Heat 4e SM Chap03

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

3-1

Solutions Manual for

Heat and Mass Transfer: Fundamentals & Applications Fourth Edition

Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2011

Chapter 3 STEADY HEAT CONDUCTION

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3-2

&

re, mperature.

Steady Heat Conduction in Plane Walls

3-1C The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with constant wall thermal conductivity.

3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient conduction.

3-3C Convection heat transfer through the wall is expressed as )( ∞−= TThA ss . In steady heat transfer, heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefoat the outer surface, the temperature will be closer to the surrounding air te

Q

3-4C The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat.

3-5C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer area of the rod is the lateral surface area of the rod,

4/2DAs π=DLA π= .

3-6C The thermal resistance of a medium represents the resistance of that medium against heat transfer.

3-7C The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a surface, and is defined as hcombined = hconvection + hradiation. It offers the convenience of incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations.

3-8C Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface area since it is defined as . )/(1 hARconv =

3-9C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat transfers occur simultaneously.


3-3

r coefficient is eh

ce will be

3-10C For a surface of A at which the convection and radiation heat transfer coefficients are hconv rad and , the single equivalent heat transfe radconvqv hh += when the medium and the surrounding surfaces are at the same

temperature. Then the equivalent thermal resistan

h

)/(1 AhR eqveqv = .

3-11C The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances connected in series.

3-12C Once the rate of heat transfer is known, the temperature drop across any layer can be determined by multiplying

heat transfer rate by the thermal resistance across that layer,

Q&

layerlayer RQT &=∆

3-13C The temperature of each surface in this case can be determined from

)(/)(

)(/)(

22222222

11111111

∞−∞∞−∞

−∞∞−∞∞

+=⎯→⎯−=

−=⎯→⎯−=

ssss

ssss

RQTTRTTQ

RQTTRTTQ&&

&&

where is the thermal resistance between the environment iR −∞ ∞ and surface i.

3-14C Yes, it is.

window, and thus the heat sfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet.

nsfer and slow down the heat gain of the drink wrapped in a lanket. Therefore, the drink left on a table will warm up faster.

3-15C The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other will probably have thermal contact resistance which serves as an additional thermal resistance to heat transfer throughtran

3-16C The blanket will introduce additional resistance to heat trab


3-4

A

3-17 The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the wall is to be determined.

Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant.

Wall

5°C

Q&

L= 0.25 m

Properties The thermal conductivity is given to be k = 0.8 W/m⋅°C.

Analysis The surface area of the wall and the rate of heat loss through the wall are 14°C

= 2m 18m) 6(m) 3( =×

W518=°−

°⋅=−

=m25.0

C)514()m C)(18 W/m8.0( 221

LTT

kAQ&

3-18 Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the bottom of the pan are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the thickness of the bottom of the pan is small relative to its diameter. 3 The thermal conductivity of the pan is constant.

Properties The thermal conductivity of the aluminum pan is given to be k = 237 W/m⋅°C.

Analysis (a) The boiling heat transfer coefficient is

222

m 0491.04

m) 25.0(4

===ππDAs

95°C 108°C

800 W C. W/m1254 2 °=

°−− ∞ C)95108)(m 0491.0()( 2TTA ss==

−= ∞

W800

)(

Qh

TThAQ ss&

&

The ou r surface temperature of the bottom of the pan is

0.5 cm(b) te

C108.3°=°⋅

°=+=

−=

)m C)(0.0491 W/m237(m) 005.0 W)(800(+C108

21,,

,,

kALQTT

LTT

kAQ

innersouters

innersouters

&

&


3-5

3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined.

Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C.

Analysis The area of the window and the individual resistances are 2m 6.3m) 4.2(m) 5.1( =×=A

L

Glass

T1

Q&

C/W 04103.001111.000214.002778.0

C/W 01111.0)m 6.3(C). W/m25( 22

22,o °=

°=== conv Ah

RR 11

C/W 002140)m 6.3(C) W/m.78.0(

m 006.0

C/W 02778.011

2,1,

21

glass

°=++=

++=

°=°

==

°====

convglassconvtal RRRR

AkLR

RR

he steady rate of heat transfer through window glass is then

)m 6.3(C). W/m10( 221

1,i°

conv Ah

.

to

RglassRi RoT T∞1 T∞2

W707=°°−−

=−

= ∞∞

C/W 04103.0C)]5(24[21

totalRTT

Q&

The inner surface temperature of the window glass can be determined from

C4.4°=°−°=−=⎯→⎯−

= ∞∞ C/W) 8 W)(0.0277707(C241,11

1,

11conv

convRQTT

RTT

Q &&


3-6

3-20 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.

Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible.

Air

Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m⋅°C and kair = 0.026 W/m⋅°C.

Analysis The area of the window and the individual resistances are

2m 6.3m) 4.2(m) 5.1( =×=A

C/W 16924.001111.012821.0)00107.0(202778.02

C/W 01111.0)m 6.3(C). W/m25(

11

C/W 12821.0)m 6.3(C) W/m.026.0( 2

2

22 °=

°=== air Ak

RR m 012.0

C/W 00107.0) 6.3(C) W/m.78.0(

m 003.0

C/W 02778.011

2,211,

o2o2

22,o

21

1glass31

1,i

°=

+++=+++=

====

°=°

====

°====

convconvtotal

conv

conv

RRRRRAh

RR

LAk

LRRR

R

The steady rate of heat transfer through window glass then becomes

R1 R2 R3Ri Ro

T∞1 T∞2R)m 6.3(C). W/m10( 22

1 °Ah

m

W154=°

°−−=

−= ∞∞

C/W16924.0C)]5(21[21

totalRTT

Q&


C16.7°°−°=−=⎯→⎯−

= ∞∞ =C/W)8 W)(0.0277154(C211,11

1,

11conv

convRQTT

RTT

Q &&


3-7

3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.

Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the glass is constant. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m⋅°C.

Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is disregarded, the heat transfer through the window will be zero. Then the answer of this problem is zero since the problem states to disregard radiation.

Vacuum

Discussion In reality, heat will be transferred between the glasses by radiation. We do not know the inner surface temperatures of windows. In order to determine radiation heat resistance we assume them to be 5°C and 15°C, respectively, and take the emissivity to be 1. Then individual resistances are

2m 6.3m) 4.2(m) 5.1( =×=A

C/W 09505.001111.005402.0)00107.0(202778.02

C/W 01111.0)m 6.3(C). W/m25(

11C/W 05402.0

]278288][278288)[m 6.3().K W/m1067.5(1 3222428

===

°=++× −

RR

K1

1

C/W 00107.0)m 6.3(C) W/m.78.0(

m 003.0

C/W 02778.011

2,11,

o2o2

22,o

21

1glass31

1,i

°=

+++=+++=

=

=

°=°

====

°====

convradconvtotal

conv

conv

RRRRRAh

AkL

RRR

R

The steady rate of heat transfer through window glass then becomes

R1 Rrad R3Ri Ro

T∞1 T∞2R)m 6.3(C). W/m10( 22

1 °Ah

))(( 22 ++=

surrssurrsrad

TTTTAR

εσ

W274=°

°−−=

−= ∞∞

C/W09505.0C)]5(21[21

totalRTT

Q&


C13.4°=°−°=−=⎯→⎯−

= ∞∞ C/W)8 W)(0.0277274(C211,11

1,

11conv

convRQTT

RTT

Q &&

Similarly, the inner surface temperatures of the glasses are calculated to be 13.1 and -1.7°C (we had assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations.


3-8

3-22 Prob. 3-20 is reconsidered. The rate of heat transfer through the window as a function of the width of air space is to be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" A=1.5*2.4 [m^2] L_glass=3 [mm] k_glass=0.78 [W/m-C] L_air=12 [mm] T_infinity_1=21 [C] T_infinity_2=-5 [C] h_1=10 [W/m^2-C] h_2=25 [W/m^2-C] "PROPERTIES" k_air=conductivity(Air,T=25) "ANALYSIS" R_conv_1=1/(h_1*A) R_glass=(L_glass*Convert(mm, m))/(k_glass*A) R_air=(L_air*Convert(mm, m))/(k_air*A) R_conv_2=1/(h_2*A) R_total=R_conv_1+2*R_glass+R_air+R_conv_2 Q_dot=(T_infinity_1-T_infinity_2)/R_total

Lair [mm]

Q [W]

2 4 6 8 10 12 14 16 18 20

414 307.4 244.5 202.9 173.4 151.4 134.4 120.8 109.7 100.5

2 4 6 8 10 12 14 16 18 20100

150

200

250

300

350

400

450

Lair [mm]

Q [

W]


3-9

A

3-23E The inner and outer surfaces of the walls of an electrically heated house remain at specified temperatures during a winter day. The amount of heat lost from the house that day and its cost are to be determined.

Assumptions 1 Heat transfer through the walls is steady since the surface temperatures of the walls remain constant at the specified values during the time period considered. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the walls is constant.

Properties The thermal conductivity of the brick wall is given to be k = 0.40 Btu/h⋅ft⋅°F.

Analysis We consider heat loss through the walls only. The total heat transfer area is

= 2ft 1530)935950(2 =×+× Wall

T2

Q&

LThe rate of heat loss during the daytime is

Btu/h 6120ft 1

F)4555()ft F)(1530ftBtu/h 40.0( 221

day =°−

°⋅⋅=−

=L

TTkAQ&

The rate of heat loss during nighttime is T1

Btu/h 240,12ft 1

C)3555()ft F)(1530ftBtu/h 40.0( 2

21night L

=°−

°⋅⋅=

−=

TTkAQ&

The amount of heat loss from the house that night will be

Btu 232,560=

+=+=∆=⎯→⎯= 1410 nightday QQtQQQ

Q &&&&∆

)Btu/h 240,12(h) 14()Btu/h 6120(h) 10(t

hen the cost of this heat loss for that day becomes T

$6.13== kWh)/09.0)($kWh 3412/560,232(Cost

3-24 A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily inheat dissipated in 24 h, the surface heat flux, and the surface temperature of the resist

a specified environment. The amount of or are to be determined.

ssumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the resistor.

pates during a 24-hour period is

h) W)(2415.0(tQQ &

A

Analysis (a) The amount of heat this resistor dissi

Resistor 0.15 W

Q&

Wh3.6==∆ =

(b) The heat flux on the surface of the resistor is

222

m 000127.0m m)(0.012 003.0(4

m) 003.0(24

2 +=+= ππDLDAs ) =ππ

2 W/m1179===2m 000127.0

W15.0

sAQ

q&

&

(c) The surface temperature of the resistor can be determined from

C166°=°⋅

+°=+=⎯→⎯−= ∞∞ )m 7C)(0.00012 W/m(9 W15.0C35)( 22

ssss hA

QTTTThAQ&

&


3-10

3-25 A very thin transparent heating element is attached to the inner surface of an automobile window for defogging purposes, the inside surface temperature of the window is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal resistance of the thin heating element is negligible.

Properties Thermal conductivity of the window is given to be k = 1.2 W/m · °C.

Analysis The thermal resistances are

Ah

Ri

i1

= Ah

Ro

o1

= and kALR =win

From energy balance and using the thermal resistance concept, the following equation is expressed:

o

oh

i

i −∞,

RRTT

AqR

TT+

−=+ ∞

win

,11 &

or )/(1)/()/(1

,11,

AhkALTT

AqAhTT

o

oh

i

i

+

−=+

− ∞∞ &

o

oh

i

i

hkLTT

qh

TT/1//1,11,

+

−=+

− ∞∞ &

C) W/m100/1(C) W/m2.1/m 005.0(C W/m15/1 2 +°⋅°⋅

)C 5( W/m1300C 22 121 °−−=+

−° TT 2 °⋅

equation:

(2 -T_1)/(1/15)+1300=(T_1-(-5))/(0.005/1.2+1/100)

Discussion In actuality, the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed. To maintain the inner surface temperature of the window, it is necessary to vary the heat flux to the heating element according to the outside condition.

Copy the following line and paste on a blank EES screen to solve the above

2

Solving by EES software, the inside surface temperature of the window is

C 14.9 °=1T


3-11

3-26 A process of bonding a transparent film onto a solid plate is taking place inside a heated chamber. The temperatures inside the heated chamber and on the transparent film surface are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal contact resistance is negligible.

Properties The thermal conductivities of the transparent film and the solid plate are given to be 0.05 W/m · °C and 1.2 W/m · °C, respectively.


hAR 1

conv =

AkL

Rf

ff =

and Aks

s

Using the thermal resistance conc

LR s=

ept, the llowing e uation is expressed: fo q

s

b

f

b

RTT

RRTT 2

conv

−=

+−∞

Rinside the cham

earranging and solving for the temperature ber yields

( ) bf

f sss kLR ⎝/

bbf

b TkL

hTTTRRTTT +⎟

⎟⎠

⎞⎜⎜⎛

+−

=++−

=12

conv2 ∞

C 127 °=°+⎟⎠⎞

⎜⎝⎛

°⋅+

°⋅°⋅°−

=∞ C 70C W/m05.0

m 001.0C W/m70

1C

W/m2.1/m 013.0

C )5270(2T

he surfac temperature of the transparent film is T e

s

b

f

b

RTT

RTT 21 −

=−

bf

f

ss

bbf

s

b TkL

kLTTTR

RTTT +⎟

⎟⎠

⎞⎜⎜⎝

⎛−=+

−=

/22

1

C 103 °=°+⎟⎠⎞

⎜⎝⎛

°⋅°⋅°−

= C 70C W/m05.0

m 001.0C W/m2.1/m 013.0

C )5270(1T

Discussion If a thicker transparent film were to be bonded on the solid plate, then the inside temperature of the heated chamber would have to be higher to maintain the temperature of the bond at 70 °C.


3-12

3-27 A power transistor dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the transistor.

Analysis (a) The amount of heat this transistor dissipates during a 24-hour period is

Air, 30°C kWh 0.0036===∆= Wh6.3h) W)(2415.0(tQQ &

(b) The heat flux on the surface of the transistor is

22

2

m 0001021.0m) m)(0.004 005.0(4

m) 005.0(2

42

=+=

+=

ππ

ππ DLDAs

Power Transistor

0.15 W

2 W/m1469=== 2m 0001021.0sAq& W15.0Q&

(c) The surface temperature of the transistor can be determined from

C111.6°=°⋅

+°=+=⎯→⎯−= ∞∞ )m 21C)(0.00010 W/m(18 W15.0C30)( 22

ssss hA

QTTTThAQ&

&

3-28 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface temperature of the chips, and the thermal resistance between the surface of the board and the cooling medium are to be determined.

1 Steady operating conditions exist. 2 Assumptionstransferred uniformly from the entire front surface.

Heat transfer from the back surface of the board is negligible. 2 Heat is

nalysis (a The heat flux on the surface of the circuit board is A )

Chips Ts

Q&

T∞

2m 0216.0m) m)(0.18 12.0( ==sA

2 W278===2m 0216.0sA

q& /m× W)06.0100(Q&

e sur ce temperature of the chips is

(b) Th fa

−= ∞ )( ss TThAQ&

C67.8°=

°⋅

×°=+= ∞

)m 0216.0)(C W/m10( W)06.0100(+C40

22s

s hAQTT&

(c) The thermal resistance is

C/W4.63°=°⋅

==)m 0216.0(C) W/m10(

1122

sconv hA

R


3-13

3-29 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces. For a given deep body temperature, the outer skin temperature is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire exposed surface of the person. 3 The surrounding surfaces are at the same temperature as the indoor air temperature. 4 Heat generation within the 0.5-cm thick outer layer of the tissue is negligible.

Qconv

Tskin

Qrad

Properties The thermal conductivity of the tissue near the skin is given to be k = 0.3 W/m⋅°C.

Analysis The skin temperature can be determined directly from

C35.5°=°⋅

−°=−=

−=

)m C)(1.7 W/m3.0(m) 005.0 W)(150(

C3721

1

kALQ

TT

LTT

kAQ

skin

skin

&

&

3-30 A double-pane window is considered. The rate of heat loss through the window and the temperature difference across

Properties The thermal conductivities of glass and air are given to be 0.78 W/m⋅K and 0.025 W/m⋅K, respectively.

Analysis (a) The rate of heat transfer through the window is determined to be

the largest thermal resistance are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficients are constant.

[ ]

[ ] W210=++++

°×°⋅°⋅°⋅°⋅°⋅ C W/m20C W/m78.0C W/m025.0C W/m78.0C W/m40

2

22

=

++++

°×=

++++

∆=

05.0000513.02.0000513.0025.0C(-20)-20)m 5.11(

1m 004.0m 005.0m 004.01C(-20)-20)m 5.11(

11

2

og

g

a

a

g

g

i hkL

kL

kL

h

TAQ&

b) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is determined from

(

C28°=×°⋅

===∆)m 5.11(C) W/m025.0(

m 005.0 W)210(2Ak

LQRQT

a

aaa

&&


3-14

3-31E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall and its R-value of insulation are to be determined.

Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant.

Properties The thermal conductivities are given to be ksheetrock = 0.10 Btu/h⋅ft⋅°F and kinsulation = 0.020 Btu/h⋅ft⋅°F.

L1 L2 L3

R1 R2 R3

Analysis (a) The surface area of the wall is not given and thus we consider a unit surface area (A = 1 ft2). Then the R-value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from.

F.h/Btu.ft 30.17 2 °=+×=+=

°=°

===

°=°

====

17.29500.022

F.h/Btu.ft 17.29F)Btu/h.ft. 020.0(

ft 12/7

F.h/Btu.ft 500.0F)Btu/h.ft. 10.0(

ft 12/6.0

21

2

2

22

2

1

131

RRR

kL

RR

kL

RRR

total

fiberglass

sheetrock

(b) Therefore, this is approximately a R-30 wall in English units.


3-15

3-32 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. Q&

Tin=20°C

Tsky = 100 K Tair =10°C Properties The thermal conductivity of the concrete is given to be

k = 2 W/m⋅°C. The emissivity of both surfaces of the roof is given to be 0.9.

L=15 cm

Analysis When the surrounding surface temperature is different than the ambient temperature, the thermal resistances network approach becomes cumbersome in problems that involve radiation. Therefore, we will use a different but intuitive approach.

In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to the heat transfer from the roof to the surroundings (by convection and radiation), that must be equal to the heat transfer through the roof by conduction. That is,

rad+conv gs,surroundin toroofcond roof,rad+conv roof, toroom QQQQ &&&& ===

Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities above can be expressed as

[ ]4,

44282

,224

,4

, rad+conv roof, toroom

K) 273(K) 27320()K W/m1067.5)(m 300)(9.0(

C))(20m C)(300 W/m5()()(

+−+⋅×+

°−°⋅=−+−=−

ins

insinsroominsroomi

T

TTTATTAhQ σε&

m 15.0)m 300)(C W/m2( ,,2,,

cond roof,outsinsoutsins TT

LTT

kAQ−

°⋅=−

=&

[ ]44,

4282m 300)(9.0( +

,2244

,, rad+conv surr, toroof

K) 100(K) 273()K W/m1067.5)(

C)10)(m C)(300 W/m12()()(

−+⋅×

°−°⋅=−+−=−

outs

outssurroutssurroutso

T

TTTATTAhQ σε&

lving the quations above simultaneously gives

he total amount of natural gas consumption during a 14-hour period is

So e

C1.2and , , out,, °−=°== sins TTQ C7.3 W37,440&

T

therms36.22kJ 1080.080.080.0 ⎜⎜

⎝===Q total

gas 5,500 therm1)s 360014)(kJ/s 440.37(

=⎟⎟⎠

⎞⎛×∆tQQ &

inally, the money lost through the roof during that period is

F

$26.8== )therm/20.1$ therms)(36.22(lostMoney


3-16

3-33 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined.

Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects.

Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m⋅°C.

Analysis The rate of heat transfer without insulation is

Insulation

L Ts

Rinsulation Ro

2m 3m) m)(1.5 2( ==A

W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=−= ∞TThAQ s&

In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be T∞

C/W 333.0C)32110(°=

°−=

∆=⎯→⎯

∆=

TRTQ total&

W234

W234 W234010.0 =×

QR

Q

total&

&

and in order to have this thermal resistance, the thickness of insulation must be

=

cm 3.4==°°⋅

m 034.0)m C)(3 W/m.038.0()m C)(3 W/m10( 222

L

°=+=

+=+=

C/W 333.01

1conv

LkAL

hARRR insulationtotal

oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is

N t hper year, and that the

therms1.807kJ 105,500

therm1h 1

s 36000.78

h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

∆=

EfficiencytQsaved

&

he money aved is

gy ==

he insulation will pay for its cost of $250 in

T s

Ener(savedMoney = year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos

T

yr 0.282===$887.8/yr

$250savedMoney spent Money periodPayback

which is equal to 3.4 months.


3-17

∞TThAQ s&

3-34 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined.

Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficients accounts for the radiation effects.

Properties The thermal conductivity of the expanded perlite insulation is given to be k = 0.052 W/m⋅°C.

Analysis The rate of heat transfer without insulation is

Insulation

L Ts

Rinsulation Ro

2m 3m) m)(1.5 2( ==A

= W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=−

In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be T∞

C/W 333.0C)32110(°=

°−=

∆=⎯→⎯

∆=

TRTQ total&

W234

W234 W234010.0 =×

QR

Q

total&

&

and in order to have this thermal resistance, the thickness of insulation must be

=

cm 4.7==°⋅°⋅

m 047.0)m C)(3 W/m052.0()m C)(3 W/m10( 222

L

°=+=

+=+=

C/W 333.01

1conv

LkAL

hARRR insulationtotal

oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is

N t hper year, and that the

therms1.807kJ 105,500

therm1h 1

s 36000.78

h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

∆=

EfficiencytQsaved

&

oney

gy

The m saved is

Ener(savedMoney year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos= ==

he insulation will pay for its cost of $250 in

T

yr 0.282===$887.8/yr

$250savedMoney spent Money periodPayback

which is equal to 3.4 months.


3-18

3-35 Prob. 3-33 is reconsidered. The effect of thermal conductivity on the required insulation thickness is to be investigated.


"GIVEN" A=2*1.5 [m^2] T_s=110 [C] T_infinity=32 [C] h=10 [W/m^2-C] k_ins=0.038 [W/m-C] f_reduce=0.90 "ANALYSIS" Q_dot_old=h*A*(T_s-T_infinity) Q_dot_new=(1-f_reduce)*Q_dot_old Q_dot_new=(T_s-T_infinity)/R_total R_total=R_conv+R_ins R_conv=1/(h*A) R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm"

kins [W/m.C]

Lins [cm]

0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08

1.8 2.25 2.7 3.15 3.6 4.05 4.5 4.95 5.4 5.85 6.3 6.75 7.2 0.02 0.03 0.04 0.05 0.06 0.07 0.08

1

2

3

4

5

6

7

8

kins [W/m-C]

L ins

[cm

]


3-19

3-36 Two of the walls of a house have no windows while the other two walls have single- or double-pane windows. The average rate of heat transfer through each wall, and the amount of money this household will save per heating season by converting the single pane windows to double pane windows are to be determined.

Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is disregarded.

Properties The thermal conductivities are given to be k = 0.026 W/m⋅°C for air, and 0.78 W/m⋅°C for glass.

Analysis The rate of heat transfer through each wall can be determined by applying thermal resistance network. The convection resistances at the inner and outer surfaces are common in all cases.

Walls without windows: Wall

C/W 06271.0001389.005775.0003571.0

C/W001389.0)m 410(C) W/m18( 2

==o

o AhR 11

C/W 05775.0)m 410(

C/Wm 31.2

C/W 003571.0)m 410(C) W/m7(

11

walltal

2

2

2wall

wall

22

°=++=++=

°=×°⋅

°=×

°⋅=

−==

°=×°⋅

==

oi

ii

RRRR

AvalueR

kAL

R

Ah

R

Q&

L

to

Then RwallRi Ro

W255.1=°°−

=−

= ∞∞

C/W06271.0C)824(21

totalRTT

Q&

Wall with single pane windows:

C/W 003063.0000694.0000583.0001786.0

C/W 000694.0)m 420(C) W/m18(

11

C/W 000583.0002968.0

15033382.0

11511

C/W 002968.0m)8.12.1)(C W/m78.0(

m 005.0

C/W 033382.0m )8.12.1(5)420(

C/Wm 31.2

C/W 001786.0)m 420(C) W/m7(

11

eqvtotal

22

o

glasswalleqv

2o2glass

glass

2

2wall

wall

22

°=++=++=

°=×°⋅

==

=→+=+=

°=×⋅

==

°=×−×

°⋅=

−==

°=×°⋅

==

oi

oo

eqv

ii

RRRRAh

R

RRRR

kAL

R

AvalueR

kAL

R

AhR

Ri Rwall Ro

Rglass

Then

W5224=°°−

=−

= ∞∞

C/W003063.0C)824(

total

21

RTT

Q&


3-20

4th wall with double pane windows:

Rwall

Rglass Rair Rglass

Ri

Ro

C/W 023197.0000694.0020717.0001786.0

C/W 020717.027303.0

1511511

eqv°=⎯→⎯+=+= R

R 033382.0

C/W 27303.0267094.0002968.022

C/W 002968.0m)12.1)(C W/m78.0(

m 005.0

C/W 033382.0m)8.12.1(5)420(

C/Wm 31.2

eqvtotal

eqvwindowwall

airglass

22glass

glass

2

2wall

wall

°=++=++=

°=+×=+=

°=×°⋅

==

°=×−×

°⋅=

−==

oi RRRR

RR

RRR

kAL

R

AvalueR

kAL

R

8.

C/W 267094.0m)8.12.1)(C W/m026.0(

m 015.02o2

airair °=

×⋅==

kAL

R

window

Then

W690=°°−

=−

= ∞∞

C/W023197.0C)824(

total

21

RTT

Q&

The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is

The amount of energy and money saved during a 7-month long heating season by switching from single pane to double pane windows become

Money savings = (Energy saved)(Unit cost of energy) = (22,851 kWh)($0.08/kWh) = $1828

W45346905224panedouble

panesinglesave =−=−= QQQ &&&

kWh 22,851=h) 2430kW)(7 534.4( ××=∆= tQQ savesave&


3-21

,− outsroomo TTAhQ&

room and the refrigerated space can be expressed as

3-37 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined.

Assumptions 1 Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the kitchen air remain constant at the specified values. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation effects.

Properties The thermal conductivities are given to be k = 15.1 W/m⋅°C for sheet metal and 0.035 W/m⋅°C for fiberglass insulation.

Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 20°C. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area,

1 mm L 1 mm

insulation

)(=

W36=C)2024)(m 1(C) W/m9( 22 °−°⋅=

Using the thermal resistance network, heat transfer between the

Ri R1 Rins R3 Ro

iinsulationmetalo

refrigroom

total

refrigroom

hkL

kL

h

TTAQ

RTT

Q

121/

+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

−=

−=

&

&

Troom Trefrig

S g,ubstitutin

C W/m41

C W/m035.0C W/m15.1m 001.02

C W/m91

+

C)224( W/m36 2 °−=

2222 °⋅+

°⋅+

°⋅

×

°⋅

L

Solv ing for L, the minimum thickness of insulation is determined to be

L = 0.00875 m = 0.875 cm


3-22

3-38 Prob. 3-37 is reconsidered. The effects of the thermal conductivities of the insulation material and the sheet metal on the thickness of the insulation is to be investigated.


"GIVEN" k_ins=0.035 [W/m-C] L_metal=0.001 [m] k_metal=15.1 [W/m-C] T_refrig=2 [C] T_kitchen=24 [C] h_i=4 [W/m^2-C] h_o=9 [W/m^2-C] T_s_out=20 [C] "ANALYSIS" A=1 [m^2] “a unit surface area is considered" Q_dot=h_o*A*(T_kitchen-T_s_out) Q_dot=(T_kitchen-T_refrig)/R_total R_total=R_conv_i+2*R_metal+R_ins+R_conv_o R_conv_i=1/(h_i*A) R_metal=L_metal/(k_metal*A) R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm" R_conv_o=1/(h_o*A)

0.02 0.03 0.04 0.05 0.06 0.07 0.080.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

kins [W/m-C]

L ins

[cm

]

kins [W/m.C]

Lins [cm]

0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08

0.4997 0.6247 0.7496 0.8745 0.9995 1.124 1.249 1.374 1.499 1.624 1.749 1.874 1.999


3-23

kmetal [W/m.C]

Lins [cm]

10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400

0.8743 0.8748 0.8749 0.8749 0.8749 0.8749 0.8749 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875 0.875

0 50 100 150 200 250 300 350 4000.8743

0.8744

0.8745

0.8746

0.8747

0.8748

0.8749

0.875

kmetal [W/m-C]

L ins

[cm

]


3-24

3-39 Heat is to be conducted along a circuit board with a copper layer on one side. The percentages of heat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the board are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since heat transfer from the side surfaces is disregarded 3 Thermal conductivities are constant. Copper

Epoxy

Ts tepoxytcopper

Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper and 0.26 W/m⋅°C for epoxy layers.

Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer board can be expressed as

[ ]LT

TkATkAQQQ ⎟⎞

⎜⎛ ∆

+⎟⎞

⎜⎛ ∆

=+

epoxycopper

&&&

eat conduction along an “equivalent” board of thickness t = tcopper + tepoxy and thermal conductivity k can be expressed as

wktkt

LL

∆+=

⎠⎝⎠⎝ epoxycopperepoxycopper

)()(

=

Heff

Q LTwttk

LTkA⎜

⎝= Q ∆

+=⎟⎠⎞⎛ ∆ )( epoxycoppereff

board

&

Setting the two relations above equal to each other and solving for the effective conductivity gives

epoxycopper

epoxycopperepoxycopperepoxycopper

)()()()()(

ttktkt

kktktttk effeff +

+=⎯→⎯+=+

Note that heat conduction is proportional to kt. Substituting, the fractions of heat onducted along the copper and epoxy layers as well as the effective thermal conductivity of the board are determined to be

c

99.2%

0.8%

====

====

°=+=+=

°=°⋅=

°=°⋅=

992.0038912.0

0386.0)(

)(

008.0038912.0000312.0

)()(

C W/038912.0000312.00386.0)()()(

C W/000312.0m) C)(0.0012 W/m26.0()(

C W/0386.0m) C)(0.0001 W/m386()(

total

coppercopper

total

epoxyepoxyf

epoxycoppertotal

epoxy

copper

ktkt

f

ktkt

ktktkt

kt

kt

and

C W/m.29.9 °=+

°×+×=

m )0012.00001.0(C W/)0012.026.00001.0386(

effk


3-25

3-40E A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since heat transfer from the side surfaces are disregarded 3 Thermal conductivities are constant. Copper

Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper and 0.15 Btu/h⋅ft⋅°F for epoxy layers.

Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a single layer that is twice as thick)

[ ]LTwktkt

LTkA

LTkA

QQQ

∆+=⎟

⎠⎞

⎜⎝⎛ ∆

+⎟⎠⎞

⎜⎝⎛ ∆

=

+=

epoxycopperepoxycopper

epoxycopper

)()(

&&&

Epoxy

Ts

½ tepoxytcopper½ tepoxy

Epoxy

Heat conduction along an “equivalent” plate of thick ness t = tcopper + tepoxy and thermal conductivity keff can be expressed as Q

LL ⎠⎝

epoxycoppereffboard

TwttkTkAQ ∆+=⎟

⎞⎜⎛ ∆ )(&

Setting the two relations above equal to each other and solving for the effective conductivity gives

=

epoxycopperepoxycopperepoxycopper tteffeff +

Note that heat conduction is proportional to kt. Substituting, the

epoxycopper )()()()()(

ktktkktktttk

+=⎯→⎯+=+

fraction of heat conducted along the copper layer and the effective th rmal conductivity of the plate are determined to be

and

e

FBtu/h. 93292.000375.09292.0)()()(

FBtu/h. 00375.0ft) F)(0.15/12Btu/h.ft. 15.0(2)(

FBtu/h. 9292.0ft) F)(0.05/12Btu/h.ft. 223()(

epoxycoppertotal

epoxy

copper

°=+=+=

°=°=

°=°=

ktktkt

kt

kt

F.Btu/h.ft 32.0 2 °=+

°=

+

+=

ft )]12/15.0(2)12/05.0[(FBtu/h. 93292.0

t)()(

epoxycopper

epoxycopper

tktkt

keff

99.6%==== 996.093292.09292.0

)()(

total

coppercopper kt

ktf


3-26

3-41 Warm air blowing over the inner surface of an automobile windshield is used for defrosting ice accumulated on the outer surface. The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the windshield is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 The automobile is operating at 1 atm.

Properties Thermal conductivity of the windshield is given to be k = 1.4 W/m · °C.


Ah

Ri

i1

=

AhR

oo

1=

and kAwin

From energy balance and using the thermal res

LR =

istance oncept, th following equation is expressed: c e

i

i

o

o

RRTT

RTT

+

−=

− ∞∞

win

,11,

win1,o∞

or

,1 RRTT

TTR o

i −−

−= ∞ i

kLTT i −⎟

⎞⎜⎛−

= ∞ 11 ,1 hTTh oo

⎟⎠

⎜⎝−∞ 1,

ection

i

For the ice to begin melting, the outer surface temperature of the windshield ( 1T ) should be at least 0 °C. The convheat transfer coefficient for the warm air is

C W/m112 2 °⋅=

⎥⎦

⎤⎢⎣

⎡°⋅

−⎟⎠

⎞⎜⎝

⎛°⋅°−−

°−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

−

−1

∞

∞

1

2

1,

,1

C W/m4.1m 005.0

C W/m2001

C )010(C )250(

1kL

hTTTT

hoo

ii

Discussion In practical situations, the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed. Therefore the convection heat transfer coefficient of the warm air necessary to melt the ice should be varied as well. This is done by adjusting the warm air flow rate and temperature.


3-27

3-42 The thermal contact conductance for an aluminum plate attached on a copper plate, that is heated electrically, is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the aluminum plate is given to be 235 W/m · °C.


kALR =cond

and hA

From energy balance and using the thermal res

R 1=

istance concept, the following equation is expressed:

conv

convcond

1elec / RRAR

TTAq

c ++−

= ∞&

or )/(1)/(/

1elec hAkALAR

TTAq

++−

= ∞& c

Rearranging the equation and solving for the contact resistance yields

C/Wm102586

CW/m671

C W/m235025.0C )20100(

2 −°−

=m

W/m5300

1

252

elec

1

°⋅×=°⋅

−°⋅

−−−

=

−

∞

.

hkL

qTT

Rc &

herma contact conductance is

Discussion By comparing the value of the thermal contact conductance with the values listed in Table 3-2, the surface conditions of the plates appear to be milled.

The t l

C W/m16000 2 °⋅== cc Rh /1


3-28

Thermal Contact Resistance

3-43C The resistance that an interface offers to heat transfer per unit interface area is called thermal contact resistance, . The inverse of thermal contact resistance is called the thermal contact conductance.

cR

3-44C The thermal contact resistance will be greater for rough surfaces because an interface with rough surfaces will contain more air gaps whose thermal conductivity is low.

3-45C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance has significance only for highly conducting materials like metals. Therefore, the thermal contact resistance can be ignored for two layers of insulation pressed against each other.

3-46C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance is significant for highly conducting materials like metals. Therefore, the thermal contact resistance must be considered for two layers of metals pressed against each other.

3-47C Heat transfer through the voids at an interface is by conduction and radiation. Evacuating the interface eliminates heat transfer by conduction, and thus increases the thermal contact resistance.

3-48C Thermal contact resistance can be minimized by (1) applying a thermally conducting liquid on the surfaces before they are pressed against each other, (2) by replacing the air at the interface by a better conducting gas such as helium or hydrogen, (3) by increasing the interface pressure, and (4) by inserting a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.


3-29

3-49 The thickness of copper plate whose thermal resistance is equal to the thermal contact resistance is to be determined.

Properties The thermal conductivity of copper is k = 386 W/m⋅°C.

Analysis Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance is determined to be

C/W.m 10143.7C. W/m000,14

11 252

c

For a unit surface area, the therm

°×=°

== −

hR

al resistance of a flat plate is defined as

c

kR = where L is the thickness oL f the plate and k is

e thermal conductivity. Setting he equivalent thickness is determined from t e relation above to be

copper. Note that e thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates.

ror involved in the total thermal resistance of the

ting conditions exist. 2 Heat transfer is one-dimensional since the plate is large. 3 Thermal

ers is given to be hc = 6000 W/m2⋅°C.

rmal resistances of different layers for unit surface a of 1 m

th h,cRR = t

cm 2.76==°⋅×°⋅=== − m 0276.0C/W)m 10C)(7.143 W/m386( 25ckRkRL

Therefore, the interface between the two plates offers as much resistance to heat transfer as a 2.76 cm thickth

3-50 A thin copper plate is sandwiched between two epoxy boards. The erplate if the thermal contact conductances are ignored is to be determined.

Assumptions 1 Steady operaconductivities are constant.

Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper plates and k = 0.26 W/m⋅°C for epoxyboards. The contact conductance at the interface of copper-epoxy lay

Analysis The theare 2 are

Epoxy

7 mm

Epoxy

7 mm

Copper plate

Q&

C° /W 00017.0)m C)(1 W/m6000(

1122

ccontact =

°⋅==

cAhR

C/W 106.2)m C)(1

W/(386kA m

m 001.0 62ate °×=

°⋅== −LR

pl

C/W 02692.0)m C)(1 W/m(0.26

m 007.02epoxy °=

°⋅==

kALR

The total thermal resistance is

22 epoxyplatecontacttotal ++= RRRR

C/W 05419.002692.02106.200017.02 6 °=×+×+×= −

nvolved in the total thermal resistance of the ntact resistances are ignored is determined to

be

Then the percent error iplate if the thermal co

Rcontact Rcontact

Repoxy

T1T2

Repoxy Rplate

0.63%=××

=×= 10005419.0

00017.021002

Error%total

contact

RR

which is negligible.


3-30

3-51 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the interface are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional in the axial direction since the lateral surfaces of both cylinders are well-insulated. 3 Thermal conductivities are constant.

Bar Bar

Interface

Properties The thermal conductivity of aluminum bars is given to be k = 176 W/m⋅°C. The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 20 atm ≈ 2 MPa pressure is hc = 11,400 W/m2⋅°C (Table 3-2).

Ri Rglass RoAnalysis (a) The thermal resistance network in this case consists of two conduction resistance and the contact resistance, and they are determined to be

T1 T2

C/W 0447.0/4]m) (0.05C)[ W/m400,11(

1122

ccontact °=

°⋅==

πcAhR

C/W 4341.0/4]m) (0.05C)[ W/m(176

m 15.0plate ==

LR2

°=°⋅ πkA

hen the rate of heat transfer is determined to be T

W142.4=°

×++ )4341.020447.0(2 barcontacttotal RRR

°−=

∆=

∆=

C)20150(TTQ&C/W

e rate of heat transfer through the bars is 142.4 W.

) The temperature drop at the interface is determined to be

Therefore, th

(b

C6.4°=°==∆ C/W) W)(0.04474.142(contactinterface RQT &


3-31

Generalized Thermal Resistance Networks

3-52C Two approaches used in development of the thermal resistance network in the x-direction for multi-dimensional problems are (1) to assume any plane wall normal to the x-axis to be isothermal and (2) to assume any plane parallel to the x-axis to be adiabatic.

3-53C The thermal resistance network approach will give adequate results for multi-dimensional heat transfer problems if heat transfer occurs predominantly in one direction.

3-54C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a surface). Series resistances indicate sequential heat transfer (such as two homogeneous layers of a wall).

3-55 A typical section of a building wall is considered. The average heat flux through the wall is to be determined.

Assumptions 1 Steady operating conditions exist.

Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K.

Analysis We consider 1 m2 of wall area. The thermal resistances are

C/Wm 1.0C) W/m0.1(

m 1.0

C/Wm 1032.10.005)C)(0.6 W/m50(

m 005.0m) 08.0(

)(

C/Wm 645.20.005)C)(0.6 W/m03.0(

m 6.0m) 08.0(

)(

C/Wm 02.0C) W/m5.0(

m 01.0

2

34

3434

25

23b

b2323

2

23a

a2323

2

12

1212

°⋅=°⋅

==

°⋅×=+°⋅

=

+=

°⋅=+°⋅

=

+=

°⋅=°⋅

==

−

kt

R

LLkL

tR

LLkL

tR

kt

R

bab

baa

The total thermal resistance and the rate of heat transfer are

C/Wm 120.01.01032.1645.2

1032.1645.202.0 25

5

342323

232312total

°⋅=+⎟⎟⎠

⎞⎜⎜⎝

⎛

×+

×+=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

−

−

RRR

RRRR

ba

ba

2 W/m125=⋅

°−=

−=

C/Wm 0.120C)2035(

2total

14

RTT

q&


3-32

3-56 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is disregarded.

Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for plaster layers, and k = 0.026 W/m⋅°C for the rigid foam.

Analysis We consider 1 m deep and 0.28 m high portion of wall which is representative of the entire wall. The thermal resistance network and individual resistances are

R7R6

R5

R4

R3

R2

R1Ri

T∞1

T∞2

C/W 737.4179.0804.0)325.0(2747.2357.02

C/W 804.045.45

1833.01

45.4511111

C/W 179.0)m 128.0(C) W/m20(

11

C/W 833.0)m 125.0(C) W/m72.0(

m 15.0

C/W45.45)m 1015.0(C) W/m22.0(

m 15.0

C/W 325.0)m 128.0(C) W/m22.0(

m 02.0====

LRRR

C/W 747.2)m 128.0(C) W/m026.0(

C/W 357.0)m 1

11

21

543

22

2,o

24

253

262

21

221,

°=++++=++++=

°=⎯→⎯++=++=

°=×°⋅

===

°=×°⋅

===

°=×°⋅

====

°=×°⋅

°=×°⋅

===

°=×

===

omiditotal

midmid

conv

brick

ocenterplaster

sideplaster

foam

convi

RRRRRR

RRRRR

AhRR

kALRR

AhLRRR

kA

kARR

RR

The steady rate of heat transfer through the wall per is

m 02.028.0(C) W/m10(1 °⋅

LAh

2m 28.0

W49.5C/W737.4

C)]4(22[( 21 =°

°−−=

−= ∞∞

totalRTT

Q&

Then steady rate of heat transfer through the entire wall becomes

W470=×

=2

2

m 28.0m)64( W)49.5(totalQ&


3-33

3-57 Prob. 3-56 is reconsidered. The rate of heat transfer through the wall as a function of the thickness of the rigid

nalysis The problem is solved using EES, and the solution is given below.

] ]

-C]

[m^2]

ter/(k_plaster*A_3)

side+R_mid+R_conv_2 ity_2)/R_total

_dot_total=Q_dot*A/A_1

]

foam is to be plotted.

A

"GIVEN" A=4*6 [m^2] L_brick=0.15 [m] L_plaster_center=0.15 [mL_plaster_side=0.02 [m"L_foam=2 [cm]" k_brick=0.72 [W/m-C] k_plaster=0.22 [W/m-C] k_foam=0.026 [W/mT_infinity_1=22 [C] T_infinity_2=-4 [C] h_1=10 [W/m^2-C] h_2=20 [W/m^2-C] A_1=0.28*1 [m^2] A_2=0.25*1 [m^2] A_3=0.015*1 "ANALYSIS" R_conv_1=1/(h_1*A_1) R_foam=(L_foam*Convert(cm, m))/(k_foam*A_1) "L_foam is in cm"R_plaster_side=L_plaster_side/(k_plaster*A_1) R_plaster_center=L_plaster_cenR_brick=L_brick/(k_brick*A_2) R_conv_2=1/(h_2*A_1) 1/R_mid=2*1/R_plaster_center+1/R_brick R_total=R_conv_1+R_foam+2*R_plaster_Q_dot=(T_infinity_1-T_infinQ

Lfo[cm

am Q total[W]

1 2 3 4 5 6 7 8 9 10 141.7

662.8 470.5 364.8 297.8 251.6 217.8 192 171.7 155.3

1 2 3 4 5 6 7 8 9 10100

200

300

400

500

600

700

Qto

tal

[W]

Lfoam [cm]


3-34

3-58 A wall is constructed of two layers of sheetrock spaced by 5 cm × 16 cm wood studs. The space between the studs is filled with fiberglass insulation. The thermal resistance of the wall and the rate of heat transfer through the wall are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer.

Properties The thermal conductivities are given to be k = 0.17 W/m⋅°C for sheetrock, k = 0.11 W/m⋅°C for wood studs, and k = 0.034 W/m⋅°C for fiberglass insulation.

Analysis (a) The representative surface area is . The thermal resistance network and the individual thermal resistances are

2m 65.065.01 =×=A

R5R4

R3

R2R1Ri

T∞1 T∞2

W40.4C/W 588.6

C)]9(20[section) m 0.65m 1 a(for 045.0090.0178.6090.0185.0

C/W 178.6843.71

091.291111

C/W 045.0)m 65.0(C) W/m34(

11

C/W 843.7)m 60.0(C) W/m034.0(

m 16.0

C/W 091.29)m 05.0(C) W/m11.0(

m 16.0

C/W 090.0)m 65.0(C) W/m17.0(

m 01.0

C/W 185.011)m 65.0(C) W/m3.8( 22 °⋅i Ah

21

41

32

2o2

23

22

241

=°

°−−=

−=

×°=++++=++++=

°=⎯→⎯+=+=

°=⋅

==

°=°⋅

===

°=°⋅

===

°=°⋅

====

°===

∞∞

total

omiditotal

midmid

oo

fiberglass

stud

sheetrock

i

RTT

Q

RRRRRR

RRRR

AhR

kALRR

kALRR

kALRRR

R

&

C/W 6.588

(b) Then steady rate of heat transfer through entire wall becomes

W406==2m 65.0m) 5(m) 12(

W)40.4(totalQ&


3-35

3-59 A wall is to be constructed of 10-cm thick wood studs or with pairs of 5-cm thick wood studs nailed to each other. The rate of heat transfer through the solid stud and through a stud pair nailed to each other, as well as the effective conductivity of the nailed stud pair are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can be approximated as being one-dimensional since it is predominantly in the x direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance between the two layers is negligible. 4 Heat transfer by radiation is disregarded.

Properties The thermal conductivities are given to be k = 0.11 W/m⋅°C for wood studs and k = 50 W/m⋅°C for manganese steel nails.

Analysis (a) The heat transfer area of the stud is A = (0.1 m)(2.5 m) = 0.25 m2. The thermal resistance and heat transfer rate through the solid stud are

W2.2=°°

=∆

=

°=

°⋅ )m 25.0(C) W/m11.0( 2kA

==

C/W 636.3C8

C/W 636.3m 1.0

studRTQ

LR

&

(b) The thermal resistances of stud pair and nails are in parallel

Stud stud

T1 T2

L

Q&

W4.7=°°

=∆

=

°=⎯→⎯+=+=

°=−°⋅

==

°=°⋅

==

=⎥⎥⎦

⎤

⎢⎢⎣

⎡==

C/W 70.1C8

C/W 70.118.365.3nailsstudtotal RRR11111

C/W 65.3)m 000628.025.0(C) W/m11.0(

m 1.0

C/W 18.3)m 000628.0(C) W/m50(

m 1.0

m 000628.04

m) 004.0(504

50

2

2

222

stud

total

stud

nails

nails

RTQ

R

kALR

kALR

DA

&

ππ

c) The effective conductivity of the nailed stud pair can be determined from

Rstud

T1 T2

(

C W/m.0.235 °=°

=∆

=⎯→⎯∆

=)m C)(0.258(

m) 1.0 W)(7.4(2TA

LQkLTAkQ effeff

&&


3-36

3-60E A wall is to be constructed using solid bricks or identical size bricks with 9 square air holes. There is a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall. The rates of heat transfer through the wall constructed of solid bricks and of bricks with air holes are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer.

Properties The thermal conductivities are given to be k = 0.40 Btu/h⋅ft⋅°F for bricks, k = 0.015 Btu/h⋅ft⋅°F for air, and k = 0.10 Btu/h⋅ft⋅°F for sheetrock.

Analysis (a) The representative surface area is . The thermal resistance network and the individual thermal resistances if the wall is constructed of solid bricks are

2ft 3906.0)12/5.7)(12/5.7( ==A

RoR5

R4

R3

R2

R1Ri

T∞1

T∞2

F/Btuh 57.308ft)]12/5.0()12/7(F)[ftBtu/h 10.0(

ft 12/9

F/Btuh 288ft)]12/5.0()12/5.7(F)[ftBtu/h 10.0(

ft 12/9

F/Btuh 0667.1)ft 3906.0(F)ftBtu/h 10.0(

ft 12/5.0

F/Btuh 7068.1)ft 3906.0(F)ftBtu/h 5.1(

11

2o3

22

251

22

°⋅=×⋅⋅

===

°⋅=×°⋅⋅

===

°⋅=°⋅⋅

====

°=°⋅⋅

==

kALRR

kALRR

kALRRR

AhR

plaster

plaster

plaster

ii

Btu/h 6971.4F/Btuh 5804.9

F)3580(F/Btuh 5804.94267.00667.13135.50667.17068.1

F/Btuh 3135.551.557.308288432

°⋅=⎯→⎯++=++= midmid

RRRRR

111111

F/Btuh )ft 3906.0(F)ftBtu/h 6(

F/Btuh 51.5ft)]12/7()12/7(F)[ftBtu/h 40.0(

ft 12/9

21

51

24

=°⋅°−

=−

=

°⋅=++++=++++=

°⋅°⋅⋅

°⋅=×°⋅⋅

===

∞∞

total

omiditotal

o

brick

RTT

Q

RRRRRR

Ah

kALRR

&

en ste y of heat transfer through entire wall becomes

4267.01122

===oR

1

Th ad rate

Btu/h 3610== 2m 3906.0ft) 10(ft) 30(Btu/h) 6971.4(totalQ&

(ba

) The thermal resistance network and the individual thermal resistances if the wall is constructed of bricks with air holes re

ft 1406.0)12/5.1()12/5.1(9 =×=airholesA

T∞1

Ri R1

R2

R3

R4

R5

R6 Ro

T∞2

22 ft 1997.01406.0ft) 12/7( =−=bricksA

2


3-37

F/Btuh 389.9

)ft F)(0.1997ftBtu/h 40.0(ft 12/9

F/Btuh 62.355)ft F)(0.1406ftBtu/h 015.0(

ft 12/9

25

24

°⋅=°⋅⋅

===

°⋅=°⋅⋅

===

kALRR

kALRR

brick

airholes

Btu/h 492.3F/Btuh 885.12

F)3580(F/Btuh 885.124267.00667.1618.80667.17068.1

F/Btuh 618.8389.91

62.3551

57.3081

288111111

21

61

5432

=°⋅°−

=−

=

°⋅=++++=++++=

°⋅=⎯→⎯+++=+++=

∞∞

total

omiditotal

midmid

RTT

Q

RRRRRR

RRRRRR

&

Then steady rate of heat transfer through entire wall becomes

Btu/h 2680== 2ft 3906.0ft) 10(ft) 30(Btu/h) 492.3(totalQ&


3-38

3-61 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of the wall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interface temperatures, and the temperature drop across the section F are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Thermal contact resistances at the interfaces are disregarded. Properties The thermal conductivities are given to be kA = kF = 2, kB = 8, kC = 20, kD = 15, kE = 35 W/m⋅°C.

Analysis (a) The representative surface area is . The thermal resistance network and the individual thermal resistances are

2m 12.0112.0 =×=A

R1

R2

R3

R4

R5

R6

R7

T2T1

C/W 16.0)m 04.0(C) W/m8(

m 05.0

C/W 06.0)m 04.0(C) W/m20(

m 05.0

C/W 04.0)m 12.0(C) W/m2(

m 01.0

23

242

21

°=°⋅

=⎟⎠⎞

⎜⎝⎛==

°=°⋅

=⎟⎠⎞

⎜⎝⎛===

°=°⋅

=⎟⎠⎞

⎜⎝⎛==

BB

CC

AA

kALRR

kALRRR

kALRR

C/W 25.0)m 12.0(C) W/m2( 2°⋅⎠⎝ FkA

m 06.0

C/W)m 06.0(C) W/m35(

C/W 11.0)m 06.0(C) W/m15(

m 1.0

7

o2

2o

°==⎟⎞

⎜⎛==

°⋅⎠⎝

°=⋅

=⎟⎠⎞

⎜⎝⎛==

F

E

DD

LRR

kA

kALRR

5

05.0m 1.06 ==⎟

⎞⎜⎛== E

LRR

section) m 1 m 0.12 a(for W 572C/W 349.0

C)100300(

C/W 349.025.0034.0025.004.0

C/W 034.005.01

11.01111

C/W 025.006.016.006.04321,mid RRRR

1111111

21

2,652,

1,

×=°

==

°=

°=⎯→⎯+=+=

°=⎯→⎯++=++=

∞∞

total

midmid

mid

RQ

RRRR

R

&


72,1,1

°−−

+++=+++= midmidtotal

TTRRRRR

W101.91 5×==2m 12.0m) 8(m) 5(

W)572(totalQ&

(b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is

C/W 065.0025.004.01,1 °=+=+= midtotal RRR

Then the temperature at the point where the sections B, D, and E meet becomes

C263°=°−°=−=⎯→⎯−

= C/W) W)(0.065572(C30011

totaltotal

RQTTR

TTQ &&

(c) The temperature drop across the section F can be determined from

C143°=°==∆→∆

= C/W) W)(0.25572(FF

RQTR

TQ &&


3-39

3-62 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of the wall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interface temperatures, and the temperature drop across the section F are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Thermal contact resistances at the interfaces are to be considered. Properties The thermal conductivities of various materials used are given to be kA = kF = 2, kB = 8, kC = 20, kD = 15, and kE = 35 W/m⋅°C. Analysis The representative surface area is 2m 12.0112.0 =×=A

R1

R2

R3

R4

R5

R6

R7 R8 (a) The thermal resistance network and the individual thermal resistances are

C/W 16.0)m 04.0(C) W/m8(

m 05.0

C/W 06.0)m 04.0(C) W/m20(

m 05.0

C/W 04.0)m 12.0(C) W/m2(

m 01.0

23

242

21

°=°⋅

=⎟⎠⎞

⎜⎝⎛==

°=°⋅

=⎟⎠⎞

⎜⎝⎛===

°=°⋅

=⎟⎠⎞

⎜⎝⎛==

BB

CC

AA

kALRR

kALRRR

kALRR

C/W 001.0C/Wm 00012.028 °=°⋅

=Rm 12.0

C/W 25.0m 12.0(C) W/m2(

C/W 05.0)m 06.0(C) W/m35(

m 1.0

C/W 11.0)m 06.0(C) W/m15(

m 1.0

2

27

o2

2o5

°⋅=⎟

⎠⎜⎝

=°⋅

=⎟⎠⎞

⎜⎝⎛==

°=⋅

=⎟⎠⎞

⎜⎝⎛==

kARR

kALRR

kALRR

FF

EE

DD

6

)m 06.0

°=⎞⎛==L

section) m 1 m 0.12 a(for W 571C/W 350.0

C)100300(

C/W 350.0001.025.0034.0025.004.0

C/W 034.0 05.011.0 2,

652,mid

mid RRR11111

21

872,1,1

4321,

×=°

°−=

−=

°=++++=++++=

°=⎯→⎯+=+=

∞∞

total

midmidtotal

mid

RTT

Q

RRRRR

R

R

&


C/W 025.006.01

16.01

06.011111

1, °=⎯→⎯++=++= midRRRR

R

W101.90 5×==2m 12.0m) 8(m) 5(

W)571(totalQ&

(b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is C/W 065.0025.004.01,1 °=+=+= midtotal RRR Then the temperature at the point where The sections B, D, and E meet becomes

C263°=°−°=−=⎯→⎯−

= C/W) W)(0.065571(C30011

totaltotal

RQTTR

TTQ &&

(c) The temperature drop across the section F can be determined from

C143°=°==∆⎯→⎯∆

= C/W) W)(0.25571(FF

RQTR

TQ &&


3-40

3-63 A coat is made of 5 layers of 0.15 mm thick synthetic fabric separated by 1.5 mm thick air space. The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer.

Properties The thermal conductivities are given to be k = 0.13 W/m⋅°C for synthetic fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric.

Analysis The thermal resistance network and the individual thermal resistances are

R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro

Ts1 T∞2

C/W 2214.00320.00462.040009.0545

C/W 0320.0)m 25.1(C) W/m25(

11

C/W 0462.0)m 25.1(C) W/m026.0(

m 0015.0

C/W 0009.0)m 25.1(C) W/m13.0(

m 00015.0

22

28642

297531

°=+×+×=++=

°=°⋅

==

°=°⋅

======

°=°⋅

=======

oairfabrictotal

o

air

fabric

RRRRhA

R

kALRRRRR

kALRRRRRR

and

W113=°C/W 2214.0totalR°−

=− ∞ C)025(21s TT

Q&

If the jacket is made of a single layer of 0.75 mm thick synthetic fabric, the rate of heat transfer would be

=

W685=°+×+× C/W )0320.00009.05(5 ofabrictotal RRR

°−=

−=

−= ∞∞ C)025(2121 ss TTTT

Q&

he thickn s of a wool fabric that has the same thermal resistance is determined from

T es

mm 8.29==⎯→⎯+°⋅

=°

+=+=

m 00829.00320.0)m 25.1(C) W/m035.0(

C/W 2214.0

1

2

fabricwooltotal

LLhAkA

LRRR o


3-41

3-64 A coat is made of 5 layers of 0.15 mm thick cotton fabric separated by 1.5 mm thick air space. The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer.

Properties The thermal conductivities are given to be k = 0.06 W/m⋅°C for cotton fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric.

Analysis The thermal resistance network and the individual thermal resistances are

R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro

T1 T∞2

C/W 2268.00320.00462.04002.0545

C/W 0320.0)m 25.1(C) W/m25(

11

C/W 0462.0)m 25.1(C) W/m026.0(

m 0015.0

C/W 002.0)m 25.1(C) W/m06.0(

m 00015.0

22

2o8642

297531cot

°=+×+×=++=

°=°⋅

==

°=⋅

======

°=°⋅

=======

oairfabrictotal

o

air

ton

RRRRhA

R

kALRRRRR

kALRRRRRR

and

W110=°C/W 2268.0totalR°−

=− ∞ C)025(21s TT

Q&

If the jacket is made of a single layer of 0.75 mm thick cotton fabric, the rate of heat transfer will be

=

W595=°+×+× C/W )0320.0002.05(5 ofabrictotal RRR

°−=

−=

−= ∞∞ C)025(2121 ss TTTT

Q&

he thickn s of a wool fabric for that case can be determined from

T es

mm 8.52==⎯→⎯+°⋅

=°

+=+=

m 00852.00320.0)m 25.1(C) W/m035.0(

C/W 2268.0

1

2

fabricwooltotal

LLhAkA

LRRR o


3-42

3-65 In an experiment, the convection heat transfer coefficients of (a) air and (b) water flowing over the metal foil are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Thermal resistance of the thin metal foil is negligible.

Properties Thermal conductivity of the slab is given to be k = 0.023 W/m · K and the emissivity of the metal foil is 0.02.


kALR =cond

hAR 1

conv =

and Ah

Rrad

rad1

=

From energy balance and using the thermal resistance concept, the following equation is expressed:

cond

21elec

rad

1surr

conv

1

RTT

AqR

TTR

T−∞T −=+

−+ &

or 1

elecrad

1surr

cond

21

conv

11TT

AqR

TTR

TTR −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−

−=

∞

&

1

elec1surr21 1q

TTTTh ⎟

⎞⎜⎛

−−

−−

= & rad/1/ TThkL −⎟

⎠⎜⎝ ∞

) For air flowing over the metal foil, the radiation heat transfer coefficient is

))((

2

surr2

surr2

rad

⋅=

++= TTTTh ssεσ

(a

K )293423(K )293423)(K W/m1067.5)(02.0( 222428 ++⋅×= −

K W/m215.0

The convection heat transfer coefficient for air flowing over the metal foil is

K W/m37.3 2 ⋅=

−⎥⎦

⎤⎢⎣

⎡ −⋅

−−

⋅−

=K )15020(

1 W/m5000 W/m215.0/1

K )15020(K W/m023.0/m 025.0

K )20150( 22 K

h

(b) For water flowing over the metal foil, the radiation heat transfer coefficient is

))((

K W/m1201.0

K )293303(K )293303)(K W/m1067.5)(02.0(2 ⋅=

++⋅×=

The convection heat transfer coefficient for water flowing over the metal foil is

222428surr

2surr

2 ++=−

TTTTh ssεσ

rad

K W/m499 2 ⋅=

−⎥⎦

⎤⎢⎣

⎡ −⋅

−−

⋅−

=K )3020(

1 W/m5000 W/m1201.0/1

K )3020(K W/m023.0/m 025.0

K )2030( 22 K

h

Discussion If heat transfer by conduction through the slab and radiation on the metal foil surface is neglected, the convection heat transfer coefficient for the case with air flow would deviate by 3.2% from the result in part (a), while the convection heat transfer coefficient for the case with water flow would deviate by 0.2% from the result in part (b).


3-43

3-66 A kiln is made of 20 cm thick concrete walls and ceiling. The two ends of the kiln are made of thin sheet metal covered with 2-cm thick styrofoam. For specified indoor and outdoor temperatures, the rate of heat transfer from the kiln is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the walls and ceiling is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. 5 Heat loss through the floor is negligible. 6 Thermal resistance of sheet metal is negligible.

Properties The thermal conductivities are given to be k = 0.9 W/m⋅°C for concrete and k = 0.033 W/m⋅°C for styrofoam insulation.

Analysis In this problem there is a question of which surface area to use. We will use the outer surface area for outer convection resistance, the inner surface area for inner convection resistance, and the average area for the conduction resistance. Or we could use the inner or the outer surface areas in the calculation of all thermal resistances with little loss in accuracy. For top and the two side surfaces:

Ri Rconcrete Ro

Tin Tout

C/W 10256.510)769.0480.40071.0(

C/W 10769.0m)] m)(13 40(C)[ W/m25(

11

C/W 10480.4m] )6.0m)(13 40(C)[ W/m9.0(

m 2.0

C/W 100071.0m] )2.1m)(13 40(C)[ W/m3000(

11

44

42

4

42

°×=×++=++=

°×=°⋅

==

°×=−°⋅

==

°×=−°⋅

==

−−

−

−

−

oconcreteitotal

ooo

aveconcrete

iii

RRRR

AhR

kALR

AhR

and W83,700C/W 10256.5 4 °× −

totalRC)]4(40[

=°−−

=−

=+outin

sidesTT

Q&

eat loss through the end surface of the kiln with styrofoam:

top

H

Ri Rstyrofoam Ro

Tin Tout

C/W 0352.00020.00332.010201.0

C/W 0020.0]m 54C)[ W/m25(

11

C/W 0332.0]m )2.05)(2.04(C)[ W/m033.0(

m 02.0

C/W 10201.0]m )4.05)(4.04(C)[ W/m3000(

11

4

22

2

422

°=++×=++=

°=×°⋅

==

°=−−°⋅

==

°×=−−°⋅

==

−

−

ostyrpfoamitotal

ooo

avestyrofoam

iii

RRRR

AhR

kALR

AhR

and W1250C)]4(40[

=°−−

=−

= outinsurfaceend

TTQ&

C/W 0352.0 °totalR

Then the total rate of heat transfer from the kiln becomes

W86,200=×+=+= + 12502700,832 sidesidestoptotal QQQ &&&


3-44

3-67 Prob. 3-66 is reconsidered. The effects of the thickness of the wall and the convection heat transfer coefficient .


.9 [W/m-C]

/m-C] ]

^2-C]

l)*length

ut)/R_total_top_sides "Heat loss from top and the two side surfaces"

am*A_5) wall)

ss from one end surface" _dot_total=Q_dot_top_sides+2*Q_dot_end

on the outer surface of the rate of heat loss from the kiln are to be investigated

A

"GIVEN" width=5 [m] height=4 [m] length=40 [m] L_wall=0.2 [m] k_concrete=0T_in=40 [C] T_out=-4 [C] L_sheet=0.003 [m] L_styrofoam=0.02 [m] k_styrofoam=0.033 [Wh_i=3000 [W/m^2-Ch_o=25 [W/m "ANALYSIS" R_conv_i=1/(h_i*A_1) A_1=(2*height+width-6*L_wall)*length R_concrete=L_wall/(k_concrete*A_2) A_2=(2*height+width-3*L_walR_conv_o=1/(h_o*A_3) A_3=(2*height+width)*length R_total_top_sides=R_conv_i+R_concrete+R_conv_o Q_dot_top_sides=(T_in-T_o R_conv_i_end=1/(h_i*A_4) A_4=(height-2*L_wall)*(width-2*L_wall) R_styrofoam=L_styrofoam/(k_styrofoA_5=(height-L_wall)*(width-L_R_conv_o_end=1/(h_o*A_6) A_6=height*width R_total_end=R_conv_i_end+R_styrofoam+R_conv_o_end Q_dot_end=(T_in-T_out)/R_total_end "Heat loQ

0.08 0.12 0.16 0.2 0.24 0.28 0.32

60000

80000

100000

120000

140000

160000

Qto

tal

[W]

Lwall [m]

L wall[m]

Q total[W]

0.1 151098 0.12 131499 0.14 116335 0.16 104251 0.18 94395 0.2 86201 0.22 79281 0.24 73359 0.26 68233 0.28 63751 0.3 59800


3-45

ho [W/m2.C]

Qtotal [W]

5 54834 10 70939 15 78670 20 83212 25 86201 30 88318 35 89895 40 91116 45 92089 50 92882

5 10 15 20 25 30 35 40 45 5050000

55000

60000

65000

70000

75000

80000

85000

90000

95000

ho [W/m2-C]Q

tota

l [W

]


3-46

3-68E The thermal resistance of an epoxy glass laminate across its thickness is to be reduced by planting cylindrical copper fillings throughout. The thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the plate is one-dimensional. 3 Thermal conductivities are constant.

Properties The thermal conductivities are given to be k = 0.10 Btu/h⋅ft⋅°F for epoxy glass laminate and k = 223 Btu/h⋅ft⋅°F for copper fillings.

Analysis The thermal resistances of copper fillings and the epoxy board are in parallel. The number of copper fillings in the board and the area they comprise are

2

222

2

2

ft 7606.007272.08333.0

ft 07272.04

ft) 12/02.0(333,334

fillings)copper of(number 333,33ft) 12/06.0(ft) 12/06.0(

ft 8333.0

ft 8333.0ft) 12/12(ft) 12/10(

=−=−=

===

==

==

coppertotalepoxy

copper

copper

total

AAA

DnA

n

A

ππ

Rcopper

Repoxy

The thermal resistances are evaluated to be

F/Btuh 0548.0)ft 7606.0(F)ftBtu/h 10.0(

ft 12/05.0

F/Btuh 000257.0)ft 07272.0(F)ftBtu/h 223( 2°⋅⋅kAcopper

ft 12/05.0

2°⋅=

°⋅⋅==

°⋅===

kALR

LR

epoxy

hen the thermal resistance of the entire epoxy board becomes

T

F/Btuh 0.000256 °⋅=⎯→⎯+=+= boardepoxycopperboard

RRRR 0548.0

1000257.0

1111


3-47

Heat Conduction in Cylinders and Spheres

3-69C When the diameter of cylinder is very small compared to its length, it can be treated as an infinitely long cylinder. Cylindrical rods can also be treated as being infinitely long when dealing with heat transfer at locations far from the top or bottom surfaces. However, it is not proper to use this model when finding temperatures near the bottom and the top of the cylinder.

3-70C No. In steady-operation the temperature of a solid cylinder or sphere does not change in radial direction (unless there is heat generation).

3-71C Heat transfer in this short cylinder is one-dimensional since there will be no heat transfer in the axial and tangential directions.


3-48

its emissivity is ε = 1.

))((

2

2242

222

=

+++++

++= surrsurrd TTTTh εσ

The individual thermal resistances are

3-72 A spherical container filled with iced water is subjected to convection and radiation heat transfer at its outer surface. The rate of heat transfer and the amount of ice that melts per day are to be determined.

Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 Thermal conductivity is constant.

Properties The thermal conductivity of steel is given to be k = 15 W/m⋅°C. The heat of fusion of water at 1 atm is kJ/kg 7.333=ifh . The outer surface of the tank is black and thus

Analysis (a) The inner and the outer surface areas of sphere are

222

222

m 57.202m) 03.8(

m 06.201m) 8(

===

===

ππ

ππ

oo

ii

DA

DA

We assume the outer surface temperature T2 to be 5°C after comparing convection heat transfer coefficients at the inner and the outer surfaces of the tank. With this assumption, the radiation heat transfer coefficient can be determined from

ra

.K W/m424.5

)]K 5273(K) 25273]()K 25273()K 5273)[(K W/m1067.5(1 28 ⋅×= −

C/W 000387.0000320.0000005.0000062.0

C/W 000320.0000910.0000494.0,

⎯→⎯+=+=radoconveqv RRR

11111

C/W 000910.0)m 57.202(C) W/m424.5(

11

C/W 000494.0)m 57.202(C) W/m10(

11

C/W 000005.0)m 0.4)(m 015.4(C) W/m15(4

m )0.4015.4(4

)m 06.201(C) W/m80(11

1,

22

22,

21

121

22,

°=++=++=

°=

°=°⋅

==

°=°⋅

==

°=°⋅

−=

−==

°=°⋅

==

eqviconvtotal

eqv

radrad

ooconv

sphere

iiconv

RRR

R

AhR

AhR

rkrrr

RR

AhR

ππ

C/W 000062.0

R

Then the steady rate of heat transfer to the iced water becomes

W64,600=°°−

=−

= ∞∞

C/W 0.000387C)025(21

totalRTT

Q&

(b) The total amount of heat transfer during a 24-hour period and the amount of ice that will melt during this period are

kg 16,730=

×==

×=×=∆= .5s) 3600kJ/s)(24 600.64(tQQ &

kJ/kg 7.333ifh

Check: The outer surface temperature of the tank is

kJ 10581.5

kJ 105816

6

Qm

ice

C3.4)m C)(202.57 W/m5.424+(10

W600,64C25

)(

221

1

°=°⋅

−°=−=→

−=

+∞

∞+

oradconvs

soradconv

AhQTT

TTAhQ&

&

which is very close to the assumed temperature of 5°C for the outer surface temperature used in the evaluation of the radiation heat transfer coefficient. Therefore, there is no need to repeat the calculations.

T∞1

Ri T1R1

Rrad

Ro

T∞2


3-49

3-73 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivities are given to be k = 15 W/m⋅°C for steel and k = 0.038 W/m⋅°C for glass wool insulation

Analysis The inner and the outer surface areas of the insulated pipe per unit length are

2

2

m 361.0m) 1(m) 06.0055.0(

m 157.0m) 1(m) 05.0(

=+==

===

ππ

ππ

LDA

LDA

oo

ii

Ri

T∞1

R2 Ro

T∞2

R1 The individual thermal resistances are

C/W 296.31259.0089.300101.008.0

C/W 1259.0)m 361.0(C) W/m22(

11

C/W 089.3)m 1

(C) W/m038.0(22 2

2 °⋅=== insulation Lk

RRππ

)75.2/75.5ln()/ln(

C/W 00101.0)m 1(C) W/m15(2

)5.2/75.2ln(2

)/ln(

C/W 08.011

21

22

23

1

121

°=+++=+++=

°=°⋅

==

°=

°====

°===

oitotal

ooo

pipe

i

RRRRRAh

rrLkrrRR

R

ππ

Then the steady rate of heat loss from the steam per m. pipe length becomes

)m 157.0(C) W/m80( 22 °⋅ii Ah

°⋅

R

W83.4=°

°−=

−= ∞∞

C/W 3.296C)5280(21

totalRTTQ&

The temperature drops across the pipe and the insulation are

C257.6

C0.084

°=°==∆

°=°==∆

C/W) W)(3.0894.83(

C/W) 1 W)(0.00104.83(

insulationinsulation

pipepipe

RQT

RQT&

&


3-50

3-74 Prob. 3-73 is reconsidered. The effect of the thickness of the insulation on the rate of heat loss from the steam and the temperature drop across the insulation layer are to be investigated.


"GIVEN" T_infinity_1=280 [C] T_infinity_2=5 [C] k_steel=15 [W/m-C] D_i=0.05 [m] D_o=0.055 [m] r_1=D_i/2 r_2=D_o/2 t_ins=3 [cm] k_ins=0.038 [W/m-C] h_o=22 [W/m^2-C] h_i=80 [W/m^2-C] L=1 [m] "ANALYSIS" A_i=pi*D_i*L A_o=pi*(D_o+2*t_ins*Convert(cm, m))*L R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_steel*L) R_ins=ln(r_3/r_2)/(2*pi*k_ins*L) r_3=r_2+t_ins*Convert(cm, m) "t_ins is in cm" R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_ins+R_conv_o Q_dot=(T_infinity_1-T_infinity_2)/R_total DELTAT_pipe=Q_dot*R_pipe DELTAT_ins=Q_dot*R_ins

Tins [cm]

Q [W]

∆Tins [C]

1 2 3 4 5 6 7 8 9 10

174.9 109 83.44 69.64 60.93 54.88 50.41 46.95 44.18 41.91

227.2 249.6 257.8 261.9 264.4 266 267.2 268.1 268.7 269.2

1 2 3 4 5 6 7 8 9 1040

60

80

100

120

140

160

180

220

230

240

250

260

270

tins [cm]

Q [

W]

∆T i

ns [

C]


3-51

3-75 A 50-m long section of a steam pipe passes through an open space at 15°C. The rate of heat loss from the steam pipe, the annual cost of this heat loss, and the thickness of fiberglass insulation needed to save 90 percent of the heat lost are

sulation. 6 The combined heat transfer coefficient on the outer surface remains

on is given to be k = 0.035 W/m⋅°C.

lysis (a) The rate of heat loss from the steam

/m 22 °−°⋅

) The amount of heat loss per year is

he amount of gas consumption from the natural gas furnace that has an efficiency of 75% is

to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivity is constant. 4 The thermal contact resistance at the interface is negligible. 5 The pipe temperature remains constant at about 150°C with or without inconstant even after the pipe is insulated.

Properties The thermal conductivity of fiberglass insulati

Ana pipe is 2m 71.15m) 50(m) 1.0( === ππDLAo

W20()( =−= airsobare TTAhQ& W42,412=C)15150)(m 71.15(C)

(b

kJ/yr 10337.1s/yr) 360024kJ/s)(365 412.42( 9×=××=∆= tQQ &

T

therms/yr903,16kJ 105,500

therm1kJ/yr 10337.1 9

⎜⎜⎛×

=gasQ75.0

=⎟⎟⎠

⎞

⎝

lost is

therm)/52.0($) therms/yr(16,903=

(c) In order to save 90% of the heat loss and thus to reduce it to 0.1×42,412 = 4241 W, the thickness of insulation needed is

The annual cost of this energy

= energy) ofcost used)(UnitEnergy (costEnergy

$8790/yr=

determined from

kLrr

Ah

TTRR

TTQ

oo

airs

insulationo

airsinsulated

π2)/ln(1 12+

−=

+−

=& Ro

TairRinsulation

Ts

Substituting and solving for r2, we get

m 0692.0

)m 50(C) W/m035.0(2)05.0/ln(

)]m 50(2(C)[ W/m20(1

C)15150( W4241 22

22

=⎯→⎯

°⋅+

°⋅

°−= r

rr ππ

Then the thickness of insulation becomes

cm 1.92=−=−= 592.612 rrtinsulation


3-52

3-76 An electric hot water tank is made of two concentric cylindrical metal sheets with foam insulation in between. The fraction of the hot water cost that is due to the heat loss from the tank and the payback period of the do-it-yourself insulation kit are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal resistances of the water tank and the outer thin sheet metal shell are negligible. 5 Heat loss from the top and bottom surfaces is negligible. Properties The thermal conductivities are given to be k = 0.03 W/m⋅°C for foam insulation and k = 0.035 W/m⋅°C for fiber glass insulation Analysis We consider only the side surfaces of the water heater for simplicity, and disregard the top and bottom surfaces (it will make difference of about 10 percent). The individual thermal resistances are 2m 885.1m) 5.1(m) 40.0( === ππ LDA ii

C/W 0106.0)m 885.1(C). W/m50(

1122 °=

°==

iii Ah

R

Tw

Ro

T∞2

RfoamRi 2m 168.2m) 5.1(m) 46.0( == =o ππ LDA o

C/W 0384.0)m 168.2(C). W/m12(

1122 °=

°==

ooo Ah

R

C/W 5433.04943.00384.00106.0

C/W 4943.0)m 5.1(C) W/m03.0(22 2 °⋅

foam kL ππ)20/23ln()/ln( 12

°=++=++=

°===

foamoi RRRR

rrR

total

The rate of heat loss from the hot water tank is

W74.60C/W 0.5433

C)2760(2 =°°−

=−

= ∞

total

w

RTTQ&

The amount and cost of heat loss per year are 5h/yr) 24kW)(365 06074.0( =×=∆= tQQ &

kWh/yr 1.32

ofost

15.2%===

==

152.0280$

57.42$57.42$kWh)/08.0($kWh) 1.(532=cost)it energy)(Un ofAmount (Energy C

f

If 3 cm thick fiber glass insulation is used to wrap the entire tank, the individual resistances becomes 2m 450.2m) 5.1(m) 52.0( === ππ LDA oo

Tw

Rfiberglass Ro

T∞2

RfoamRi C/W0340.0)m 450.2(C) W/m12( 2o2 ⋅oo

o Ah11

°===R

C/W 9106.03717.04943.00340.00106.0

C/W 3717.0)m 5.1(C) W/m035.0(22 2

2°=

°⋅==fiberglass Lk

Rππ

)23/26

C/W 4943.0)m 5.1(C) W/m03.0(2

)20/23ln(2

)/ln(2

1

12

°=++=+++=

°=°⋅

=

fiberglassfoamoitotal RRRRR

Lkrr

ππ

The rate of heat loss from the hot water heater in this case is

=foamR

ln()/ln( 23 rr

+

W24.36C/W 0.9106

C)2760(2 =°°−

=−

= ∞

total

w

RTT

Q&

The energy saving is saving = 60.74 - 36.24 = 24.5 W The time necessary for this additional insulation to pay for its cost of $30 is then determined to be

months 21≈==⎯→⎯

==

days 638hours 306,15 period Time

30$kWh)/08.0$period)( kW)(Time 0245.0(Cost


3-53

3-77 Prob. 3-76 is reconsidered. The fraction of energy cost of hot water due to the heat loss from the tank as a


m]

=280 [$/year]

o

HeatLoss=Cost_HeatLoss/Cost_heating*Convert(, %)

function of the hot-water temperature is to be plotted.

A

"GIVEN" L=1.5 [m] D_i=0.40 [m] D_o=0.46 [r_1=D_i/2 r_2=D_o/2 T_w=60 [C] T_infinity_2=27 [C] h_i=50 [W/m^2-C] h_o=12 [W/m^2-C] k_ins=0.03 [W/m-C] Price_electric=0.08 [$/kWh]Cost_heating "ANALYSIS" A_i=pi*D_i*L A_o=pi*D_o*L R_conv_i=1/(h_i*A_i) R_ins=ln(r_2/r_1)/(2*pi*k_ins*L) R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_ins+R_conv_Q_dot=(T_w-T_infinity_2)/R_total Q=(Q_dot*Convert(W, kW))*time time=365*24 [h/year] Cost_HeatLoss=Q*Price_electric f_

T w[C]

fHeatLoss[%]

40 45 50 55 60 65 70 75 80 85 90

29.02

5.988 8.291 10.5912.9 15.2 17.5 19.81 22.11 24.41 26.72

40 50 60 70 80 905

10

15

20

25

30

f Hea

tLos

s [%

]

Tw [C]


3-54

3-78 Chilled water is flowing inside a pipe. The thickness of the insulation needed to reduce the temperature rise of water to one-fourth of the original value is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivity is given to be k = 0.05 W/m⋅°C for insulation.

Water

L

Insulation

r1r2

Analysis The rate of heat transfer without the insulation is

W4096C7)-C)(8J/kg kg/s)(4180 (0.98old =°°⋅=∆= TcmQ p&&

The total resistance in this case is

C/W0.005493C)5.730(

W4096 totaltotal

totalold R

°=⎯→⎯°−

=

−= ∞

RR

TTQ w&

R1

∞1

RinsT∞2

Ro

T

The convection resistance on the outer surface is

C/W 004421.0m) m)(200 04.0(C) W/m9(

11=oR 2 °=

°⋅=

πoo Ah

ction resistance on the inner surface and the resistance of the pipe and it is The rest of thermal resistances are due to convedetermined from

C/W 001072.0004421.0005493.0ototal1 °=−=−= RRR

The rate of heat transfer with the insulation is

new W1024C)C)(0.25J/kg kg/s)(4180 (0.98 =°°⋅=∆= TcmQ p&&

ce with the insulation is The total thermal resistan

C/W0.02234C)]2/)25.77(30[ W1024 newtotal,newtotal,newtotal,

new °=⎯→⎯°+−

=⎯→⎯−

= ∞ RRR

TTQ w&

It is expressed by

)m 200(C) W/m05.0(2)04.0/ln(

m) 200(C) W/m9(1

001072.0C/W02234.0 =°

2

2

22

ins

°⋅+

°⋅+

ππ

πD

D

LkAh oo

v ain

The following line in EES is used:

0.02234=0.001072+1/(9*pi*D2*200)+ln(D2/0.04)/(2*pi*0.05*200)

Then the required thickness of the insulation becomes

)/ln(1 121insnewo,1newtotal, ++=++=

DDRRRRR

Sol ing this equation by trial-error or by using an equation solver such as EES, we obt

m 1406.0=D 2

cm 5.03==−=−= m 0503.02/)04.01406.0(2/)( 12ins DDt


3-55

3-79 Steam flows in a steel pipe, which is insulated by gypsum plaster. The rate of heat transfer from the steam and the temperature on the outside surface of the insulation are be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.

Properties (a) The thermal conductivities of steel and gypsum plaster are given to be 50 and 0.5 W/m⋅°C, respectively.

Steam

L

Insulation


Ri

Ti

Rins

To

Rsteel Ro

C/W0004974.0m) m)(20 (0.16C) W/m200(

11

C/W011032.0m) C)(20 W/m5.0(22 ins

ins °⋅ππ Lk)8/16ln()/ln(

C/W0000458.0m) C)(20 W/m50(2

)6/8ln(2

)/ln(

C/W0003316.011

2

23

steel

12steel

°=°⋅

==

°===

°=°⋅

==

°===

π

ππ

ooo Ah

R

DDR

LkDD

R

he total thermal resistance and the rate of heat transfer are

Rm) m)(20 (0.06C) W/m800( 2 °⋅ πii

i Ah

T

C/W011907.00004974.0011032.00000458.00003316.0inssteeltotal °=+++=+++= oi RRRRR

W15,957=⋅

°−=

−=

C/Wm 0.011907C)10200(

2totalR

TTQ oi&

(b) The temperature at the outer surface of the insulation is determined from

C17.9°=⎯→⎯°⋅

°−=⎯→⎯

−= s

s

o

os TT

RTT

QC/Wm 0.0004974

C)10( W957,15

2&


3-56

ft 916.0ft) 1(ft) 12/5.3(

=

=== ππ LDA i

The individual resistances are

3-80E A steam pipe covered with 2-in thick fiberglass insulation is subjected to convection on its surfaces. The rate of heat loss from the steam per unit length and the error involved in neglecting the thermal resistance of the steel pipe in calculations are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivities are given to be k = 8.7 Btu/h⋅ft⋅°F for steel and k = 0.020 Btu/h⋅ft⋅°F for fiberglass insulation.

Analysis The inner and outer surface areas of the insulated pipe are

2

2

ft 094.2ft) 1(ft) Ri

T∞1

Rinsulation Ro

T∞2

Rpipei

12/8(== ππ LDA oo

F/Btuh 65.5096.0516.5002.0036.021 ⋅=+++=+++= oitotal RRRRR

F/Btuh 096.0)ft 094.2(F).Btu/h.ft 5(

11

F/Btuh 516.5)ft 1(F)Btu/h.ft. 020.0(2

)2/4ln(2

)/ln(

F/Btuh 002.0)ft 1(F)Btu/h.ft. 7.8(2

)75.1/2ln(2

)/ln(

F/Btuh 036.0)ft 916.0(F).Btu/h.ft 30(

11

2o2

2

232

1

121

22

°

°⋅===

°⋅=°

===

°⋅=°

===

°⋅=°

==

ooo

insulation

pipe

iii

AhR

Lkrr

RR

Lkrr

RR

AhR

ππ

ππ

hen the st ady rate of heat loss from the steam per ft. pipe length becomes T e

Btu/h 69.91=°−

=−

= ∞∞ F)55450(21 TTQ&

°⋅ F/Btuh 5.65totalR

alue of total thermal resistance will be If the thermal resistance of the steel pipe is neglected, the new v

F/Btuh 648.5096.0516.5036.02 °=++=++= oitotal RRRR

or involved in calculations becomes Then the percentage err

0.035%=×°

°−= 100

F/Btuh 65.5F/Btuh)648.565.5(%error

which is insignificant.


3-57

3-81 Hot water is flowing through a 15-m section of a cast iron pipe. The pipe is exposed to cold air and surfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal properties are constant.

Properties The thermal conductivity and emissivity of cast iron are given to be k = 52 W/m⋅°C and ε = 0.7.

Analysis The individual resistances are

2

2

m 168.2m) 15(m) 046.0(

m 885.1m) 15(m) 04.0(

===

===

ππ

ππ

LDA

LDA

oo

ii

C/W 00003.0)2/3.2ln()/ln(

)m 885.1(C). W/m120(

12

22

°===

°rr

R

Ah ii

)m 15(C) W/m.52(22

C/W 00442.011

1 °

°===

ππ Lk

R

pipe

i

The outer surface temperature of the pipe will be somewhat below the water temperature. Assuming the outer surface temperature of the pipe to be 80°C (we will check this assumption later), the radiation heat tran efficient is determined to be

222428 =+×= −

mperature, the radiation and convection heat transfer coefficients can be added and the result can be taken as the combined heat transfer coefficient. Then,

Ri Rpipe Ro

T∞1 T∞2

sfer co

.K W/m167.5283)+353]()K 283()K 353)[(.K W/m1067.5)(7.0(

))(( 222

2 ++= surrsurrrad TTTTh εσ

Since the surrounding medium and surfaces are at the same te

C/W 02732.002287.000003.000442.0

C/W 02287.0)m 168.2(C). W/m17.20(

1122 °=

°==o Ah

R

C. W/m.2015167.5 22,

°=++=++=

°=+=+=

opipeitotal

ocombined

convradcombined

RRRR

hhh

es

17

The rate of heat loss from the hot water pipe then becom

W2928=°°−

=−

= ∞∞

C/W 0.02732C)1090(21

totalRTTQ&

For a temperature drop of 3°C, the mass flow rate of water and the average velocity of water must be

m/s 0.186===⎯→⎯=

=°°

=∆

=⎯→⎯∆=

4m) 04.0()kg/m 1000(

kg/s 2335.0

kg/s 2335.0C) C)(3J/kg. (4180

J/s 2928

23 πρ

ρc

c

pp

AmVVAm

TcQmTcmQ

&&

&&&&

Discussion The outer surface temperature of the pipe is

C77.0= C/W0.00003)+(0.00442

C)90( W928 2 1 °→°

°−=→

+−

= ∞s

s

pipei

s TTRR

TTQ&

which is close to the value assumed for the surface temperature in the evaluation of the radiation resistance. Therefore, there is no need to repeat the calculations.


3-58

3-82 Hot water is flowing through a 15 m section of a copper pipe. The pipe is exposed to cold air and surfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant.

Properties The thermal conductivity and emissivity of copper are given to be k = 386 W/m⋅°C and ε = 0.7.


i2

2

m 168.2m) 15(m) 046.0(

m 885.1m) 15(m) 04.0(

===

===

ππ

ππ

LDA

LDA

oo

i

C/W 0000038.0)m 15(C) W/m.386(22

°=°

==ππkL

R pipe

The outer surface temperature of the pipe will be somewhat below the water temperature. Assuming the outer surface

)2/3.2ln()/ln(

C/W 00442.0)m 885.1(C). W/m120(

11

12

22°=

°==

rrAh

Rii

i

temperature of the pipe to be 80°C (we will check this assumption later), the radiation heat transfer coefficient is determined to be

Since the surrounding medium and surfaces are at the same temperature, the radiation and convection heat transfer oefficients can be added and the result can be taken as the combined heat transfer coefficient. Then,

Ri Rpipe Ro

T∞1 T∞2

.K W/m167.5283)+353]()K 283()K 353)[(.K W/m1067.5)(7.0(

))((222428

222

2

=+×=

++=−

surrsurrrad TTTTh εσ

c

C/W 02733.002287.00000038.000442.0

C/W 02287.0)m 168.2(C). W/m17.20( 2 °

==ocombined

o AhR 11

C. W/m17.2015167.5

2

22,

°=++=++=

°=

°=+=+=

opipeitotal

convradcombined

RRRR

hhh

The rate of heat loss from the hot water pipe then becomes

W2927=°°−

=−

= ∞∞

C/W 0.02733C)1090(21

totalRTTQ&

For a temperature drop of 3°C, the mass flow rate of water and the average velocity of water must be

m/s 0.186===⎯→⎯

=

=°

==⎯→⎯∆=

m) 04.0(kg/s 2334.0

kg/s 2334.0C)

J/s 2927

23 πρ

ρc

c

p

AmVVAm

QmTcmQ

&&

&&&&

°∆ C)(3J/kg. (4180p Tc

4)kg/m 1000(

Discussion The outer surface temperature of the pipe is

C77.1= C/W0.0000038)+(0.00442

C)90( W2927 1 °→

°°−

=→+−

= ∞s

s

pipei

s TT

RRTT

Q&

which is close to the value assumed for the surface temperature in the evaluation of the radiation resistance. Therefore, there is no need to repeat the calculations.


3-59

3-83E Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper tubes. For specified heat transfer coefficients, the length of the tube required to condense steam at a rate of 250 lbm/h is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal properties are constant. 4 Heat transfer coefficients are constant and uniform over the surfaces.

Properties The thermal conductivity of copper tube is given to be k = 223 Btu/h⋅ft⋅°F. The heat of vaporization of water at 100°F is given to be 1037 Btu/lbm.


2

2

ft 157.0ft) 1(ft) 12/6.0(

ft 105.0ft) 1(ft) 12/4.0(

===

===

ππ

ππ

LDA

LDA

oo

ii

F/Btuh 27505.000265.000029.027211.0

F/Btuh 00265.0)ft 157.0(F).Btu/h.ft 2400( 22

=

°=°

==

total

ooo

RRAh

R 11

F/Btuh 00029.0)ft 1(F)Btu/h.ft. 223(2

)2/3ln(2

)/ln(

F/Btuh 27211.0)

11

12

2

°=++=++

°=°

==

°===

opipei

pipe

i

RR

kLrrR

R

ππ

he heat tr sfer rate per ft length of the tube is

RiRpipe Ro

T∞1 T∞2

ft 105.0(F).Btu/h.ft 35( 2 °ii Ah

T an

Btu/h 109.1F/Btu 0.27505

F)70100(21 =°

°−=

−= ∞∞

totalRTTQ&

The total rate of heat transfer required to condense steam at a rate of 250 lbm/h and the length of the tube required is determined to be

ft 2380≅===

===

ft 2376 1.109

250,259length Tube

Btu/h 250,259Btu/lbm) 7lbm/h)(103 250(

QQ

hmQ

total

fgtotal

&

&

&&


3-60

3-84E Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper tubes. For specified heat transfer coefficients and 0.01-in thick scale build up on the inner surface, the length of the tube required to condense steam at a rate of 120 lbm/h is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 Heat transfer coefficients are constant and uniform over the surfaces.

Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper tube and be k = 0.5 Btu/h⋅ft⋅°F for the mineral deposit. The heat of vaporization of water at 100°F is given to be 1037 Btu/lbm.

Analysis When a 0.01-in thick layer of deposit forms on the inner surface of the pipe, the inner diameter of the pipe will reduce from 0.4 in to 0.38 in. The individual thermal resistances are

2

2

ft 157.0ft) 1(ft) 12/6.0(

ft 099.0ft) 1(ft) 12/38.0(

===

===

ππ

ππ

LDA

LDA

oo

ii Ri

T∞1

Rpipr Ro

T∞2

Rdeposit

F/Btuh 3095.000425.001633.000029.02886.0

F/Btuh 00425.0)ft 157.0(F).Btu/h.ft 1500(

11

F/Btuh. 01633.0)ft 1(F)Btu/h.ft. 5.0(22 2

1 °=°

== depdeposit Lk

Rππ

)19.0/2.0ln()/ln(

F/Btuh 00029.0f 1(F)Btu/h.ft. 223(2

)2/3ln(2

)/ln(

F/Btuh 2886.0)ft 9

11

22

12

2

°=+++=+++=

°=°

==

°=°

==

°===

odepositpipeitotal

ooo

pipe

RRRRRAh

R

rrkL

rrR

R

ππ

The heat transfer rate per ft length of the tube is

09.0(F).Btu/h.ft 35( 2 °iii Ah

)t

Btu/h 9.96F/Btu 0.3095

F)70100(21 =°

°−=

−= ∞∞

totalRTT

Q&

The total rate of heat transfer required to condense steam at a rate of 120 lbm/h and the length of the tube required can be determined to be

ft 1284===

===

9.96 440,124length Tube

Btu/h 440,124Btu/lbm) 7lbm/h)(103 120(

QQ

hmQ

total

fgtotal

&

&

&&


3-61

3-85E Prob. 3-83E is reconsidered. The effects of the thermal conductivity of the pipe material and the outer diameter of the pipe on the length of the tube required are to be investigated.


"GIVEN" T_infinity_1=100 [F] T_infinity_2=70 [F] k_pipe=223 [Btu/h-ft-F] D_i=0.4 [in] D_o=0.6 [in] r_1=D_i/2 r_2=D_o/2 h_fg=1037 [Btu/lbm] h_o=1500 [Btu/h-ft^2-F] h_i=35 [Btu/h-ft^2-F] m_dot=120 [lbm/h] "ANALYSIS" L=1 [ft] “for 1 ft length of the tube" A_i=pi*(D_i*Convert(in, ft))*L A_o=pi*(D_o*Convert(in, ft))*L R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_pipe*L) R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_conv_o Q_dot=(T_infinity_1-T_infinity_2)/R_total Q_dot_total=m_dot*h_fg L_tube=Q_dot_total/Q_dot

kpipe [Btu/h.ft.F]

Ltube [ft]

10 1176 30.53 1158 51.05 1155 71.58 1153 92.11 1152 112.6 1152 133.2 1151 153.7 1151 174.2 1151 194.7 1151 215.3 1151 235.8 1150 256.3 1150 276.8 1150 297.4 1150 317.9 1150 338.4 1150 358.9 1150 379.5 1150 400 1150

0 50 100 150 200 250 300 350 4001145

1150

1155

1160

1165

1170

1175

1180

kpipe [Btu/h-ft-F]

L tub

e [f

t]


3-62

Do [in]

Ltube [ft]

0.5 1154 0.525 1153 0.55 1152 0.575 1151 0.6 1151 0.625 1150 0.65 1149 0.675 1149 0.7 1148 0.725 1148 0.75 1148 0.775 1147 0.8 1147 0.825 1147 0.85 1146 0.875 1146 0.9 1146 0.925 1146 0.95 1145 0.975 1145 1 1145

0.5 0.6 0.7 0.8 0.9 11145.0

1147.5

1150.0

1152.5

1155.0

D o [in]L t

ube

[ft]


3-63

m 27.28=A

3-86 A spherical tank filled with liquid nitrogen at 1 atm and -196°C is exposed to convection and radiation with the surrounding air and surfaces. The rate of evaporation of liquid nitrogen in the tank as a result of the heat gain from the surroundings for the cases of no insulation, 5-cm thick fiberglass insulation, and 2-cm thick superinsulation are to be determined.

Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 The combined heat transfer coefficient is constant and uniform over the entire surface. 4 The temperature of the thin-shelled spherical tank is said to be nearly equal to the temperature of the nitrogen inside, and thus thermal resistance of the tank and the internal convection resistance are negligible.

Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg/m3, respectively. The thermal conductivities are given to be k = 0.035 W/m⋅°C for fiberglass insulation and k = 0.00005 W/m⋅°C for super insulation.

Analysis (a) The heat transfer rate and the rate of evaporation of the liquid without insulation are

2 m) 3(== ππD 22

C/W 00101.0)m 27.28(C). W/m35(

1122

°=°

==Ah

Ro

o

kg/s 1.055===⎯→⎯=

=°−−

=−

= ∞ W910,208C)]196(15[21s TT

Q&°

kJ/kg 198kJ/s 910.208

C/W 0.00101

fgfg

o

hQ

mhmQ

R&

&&&

(b) The heat transfer rate and the rate of evaporation of the liquid with a 5-cm thick layer of fiberglass insulation are

ππDA

Ro Ts1 T∞2

222 m 19.30m) 1.3( == =Ro T∞2

Rinsulation

C/W 000946.01122

°===RoTs1

)m 19.30(C). W/m35( °Aho

C/W 0498.00489.0000946.0

C/W 0489.0)m 5.1)(m 55.1(C) W/m.035.0(4

m )5.155.1(4 21

12

°=+=+=

°=°−

=−

=

insulationototal

insulation

RRRrkrrr

Rππ

kg/s 0.0214===⎯→⎯=

=°

°−−=

−= ∞

kJ/kg 198kJ/s 233.4

W4233C/W 0.0498

C)]196(15[21

fgfg

total

s

hQ

mhmQ

RTT

Q

&&&&

&

(c) The heat transfer rate and the rate of evaporation of the liquid with 2-cm thick layer of superinsulation is 222 m 03.29m) 04.3( === ππDA

C/W 000984.0)m 03.29(C). W/m35(

1122

°=°

==Ah

Ro

o

C/W 96.1396.13000984.0

C/W 96.13)m 5.1)(m 52.1(C) W/m.00005.0(4

m )5.152.1(4 21

12

°=+=+=

°=°−

=−

=

insulationototal

insulation

RRRrkrrr

Rππ

Ro T∞2

RinsulationTs1

kg/s 0.000076===⎯→⎯=

=°

°−−=

−= ∞

kJ/kg 198kJ/s 01511.0

W11.15C/W 13.96

C)]196(15[21

fgfg

total

s

hQ

mhmQ

RTT

Q

&&&&

&


3-64

3-87 A spherical tank filled with liquid oxygen at 1 atm and -183°C is exposed to convection and radiation with the surrounding air and surfaces. The rate of evaporation of liquid oxygen in the tank as a result of the heat gain from the surroundings for the cases of no insulation, 5-cm thick fiberglass insulation, and 2-cm thick superinsulation are to be determined.

Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 The combined heat transfer coefficient is constant and uniform over the entire surface. 4 The temperature of the thin-shelled spherical tank is said to be nearly equal to the temperature of the oxygen inside, and thus thermal resistance of the tank and the internal convection resistance are negligible.

Properties The heat of vaporization and density of liquid oxygen at 1 atm are given to be 213 kJ/kg and 1140 kg/m3, respectively. The thermal conductivities are given to be k = 0.035 W/m⋅°C for fiberglass insulation and k = 0.00005 W/m⋅°C for super insulation.

Analysis (a) The heat transfer rate and the rate of evaporation of the liquid without insulation are

222 m 27.28m) 3( === ππDA

C/W 00101.0)m 27.28(C). W/m35(

1122

°=°

==Ah

Ro

o

kg/s 0.920===⎯→⎯=

kJ/kg 213kJ/s 040.196

fg hQ

mhmQ&

&&&

=°

°−−=

−= ∞ W040,196

C/W 0.00101C)]183(15[21

fg

o

s

RTT

Q&

(b) The heat transfer rate and the rate of evaporation of the liquid with a 5-cm thick layer of fiberglass insulation are

T∞2

Ro Ts1

222 m 19.30m) 1.3( === ππDA

C/W 000946.0)m 19.30(C). W/m35(

1122

°=°

==Ah

Ro

o

C/W 0498.00489.0000946.0

C/W 0489.0)m 5.1)(m 55.1(C) W/m.035.0(4

m )5.155.1(4 21

12

°=+=+=

°=°−

=−

=

Ro T∞2

RinsulationTs1

insulationototal

insulation

RRRrkrrr

Rππ

kg/s 0.0187===⎯→⎯=

=°

°−−== ∞21s

RTT −

kJ/kg 213kJ/s 976.3

W3976C/W 0.0498

C)]183(15[

fgfg

total

hQ

mhmQ

Q

&&&&

&

(c) The heat transfer rate and the rate of evaporation of the liquid with a 2-cm superinsulation is

222 m 03.29m) 04.3( === ππDA

C/W 000984.0)m 03.29(C). W/m35(

1122

°=°

==Ah

Ro

o

C/W 96.1396.13000984.0

C/W 96.13)m 5.1)(m 52.1(C) W/m.00005.0(4

m )5.152.1(4 21

12

°=+=+=

°=°−

=−

=

insulationototal

insulation

RRRrkrrr

Rππ

Ro T∞2

RinsulationTs1

kg/s 0.000067===⎯→⎯=

=°

°−−=

−= ∞

kJ/kg 213kJ/s 01418.0

W18.14C/W 13.96

C)]183(15[21

fgfg

total

s

hQ

mhmQ

RTT

Q

&&&&

&


3-65

3-88 An electric wire is tightly wrapped with a 1-mm thick plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient accounts for the radiation effects, if any.

Properties The thermal conductivity of plastic cover is given to be k = 0.15 W/m⋅°C.

Analysis In steady operation, the rate of heat transfer from the wire is equal to the heat generated within the wire,

W104)A 13)(V 8( ==== IWQ e V&& Rconv

T∞2

Rplastic

T1The total thermal resistance is

C/W 2746.00490.02256.0

C/W 0490.0)m 14(C) W/m.15.0(2

)1.1/1.2ln(2

)/ln(

C/W 2256.0m)] m)(14 (0.0042C)[. W/m24(

11

plasticconvtotal

12plastic

2conv

°=+=+=

°=°

==

°=°

==

RRRkL

rrR

AhR

oo

ππ

π

Then the interface temperature becomes

C58.6°=°+°=+=⎯→⎯−

= ∞ total1totalR

∞ )C/W 2746.0)( W104(C3021 RQTTTTQ &&

The critical radius of plastic insulation is

mm 25.6m 00625.0C. W/m24 2 ==°

==h

rcr

Doubling the thickness of the plastic cover will increase the outer radius of the wire to 3 mm, which is less than the criticalradius of insulation. Therefore, doubling

C W/m.15.0 °k

the thickness of plastic cover will increase the rate of heat loss and decrease the

interface temperature.


3-66

3-89 To avoid condensation on the outer surface, the necessary thickness of the insulation around a copper pipe that carries liquid oxygen is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Thermal contact resistance is negligible. Properties The thermal conductivities of the copper pipe and the insulation are given to be 400 W/m · °C and 0.05 W/m · °C, respectively.

Analysis From energy balance and using the thermal resistance concept, the following equation is expressed:

combined

,

convccond,icond,combined

,,

RTT

RRRRTT soio −

=+++

− ∞∞∞

AhhALkDD

LkDD

Ah1223 11

2)/ln(

2)/ln(1

+++ππ

TTTT soio ,,, −=

− ∞∞∞

ci combinedcombined

LDh

TT

L

DπhLkLkLDh

TT

ci 3combined1

12

3combined 22 ππππ

−

+++

−DDDD

soio ,

23

,,

11)/ln()/ln(1= ∞∞∞

Rearranging yields

⎥⎦

⎤⎢⎣

⎡+=

− ∞∞3combined

,, ln(1 Dh

TT io ++−∞ 1

1223

,

12

)/ln(2

)/hDk

DDk

DDTT ciso

⎥⎥⎦

⎤

°⋅+

°⋅+

⎢⎣

⎡°⋅

°⋅+=°−°+

)m 020.0)(C W/m120(1

)C W/m400(2)m 020.0m/ 025.0ln(

)C W/m05.0(2)m 025.0/ln(

)C W/m20(1C )1020(C )20020(

2

33

2 DD

Copy the following line and paste on a blank EES screen to solve the above equation: (20+200)/(20-10)=1+20*D_3*(ln(D_3/25e-3)/(2*0.05)+ln(25/20)/(2*400)+1/(120*20e-3))

Solving by EES software, the outer diameter of the insulation is

The thickness of the insulation necessary to avoid condensation on the outer surface is

m 0839.03 =D

m 0.0295=−

=−

>2

m 025.0m 0839.02

23 DDt

Discussion If the insulation thickness is less than 29.5 mm, the outer surface temperature would decrease to the dew point at 10 °C where condensation would occur.


3-67

n ent.

Critical Radius of Insulation

3-90C In a cylindrical pipe or a spherical shell, the additional insulation increases the conduction resistance of insulation, but decreases the convection resistance of the surface because of the increase in the outer surface area. Due to these opposite effects, a critical radius of insulation is defined as the outer radius that provides maximum rate of heat transfer. For a cylindrical layer, it is defined as hkr /= where k is the thermal conductivity of insulation and h is the external convectioheat transfer coeffici

rc

3-91C For a cylindrical pipe, the critical radius of insulation is defined as . On windy days, the external convection heat transfer coefficient is greater compared to calm days. Therefore critical radius of insulation will be greater on calm days.

hkrcr /=

3-92C Yes, the measurements can be right. If the radius of insulation is less than critical radius of insulation of the pipe, the rate of heat loss will increase.

3-93C No.

3-94C It will decrease.

3-95E An electrical wire is covered with 0.02-in thick plastic insulation. It is to be determined if the plastic insulation on the wire will increase or decrease heat transfer from the wire.

Assumptions 1 Heat transfer from the wire is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivity of plastic cover is given to be k = 0.075 Btu/h⋅ft⋅°F.

Analysis The critical radius of plastic insulation is

in) 0615.0(in 36.0ft 03.0F.Btu/h.ft 5.2

22 °hcr

Since the outer radius of the

FBtu/h.ft. 075.0=>==

°== rkr

wire with insulation is smaller than critical radius of insulation, plastic insulation will increase eat transfer from the wire.

Wire Insulation

h


3-68

3-96E An electrical wire is covered with 0.02-in thick plastic insulation. By considering the effect of thermal contact resistance, it is to be determined if the plastic insulation on the wire will increase or decrease heat transfer from the wire.

Assumptions 1 Heat transfer from the wire is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant

Properties The thermal conductivity of plastic cover is given to be k = 0.075 Btu/h⋅ft⋅°F. Wire

Insulation Rplastic Rinterface Rconv

Ts T∞

Analysis Without insulation, the total thermal resistance is (per ft length of the wire)

F/Btuh. 4.18ft)] ft)(1 (0.083/12F)[.Btu/h.ft 5.2(

112convtot °=°

===πoo Ah

RR

With insulation, the total thermal resistance is

F/Btuh. 30.13046.0835.042.12

F/Btuh. 046.0ft)] ft)(1 (0.083/12[

F/Btu.h.ft 001.0

F/Bh. 835.0)083.0/123.0ln()/ln(ft)] ft)(1 (0.123/12F)[.Btu/h.ft 5.2(

12plastic

2

°===

°rrR

Ah oo π

tu)ft 1(F)Btu/h.ft. 075.0(22

F/Btuh. 42.1211

interfaceplasticconvtotal

2

interface

conv

°=++=++=

°=°

==

°

°===

RRRRAh

R

kL

c

c

π

ππ

o be determined if the plastic insulation on the ball will increase or decrease heat transfer from it.

1 Heat transfer from the ball is steady since there is no indication of any change with time. 2 Heat transfer is 3 Thermal properties are constant. 4 The thermal

conductivity of plastic cover is given to be k = 0.13 W/m⋅°C.

nalysis The critical radius of plastic insulation for the spherical ball is

R

Since the total thermal resistance decreases after insulation, plastic insulation will increase heat transfer from the wire. The thermal contact resistance appears to have negligible effect in this case.

3-97 A spherical ball is covered with 1-mm thick plastic insulation. It is t

Assumptionsone-dimensional since there is thermal symmetry about the midpoint. contact resistance at the interface is negligible.

Properties The thermal Insulation

A

mm) 3(mm 13m 013.0C. W/m20C) W/m.13.0(22

22=>==

°

°== r

hkrcr

Since the outer radius of the ball with insulation is smaller than critical radius of insulation, plastic insulation will increase heat transfer from the wire.


3-69

3-98 Prob. 3-97 is reconsidered. The rate of heat transfer from the ball as a function of the plastic insulation thickness


C]

^2-C]

rt(mm, m)

r_1)/(4*pi*r_1*r_2*k_ins)

_dot=(T_ball-T_infinity)/R_total

]

is to be plotted.

A

"GIVEN" D_1=0.004 [m] t_ins=1 [mm] k_ins=0.13 [W/m-T_ball=50 [C] T_infinity=15 [C] h_o=20 [W/m "ANALYSIS" D_2=D_1+2*t_ins*ConveA_o=pi*D_2^2 R_conv_o=1/(h_o*A_o) R_ins=(r_2-r_1=D_1/2 r_2=D_2/2 R_total=R_conv_o+R_ins Q

t ins[mm

Q [W]

0.5 1.526 2.553 3.579 4.605 5.632 6.658 7.684 8.711 9.737 10.76 11.79 12.82 13.84 14.87 15.89 16.92 17.95 18.97 20

0.1222

0.05016 0.07736 0.096260.108 0.1149 0.119 0.1213 0.1227 0.1234 0.1238 0.1239 0.1238 0.1237 0.1236 0.1233 0.1231 0.1229 0.1226 0.1224

0 4 8 12 16 200.05

0.06

0.07

0.08

0.09

0.1

0.11

0.12

0.13

Q [

W]

tins [mm]


3-70

Heat Transfer from Finned Surfaces

3-99C Fins should be attached to the outside since the heat transfer coefficient inside the tube will be higher due to forced convection. Fins should be added to both sides of the tubes when the convection coefficients at the inner and outer surfaces are comparable in magnitude.

3-100C Increasing the rate of heat transfer from a surface by increasing the heat transfer surface area.

3-101C The fin efficiency is defined as the ratio of actual heat transfer rate from the fin to the ideal heat transfer rate from the fin if the entire fin were at base temperature, and its value is between 0 and 1. Fin effectiveness is defined as the ratio of heat transfer rate from a finned surface to the heat transfer rate from the same surface if there were no fins, and its value is expected to be greater than 1.

3-102C Heat transfer rate will decrease since a fin effectiveness smaller than 1 indicates that the fin acts as insulation.

3-103C Fins enhance heat transfer from a surface by increasing heat transfer surface area for convection heat transfer. However, adding too many fins on a surface can suffocate the fluid and retard convection, and thus it may cause the overall heat transfer coefficient and heat transfer to decrease.

3-104C Effectiveness of a single fin is the ratio of the heat transfer rate from the entire exposed surface of the fin to the heat transfer rate from the fin base area. The overall effectiveness of a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same surface if there were no fins.

3-105C Fins should be attached on the air side since the convection heat transfer coefficient is lower on the air side than it is on the water side.

3-106C Welding or tight fitting introduces thermal contact resistance at the interface, and thus retards heat transfer. Therefore, the fins formed by casting or extrusion will provide greater enhancement in heat transfer.

3-107C If the fin is too long, the temperature of the fin tip will approach the surrounding temperature and we can neglect heat transfer from the fin tip. Also, if the surface area of the fin tip is very small compared to the total surface area of the fin, heat transfer from the tip can again be neglected.

3-108C Increasing the length of a fin decreases its efficiency but increases its effectiveness.


3-71

3-109C Increasing the diameter of a fin increases its efficiency but decreases its effectiveness.

3-110C The thicker fin has higher efficiency; the thinner one has higher effectiveness.

3-111C The fin with the lower heat transfer coefficient has the higher efficiency and the higher effectiveness.

3-112 A relation is to be obtained for the fin efficiency for a fin of constant cross-sectional area , perimeter p, length L, and thermal conductivity k exposed to convection to a medium at ∞ with a heat transfer coefficient h. The relation is to be simplified for circular fin of diameter D and for a rectangular fin of thickness t.

cAT

Assumptions 1 The fins are sufficiently long so that the temperature of the fin at the tip is nearly T . 2 Heat transfer from the fin tips is negligible.

∞

Analysis Taking the temperature of the fin at the base to be T and using the heat transfer relation for a long fin, fin efficiency for long fins can be expressed as

b

phkA

LhpLhpkA

TThATThpkA cc

bfin

bc 1)(

)(

re temperatubaseat fin were entire theiffin thefrom ratefer heat trans Ideal

fin thefrom ratefer heat trans Actualfin

==−

−=

=

∞

∞

η

h, T∞

D p= πD Ac = πD2/4

Tb

This relation can be simplified for a circular fin of diameter D and rectangular fin of thickness t and width w to be

hkt

Lwhwtk

Lhtwwtk

LphkA

L

hkD

LhDDk

LphkA

L

c

c

21

2)(1

)(2)(11

21

)()4/(11

rrectangulafin,

2

circularfin,

=≅+

==

===

η

ππ

η

3-113 The maximum power rating of a transistor whose case temperature is not to exceed 80 is to be determined.

s isothermal at 80

operties bient thermal resistance is given to be

Analysis The maximum power at which this transistor can be operated safely is

°C

Assumptions 1 Steady operating conditions exist. 2 The transistor case i °C.

Pr The case-to-am20

T∞

RTs

°C / W.

W1.8=°

°−=

−=

∆=

−

∞

− C/W 25C )3580(

ambientcase

case

ambientcase RTT

RTQ&


3-72

3-114 A fin is attached to a surface. The percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 The heat transfer coefficient is constant and uniform over the entire fin surface. 4 The thermal properties of the fins are constant. 5 The heat transfer coefficient accounts for the effect of radiation from the fins.

Properties The thermal conductivity of the aluminum fin is given to be k = 237 W/m⋅°C.

Analysis The expressions for the heat transfer from a fin under infinitely long fin and adiabatic fin tip assumptions are

)tanh()(

)(

tipins.

fin long

mLTThpkAQ

TThpkAQ

bc

bc

∞

∞

−=

−=

&

&

L = 10 cm D = 4 mm

The percent error in using long fin assumption can be expressed as

1)tanh(

1)tanh()(

)tanh()()(Error %

tipins.

tipins.fin long −=−

−−−=

−=

∞

∞∞

mLmLTThpkA

mLTThpkATThpkA

Q

QQ

bc

bcbc&

&&

where

1-2

2m 116.7

4/m) 004.0()C W/m.237(m) 004.0(C). W/m12(

=°

°==m

ππ

ckAhp

ubstitutin S g,

[ ] 63.5%==−=−= 635.01m) 10.0)(m 116.7(tanh

11 1)tanh(

Error %1-mL

his result shows that using infinitely long fin assumption may yield results grossly in error.

fins varies in one direction only (normal to the ce. 4 The thermal properties of the fins he fins.

Properties The thermal conductivity of the fin is given to be k = 200 W/m⋅°C.

Analysis The fin temperature at a distance of 5 cm from the base is determined from

T

3-115 A very long fin is attached to a flat surface. The fin temperature at a certain distance from the base and the rate of heat loss from the entire fin are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the plate). 3 The heat transfer coefficient is constant and uniform over the entire fin surfaare constant. 5 The heat transfer coefficient accounts for the effect of radiation from t

C29.8°=⎯→⎯=−−− −−∞ TTT mx

c

14(20⎯→⎯

=×°

×+×°==

TeT

kAhp

m

)05.0)(3.

1-2

2

2040

m 3.14)m001.005.0)(C W/m.200(

0.001)m20.05C)(2. W/m20(

The rate of heat loss from this very long fin is

20°C

40°C

=− ∞

eTb

W2.9=

−××+×=

−= ∞

)2040()001.005.0(200(0.001)20.05)(220(

)(fin long TThpkAQ bc&


3-73

3-116 A DC motor draws electrical power and delivers mechanical power to rotate a stainless steel shaft. The surface temperature of the motor housing is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 The surface temperature of the motor housing is uniform. 5 The base temperature of the shaft is equal to the surface temperature of the motor housing. Properties The thermal conductivity of the stainless steel shaft is given as 15.1 W/m · °C.

Analysis From energy balance, the following equation is expressed: W&= melec sh QQW &&& ++ech or

&

sh QQWW &&&& ++= elecelec 55.0 The heat transfer rate from the motor housing surface is h )( ∞−= TThAQ hs

The motor shaft can be treated as a circular fin with a specified fin tip temperature. The heat transfer rate from the motor shaft can be written as

mLTTTTmL

TThkD

mLTTTTmL

TThpkAQ

hLh

hLhcs

sinh)/()(cosh

)(4

sinh)/()(cosh

)(

23 ∞∞

∞

∞∞∞

−−−−=

−−−−=

π

&

where

069.4)m 25.0()m 025.0)(C W/m1.15(

)C W/m25(445.025.05.0

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

°⋅°⋅

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= L

kDhL

kAhpmL

c

C W/1206.04

)m 025.0)(C W/m1.15)(C W/m25( 3232

°⋅°⋅=ππ Dhk

4

2°=

Substituting the listed terms into the energy balance equation we get

mLTTTTmL

TTDhkTThAW hLhhs sinh

)/()(cosh)(

4)(45.0 3

2

elec∞∞

∞∞−−−

−+−=π&

Rearranging the equation, the surface temperature of the motor housing is

C 87.7 °=°+

+°= 069.4sinh)C W/1206.0() W300(45.0

C 202 ⎟

⎠⎞

⎜⎝⎛°+°⋅

°−

⎟⎠⎞

⎜⎝⎛+

−+

+=

∞

∞

069.4sinh069.4cosh)C W/1206.0()m 075.0)(C W/m25(

C )2022(sinhcosh

4

sinh)(

445.0

2

32

32

elec

mLmLDhkhA

mLTT

DhkWTT

s

L

hπ

π&

Discussion If the surface of the motor housing has a high emissivity, heat transfer by radiation from the motor housing would decrease the surface temperature.


3-74

3-117 Two cast iron steam pipes are connected to each other through two 1-cm thick flanges exposed to cold ambient air. The average outer surface temperature of the pipe, the fin efficiency, the rate of heat transfer from the flanges, and the equivalent pipe length of the flange for heat transfer are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the flanges (fins) varies in one direction only (normal to the pipe). 3 The heat transfer coefficient is constant and uniform over the entire fin surface. 4 The thermal properties of the fins are constant. 5 The heat transfer coefficient accounts for the effect of radiation from the fins.

Properties The thermal conductivity of the cast iron is given to be k = 52 W/m⋅°C.

Analysis (a) We treat the flanges as fins. The individual thermal resistances are

i2

2

m 513.2m) 8(m) 1.0(

m 312.2m) 8(m) 092.0(

===

===

ππ

ππ

LDA

LDA

oo

i Ri Rcond Ro

T∞1 T∞2

T1 T2

C/W 01592.0)m 513.2(C). W/m25(

11

C/W 00003.0)m 8(C) W/m.52(2

)6.4/5ln(2

)/ln(

C/W 00240.0)m 312.2(C). W/m180(

11

22o

12cond

22i

°=°

==

°=°

==

°=°

==

oo

ii

AhR

kLrrR

AhR

ππ

C/W 01835.001592.000003.000240.0oconditotal °=++=++= RRRR

The rate of heat transfer and average outer surface temperature of the pipe are

C175.1°=°+°=+=⎯→⎯= ∞ ,10(C 12o22

oRQTT

RQ −

=°°−

=−

=

∞

∞∞

C/W) 01592.0)( W245

W245,10C 0.01835C)12200(

22

total

21

TTR

TTQ

&&

&

fin fficiency can be determined from (Fig. 3-44) (b) The e

0.88=

⎪⎪⎭

=⎟⎠

⎜⎝

+=⎟⎠

⎜⎝

+=⎟⎟⎠

⎜⎜⎝

=o

2/3 29.0m) C)(0.02 W/m52(

m 2

m 05.02

ξkt

LkA

Lp

c

⎪

⎪⎪⎪

⎬

⎫

⎞⎛⎞⎛⎞⎛

++

fino22/1

1

2

C W/m2502.0

02.009.0

ηhth

tr

22

21

22fin m 0465.0m) m)(0.02 09.0(2]m 05.0(m) 09.0[(22)(2 =+−=+−= ππππ trrrA

the flanges is

(c) An 8-m long section of the steam pipe is losing heat at a rate of 10,245 W or 10,245/8 = 1280 W per m length. Then for heat transfer purposes the flange section is equivalent to

== 0.205.0

22r

2) 2

The heat transfer rate from

W167=°−°=

−== ∞

C)121.175)(m 0465.0)(C. W/m25(88.0

)(22

finfinmaxfin,finfinned TThAQQ bηη &&

cm 13.0= m 130.0 W/m1280 W167length Equivalent ==


3-75

3-118 A commercially available heat sink is to be selected to keep the case temperature of a transistor below 90°C in an environment at 20°C.

Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90°C. 3 The contact resistance between the transistor and the heat sink is negligible.

Analysis The thermal resistance between the transistor attached to the sink and the ambient air is determined to be

C/W 1.75 °=°−

=−

=⎯→⎯∆

= ∞−

C)2090(transistorambientcase

TTRTQ&

le 3-6 reveals that HS6071 in vertical position, HS5030 nd HS6115 in both horizontal and vertical position can be selected.

available heat sink is to be selected to keep the case temperature of a transistor below 55°C in an

hermal transistor and the heat sink is

between the transistor attached to the sink and the ambient air is determined to be

T∞

RTs

− W40ambientcase QR &

The thermal resistance of the heat sink must be below 1.75°C/W. Taba

3-119 A commerciallyenvironment at 18°C.

Assumptions 1 Steady operating conditions exist. 2 The transistor case is isotat 55°C. 3 The contact resistance between the negligible.

Analysis The thermal resistance

C/W 1.5 °=°−

=−

=⎯→⎯∆

= ∞−

− W25C)1855(transistor

ambientcaseambientcase Q

TTR

RTQ

&&

T∞

RTs

The thermal resistance of the heat sink must be below 1.5°C/W. Table 3-6 reveals that HS5030 in both horizontal and ertical positions, HS6071 in vertical position, and HS6115 in both horizontal and vertical positions can be selected.

v


3-76

3-120 A turbine blade is exposed to hot gas from the combustion chamber. The heat transfer rate to the turbine blade and the temperature at the tip are to be determined.

Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 The cross-sectional area of the turbine blade is uniform.

Properties The thermal conductivity of the turbine blade is given as 17 W/m · °C.

Analysis The turbine blade can be treated as a uniform cross section fin with adiabatic tip. The heat transfer rate to the turbine blade can be expressed as

mLTThpkAQ bc tanh)(blade −= ∞&

where

366.4)m 053.0()m 1013.5)(C W/m17(

)m 11.0)(C W/m538(5.0

24

25.0

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×°⋅

°⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−L

kAhpmL

c

C W/7184.0)m 1013.5)(C W/m17)(m 11.0)(C W/m538( 242 °=×°⋅°⋅= −chpkA

to the turbine blade is

&

perature distribution is expressed as

The heat transfer rate

W376=°−°= )366.4tanhC( )450973)(C W/7184.0( bladeQ

For adiabatic tip, the tem

mL

xLmTTTxT

cosh)(cosh)( −

=−− ∞

b ∞

he temperature at the tip of the turbine blade is T

C 960 °=°+°−

=+−

= ∞∞ C 973

366.4coshC )973450(

coshT

mLTT

T bL

Discussion The tolerance of the turbine blade to high temperature can be increased by applying Zirconia based thermal barrier coatings (TBCs) on the blade surface.


3-77

3-121 Pipes used for transporting superheated vapor are connected together by flanges. The temperature at the base of the flange and the rate of heat loss through the flange are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 The flanges profile is similar to circular fins of rectangular profile. Properties The thermal conductivity of the pipes is given as 16 W/m · °C.

Analysis The heat transfer rate through the pipe wall is equal to the heat transfer rate through the flanges:

or fQQ && =pipe )()/ln(

2 ∞−=−

TThADD

TTtk bff

io

bi ηπ

Rearranging the equation yields

)/ln(2

ff DDtkhA πη +

)ln(2

io

iio

ff

b

TD

tkThAT

πη +=

∞

From Table 3-3, for circular fins of rectangular profile we have

/ D

m 055.02

m 02.02

m 09.02/22 =+=+= trr c

12

2 m 055.0[(2)(2 =−= ππ rrA cf2222 m 0151.0])m 2/05.0() =−

m 025.022c

m 02.0m 06.0m 09.02/ =+−

=+= tLL

Hence,

2m 0005.0)m 02.0)(m 025.0( === tLA cp

1398.0)m 0005.0)(C W/m16(

C W/m10)m 025.0(2/1

2

22/3

2/12/3 =

⎥⎥⎦

⎤

⎢⎢⎣

⎡

°⋅

°⋅=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

pc kA

hLξ

83.1m 030.0m 055.0/ 12 ==rr c

Using Figure 3-44, the fin efficiency is 97.0≈fη . The temperature at the base of the flange is

C 148 °=°⋅

+°⋅

°°⋅

+°°⋅=

)50/60ln()C W/m16)(m 020(2)m 0151.0)(C W/m10)(97.0(

)C 150()50/60ln(

)C W/m16)(m 020(2)C 25)(m 0151.0)(C W/m10)(97.0(

22

22

π

π

.

.

Tb

The rate of heat loss through the flange is

Discussion The flanges act as extended surfaces, which enhanced heat transfer from the pipes.

W18=°−°⋅=−= ∞ C )25148)(m 0151.0)(C W/m10)(97.0()( 22TThAQ bfff η&


3-78

3-122 Using Table 3-3 and Figure 3-43, the efficiency, heat transfer rate, and effectiveness of a straight rectangular fin are to be determined.

Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the fin is given as 235 W/m · °C.

Analysis (a) From Table 3-3, for straight rectangular fins, we have

12

m 19.16)m 005.0)(C W/m235(

)C W/m154(22 -

kthm =

°⋅°⋅

==

c m 0525.02/)m 005.0()m 05.0(2/ =+=+= tLL

2fin m 0105.0)m 0525.0)(m 1.0(22 === cwLA

The fin efficiency is

[ ] 0.813===)m 0525.0)(m 19.16(fin

cmL)m 0525.0)(m 19.16(tanhtanh

1

1

-

-cmL

η

he heat transfer rate for a single fin is

∞ C )25350)(m 0105.0)(C W/m154)(813.0()( 22TTb

he fin eff tiveness is

T

= finfinfin hAQ η& W427=°−°⋅=−

T ec

17.1=°−°⋅

=−

= − ∞ )(() hT

=∞ C )25350)(m 1.0)(m 005.0)(C W/m154(

W427)( 2

finfin TTtw

QThA bbb

&ε

se Figure 3-43, we need

and

Hence,

finQ&

(b) To u

m 0525.0=cL tLA cp =

60.0)m 005.0)(m 0525.0)(C W/m235(

C W/m154)m 0525.0(2/12

2/32/1

2/3 ⎟⎞

⎜⎛ h

≈⎥⎥⎦

⎤

⎢⎢⎣

⎡

°⋅°⋅

=⎟⎠

⎜⎝ pkA

L c

Using Figure 3-43, the fin efficiency is

0.81≈fη

The heat transfer rate for a single fin is

n TThAQ bη& W426=°−°⋅=−= ∞ C )25350)(m 0105.0)(C W/m154)(81.0()( 22finfin fi

The fin effectiveness is

17.0=°−°⋅

=−

=−

=∞∞ C )25350)(m 1.0)(m 005.0)(C W/m154(

W426))(()( 2

finfinfin TTtwh

QTThA

Q

bbb

&&ε

Discussion The results determined using Table 3-3 and Figure 3-43 are very comparable. However, it should be noted that results determined using Table 3-3 are more accurate.


3-79

3-123 Circular aluminum fins are to be attached to the tubes of a heating system. The increase in heat transfer from the tubes per unit length as a result of adding fins is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire fin surfaces. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the fins is given to be k = 186 W/m⋅°C.

Analysis In case of no fins, heat transfer from the tube per meter of its length is

W660C)25130)(m 1571.0)(C. W/m40()(

m 1571.0)m 1)(m 05.0(22

fin nofin no

21fin no

=°−°=−=

===

∞TThAQ

LDA

b&

ππ

The efficiency of these circular fins is, from the efficiency curve, Fig. 3-43

97.0

08.0m) C)( W/m186(2 o⎠⎝ 0.001

C W/m40001.0005.0

2

22.10.025

)2/001.0(m 005.02/)05.006.0(2/)(

fin

o2

2/12/3

1

12

=

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

=⎟⎞

⎜⎛ +=

⎟⎠⎞

⎜⎝⎛ +=⎟

⎟⎠

⎞⎜⎜⎝

⎛

=

=−=−=

ηkthtL

kAhL

r

DDL

pc

eat transf from a single fin is

22

finfinmaxfin,finfin

°−°=

−= ∞TThA bη

eat transf from a single unfinned portion of the tube is

There are 250 fins and thus 250 interfin spacings per meter length of the tube. The total heat transfer from the finned tube is then determined from

Therefore the increase in heat transfer from the tube per meter of its length as a result of the addition of the fins is

25°C

130°C

03.0)2/(2 +=

+ tr

H er

C)25130)(m 001916.0)(C. W/m40(97.0

)(

m 001916.0)001.0)(03.0(2)025.003.0(22)(2 2222

21

22fin

=

=+−=+−=

QQ

trrrA

η

ππππ&&

W81.7=

H er

W98.1C)25130)(m 0004712.0)(C. W/m40()(

m 0004712.0m) m)(0.003 05.0(22

unfinunfin

21unfin

=°−°=−=

===

∞TThAQ

sDA

b&

ππ

W2448)98.181.7(250)( unfinfinfintotal, =+=+= QQnQ &&&

W1788=−=−= 6602448fin nofintotal,increase QQQ &&&


3-80

3-124E The handle of a stainless steel spoon partially immersed in boiling water extends 7 in. in the air from the free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be determined.

Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the spoon.

Properties The thermal conductivity of the spoon is given to be k = 8.7 Btu/h⋅ft⋅°F.

Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water, the variation of temperature along the spoon can be expressed as

cosh

)(cosh)(mL

xLmTTTxT

b

−=

−−

∞

∞ h, T∞

Tb

L = 7 in

0.08 in

0.5 in

where

2ft 000278.0)ft 12/08.0ft)( 12/5.0(

ft 0967.0)ft 12/08.0ft 12/5.0(2

==

=+=

cA

p

1-2

2ft 95.10

)ft 000278.0)(FBtu/h.ft. 7.8( °ckA

Noting that x = L

)ft 0967.0)(F.Btu/h.ft 3(=

°==

hpm

= 7/12=0.583 ft at the tip and substituting, the tip temperature of the spoon is etermined to be

d

F75.4=2961)75(200+F75=

)583.095.10cosh(0cosh)75(200+F75=

cosh)(cosh

)()(−

−+= ∞∞LLm

TTTLT b

°

−°

×−°

mL

Therefore, the temperature difference across the exposed section of the spoon handle is

F124.6°=°−=−=∆ F)4.75200(tipTTT b


3-81

3-125E The handle of a silver spoon partially immersed in boiling water extends 7 in. in the air from the free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be determined.

Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the spoon..

Properties The thermal conductivity of the spoon is given to be k = 247 Btu/h⋅ft⋅°F.

Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water, the variation of temperature along the spoon can be expressed as

cosh

)(cosh)(mL

xLmTTTxT

b

−=

−−

∞

∞

where

A

p

h, T∞

Tb

L = 7 in

0.08 in

0.5 in2ft 000278.0)ft 12/08.0ft)( 12/5.0(

ft 0967.0)ft 12/08.0ft 12/5.0(2

==

=+=

c

1-2

ft 055.2)ft 0967.0)(F.Btu/h.ft 3(=

°==

kAhpm

2 )ft 000278.0)(FBtu/h.ft. 247( °c

Noting that x = L = 0.7/12=0.583 ft at the tip and substituting, the tip temperature of the spoon is determined to be

F144.1=81.11)75(200+F75= −°

)583.0055.2cosh(0cosh)75(200+F75=

cosh)(cosh

)()(

°

×−°

−−+= ∞∞ mL

LLmTTTLT b

erefore, the temperature difference across the exposed section of the spoon handle is

Th

F55.9°=°−=−=∆ C)1.144200(tipTTT b


3-82

3-126E Prob. 3-124E is reconsidered. The effects of the thermal conductivity of the spoon material and the length of its extension in the air on the temperature difference across the exposed surface of the spoon handle are to be investigated.


"GIVEN" k_spoon=8.7 [Btu/h-ft-F] T_w=200 [F] T_infinity=75 [F] A_c=(0.08/12*0.5/12) [ft^2] L=7 [in] h=3 [Btu/h-ft^2-F] "ANALYSIS" p=2*(0.08/12+0.5/12) a=sqrt((h*p)/(k_spoon*A_c)) (T_tip-T_infinity)/(T_w-T_infinity)=cosh(a*(L-x)*Convert(in, ft))/cosh(a*L*Convert(in, ft)) x=L "for tip temperature" DELTAT=T_w-T_tip

kspoon [Btu/h.ft.F]

∆T [F]

5 124.9 16.58 122.6 28.16 117.8 39.74 112.5 51.32 107.1 62.89 102 74.47 97.21 86.05 92.78 97.63 88.69 109.2 84.91 120.8 81.42 132.4 78.19 143.9 75.19 155.5 72.41 167.1 69.82 178.7 67.4 190.3 65.14 201.8 63.02 213.4 61.04 225 59.17

0 45 90 135 180 22550

60

70

80

90

100

110

120

130

kspoon [Btu/h-ft-F]

∆T

[F]


3-83

L [in]

∆T [F]

5 122.4 5.5 123.4 6 124 6.5 124.3 7 124.6 7.5 124.7 8 124.8 8.5 124.9 9 124.9 9.5 125 10 125 10.5 125 11 125 11.5 125 12 125 5 6 7 8 9 10 11 12

122

122.5

123

123.5

124

124.5

125

125.5

L [in]

∆T

[F]


3-84

3-127 A circuit board houses 80 logic chips on one side, dissipating 0.04 W each through the back side of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 864 aluminum pin fins on the back surface. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 30 W/m⋅°C for the circuit board, k = 237 W/m⋅°C for the aluminum plate and fins, and k = 1.8 W/m⋅°C for the epoxy adhesive. Analysis (a) The total rate of heat transfer dissipated by the chips is W2.3 W)04.0(80 =×=Q& The individual resistances are 2m 0216.0m) 18.0(m) 12.0( ==A

C/W 89031.0

)m 021.0(C). W/m52( °hA 6

)m 0216.0(C) W/m.30(m 004.0

22

2board

°===

°=°

==

R

kALR

11

C/W 00617.0

conv

C/W 8965.089031.000617.0convboardtotal °=+=+= RRR The temperatures on the two sides of the circuit board are

C42.9°≅°=°−°=−=⎯→⎯−

= C85.42C/W) 00617.0)( W2.3(C87.42board12board

21 RQTTR

TTQ &&

C42.9°≅°=°+°=+=⎯→⎯−

= ∞∞ C42C/W) 8965.0)( W2.3(C40total21

total

21 RQTTR

TTQ &&

Therefore, the board is nearly isothermal. (b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be

2 cm Rboard

T1

RAluminum Rconv

T∞2

Repoxy

T2

87.

1-2

2 m 74.18)m 0025.0)(C W/m.237(

)C. W/m52(444/

=°

°====

kDh

DkDh

kAhpm

c ππ

956.0m02.0m 74.18

)m 02.0m 74.18tanh(tanh1-

-1

fin =××

==mL

mLη

The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.956. Then the various thermal resistances are

C/W 00514.0)m 0216.0(C) W/m.8.1( 2epoxy

°kAm 0002.0

°===LR

C/W 00039.0)m 0216.0(C) W/m.237(

m 002.02Al °=

°==

kALR

2unfinnedfinnedfinswith total,

222

unfinned

2finfinned

m 1471.00174.01297.0

m 0174.04

)0025.0(8640216.04

8640216.0

m 1297.0m) 02.0(m) 0025.0(864956.0

=+=+=

=×−=−=

=×==

AAA

DA

DLnA

ππ

ππη

C/W 1307.0)m 1471.0(C). W/m52(

1122

finswith total,conv °=

°==

hAR

C/W 1424.01307.000039.000514.000617.0convaluminumepoxyboardtotal °=+++=+++= RRRRR Then the temperatures on the two sides of the circuit board becomes

C40.4

C40.5

°≅=°−°=−=⎯→⎯−

=

°≅°=°+°=+=⎯→⎯−

= ∞∞

44.40C/W) 00617.0)( W2.3(C46.40

C40.46C/W) 1424.0)( W2.3(C40

board12board

21

total21total

21

RQTTR

TTQ

RQTTR

TTQ

&&

&&


3-85

A

3-128 A circuit board houses 80 logic chips on one side, dissipating 0.04 W each through the back side of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 864 copper pin fins on the back surface. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 20 W/m⋅°C for the circuit board, k = 386 W/m⋅°C for the copper plate and fins, and k = 1.8 W/m⋅°C for the epoxy adhesive. Analysis (a) The total rate of heat transfer dissipated by the chips is 2 cm W2.3 W)04.0(80 =×=Q&

The individual resistances are = 2m 0216.0m) 18.0(m) 12.0( =

C/W 89031.0

)m 0216.0(C). W/m52(11

C/W 00617.0)m 0216.0(C) W/m.30(

m 004.0

22conv

2board

°=°

==

°=°

==

hAR

kALR

Rboard

T1

RAluminum Rconv

T∞2

Repoxy

T2

C/W 8965.089031.000617.0convboardtotal °=+=+= RRR The temperatures on the two sides of the circuit board are

C42.9

C42.9

°≅°=°−°=−=⎯→⎯−

=

°≅°=°+°=+=⎯→⎯−

= ∞∞

C85.42C/W) 00617.0)( W2.3(C87.42

C87.42C/W) 8965.0)( W2.3(C40

board12board

21

total21total

21

RQTTR

TTQ

RQTTR

TTQ

&&

&&

Therefore, the board is nearly isothermal. (b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be

1-2

2 m 74.18)m 0025.

0()C W/m.237(

)C. W/m52(444/

=°

°====

kDh

DkDh

kAhpm

c ππ

956.0m02.0m 74. 1- ×

The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.956. Then the various thermal resistances are

18)m 02.0m 74.18tanh(tanh -1

fin =×

==mL

mLη

C/W 00514.0)m 0216.0(C) W/m.8.1(

m 0002.02epoxy °=

°==

kALR

C/W 00024.0)m 0216.0(C) W/m.386(

m 002.02Al °=

°==

kALR

2unfinnedfinnedfinswith total,

222

unfinned

2finfinned m) 02.0(m) 0025.0(864956.0 =×== DLnA ππη

m 1471.00174.01297.0

m 0174.04

)0025.0(8640216.04

8640216.0

m 1297.0

=+=+=

=×−=−=

AAA

DA ππ

C/W 1307.0)m 1471.0(C). W/m52(

1122

finswith total,conv °=

°==

hAR

C/W 1423.01307.000024.000514.000617.0convaluminumepoxyboardtotal °=+++=+++= RRRRR Then the temperatures on the two sides of the circuit board becomes

C40.4

C40.5

°≅=°−°=−=⎯→⎯−

=

°≅°=°+°=+=⎯→⎯−

= ∞∞

44.40C/W) 00617.0)( W2.3(C46.40

C40.46C/W) 1423.0)( W2.3(C40

board12board

21

total21total

21

RQTTR

TTQ

RQTTR

TTQ

&&

&&


3-86

3-129 A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heat transfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 Heat transfer from the fin tips is negligible. 4 The heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the fins.

Properties The thermal conductivity of the aluminum plate and fins is given to be k = 237 W/m⋅°C.

Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be

3 cm

0.6 cmD=0.25 cm

1-2

2m 37.15

)m 0025.0)(C W/m.237()C. W/m35(44

4/=

°°

====kD

hDk

DhkAhpm

c ππ

935.0m 03.0m 37.15 1-fin =

×==

mLη

The number of fins, finned and unfinned surface areas, and heat transfer

)m 03.0m 37.15tanh(tanh -1 ×mL

rates from those areas are

777,27m) 006.0(m) 006.0(

m 1 2==n

W2107C)30100)(m 86.0)(C. W/m35()( 2

unfinnedunfinned °=−= ∞TThAQ b&

W300,15C)30100)(m 68.6)(C. W/m35(935.0

)(

m 86.04

)0025.0(277771

4277771

m 68.6

4)0025.0()03.0)(0025.0(27777

427777

2

22

finfinmaxfin,finfinned

222

unfinned

2

22

fin

=°−

=°−°=

−==

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

∞TThAQQ

DA

DDLA

b&& ηη

ππ

ππππ

The rate of heat transfer if there were no fin attached to the plate would be

22fin nofin =°−°=−= ∞TThAQ b

&

hen the fin effectiveness becomes

Then the total heat transfer from the finned plate becomes

kW 17.4=×=+=+= W1074.12107300,15 4unfinnedfinnedfintotal, QQQ &&&

no

m 1)m 1)(m 1( 2fin no ==A

W2450C)30100)(m 1)(C. W/m35()(

T

7.10===2450

400,17

fin no

finfin Q

Q&

&ε


3-87

3-130 A hot plate is to be cooled by attaching copper pin fins on one side. The rate of heat transfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to be determined.


Properties The thermal conductivity of the copper plate and fins is given to be k = 386 W/m⋅°C.

Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be

3 cm

0.6 cmD=0.25 cm

1-2

2m 04.12

)m 0025.0)(C W/m.386()C. W/m35(44

4/=

°°

====kD

hDk

DhkAhpm

c ππ

959.0m 03.0m 04.12 1-fin

×mL

The number of f

)m 03.0m 04.12tanh(tanh -1=

×==

mLη

ins, finned and unfinned surface areas, and heat transfer rates from those areas are

27777m) 006.0(m) 006.0(

m 1 2==n

W2107C)30100)(m 86.0)(C W/m35()(

W700,15C)30100)(m 68.6)(C.

W/m35(959.0 2

maxfin,finfinned

=

)(

m 68.64

)0025.0()03.0)(4

2o2unfinnedunfinned

2

finfin

22

22

fin

=°−=−=

=°−°

−==

⎥⎦⎢⎣⎠⎝

=⎥⎤

⎢⎣+

⎥⎦⎢⎣

∞

∞

TThAQ

TThAQ

b

b

&

& ηη

π

Then the total heat transfer from the finned plate becomes

The rate of heat transfer if there were no fin attached to the plate would be

Then the fin effectiveness becomes

0025.0(27777277772

⎢⎡

=⎥⎤

⎢⎡

+=DDLA πππ

⎥⎦

m 86.04

)0025.0(277771

4277771 2

unfinned =⎥⎤

⎢⎡

−=⎟⎟⎞

⎜⎜⎛

−=DA

ππ

Q&

kW 17.8=×=+=+= W1078.12107700,15 4unfinnedfinnedfintotal, QQQ &&&

W2450C)30100)(m 1)(C. W/m35()(

m 1)m 1)(m 1(22

fin nofin no

2fin no

=°−°=−=

==

∞TThAQ

A

b&

7.27===2450

17800

fin no

finfin Q

Q&

&ε


3-88

3-131 Prob. 3-129 is reconsidered. The effect of the center-to center distance of the fins on the rate of heat transfer from the surface and the overall effectiveness of the fins is to be investigated.


1 [m^2]

2)) "number of fins"

ned )

psilon_fin=Q_dot_total_fin/Q_dot_nofin

A

"GIVEN" T_b=100 [C] L=0.03 [m] D=0.0025 [m] k=237 [W/m-C] S=0.6 [cm] T_infinity=30 [C] h=35 [W/m^2-C] A_surface=1* "ANALYSIS" p=pi*D A_c=pi*D^2/4 a=sqrt((h*p)/(k*A_c)) eta_fin=tanh(a*L)/(a*L) n=A_surface/(S^2*Convert(cm^2, m^A_fin=n*(pi*D*L+pi*D^2/4) A_unfinned=A_surface-n*(pi*D^2/4) Q_dot_finned=eta_fin*h*A_fin*(T_b-T_infinity) Q_dot_unfinned=h*A_unfinned*(T_b-T_infinity) Q_dot_total_fin=Q_dot_finned+Q_dot_unfinQ_dot_nofin=h*A_surface*(T_b-T_infinitye

S [cm]

Qtotal fin[W]

εfin

0.4 36123 14.74 0.5 24001 9.796 0.6 17416 7.108 0.7 13445 5.488 0.8 10868 4.436 0.9 9101 3.715 1 7838 3.199 1.1 6903 2.817 1.2 6191 2.527 1.3 5638 2.301 1.4 5199 2.122 1.5 4845 1.977 1.6 4555 1.859 1.7 4314 1.761 1.8 4113 1.679 1.9 3942 1.609

0.25 0.6 0.95 1.3 1.65 20

5000

10000

15000

20000

25000

30000

35000

40000

0

2

4

6

8

10

12

14

16

18

20

Qto

tal,f

in [

W]

ε fin

S [cm]

2 3797 1.55


3-89

3-132 Circular fins made of copper are considered. The function θ(x) = T(x) - T∞ along a fin is to be expressed and the temperature at the middle is to be determined. Also, the rate of heat transfer from each fin, the fin effectiveness, and the total rate of heat transfer from the wall are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 The heat transfer coefficient is constant and uniform over the entire finned and unfinned wall surfaces. 4 The thermal properties of the fins are constant. 5 The heat transfer coefficient accounts for the effect of radiation from the fins.

Properties The thermal conductivity of the copper fin is given to be k = 380 W/m⋅°C.

Analysis (a) T∞ , h

L

Ts2

Ts1

x D

For x = L/2:

1-2 m 44.32

/4(0.001))380((0.001))100(

===ππ

ckAhpm

Noting that Tb-T∞ = Ts1 and TL - T∞ = 0,

C58.9°=×

−=

−=

−−

−=

−+⎟⎠

⎜⎝ −

=− ∞∞)( TTTxT b

⎟⎞

⎜⎛ −

−

∞

∞

)030.044.32sinh()]015.0030.0(44.32sinh[132)2/(

sinh)](sinh[

01320)2/(

sinh)](sinh[

sinh

)(sinh)sinh(

LT

mLxLmLT

mLxLm

mL

xLmmxTT

TT

L

b

(b) The rate of heat transfer from a single fin is

W1.704=×

−×−= 001.0()100()0132( π

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−∞

∞

∞

)030.044.32sinh(0)030.044.32cosh(4/)001.0()380)(

)sinh(

)cosh()

2π

mLTTTT

mLhpkAQ b

L

c&

−= (fin one TTb

The effectiveness of the fin is

164.4=−

=−

=∞ )0132()001.0(25.0)100(

704.1)( 2π

εTThA

Q

bc

&

ince ε >> 2, the fins are well justified.

(c) The total rate of heat transfer is

S

W1191=×−×+=

−−+=

+=

∞

)132)(100]()001.0(25.06251.01.0[)704.1)(625(

)()(2

finwallfin onefin

basefinstotal

π

TThAnAQn

QQQ

bc&

&&&


3-90

&

by S=1/(kR).

Heat Transfer in Common Configurations

3-133C Under steady conditions, the rate of heat transfer between two surfaces is expressed as )( 21 TTSk −= where S is the conduction shape factor. It is related to the thermal resistance

Q

3-134C It provides an easy way of calculating the steady rate of heat transfer between two isothermal surfaces in common configurations.

3-135 Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant.

Properties The thermal conductivity of concrete is given to be k = 0.75 W/m⋅°C.

Analysis The shape factor for this configuration is given in Table 3-7 to be

m 59.14

)m 06.0)(m 06.0(2)m 06.0()m 06.0()m 4.0(4cosh

)m 12(2

24cosh

2

2221

21

22

21

21

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

−

−

π

π

DDDDz

LS

T2 = 15°C

D = 6 cm

T1 = 60°C

L = 12 mz = 40 cm

Then the steady rate of heat transfer between the pipes becomes

W492=°−°=−= C)1560)(C W/m.75.0)(m 59.14()( 21 TTSkQ&


3-91

3-136 Prob. 3-135 is reconsidered. The rate of heat transfer between the pipes as a function of the distance between the centerlines of the pipes is to be plotted.


"GIVEN" L=12 [m] D_1=0.06 [m] D_2=D_1 z=0.40 [m] T_1=60 [C] T_2=15 [C] k=0.75 [W/m-C] "ANALYSIS" S=(2*pi*L)/(arccosh((4*z^2-D_1^2-D_2^2)/(2*D_1*D_2))) Q_dot=S*k*(T_1-T_2)

z [m]

Q [W]

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1158 679 555 492.3 452.8 425.1 404.2 387.7 374.2 362.9

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1300

400

500

600

700

800

900

1000

1100

1200

z [m]

Q [

W]


3-92

3-137E A row of used uranium fuel rods are buried in the ground parallel to each other. The rate of heat transfer from the fuel rods to the atmosphere through the soil is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant.

Properties The thermal conductivity of the soil is given to be k = 0.6 Btu/h⋅ft⋅°F.


ft 5298.0

)ft 12/8()ft 15(2

sinh)ft 12/1()ft 12/8(2

ln

)ft 3(24

2sinh2ln ⎟⎞

⎜⎛ πzw

24

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×=

⎠⎝

×=

ππ

ππ

π

wD

LS

hen the steady rate of heat transfer from the fuel rods becomes

pe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be

ns exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant.

Analysis Since z >1.5D, the shape factor for this onfiguration is given in Table 3-7 to be

T2 = 60°F

8 in

T1 = 350°F

15 ft

D = 1 in

L = 3 ft

total

T

Btu/h 92.2=°−°=−= F)60350)(FBtu/h.ft. 6.0)(ft 5298.0()( 21total TTkSQ&

3-138 The hot water pidetermined.

Assumptions 1 Steady operating conditio

Properties The thermal conductivity of the soil is given to be k = 0.9 W/m⋅°C.

c

m 44.20)]m 08.0/()m 8.0(4ln[

)m 12(2)/4ln(

2===

ππDz

LS

Then the steady rate of heat transfer from the pipe becomes

W1067=°−=−= C)260)(C W/m.9.0)(m 44.20()( o21 TTSkQ&

60°C

L = 12 m

D = 8 cm

2°C

80 cm


3-93

3-139 Prob. 3-138 is reconsidered. The rate of heat loss from the pipe as a function of the burial depth is to be plotted.


_dot=S*k*(T_1-T_2)

A

"GIVEN" L=12 [m] D=0.08 [m] z=0.80 [m] T_1=60 [C] T_2=2 [C] k=0.9 [W/m-C] "ANALYSIS" S=(2*pi*L)/ln(4*z/D) Q

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2800

1000

1200

1400

1600

1800z [m]

Q [W]

0.2 0.38 0.56 0.74 0.92 1.1 1.28 1.46 1.64 1.82 2 854.6

1709 1337 1181 1090 1028 982.1 946.4 917.3 893.1 872.5

Q [

W]

z [m]


3-94

3-140 Hot water flows through a 5-m long section of a thin walled hot water pipe that passes through the center of a 14-cm thick wall filled with fiberglass insulation. The rate of heat transfer from the pipe to the air in the rooms and the temperature drop of the hot water as it flows through the pipe are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the fiberglass insulation is constant. 4 The pipe is at the same temperature as the hot water.

Properties The thermal conductivity of fiberglass insulation is given to be k = 0.035 W/m⋅°C.

53°C

L= 5 m

D =2.5 cm

18°C

Analysis (a) The shape factor for this configuration is given in Table 3-7 to be

m 16

)m 025.0()m 07.0(8

ln

)m 5(28ln

2=

⎥⎦

⎤⎢⎣

⎡=

⎟⎠⎞

⎜⎝⎛

=

π

π

π

π

DzLS

Then the steady rate of heat transfer from the pipe becomes

W19.6=°−°=−= C)1853)(C W/m.035.0)(m 16()( 21 TTSkQ&

(b) Using the water properties at the room temperature, the temperature drop of the hot water as it flows through this 5-m section of the wall becomes

C0.024°=

°⎥⎥⎦

⎤

⎢⎢⎣

⎡====∆

∆=

)CJ/kg. 4180(4

)m 025.0()m/s 4.0)(kg/m 1000(

J/s 6.192

3 πρρ pcpp

p

cVAQ

cQ

cmQT

TcmQ&

&

&

&

&

&&

V


3-95

m 3142.0)m 2)(m 05.0(

2

2

=°−

=== DLA ππ

idering the shape factor, the heat loss for vertical part of the tube can be determined from

3-141 Hot water is flowing through a pipe that extends 2 m in the ambient air and continues in the ground before it enters the next building. The surface of the ground is covered with snow at 0°C. The total rate of heat loss from the hot water and the temperature drop of the hot water in the pipe are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the ground is constant. 4 The pipe is at the same temperature as the hot water.

Properties The thermal conductivity of the ground is given to be k = 1.5 W/m⋅°C.

Analysis (a) We assume that the surface temperature of the tube is equal to the temperature of the water. Then the heat loss from the part of the tube that is on the ground is

0)(C. W/m22(

)(2 °=

−= ∞TThAQ ss&

W518C)580)(m 3142.

s

Cons

m 44.3

)m 05.0( ⎦⎣⎠⎝ D)m 3(4ln

)m 3(24ln

2=

⎥⎤

⎢⎡

=⎟⎞

⎜⎛

=ππ

LLS

W/m.5.1)(m 44.3()( 21 =°=−= TTSkQ&

The shape factor, and the rate of heat loss on the horizontal part that is in the ground are

5°C

3 m

80°C

-3°C

20 m

C)]3(80)[C °−− W428

m 9.22

)m 05.0( ⎥⎦

⎢⎣⎠⎝ D

)m 3(4ln

)m 20(24ln

2=

⎤⎡=

⎟⎞

⎜⎛

ππz

LS

5.1)(m 9.22()( 21 =°−−=−= TTSkQ&

talQ

(b) Using the water properties at the room temperature, the temperature drop of the hot water as it flows through this 25-m section of the wall becomes

=

W2851C)]3(80)[C W/m.°

and the total rate of heat loss from the hot water becomes

& W3797=++= 2851428518 to

C0.31°=

°⎥⎥⎦

⎤

⎢⎢⎣

⎡====∆

∆=

)CJ/kg. 4180(4

)m 05.0()m/s 5.1)(kg/m 1000(

J/s 3797)()( 2

3 πρρ pcpp

p

cVAQ

cQ

cmQT

TcmQ&

&

&

&

&

&&

V


3-96

3-142 The walls and the roof of the house are made of 20-cm thick concrete, and the inner and outer surfaces of the house are maintained at specified temperatures. The rate of heat loss from the house through its walls and the roof is to be determined, and the error involved in ignoring the edge and corner effects is to be assessed.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer at the edges and corners is two-or three-dimensional. 3 Thermal conductivity of the concrete is constant. 4 The edge effects of adjoining surfaces on heat transfer are to be considered.

Properties The thermal conductivity of the concrete is given to be k = 0.75 W/m⋅°C.

Analysis The rate of heat transfer excluding the edges and corners is first determined to be

2total m 7.403)2.06)(4.012(4)4.012)(4.012( =−−+−−=A

W167,18C)315(m 2.021L

The heat transfer rate th

)m 7.403)(C W/m.75.0()(2

total =°−°

=−TTkA

Q&

rough the edges can be determined using the shape factor relations in Table 3-7,

=

m 04.26m) 0.54(124+m) 2.0(15.04

54.0415.04edges4corners4edges+corners

=××=

×+×=×+×= wLS

m 04.26()( 21edges+cornersedges+corners =°−°=−= TTkSQ&

Q&

e rate of heat transfer is determined from

W234C)315)(C W/m.75.0

3°C

15°C

L

L

)(

and

kW 18.4=×=+= W10840.1234167,18 4total

Ignoring the edge effects of adjoining surfaces, th

2total m 432)6)(12(4)12)(12( =+=A

kW 4.19101.94C)315(m 2.0

C W/m.75.0()( 21total °

=− )m 432)( 42

=×=°−TTL

kAQ&

The percentage error involved in ignoring the effects of the edges then becomes

=

5.4%=×−

= 1004.18

4.184.19%error


3-97

3-143 The inner and outer surfaces of a long thick-walled concrete duct are maintained at specified temperatures. The rate of heat transfer through the walls of the duct is to be determined.




m 7.89625.1ln785.0)m 25(2

ln785.0

241.125.11620

==⎟⎠⎞

⎜⎝⎛

=⎯→⎯<==ππ

ba

LSba

Then the steady rate of heat transfer through the walls of the duct becomes

kW 47.1=×=°−°=−= W10.714C)30100)(C W/m.75.0)(m 7.896()( 421 TTSkQ&

100°C

30°C

20 cm

16 cm

3-144 A spherical tank containing some radioactive material is buried in the ground. The tank and the ground surface are maintained at specified temperatures. The rate of heat transfer from the tank is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the ground is constant.

Properties The thermal conductivity of the ground is given to be k = 1.4 W/m⋅°C.

T2 =15°C

T1 = 140°Cz = 5.5 m

D = 3 m


m 83.21

m 5.5m 325.01

)m 3(2

25.01

2=

−=

−=

ππ

zD

DS

Then the steady rate of heat transfer from the tank becomes

W3820=°−°=−= C)15140)(C W/m.4.1)(m 83.21()( 21 TTSkQ&


3-98

3-145 Prob. 3-144 is reconsidered. The rate of heat transfer from the tank as a function of the tank diameter is to be plotted.


"GIVEN" D=3 [m] k=1.4 [W/m-C] h=4 [m] T_1=140 [C] T_2=15 [C] "ANALYSIS" z=h+D/2 S=(2*pi*D)/(1-0.25*D/z) Q_dot=S*k*(T_1-T_2)

0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1000

2000

3000

4000

5000

6000

7000

D [m]

Q [

W]

D [m]

Q [W]

0.5 566.4 1 1164 1.5 1791 2 2443 2.5 3120 3 3820 3.5 4539 4 5278 4.5 6034 5 6807


3-99

3-146 Hot water passes through a row of 8 parallel pipes placed vertically in the middle of a concrete wall whose surfaces are exposed to a medium at 32°C with a heat transfer coefficient of 8 W/m2.°C. The rate of heat loss from the hot water, and the surface temperature of the wall are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of concrete is constant.



90°C

L = 4 m

D

32°C

z

z

m 58.13

m) 03.0(m) 075.0(8

ln8

ln ⎟⎟⎞

⎜⎜⎛

⎟⎞

⎜⎛ z

)m 4(22=

⎠⎝

=

⎠⎝ π

π

π

π

D

LS

Then rate of heat loss from the hot water in 8 parallel pipes

e surface temperature of the wall can be determined from

=

becomes

W4726=°−°=−= C)3290)(C W/m.75.0)(m 58.13(8)(8 21 TTSkQ&

Th

C38.2°=°

+°=+=⎯→⎯−=

==

∞∞ )m 64)(C. W/m12( W4726C32)(

sides)both (from m 64)m 8)(m 4(2

22

2

ssss

s

hAQTTTThAQ

A&

&


3-100

Special Topic: Heat Transfer through the Walls and Roofs

3-147C The R-value of a wall is the thermal resistance of the wall per unit surface area. It is the same as the unit thermal resistance of the wall. It is the inverse of the U-factor of the wall, R = 1/U.

3-148C The effective emissivity for a plane-parallel air space is the “equivalent” emissivity of one surface for use in the relation that results in the same rate of radiation heat transfer between the two surfaces across the air space. It is determined from

)( 41

42effectiverad TTAQ s −= σε&

1111

21effective εεε

where ε

−+=

ffective emissivity is known, the radiation eat transfer through the air space is determined from the relation above.

ng since the convection currents that set in in the thicker air space offset any additional resistance due to a thicker air ace.

the attic (the roof or the ceiling side) nce they reduce radiation heat transfer between the ceiling and the roof considerably.

eat transfer through the ceiling since the roof in this case ill act as a radiation shield, and reduce heat transfer by radiation.

1 and ε2 are the emissivities of the surfaces of the air space. When the e

radQ&h

3-149C The unit thermal resistances (R-value) of both 40-mm and 90-mm vertical air spaces are given to be the same, which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. This is not surprisisp

3-150C Radiant barriers are highly reflective materials that minimize the radiation heat transfer between surfaces. Highly reflective materials such as aluminum foil or aluminum coated paper are suitable for use as radiant barriers. Yes, it is worthwhile to use radiant barriers in the attics of homes by covering at least one side of si

3-151C The roof of a house whose attic space is ventilated effectively so that the air temperature in the attic is the same as the ambient air temperature at all times will still have an effect on hw


3-101

3-152 The R-value and the U-factor of a wood frame wall are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant.

Properties The R-values of different materials are given in Table 3-8.

Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from

Roverall = 1/Uoverall where Uoverall = (Ufarea )insulation + (Ufarea )stud

and the value of the area fraction farea is 0.80 for insulation section and 0.20 for stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available R-values from Table 3-8 and calculating others, the total R-values for each section is determined in the table below.

R -value, m2.°C/W

Construction Between studs

At studs

1. Outside surface, 12 km/h wind 0.044 0.044

2. Wood bevel lapped siding 0.14 0.14

3. Fiberboard sheathing, 13 mm 0.23 0.23

4a. Mineral fiber insulation, 140 mm

4b. Wood stud, 38 mm by 140 mm

3.696

--

--

0.98

5. Gypsum wallboard, 13 mm 0.079 0.079

6. Inside surface, still air 0.12 0.12

1

3 4a5

6

4b

Total unit thermal resistance of each section, R (in m2.°C/W) 4.309 1.593

The U-factor of each section, U = 1/R, in W/m2.°C 0.232 0.628

Area fraction of each section, farea 0.80 0.20

Overall U-factor, U = Σfarea,iUi = 0.80×0.232+0.20×0.628 0.311 W/m2.°C

Overall unit thermal resistance, R = 1/U 3.213 m2.°C/W

Therefore, the R-value and U-factor of the wall are R = 3.213 m2.°C/W and U = 0.311 W/m2.°C.


3-102

3-153 The change in the R-value of a wood frame wall due to replacing fiberwood sheathing in the wall by rigid foam sheathing is to be determined.



Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from

Roverall = 1/Uoverall where Uoverall = (Ufarea )insulation + (Ufarea )stud

and the value of the area fraction farea is 0.80 for insulation section and 0.20 for stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available R-values from Table 3-6 and calculating others, the total R-values for each section of the existing wall is determined in the table below.

R -value, m2.°C/W

Construction Between studs

At studs

1. Outside surface, 12 km/h wind 0.044 0.044

2. Wood bevel lapped siding 0.14 0.14

3. Rigid foam, 25 mm 0.98 0.98

4a. Mineral fiber insulation, 140 mm

4b. Wood stud, 38 mm by 140 mm

3.696

--

--

0.98

5. Gypsum wallboard, 13 mm 0.079 0.079


Total unit thermal resistance of each section, R (in m2.°C/W) 5.059 2.343





The R-value of the existing wall is R = 3.213 m2.°C/W. Then the change in the R-value becomes

21.7%)(or 217.0105.4

213.3105.4oldvalue,

valueChange % =−

=−

−∆=

RR

1 2

3 4a5

6

4b


3-103

++= RRRR

hen the U-value of the wall after modification becomes 2 °⋅

rate of eat transfer through the modified wall is

3-154 The U-value of a wall is given. A layer of face brick is added to the outside of a wall, leaving a 20-mm air space between the wall and the bricks. The new U-value of the wall and the rate of heat transfer through the wall is to be determined.


Properties The U-value of a wall is given to be U = 2.25 W/m2.°C. The R - values of 100-mm face brick and a 20-mm air space between the wall and the bricks various layers are 0.075 and 0.170 m2.°C/W, respectively.

Analysis The R-value of the existing wall for the winter conditions is

Face brick

Existing wall

C/Wm 444.025.2/1/1 2 wallexisting wallexisting °⋅=== UR

Noting that the added thermal resistances are in series, the overall R-value of the wall becomes

layerair brick wallexisting wallmodified C/Wm 689.0170.0075.044.0 2 °⋅=++=

T

1.45=== 689.0/1/1 wallmodified wallmodified UR C/Wm

The h

W1431=°−−×°⋅=−= C])25(22)[m 73)(C W/m45.1()()( 22wallwall oi TTUAQ&


3-104

3-155 The winter R-value and the U-factor of a flat ceiling with an air space are to be determined for the cases of air space with reflective and nonreflective surfaces. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the ceiling is one-dimensional. 3 Thermal properties of the ceiling and the heat transfer coefficients are constant. Properties The R-values are given in Table 3-8 for different materials, and in Table 3-11 for air layers. Analysis The schematic of the ceiling as well as the different elements used in its construction are shown below. Heat transfer through the air space and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the air space and stud sections are available, the overall average thermal resistance for the entire wall can be determined from Roverall = 1/Uoverall where Uoverall = (Ufarea )air space + (Ufarea )stud

and the value of the area fraction farea is 0.82 for air space and 0.18 for stud section since the headers which constitute a small part of the wall are to be treated as studs.

(a) Nonreflective surfaces, 9.021 == εε and thus 82.019.0/19.0/1

11/1/1 1 +ε

1

2effective =

−+=

−=

εε

R -value, m2.°C/W Construction en At studs Betwe

studs 1. Still air above ceiling 0.12 0.044 2. Linoleum (R = 0.009 m2.°C/W) 0.009 0.14 3. Felt (R = 0.011 m2.°C/W) 23 0.011 0.4. Plywood, 13 mm 0.11 5. Wood subfloor (R = 0.166 m2.°C/W) 0.166 6a. Air space, 90 mm, nonreflective

m

6b. Wood stud, 38 mm by 90 m0.16---

--- 0.63

7. Gypsum wallboard, 13 mm 0.079 0.0798. Still air below ceiling 0.12 0.12

2.°C/W) Total unit thermal resistance of each section, R (in m 0.775 1.243

The U-factor of each section, U = 1/R, in W/m2.°C 1.290 0.805Area fraction of each section, farea 0.82 0.18 Overall U-factor, U = Σfarea,iUi = 0.82×1.290+0.18×0.805 1.203 W/m2.°C Overall unit thermal resistance, R = 1/U 0.831 m2.°C/W

(b) One-reflective surface, 9.0 and 05.0 21 == εε → 05.011effective ===ε

19.0/105.0/11/1/1 21 −+−+ εε

this case we replace item 6a from 0.16 to 0.47 m2.°C/W. It gives R = 1.085 m2.°C/W and U = 0.922 W/ m2.°C for the air sp

ea,iUi = 0.82×1.085+0.18×0.805 1.035 W/m .°C

Inace. Then,

2Overall U-factor, U = Σfar

Overall unit thermal resistance, R = 1/U 0.967 m .°C/W 2

(c) Two-reflective surface, 05.021 == εε → 03.0105.0/105.0/1

11/1/1

1

21effective =

−+=

−+=

εεε

In t g 2.°C/W and U = 0.905 W/ m2.°C for the air sp


this case we replace item 6a from 0.16 to 0.49 m2.°C/W. I ives R = 1.105 mace. Then,


1 2 3 4 5 6 7 8


3-105

3-156 The winter R-value and the U-factor of a masonry cavity wall are to be determined.



Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Heat transfer through the air space and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the air space and stud sections are available, the overall average thermal resistance for the entire wall can be determined from

Roverall = 1/Uoverall where Uoverall = (Ufarea )air space + (Ufarea )stud

and the value of the area fraction farea is 0.84 for air space and 0.16 for the ferrings and similar structures. Using the available R-values from Tables 3-8 and 3-11 and calculating others, the total R-values for each section of the existing wall is determined in the table below.

R -value, m2.°C/W

Construction Between furring

At furring

1. Outside surface, 24 km/h 0.030 0.030

2. Face brick, 100 mm 0.12 0.12

3. Air space, 90-mm, nonreflective 0.16 0.16

4. Concrete block, lightweight, 100-mm

0.27 0.27

5a. Air space, 20 mm, nonreflective

5b. Vertical ferring, 20 mm thick

0.17

---

---

0.94

6. Gypsum wallboard, 13 0.079 0.079


6 5a

3 4

5b

2 1

7

Total unit thermal resistance of each section, R 0.949 1.719





Therefore, the overall unit thermal resistance of the wall is R = 1.02 m2.°C/W and the overall U-factor is U = 0.978 W/m2.°C. These values account for the effects of the vertical ferring.


3-106

3-157 The winter R-value and the U-factor of a masonry cavity wall with a reflective surface are to be determined.


Properties The R-values of different materials are given in Table 3-8. The R-values of air spaces are given in Table 3-11.



and the value of the area fraction farea is 0.84 for air space and 0.16 for the ferrings and similar structures. For an air space with one-reflective surface, we have 9.0 and 05.0 21 == εε , and thus

05.019.0/105.0/11/1/1 21 −+−+ εε

Using the available R-values from Tables 3-8 an

11effective ===ε

d 3-11 and calculating others, the total R-values for each section of the xisting wall is determined in the table below.

/W

e

R -value, m2.°C

Construction

1 2 3

4 5a

6

Betweenfurring

At furring

1. Outside surface, 24 km/h 0.030 0.0302. Face brick, 100 mm 0.12 0.12 3. Air space, 90-mm, reflective with ε = 0.45 0.45 0.05 4. Concrete block, lightweight, 100-mm 0.27 0.275a. Air space, 20 mm, reflective with ε

thick

--- 0.94 =0.05

5b. Vertical ferring, 20 mm

0.49 ---

6. Gypsum wallboard, 13 0.079 0.0797. Inside surface, still air 0.12 0.12






Therefore, the overall unit thermal resistance of the wall is R = 1.62 m2.°C/W and the overall U-factor is U = 0.618 W/m2.°C.

Discussion The change in the U-value as a result of adding reflective surfaces is

These values account for the effects of the vertical ferring.

368.0978.0ivenonreflect value,−U

618.0978.0valueChange =−

=−∆

=U

Therefore, the rate of heat transfer through the wall will decrease by 36.8% as a result of adding a reflective surface.


3-107

3-158 The winter R-value and the U-factor of a masonry wall are to be determined.



Analysis Using the available R-values from Tables 3-8, the total R-value of the wall is determined in the table below.

R-value,

Construction m2.°C/W

1. Outside surface, 24 km/h 0.030

2. Face brick, 100 mm 0.075

3. Common brick, 100 mm 0.12

4. Urethane foam insulation, 25-mm 0.98

5. Gypsum wallboard, 13 mm 0.079

6. Inside surface, still air 0.12

52 4 1 3 6

Total unit thermal resistance of each section, R 1.404 m2.°C/W

The U-factor of each section, U = 1/R 0.712 W/m2.°C

Therefore, the overall unit thermal resistance of the wall is R = 1.404 m2.°C/W and the overall U-factor is U = 0.712 W/m2.°C.

3-159 The U-value of a wall under winter design conditions is given. The U-value of the wall under summer design conditions is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant except the one at the outer surface.

Properties The R-values at the outer surface of a wall for summer (12 km/h winds) and winter (24 km/h winds) conditions are given in Table 3-8 to be Ro, summer = 0.044 m2.°C/W and Ro, winter = 0.030 m2.°C/W.

Analysis The R-value of the existing wall is

WALL Ro, winter Winter wi C/Wm 714.040.1/1/1 2

winternter °⋅=== UR

Noting that the added and removed thermal resistances are in series, the overall R-value of the wall under summer conditions becomes

C/Wm 728.0

044.0030.0714.02

summero,wintero,wintersummer

°⋅=

+−=

+−= RRRR

WALL Ro, summer Summer

Then the summer U-value of the wall becomes

C/Wm 1.37 2 °⋅=== 728.0/1/1 summersummer UR


3-108

3-160E The R-value and the U-factor of a masonry cavity wall are to be determined.





and the value of the area fraction farea is 0.80 for air space and 0.20 for the ferrings and similar structures. Using the available R-values from Table 3-8 and calculating others, the total R-values for each section of the existing wall is determined in the table below.

R -value, h.ft2.°F/Btu

Construction Between furring

At furring

1. Outside surface, 15 mph wind 0.17 0.17

2. Face brick, 4 in 0.43 0.43

3. Cement mortar, 0.5 in 0.10 0.10

4. Concrete block, 4-in 1.51 1.51

5a. Air space, 3/4-in, nonreflective

5b. Nominal 1 × 3 vertical furring

2.91

--

--

0.94

6. Gypsum wallboard, 0.5 in 0.45 0.45


1 2 3

4 5a

6 7

5b


The U-factor of each section, U = 1/R, in Btu/h.ft2.°F 0.160 0.234


Overall U-factor, U = Σfarea,iUi = 0.80×0.160+0.20×0.234 0.175 Btu/h.ft2.°F

Overall unit thermal resistance, R = 1/U 5.72 h.ft2.°F/Btu

Therefore, the overall unit thermal resistance of the wall is R = 5.72 h.ft2.°F/Btu and the overall U-factor is U = 0.175 Btu/h.ft2.°F. These values account for the effects of the vertical ferring.


3-109

3-161 The summer and winter R-values of a masonry wall are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant. 4 The air cavity does not have any reflecting surfaces.


Analysis Using the available R-values from Tables 3-8, the total R-value of the wall is determined in the table below.

R -value, m2.°C/W

Construction Summer Winter

1a. Outside surface, 24 km/h (winter)

1b. Outside surface, 12 km/h (summer)

---

0.044

0.030

---

2. Face brick, 100 mm 0.075 0.075

3. Cement mortar, 13 mm 0.018 0.018

4. Concrete block, lightweight, 100 mm 0.27 0.27

5. Air space, nonreflecting, 40-mm 0.16 0.16

5. Plaster board, 20 mm 0.122 0.122


6 5

4 3

2 1

7

Total unit thermal resistance of each section (the R-value) , m2.°C/W 0.809 0.795

Therefore, the overall unit thermal resistance of the wall is R = 0.809 m2.°C/W in summer and R = 0.795 m2.°C/W in winter.


3-110

mph 15 o,mph 7.5 o,mph 7.5 wall,mph 15 all,

3-162E The U-value of a wall for 7.5 mph winds outside are given. The U-value of the wall for the case of 15 mph winds outside is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant except the one at the outer surface.

Properties The R-values at the outer surface of a wall for summer (7.5 mph winds) and winter (15 mph winds) conditions are given in Table 3-8 to be

WALL Outside 7.5 mph

Inside

Ro, 7.5 mph = Ro, summer = 0.25 h.ft2.°F/Btu

and Ro, 15 mph = Ro, winter = 0.17 h.ft2.°F/Btu

Analysis The R-value of the wall at 7.5 mph winds (summer) is

F/Btuh.ft 33.13075.0/1/1 2mph 7.5 wall,mph 7.5 wall, °⋅=== UR

Noting that the added and removed thermal resistances are in series, the overall R-value of the wall at 15 mph (winter) conditions is obtained by replacing the summer value of outer convection resistance by the winter value,

w

F/Btuh.ft 25.1317.025.033.13 2 °⋅=+−=

+−= RRRR

n the U-value of the wall at 15 mph winds becomes

on the U-value of the wall is less than 1 percent since

WALL Outside 15 mph

Inside

The

FBtu/h.ft 0.0755 2 °⋅=== 25.13/1/1 mph 15 wal,mph 15 wall, UR

Discussion Note that the effect of doubling the wind velocity

0.67%)(or 0067.0075.0

075.00755.0valuevalueChange =

−=

−−∆

=UU


3-111

3-163 Two homes are identical, except that their walls are constructed differently. The house that is more energy efficient is to be determined.

Assumptions 1 The homes are identical, except that their walls are constructed differently. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant.


Analysis Using the available R-values from Tables 3-8, the total R-value of the masonry wall is determined in the table below.

R -value,


1. Outside surface, 24 km/h (winter) 0.030

2. Concrete block, light weight, 200 mm 2×0.27=0.54

3. Air space, nonreflecting, 20 mm 0.17

5. Plasterboard, 20 mm 0.12

6. Inside surface, still air 0.12

5 2 4 1 3 6

Total unit thermal resistance (the R-value) 0.98 m2.°C/W

which is less than 2.4 m2.°C/W. Therefore, the standard R-2.4 m2.°C/W wall is better insulated and thus it is more energy efficient.

3-164 A ceiling consists of a layer of reflective acoustical tiles. The R-value of the ceiling is to be determined for winter conditions.

Assumptions 1 Heat transfer through the ceiling is one-dimensional. 3 Thermal properties of the ceiling and the heat transfer coefficients are constant.

Properties The R-values of different materials are given in Tables 3-8 and 3-9.

Analysis Using the available R-values, the total R-value of the ceiling is determined in the table below.

R -value,


1. Still air, reflective horizontal surface facing up

R = 1/h = 1/4.32

= 0.23

2. Acoustic tile, 19 mm 0.32

3. Still air, horizontal surface, facing down

R = 1/h = 1/9.26

= 0.11

Highly Reflective

foil

19 mmAcoustical tiles

Total unit thermal resistance (the R-value) 0.66 m2.°C/W

Therefore, the R-value of the hanging ceiling is 0.66 m2.°C/W.


3-112

Review Problems

3-165 A nuclear fuel rod is encased in a concentric hollow ceramic cylinder, which created an air gap between the rod and the hollow cylinder. The surface temperature of the fuel rod is to be determined.

Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat generation in the fuel rod is uniform. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of ceramic is given to be 0.07 W/m · °C.

Analysis The combined thermal resistance between the nuclear fuel rod surface and the outer surface of the ceramic cylinder is

LkDD

LhDLhD

RRRR

πππ 2)/ln(11 23

21

cyl cond,cyl conv,rod conv,co

mbined

++=

++=

or

C/Wm635.5)C W/m07.0(2)C W/m10)(m 035.0()C W/m10)(m 015.0( 22 °⋅°⋅°⋅ πππ

)035.0/110.0ln(112

)/ln(11 23

21combined

°⋅=

++=

++=πππ k

DDhDhD

LR

he heat g y the fuel rod is dissipated through the air gap and the ceramic cylinder, and can be expressed as

T enerated b

LRTT

LQgen =&

combined

31gen

TTQ

−=& or

R combined

31 −

The surface temperature of the fuel rod is

3combined1 TLRL

T +⎟⎠

⎜⎝

= genQ ⎟⎞

⎜⎛ &

C 1026 °=°+°⋅×= C 30)C/Wm 635.5()m 015.0(4

) W/m101( 2361

πT

Discussion The air gap between the fuel rod and the hollow ceramic cylinder contributed about 54% to the combined thermal resistance between the nuclear fuel rod surface and the outer surface of the ceramic cylinder.


3-113

m 0942.0)m 1)(m 03.0( 21fin === LDA ππ

The efficiency of these circular fins is, from the efficiency curve, Fig. 3-44

3-166 Circular aluminum alloy fins are to be attached to the tubes of a heating system. The increase in heat transfer from the tubes per unit length as a result of adding fins is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire fin surfaces. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the fins is given to be k = 180 W/m⋅°C.

Analysis In case of no fins, heat transfer from the tube per meter of its length is

no

W537C)25120)(m 0942.0)(C. W/m60()( 22fin nofin no =°−°=−= ∞TThAQ b

&

96.0

207.0m) C)(0.002 W/m180(

C W/m602⎝

002.0015.0

2

07.20.015

)2/002.0(03.0)2/(m 015.02/)03.006.0(2/)(

fin

o

o2

2/12/3

1

2

12

=

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

=⎟⎠⎞

⎜⎛ +=

⎟⎠⎞

⎜⎝⎛ +=⎟

⎟⎠

⎞⎜⎜⎝

⎛

=+

=+

=−=−=

ηkthtL

kAhL

rtr

DDL

pc

eat transf from a single fin is

22

finfinmaxfin,finfin

2

°−°=

−==

=+

∞TThAQQ bηη

π&&

gle unfinned portion of the tube is 2

1fin === sDA ππ

here are 200 fins and thus 200 interfin spacings per meter length of the tube. The total heat transfer from the finned tube is

Therefore, the increase in heat transfer from the tube per meter of its length as a result of the addition of the fins is

Discussion The The overall effectiveness of the finned tube is 5380/537 = 10. That is, the rate of heat transfer from the steam tube increases by a factor of 10 as a result of adding fins. This explains the widespread use of finned surfaces.

H er

C)25120)(m 004624.0)(C. W/m60(96.0

)(

m 004624.0)002.0)(03.0(2)015.003.0(22)(2 222

21

22fin −=+−= trrrA πππ

W3.25=

Heat transfer from a sin

W6.1C)25120)(m 000283.0)(C. W/m60()( 22

unfinunfin =°−°=−= ∞TThAQ b&

m 000283.0m) m)(0.003 03.0(un

Tthen determined from

W5380)6.13.25(200)( unfinfinfintotal, =+=+= QQnQ &&&

W4843=−=−= 5375380fin nofintotal,increase QQQ &&&


3-114

3-167E Steam is produced in copper tubes by heat transferred from another fluid condensing outside the tubes at a high temperature. The rate of heat transfer per foot length of the tube when a 0.01 in thick layer of limestone is formed on the inner surface of the tube is to be determined.


Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper tubes and k = 1.7 Btu/h⋅ft⋅°F for limestone.

Rtotal, new HX

T∞1T∞2Analysis The total thermal resistance of the new heat exchanger is

F/Btuh. 0035.0Btu/h 102 4

newnewtotal, ×QR &

After 0.01 in thick layer of limestone forms, the new v

F)280350(21newtotal,

21 °=°−

=−

=⎯→⎯−

= ∞∞∞∞ TTR

TTQ&

alue of thermal resistance and heat transfer rate are determined to be

new

Rlimestone

T∞2

Rtotal, new HX

T∞1

F/Btuh 00539.000189.00035.0

F/Btuh 00189.0)ft 1(F)Btu/h.ft. 7.1(2

)49.0/5.0ln(2

)/ln(

ilimestone,newtotal,w/limetotal,

1ilimestone,

°=+=+=

°=°

==

RRRkL

rrR i

ππ

35%) of decline (a F/Btuh 0.00539

F)280350(

w/limetotal,

21w/lime Btu/h 101.3 4×=

°°−

=−

= ∞∞

RTTQ&

Discussion Note that the limestone layer will change the inner surface area of the pipe and thus the internal convection resistance slightly, but this effect should be negligible.


3-115

3-168E Steam is produced in copper tubes by heat transferred from another fluid condensing outside the tubes at a high temperature. The rate of heat transfer per foot length of the tube when a 0.01 in thick layer of limestone is formed on the inner and outer surfaces of the tube is to be determined.


Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper tubes and k = 1.7 Btu/h⋅ft⋅°F for limestone. Rtotal, new HX

T∞1Analysis The total thermal resistance of the new heat exchanger is T∞2

F/Btuh. 0035.0Btu/h 102

F)280350(4

new

21newtotal,

newtotal,

21new °=

×°−

=−

=⎯→⎯−

= ∞∞∞∞

QTT

RR

TTQ

&&

After 0.01 in thick layer of limestone forms, the new value of thermal resistance and heat transfer rate are determined to be

Rlimestone, oRtotal, new HXRlimestone, i

T∞1 T∞2

F/Btuh. 00143.0)ft 1(F)Btu/h.ft. 7.1(2

)65.0/66.0ln(2

)/ln(

F/Btuh. 00189.0)ft 1(F)Btu/h.ft. 7.1(2

)49.0/5.0ln(2

)/ln(

2olimestone,

1ilimestone,

°=°

==

°=°

==

kLrr

R

kLrr

R

o

i

ππ

ππ

F/Btuh. 00682.000143.000189.00035.0olimestone,ilimestone,newtotal,w/limetotal, °=++=++= RRRR

49%) of decline (a F/Btuh 0.00682

F)280350(

w/limetotal,

21w/lime Btu/h 101.03 4×=

°°−

=−

= ∞∞

RTTQ&

Discussion Note that the limestone layer will change the inner surface area of the pipe and thus the internal convection resistance slightly, but this effect should be negligible.


3-116

3-169 Hot water is flowing through a 15-m section of a cast iron pipe. The pipe is exposed to cold air and surfaces in the basement, and it experiences a 3°C-temperature drop. The combined convection and radiation heat transfer coefficient at the outer surface of the pipe is to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of any significant change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no significant variation in the axial direction. 3 Thermal properties are constant.

Properties The thermal conductivity of cast iron is given to be k = 52 W/m⋅°C.

Analysis Using water properties at room temperature, the mass flow rate of water and rate of heat transfer from the water are determined to be

[ ]

W13,296=C)6770)(CJ/kg. kg/s)(4180 06.1( −°=∆= TcmQ p&&

The thermal resistances for convection in the pipe and

kg/s 06.1m4/(0.03)m/s) )(1.5kg/m 1000( 223

°

=== VAm cc&& πρρV

the pipe itself are

=

C/W 001768.0m)]15((0.03)C)[. W/m400( 22iconv,

° πii Ah11

C/W 000031.0)m 15(C) W/m.52(2

)5.1/75.1ln(2

)/ln( 12pipe

°===

°=°

=

=

π

π

R

kLrr

R

sing arith etic mean temperature (70+67)/2 = 68.5°C for water, the heat transfer can be expressed as

Rconv ,i Rpipe Rcombined ,o

T∞1 T∞2

U m

o

aveaveave TTQ ,1,

AhRR

TTRRRTT

Rcombined

pipeiconv,

2,1,

ocombined,pipeiconv,

2,1,

total

2

1++

−=

++

−= ∞∞∞∞∞

Substituting,

−= ∞&

2combined )]m(0.035)(15[

1+C/W) 001768.0(+C/W) 000031.0(

C)155.68( W296,13 −=

πh°°

°

Solving for the combined heat transfer coefficient gives

C. W/m272.5 2 °=combinedh


3-117

3-170 An 8-m long section of a steam pipe exposed to the ambient is to be insulated to reduce the heat loss through that section of the pipe by 90 percent. The amount of heat loss from the steam in 10 h and the amount of saved per year by insulating the steam pipe.

Assumptions 1 Heat transfer through the pipe is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficients accounts for the radiation effects. 5 The temperatures of the pipe surface and the surroundings are representative of annual average during operating hours. 6 The plant operates 110 days a year.

Analysis The rate of heat transfer for the uninsulated case is Tair =8°C Ts =90°C2

o m 016.3m) 8(m) 12.0( === ππ LDA o Steam pipe

W8656C)890)(m 016.3)(C. W/m35()( 22 =°−°=−= airso TThAQ& The amount of heat loss during a 10-hour period is

day)(per )s 3600kJ/s)(10 656.8( kJ 103.116 5×=×=∆= tQQ &

The steam generator has an efficiency of 85%, and steam heating is used for 110 days a year. Then the amount of natural gas consumed per year and its cost are

therms/yr2.382days/yr) 110(kJ 105,500

therm10.85

kJ 10116.3used Fuel5

=⎟⎟⎠

⎞⎜⎜⎝

⎛×=

fuel) ofcost fuel)(Unit ofAmount (fuel ofCost =

$458.7/yr=erm))($1.20/th therms/yr2.382(=

hen the money saved by reducing the heat loss by 90% by insulation becomes

T

$413=$458.7/yr0.9fuel) of(Cost 0.9=savedMoney ×=×


3-118

3-171 A multilayer circuit board dissipating 27 W of heat consists of 4 layers of copper and 3 layers of epoxy glass sandwiched together. The circuit board is attached to a heat sink from both ends maintained at 35°C. The magnitude and location of the maximum temperature that occurs in the board is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be approximated as being one-dimensional. 3 Thermal conductivities are constant. 4 Heat is generated uniformly in the epoxy layers of the board. 5 Heat transfer from the top and bottom surfaces of the board is negligible. 6 The thermal contact resistances at the copper-epoxy interfaces are negligible.

Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper layers and k = 0.26 W/m⋅°C for epoxy glass boards.

Analysis The effective conductivity of the multilayer circuit board is first determined to be Copper

C W/m.48.58m)0015.0(3)0002.0(4[CW/)00117.03088.0()()(

C W/00117.0)]m 0015.0)(C W/m.26.0[(3)(

C W/3088.0)]m 0002.0)(C W/m.386[(4)(

epoxycopper

epoxycopper

epoxy

copper

°=+

°+=

+

+=

°=°=

°=°=

ttktkt

k

kt

kt

eff

Epoxy

The maximum temperature will occur at the midplane of the board that is the farthest to the heat sink. Its value is

C56.8°=°

+°=+==

−=

=+=

)m 000954.0)(C W/m.48.58()m 2/18.0)( W2/27(C35

)(

m 000954.0)]0015.0(3)0002.0(4[18.0

2eff

21max

21eff

2

AkLQTTT

TTL

AkQ

A

&

&


3-119

3-172 The plumbing system of a house involves some section of a plastic pipe exposed to the ambient air. The pipe is initially filled with stationary water at 0°C. It is to be determined if the water in the pipe will completely freeze during a cold night.

Assumptions 1 Heat transfer is transient, but can be treated as steady since the water temperature remains constant during freezing. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties of water are constant. 4 The water in the pipe is stationary, and its initial temperature is 0°C. 5 The convection resistance inside the pipe is negligible so that the inner surface temperature of the pipe is 0°C.

Properties The thermal conductivity of the pipe is given to be k = 0.16 W/m⋅°C. The density and latent heat of fusion of water at 0°C are ρ = 1000 kg/m3 and hif = 333.7 kJ/kg (Table A-9).

Analysis We assume the inner surface of the pipe to be at 0°C at all times. The thermal resistances involved and the rate of heat transfer are

C/W 0258.16631.03627.0

C/W 6631.0m)] 5.0)(m 024.0(C)[. W/m40(

11

C/W 3627.0)m 5.0(C) W/m.16.0(2

)1/2.1ln(2

)/ln(

oconv,pipetotal

2oconv,

12pipe

°=+=+=

°=°

==

°=°

==

RRRAh

R

kLrr

R

o π

ππ

W874.4C/W 1.0258total °R

C)]5(0[21 =°−−

=− ∞TT

Q s&

he total amount of heat lost by the water during a 14-h period that night is

he amount of heat required to freeze the water in the pipe com

he water in the pipe will freeze completely that night since the amount heat loss is greater than the amount it takes to freeze the water completely .

Water pipe

Tair = -5°C

Soil=

T

kJ 7.245)s 3600J/s)(14 874.4( =×=∆= tQQ &

T pletely is

kg 157.0)m 5.0()m 01.0()kg/m 1000( 232 ==== πρπρ Lrm V

kJ 4.52kJ/kg) 7.333kg)( 157.0 === fgmhQ (

T)4.527.245( >


3-120

3-173 The plumbing system of a house involves some section of a plastic pipe exposed to the ambient air. The pipe is initially filled with stationary water at 0°C. It is to be determined if the water in the pipe will completely freeze during a cold night.

Assumptions 1 Heat transfer is transient, but can be treated as steady since the water temperature remains constant during freezing. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties of water are constant. 4 The water in the pipe is stationary, and its initial temperature is 0°C. 5 The convection resistance inside the pipe is negligible so that the inner surface temperature of the pipe is 0°C.

Properties The thermal conductivity of the pipe is given to be k = 0.16 W/m⋅°C. The density and latent heat of fusion of water at 0°C are ρ = 1000 kg/m3 and hif = 333.7 kJ/kg (Table A-9).

Analysis We assume the inner surface of the pipe to be at 0°C at all times. The thermal resistances involved and the rate of heat transfer are

C/W 8364.14737.13627.0

C/W 4737.1m)] 5.0)(m 024.0(C)[. W/m18(

11

C/W 3627.0)m 5.0(C) W/m.16.0(2

)1/2.1ln(2

)/ln(

oconv,pipetotal

2oconv,

212

pipe

°=+=+=

°=°

==

°=°

==

RRRAh

R

kLrrR

o π

ππ

W723.2C/W 1.8364

C)]5(0[

total

21 =°°−−

=−

= ∞∞

RTTQ&

he amount of heat required to freeze the water in the pipe completely is

()m 01.0()kg/m 1000( 232 ==== πρπρ Lrm V

e completely that night since the amount heat loss is greater than the amount it takes to freeze e water completely .

Water pipe

Tair = -5°C

Soil kJ 2.137J 240,137)s 3600J/s)(14 723.2( ==×=∆= tQQ &

T

0 kg 157.0)m 5.

kJ 4.52kJ/kg) 7.333kg)( 157.0( === fgmhQ

The water in the pipe will freezth )4.5257.83( >


3-121

3-174E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is wrapped completely in a towel are to be determined.

Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2 The thermal contact resistance at the interface is negligible. 3 The heat transfer coefficients for wrapped and unwrapped potatoes are the same.

Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h⋅ft⋅°F. We take the properties of potato to be those of water at room temperature, ρ = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm⋅°F.

Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the potato cools down and the temperature difference between the potato and the surroundings decreases. However, we can solve this problem approximately by assuming a constant average temperature of (300+200)/2 = 250°F for the potato during the process. The mass of the potato is

lbm 5089.0

)ft 12/5.1(34)lbm/ft 2.62(

34

33

3

=

=

==

π

πρρ rm V

Ts

Potato

Rtowel RconvT∞

The amount of heat lost as the potato is cooled from 300 to 200°F is

Btu 8.50F200)-F)(300Btu/lbm. lbm)(0.998 5089.0( =°°=∆= TmcQ p

The rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings are

Btu/h 6.609h) 60/5(

Btu 8.50==

∆=

tQQ&

F.Btu/h.ft 17.2 °=°−

=−

=⎯→⎯−=∞

∞F)70250()ft (3/12)(

)(2πTTA

hTThAQso

so 2Btu/h 6.609Q&&

When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be

F/Btuh 6012.12539.03473.1

F/Btuh. 2539.0ft )12/24.3(F).Btu/h.ft 2.17(

11

F/Btuh 3473.1ft)12/5.1(ft]12/)12.05.1(F)[Btu/h.ft. 035.0(4

ft)12/5.1(ft]12/)12.05.1[(4

convtoweltotal += R

222conv

21

12towel

°=+=

°=°

==

°=+°−+

=−

=

RRhA

R

rkrrr

R

π

ππ

Btu/h 4.112F/Btuh 1.6012

F)70250(

total=

°°−

=−

= ∞

RTT

Q s&

min 27.1====∆ h 452.0Btu 8.50Qt Btu/h 4.112Q&

This result is conservative since the heat transfer coefficient will be lower in this case because of the smaller exposed surface temperature.


3-122

3-175E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is loosely wrapped completely in a towel are to be determined.

Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2 The heat transfer coefficients for wrapped and unwrapped potatoes are the same.

Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h⋅ft⋅°F. The thermal conductivity of air is given to be k = 0.015 Btu/h⋅ft⋅°F. We take the properties of potato to be those of water at room temperature, ρ = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm⋅°F.

Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the potato cools down and the temperature difference between the potato and the surroundings decreases. However, we can solve this problem approximately by assuming a constant average temperature of (300+200)/2 = 250°F for the potato during the process. The mass of the potato is

lbm 5089.0=

)ft 12/5.1(3

)lbm/ft 2.62(

4

33

3

=

=

π

πρρ rm V

he amount of heat lost as the potato is cooled from 300 to 200°F is

he rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings are

= Ts

Potato

Rair RconvRtowel

43

T∞

T

Btu 8.50F200)F)(300Btu/lbm. lbm)(0.998 5089.0( =°−°=∆= TmcQ p

T

Btu/h 6.609Btu 8.50===

QQ& h) 60/5(∆t

F.Btu/h.ft 17.2 2 °=°

− 70250()ft (3/12 2π

=−

=⎯→⎯−=∞

∞F)

Btu/h 6.609)(

)(TTA

QhTThAQso

so

&&

When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be

F/Btuh 1195.22477.03134.15584.0

F/Btuh. 2477.0ft )12/28.3(F).Btu/h.ft 2.17(

11

F/Btuh 3134.1ft)12/52.1(ft]12/)12.052.1(F)[Btu/h.ft. 035.0(4

ft)12/52.1(ft]12/)12.052.1[(4

F/Btuh. 5584.0ft)12/50.1(ft]12/)02.050.1(F)[Btu/h.ft. 015.0(4

ft)12/50.1(ft]12/)02.050.1[(4

12air °=

+°−+

=−

=rkrrr

Rππ

convtowelairtotal

222conv

32

23towel

21

°=++=++=

°=°

==

°=+°−+

=−

=

RRRRhA

R

rkrrr

R

π

ππ

Btu/h 9.84F/Btuh. 2.1195

F)70250(

total=

°°−

=−

= ∞

RTT

Q s&

min 35.9====∆ h 598.0Btu/h 9.84Btu 8.50

QQt&

This result is conservative since the heat transfer coefficient will be lower because of the smaller exposed surface temperature.


3-123

3-176 A wall constructed of three layers is considered. The rate of hat transfer through the wall and temperature drops across the plaster, brick, covering, and surface-ambient air are to be determined.

Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is accounted for in the heat transfer coefficient.

Properties The thermal conductivities of the plaster, brick, and covering are given to be k = 0.72 W/m⋅°C, k = 0.36 W/m⋅°C, k = 1.40 W/m⋅°C, respectively.

Analysis The surface area of the wall and the individual resistances are

2m 8.16m) 8.2(m) 6( =×=A

C/W 02253.000350.000085.001653.000165.0

C/W00350.0)m 8.16(C). W/m17(

11

C/W 00085.0)m 8.16(C) W/m.4.1(

m 02.0

C/W 01653.0)m 8.16(C) W/m.72.0(

m 20.0

C/W 00165.0m 01.0

conv,2321total

222

conv,2o

23

3

22

2brick2

1

°=+++=

+++=

°=°

===

°=°

=

°=°

===

°====

RRRRRAh

R

AkLAk

LRR

LR

The steady rate of heat transfer through the wall then becomes

)m 8.16(C) W/m.36.0( 21

plaster1°Ak

R

covering3 == RR

RT1 T∞2

R1 R2 R3 Ro

W665.8=°°−

=−

= ∞

C/W02253.0C)823(

total

21

RTT

Q&

The temperature drops are

C 2.3

C 0.6

C 11.0

C 1.1

°=°==∆

°=°==∆

°=°==∆

°=°==∆

)C/W00350.0)( W8.665(

)C/W00085.0)( W8.665(

)C/W01653.0)( W8.665(

)C/W00165.0)( W8.665(

convconv

coveringcovering

brickbrick

plasterplaster

RQT

RQT

RQT

RQT

&

&

&

&


3-124

3-177 An insulation is to be added to a wall to decrease the heat loss by 90%. The thickness of insulation and the outer surface temperature of the wall are to be determined for two different insulating materials. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is accounted for in the heat transfer coefficient. Properties The thermal conductivities of the plaster, brick, covering, polyurethane foam, and glass fiber are given to be 0.72 W/m⋅°C, 0.36 W/m⋅°C, 1.40 W/m⋅°C, 0.025 W/m⋅°C, 0.036 W/m⋅°C, respectively. Analysis The surface area of the wall and the individual resistances are 2m 8.16m) 8.2(m) 6( =×=A

C/W 02253.0 °=The rate of heat loss without the insulation is

00350.000085.001653.000165.0conv,2321ins no total,

+++

RRRRR

C/W00350.0)m 8.16(C). W/m17(

00085.0)m 8.16(C) W/m.4.1(

C/W 01653.0)m 8.16(C) W/m.72.0(

m 20.0

C/W 00165.0)m 8.6

m 01.0

222

conv,2o

2covering3

2

2brick2

21

+++=

°=°

===

°=°

==

°=°

===

°=

AhRR

AkRR

AkL

RR

L

1(C) W/m.36.0(1plaster1

°===

AkRR

2

m 02.03L= C/W

113

=

W666C/W02253.0ins no total, °R

(a) The rate of heat transfer after insulation is

C)823(21 =°−

=−

= ∞TTQ&

al resistance with the foam insulation is

W6.6666610.015.0 ins noins =×== QQ && The total therm

C) W.m/42.0(C/W 02253.0

)m C)(16.8 W/m.025.0(C/W 02253.0 4

24

conv,2foam321total °+°=

°+°=++++=

LLRRRRRR

The thickness of insulation is determined from

cm 8.51m 0.0851 ==⎯→⎯+°

=⎯→⎯= 44total

insC/W 02253.0

W6.66 LLR

Q&

°

°−− ∞21

C) W.m/42.0(

C)823(TT

The outer surface temperature of the wall is determined from

C8.23°=⎯→⎯°°−

=⎯→⎯−

= ∞2

2

conv

22ins C/W 00350.0

C)8( W6.66 T

TR

TTQ&

(b) The total thermal resistance with the fiberglass insulation is

C) W.m/6048.0(C/W 02253.0

)m C)(16.8 W/m.036.0(C/W 02253.0 4

24

conv,2glassfiber 321total

°+°=

°+°=

++++=

LL

RRRRRR

The thickness of insulation is determined from

cm 12.3m 0.123 ==⎯→⎯

°+°

°−=⎯→⎯

−= ∞

44total

21ins

C W.m/6048.0(C/W 02253.0

C)823( W6.66 LLR

TTQ&

The outer surface temperature of the wall is determined from

C8.23°=⎯→⎯°°−

=⎯→⎯−

= ∞2

2

conv

22ins C/W00350.0

C)8(6.66 T

TR

TTQ&

Discussion The outer surface temperature is same for both cases since the rate of heat transfer does not change.

T1

R2 R3 R1 Ro Rins

T∞2


3-125

3-178 A circuit board houses electronic components on one side, dissipating a total of 15 W through the backside of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 20 aluminum fins of rectangular profile on the backside. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the backside of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 12 W/m⋅°C for the circuit board, k = 237 W/m⋅°C for the aluminum plate and fins, and k = 1.8 W/m⋅°C for the epoxy adhesive. Analysis (a) The thermal resistance of the board and the convection resistance on the backside of the board are

C/W 492.1481.1011.0

C/W 481.1)m 15.0)(m 1.0(C) W/m.45(

11

C/W 011.0)m 15.0)(m 1.0(C) W/m.12(

m 002.0

convboardtotal

conv

board

°=+=+=

°=°

==

°=°

==

RRRhA

R

kALR

RconvRboard

T2

T1 T∞

Then surface temperatures on the two sides of the circuit board becomes

C59.2

C59.4

°=°−°=−=⎯→⎯−

=°+°=+=⎯→⎯ ∞ C/W) W)(1.492(15C37

21

total1total

TT

RQTTR

Q

=

°− ∞

C/W) W)(0.011(15C4.59board12board

1

RQTTR

Q

TT

&&

&&

(b) Noting that the cross-sectional areas of the fins are constant, the efficiency of these rectangular fins is determined to be

=

2 cm

1-2

m 78.13)m 002.0)(C W/m.237(

)C. W/m45(22)2(==≅=

hwhhpm)(

=°

°kttwkkAc

975.0m 02.0m 78.13


-1

fin =×

×==

mLmLη

The finned and unfinned surface areas are

2unfinned

2finned m 0.126=

2002.002.0)15.0(2)20(

22)20( ⎟

⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

tLwA

m 0090.0)15.0)(002.0(20)15.0)(1.0( =−=A

unfinnedfinfinbasefinnedunfinnedtotal

baseunfinnedunfinned

basefinfinmaxfin,finfinned

AATThQQQ

TThAQ

+−=+=

−=

∞

∞

∞

η&&&

&

stituting, the base temperature of the finned surfaces is determined to be

Then, )( TThAQQ −== ηη &&

Raluminum

T1

Rboard

T∞

Repoxy

))((

)(

Sub

C39.5°=+°

°=+

+= ∞)]m 0090.0()m 126.0)(975.0)[(C. W/m45(

W15+C37 )( 222

unfinnedfinfin

totalbase AAh

QTT

η

&

Then the temperatures on both sides of the board are determined using the thermal resistance network to be

C/W 01111.0

m) 15.0)(m 1.0(C) W/m.8.1(m 0003.0

C/W 00028.0m) 15.0)(m 1.0(C) W/m.237(

m 001.0

epoxy

aluminum

°=°

==

°=°

==

kALR

kALR

C39.6

C39.8

°=°−°=−=⎯→⎯−

=

°=°+°=⎯→⎯°++

°−=

++−

=

C/W) W)(0.011(15C8.39

C/W) 9 W)(0.0223(15C5.39C/W )011.001111.000028.0(

C)5.39(

board12board

21

11

boardepoxyaluminum

base1

RQTTR

TTQ

TT

RRRTT

Q

&&

&


3-126

3-179 A circuit board houses electronic components on one side, dissipating a total of 15 W through the backside of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 20 copper fins of rectangular profile on the backside. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the backside of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 12 W/m⋅°C for the circuit board, k = 386 W/m⋅°C for the copper plate and fins, and k = 1.8 W/m⋅°C for the epoxy adhesive. Analysis (a) The thermal resistance of the board and the convection resistance on the backside of the board are

C/W 492.1481.1011.0

C/W 481.1)m 15.0)(m 1.0(C) W/m.45(

11

C/W 011.0)m 15.0)(m 1.0(C) W/m.12(board °=

°==

kAR m 002.0

convboardtotal

conv

°=+=+=

°=°

==

RRRhA

R

L

Then surface temperatures on the two sides of the circuit board becomes

RconvRboard

T1 T∞

T2

C59.2

C59.4°=°+°=+=⎯→⎯−

= ∞ C/W) W)(1.492(15C371 RQTTTT

Q &&

°=°−°=−=⎯→⎯−

=

∞

C/W) )(0.011(15C4.59board12board

21

total1total

RQTTR

TTQ

R

&&

the fins are constant, the efficiency of these rectangular fins is determined to be

W

(b) Noting that the cross-sectional areas of

1-2

m 80.10)m 002.0)(C W/m.386(

)C. W/m45(22)()2(

=°

°==≅=

kth

twkwh

kAhpm

c 2 cm

985.0m 02.0m 80.10


-1

fin =×

×==

mLmLη

The finned and unfinned surface areas are

2

unfinned m 0090.0)15.0)(002.0(20)15.0)(1.0( =−=A

2finned m

22 ⎠⎝⎠⎝

Then,

Substituting, the base temperature of the finned surfaces determine to be

0.126=002.002.0)15.0(2)20(2)20( ⎟⎞

⎜⎛ +=⎟

⎞⎜⎛ +=

tLwA

))((

)(

)(

unfinnedfinfinbasefinnedunfinnedtotal

baseunfinnedunfinned

basefinfinmaxfin,finfinned

AATThQQQ

TThAQ

TThAQQ

+−=+=

−=

−==

∞

∞

∞

η

ηη

&&&

&

&&

C39.5°=+°

°=+

+= ∞)]m 0090.0()m 126.0)(985.0)[(C. W/m45(

W15+C37 )( 222

unfinnedfinfin

totalbase AAh

QTT

η

&

Then the temperatures on both sides of the board are determined using the thermal resistance network to be

C/W 01111.0

m) 15.0)(m 1.0(C) W/m.8.1(m 0003.0

C/W 00017.0m) 15.0)(m 1.0(C) W/m.386(

m 001.0

epoxy

copper

°=°

==

°=°

==

kALR

kALR

Rcopper

T1

Rboard

T∞

Repoxy

C39.6

C39.8

°=°−°=−=⎯→⎯−

=

°=°+°=⎯→⎯°++

°−=

++−

=

C/W) W)(0.011(15C8.39

C/W) 8 W)(0.0222(15C5.39C/W )011.001111.000017.0(

C)5.39(

board12board

21

11

boardepoxycopper

1

RQTTR

TTQ

TT

RRRTT

Q base

&&

&


3-127

3-180 Steam passes through a row of 10 parallel pipes placed horizontally in a concrete floor exposed to room air at 24° with a heat transfer coefficient of 12 W/m

C2.°C. If the surface temperature of the concrete floor is not to exceed 38° , the

minimum burial depth of the steam pipes below the floor surface is to be determined. C



Analysis In steady operation, the rate of heat loss from the steam through the concrete floor by conduction must be equal to the rate of heat transfer from the concrete floor to the room by combined convection and radiation, which is determined to be

Room24°C10 m

38°C

W8400C)2438)](m 5)(m 10)[(C. W/m12(

)(2 =°−°=

−= ∞TThAQ ss&

Then the depth the steam pipes should be buried can be determined with the aid of shape factor for this configuration from Table 3-7 to be

pipe)(per m 47.10C)38145)(C W/m.75.0(10

W8400)(

)(21

21 =°−°

=−

=⎯→⎯−=TTnk

QSTTnSkQ&

&

pipes) of distancecenter -to-(center m 110

m 10===

naw

cm 22.2==⎯→⎯

⎥⎦

⎤⎢⎣

⎡=

⎟⎠⎞

⎜⎝⎛

=

m 222.0

)m 1(2sinh

m) (0.06m) 2(1ln

m) 5(2m 47.10

2sinh2ln

2

zz

wz

Dw

LS

ππ

π

ππ

π


3-128

3-181 Two persons are wearing different clothes made of different materials with different surface areas. The fractions of heat lost from each person’s body by perspiration are to be determined.

Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is accounted for in the heat transfer coefficient. 5 The human body is assumed to be cylindrical in shape for heat transfer purposes.

Properties The thermal conductivities of the leather and synthetic fabric are given to be k = 0.159 W/m⋅°C and k = 0.13 W/m⋅°C, respectively.

Analysis The surface area of each body is first determined from

2

12

21

m 335.16675.022

m 6675.0m)/2 m)(1.7 25.0(2/

=×==

===

AA

DLA ππ

The sensible heat lost from the first person’s body is

C/W 10930.009988.000942.0

C/W 09988.0)m 6675.0(C). W/m15(

11

C/W 00942.0)m 6675.0(C) W/m.159.0( 2°kA

m 001.0

convleathertotal

22conv

°=+=+=

°=°

==

°===

RRRhA

R

LR

is the sum of heat transferred through the clothes and the skin

leather Rleather RconvT1 T∞2

The total sensible heat transfer

W3.38203.18skinclothessensible =+=+= QQQ &&&

W0.20C/W09988.0

C)3032(

W3.18C)3032(

conv

21in

21

=°°−

=−

=

=°−

=−

=

∞

∞

RTT

Q

TTQ

&

&

Then the fraction of heat lost by respiration becomes

C/W10930.0totalclothes °R

sk

0.362=−

=−

==60

3.3860

total

sensibletotal

total

nrespiratio

QQQ

Q

Qf

&

&&

&

&&

Repeating similar calculations for the second person’s body

C/W 05570.004994.000576.0

C/W° 04994.0)m 335.1(C). W/m15(

11

C/W 00576.0)m 335.1(C) W/m.13.0(

m 001.0

convleathertotal

22conv

2

°=+=+=

=°

==

°=°

==

RRRhA

R

kALR

synthetic

W9.35C/W05570.0

C)3032(

total

21sensible =

°°−

=−

= ∞

RTT

Q&

T1

Rsynthetic Rconv

T∞2

0.402=−

=−

==60

9.3560

total

sensibletotal

total

nrespiratio

QQQ

Q

Qf

&

&&

&

&&


3-129

3-182 A wall is constructed of two large steel plates separated by 1-cm thick steel bars placed 99 cm apart. The remaining space between the steel plates is filled with fiberglass insulation. The rate of heat transfer through the wall is to be determined, and it is to be assessed if the steel bars between the plates can be ignored in heat transfer analysis since they occupy only 1 percent of the heat transfer surface area.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall can be approximated to be one-dimensional. 3 Thermal conductivities are constant. 4 The surfaces of the wall are maintained at constant temperatures.

Properties The thermal conductivities are given to be k = 15 W/m⋅°C for steel plates and k = 0.035 W/m⋅°C for fiberglass insulation.

Analysis We consider 1 m high and 1 m wide portion of the wall which is representative of entire wall. Thermal resistance network and individual resistances are

R4

R3

R2R1

T1

T2

C/W 1926.100053.01915.100053.0

3492.64667.1

4eqv1total

32eqv

°=++=++= RRRR

RRRC/W 1915.11111

m 22.0

C/W 4667.1)m .0(C) W/m.15(

m 22.0

C/W 00053.0m 008.0

2steel2

steel41

°=⎯→⎯+=+=

°=°

===

°=====

R

LkALRR

LRR

eqv

he rate of heat transfer per m2 surface area of the wall is

R

1

C/W 3492.6)m 99.0(C) W/m.035.0( 2insulation3 °=

°===

kARR

01

)m 1(C) W/m.15( 2°kA

T

W45.18C/W 1.1926total

=°

==R

Q

The total rate of heat transfer through the entire wall is

C 22 °∆T&

then determined to be

If the steel bars were ignored since they constitute only 1% of the wall section, the Requiv would simply be equal to the thermal resistance of the insulation, and the heat transfer rate in this case would be

99 cm

1 cm

0.8 cm 22 cm 0.8 cm

W442.7==×= W)45.18(24)64(total QQ &&

W46.3C/W)00053.03492.600053.0(

C 22

4insulation1total=

°++°

=++

∆=

∆=

RRRT

RTQ&

which is mush less than 18.45 W obtained earlier. Therefore, (18.45-3.46)/18.45 = 81.2% of the heat transfer occurs through the steel bars across the wall despite the negligible space that they occupy, and obviously their effect cannot be neglected. The connecting bars are serving as “thermal bridges.”


3-130

3-183 Cold conditioned air is flowing inside a duct of square cross-section. The maximum length of the duct for a specified temperature increase in the duct is to be determined.

Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Steady one-dimensional heat conduction relations can be used due to small thickness of the duct wall. 5 When calculating the conduction thermal resistance of aluminum, the average of inner and outer surface areas will be used.

Properties The thermal conductivity of aluminum is given to be 237 W/m⋅°C. The specific heat of air at the given temperature is cp = 1006 J/kg⋅°C (Table A-15).

Analysis The inner and the outer surface areas of the duct per unit length and the individual thermal resistances are

Ri Ralum

T∞1

Ro m 0.1m) 1(m) 25.0(44

m 88.0m) 1(m) 22.0(44

===

===

LaA

LaA

12

22

21

T∞2

[ ]

C/W 09214.007692.0000.001515.0oalumitotal +=++= RRRR 07

C/W07692.0)m 0.1(C). W/m13(

11

C/W 00007.0m 2/)188.0(C) W/m.237(

m 015.0

C/W01515.0)m 88.0(C). W/m75(

11

222

o

2alum

221

i

°=+

°=°

==

°=+°

==

°=°

==

AhR

kALR

AhR

he rate of eat loss from the air inside the duct is T h

W228C/W09214.0

C)1233(12 =°−

=−

= ∞∞ TTQ&

total °R

uct should gain heat at a rate of

tal =°°=∆= TcmQ p&&

Then the maximum length of the duct becomes

For a temperature rise of 1°C, the air inside the d

to W805C)1)(CJ/kg. kg/s)(1006 8.0(

m 3.53=== W228 W805total

QQ

L&

&


3-131

3-184 Heat transfer through a window is considered. The percent error involved in the calculation of heat gain through the window assuming the window consist of glass only is to be determined. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Radiation is accounted for in heat transfer coefficients. Properties The thermal conductivities are given to be 0.7 W/m⋅°C for glass and 0.12 W/m⋅°C for pine wood. Analysis The surface areas of the glass and the wood and the individual thermal resistances are

2

wood

2glass

m 45.0m) 2(m) 5.1(15.0

m 55.2m) 2(m) 5.1(85.0

==

==

A

A

Ri Rglass

T∞1

Ro

C/W 41433.117094.092593.031746.0

C/W 08787.003017.000168.005602.0

C/W17094.0)m 45.0(C). W/m13(

11

C/W03017.0)m 55.2(C). W/m13(

11

C/W 92593.0

T∞2

)m 45.0(C) W/m.12.0( 2woodwood °Ak

m 05.0

C/W0016.0)m 55.2(C) W/m.7.0(

m 003.0

C/W31746.0)m 45.0(C). W/m7(

11

C/W05602.011

woodo,woodwoodi,woodtotal,

glasso,glassglassi,glass total,

22wood2

woodo,

22glass2

glasso,

woodwood

2glassglass

glassglass

22wood1

woodi,

°=++=++=

°=++=++=

°=°

==

°=°

==

°===

=°

==

°=°

==

°===

RRRR

RRRRAh

R

AhR

LR

AkL

R

AhR

R

The rate of heat gain through the glass and the wood and their total are

)m 55.2(C). W/m7( 22glass1

glassi,°Ah

8 °

Ri Rwood

T∞1 T∞2

Ro

W4.1933.111.182

W3.11C/W41433.1

C)2440(

W 1.182C/W08787.0

C)2440(

woodglasstotal

woodtotal,

12wood

glasstotal,

°−−

°

TT

R12

ass

=+=+=

=°

==

=°−

=−

=

∞∞

∞∞

QQQ

RQ

TTQ

&&&

&

&

If the window consists of glass only the heat gain through the window is

gl

2glass m 0.3m) 2(m) 5.1( ==A

C/W 07469.002564.000143.004762.0

C/W02564.0)m 0.3(C). W/m13(

11

C/W 00143.0)m 0.3(C) W/m.7.0(

m 003.0

C/W04762.0)m 0.3(C). W/m7(

11

glasso,glassglassi,glass total,

22glass2

glasso,

2glassglass

glassglass

22glass1

glassi,

°=++=++=

°=°

==

°=°

==

°=°

==

RRRR

AhR

AkL

R

AhR

W2.214C/W07469.0

C)2440(

glasstotal,

12glass =

°°−

=−

= ∞∞

RTT

Q&

Then the percentage error involved in heat gain through the window assuming the window consist of glass only becomes

10.8%=×−

=−

= 1004.193

4.1932.214Error %with wood

with woodonly glass

Q

QQ&

&&


3-132

m 1416.3m) 1(

m 3770.0

m 3142.0m) 1(m) 10.0(

DDLDA

LDA i

===

===

ππ

ππ

The individual thermal resistances are

3-185 Steam is flowing inside a steel pipe. The thickness of the insulation needed to reduce the heat loss by 95 percent and the thickness of the insulation needed to reduce outer surface temperature to 40°C are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 61 W/m⋅°C for steel and k = 0.038 W/m⋅°C for insulation. Analysis (a) Considering a unit length of the pipe, the inner and the outer surface areas of the pipe and the insulation are

m) 1(m) 12.0(LDA o === ππ2

3333

22

21

Ri

T∞1

R2 Ro

T∞2

R1

C/W 02274.0)12.0/ln(0.23876 3

DD

0.238760270.0

02274.0)12.0/ln(00048.002652.0

C/W 2165.018947.000048.002652.0

C/W 02274.0)m 1416.3(C). W/m14(

11

C/W 18947.0)m 3770.0(C). W/m14(

11

C/W 0.23876

)12.0/ln()m 1(C) W/m.038.0(2

)12.0/ln(2

)/ln(

C/W 00048.0)m 1(C) W/m.61(2

)5/6ln(2

)/ln(

C/W 02652.0)m 3142.0(C). W/m120(

11

3

3insulationo,21insulation total,

steelo,1insulation no total,

32

32insulationo,

22steelo,

33

2

23insulation2

1

12pipe1

22

°++=

+++=+++=

°=++=++=

°=°

==

°=°

==

°=°

===

°=°

===

°=°

==

D

DRRRRR

RRRR

DDAhR

AhR

DDLkrr

RR

Lkrr

RR

AhR

i

i

oo

oo

iii

ππ

ππ

3

Then the steady rate of heat loss from the steam per meter pipe length for the case of no insulation becomes

W1109C/W 0.2165

C)20260(21 =°

°−=

−= ∞∞

totalRTT Q&

The thickness of the insulation needed in order to save 95 percent of this heat loss can be determined from

C/W

0.238760270.0

3

3insulationtotal, °⎟⎟⎠

⎜⎜⎝

++D

02274.0)12.0/ln(C)20260( W)110905.0(21

⎞⎛°−

=×⎯→⎯−

= ∞∞

DRTT

Q

whose solution is

insulation&

cm 10.5===⎯→⎯=2

12-32.962-

thicknessm 3296.0 233

DDD

(b) The thickness of the insulation needed thof 40°C can be det

at would maintain the outer surface of the insulation at a maximum temperature ermined from

C/W 02274.0C)2040(

C/W 02274.00.23876

)12.0/ln(0270.0

C)20260(

33

3insulation o,

22

insulationtotal,

21insulation

°

°−=

°⎟⎟⎠

⎞⎜⎜⎝

⎛++

°−⎯→⎯

−=

−= ∞∞∞

DDDR

TTR

TTQ&

whose solution is

cm 2.48===⎯→⎯=2

12-16.962-

thicknessm 1696.0 233

DDD


3-133

3-186 A 4-m-diameter spherical tank filled with liquefied natural gas (LNG) at -160°C is exposed to ambient air. The time for the LNG temperature to rise to -150°C is to be determined.

Assumptions 1 Heat transfer can be considered to be steady since the specified thermal conditions at the boundaries do not change with time significantly. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 Radiation is accounted for in the combined heat transfer coefficient. 3 The combined heat transfer coefficient is constant and uniform over the entire surface. 4 The temperature of the thin-shelled spherical tank is said to be nearly equal to the temperature of the LNG inside, and thus thermal resistance of the tank and the internal convection resistance are negligible.

Properties The density and specific heat of LNG are given to be 425 kg/m3 and 3.475 kJ/kg⋅°C, respectively. The thermal conductivity of super insulation is given to be k = 0.00008 W/m⋅°C.

Analysis The inner and outer surface areas of the insulated tank and the volume of the LNG are

33311

22222

22211

m 51.336/m) 4(6/

m 81.52m) 10.4(

m 27.50m) 4(

===

===

===

ππ

ππ

ππ

D

DA

DA

V

LNG tank

-160°C Ro

T∞2

RinsulationT1

The rate of heat transfer to the LNG is

C/W 13157.1213071.1200086.0 °=+=+= insulationototal

o

RRR

C/W 00086.0)m 81.52(C). W/m22(

11

C/W 13071.12)m 05.2)(m 0.2(C) W/m.00008.0(4

m )0.205.2(4

22

21

12

°=°

==

°=°−

=−

=

o AhR

rkrrr

Rππ

insulation

W75.14C/W 12.13157

C)]155(24[

total

LNG2 =°

°−−=

−= ∞

RTT

Q&

We used average LNG temperature in heat transfer rate calculation. The amount of heat transfer to increase the LNG

temperature from -160°C to -150°C is

== Vρm kg 242,14)m )(33.51kg/m 425( 331 =

[ ] kJ 1095.4C160)(150)(C)kJ/kg. 475.3(kg) 242,14( 5×=°−−−°=∆= TmcQ p

Assuming that heat will be lost from the LNG at an average rate of 15.17 W, the time period for the LNG temperature to rise to -150°C becomes

days 388==×=×

==∆ h 9320s 10355.3kW 0.01475

kJ 1095.4 75

QQt&


3-134

3-187 A hot plate is to be cooled by attaching aluminum fins of square cross section on one side. The number of fins needed to triple the rate of heat transfer is to be determined.


Properties The thermal conductivity of the aluminum fins is given to be k = 237 W/m⋅°C.

Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the square cross-section fins can be determined to be

1-2

2

2m 99.12

m) 002.0)(C W/m.237(m) 002.0C)(. W/m20(44

=°

°===

kaha

kAhpm

c

919.0m 04.0m 99.12 1-fin

×mL

The finned and

)m 04.0m 99.12tanh(tanh -1=

×==

mLη

unfinned surface areas, and heat transfer rates from these areas are

2fin

2unfinnedunfinned

fin

2fin

2

finfinmaxfin,finfinned

2fin

finunfinned

2finfinfin

nTThAQ

nn

TThAQQ

n

nAnnA

b

b

°−−°=−=

=°−°=

−==

−=

−==××=

∞

∞

&

&& ηη

hen the to l heat transfer from the finned plate becomes

he rate of heat transfer if there were no fin attached to the plate would be 2

fin no ==A

he number of fins can be determined from the overall fin effectiveness equation

C)2585)(m 000004.00.03)(C. W/m20()(

W35328.0C)2585)(m 0.00032)(C. W/m20(919.0

)(

m 000004.00.03

m) m)(0.002 002.0(m) m)(0.20 15.0(m 0.00032m) m)(0.04 002.0(4

W0048.036 finn−=

T ta

W0048.03635328.0 finfinunfinnedfinnedfintotal, nnQQQ −+=+= &&&

T

W36C)2585)(m 03.0)(C. W/m20()( 22

fin nofin no =°−°=−= ∞TThAQ b&

m 03.0)m 20.0)(m 15.0(

T

207=⎯→⎯−+

=⎯→⎯= finfinfin

fin no

finfin 36

0048.03635328.03 n

nnQQ&

&ε

4 cm 2 mm × 2

Tb = 85°C

T∞ = 25°C


3-135

3-188 Prob. 3-187 is reconsidered. The number of fins as a function of the increase in the heat loss by fins relative to no fin case (i.e., overall effectiveness of the fins) is to be plotted.


"GIVEN" A_surface=0.15*0.20 [m^2] T_b=85 [C]; k=237 [W/m-C] side=0.002 [m]; L=0.04 [m] T_infinity=25 [C] h=20 [W/m^2-C] epsilon_fin=3 "ANALYSIS" A_c=side^2 p=4*side a=sqrt((h*p)/(k*A_c)) eta_fin=tanh(a*L)/(a*L) A_fin=n_fin*4*side*L A_unfinned=A_surface-n_fin*side^2 Q_dot_finned=eta_fin*h*A_fin*(T_b-T_infinity) Q_dot_unfinned=h*A_unfinned*(T_b-T_infinity) Q_dot_total_fin=Q_dot_finned+Q_dot_unfinned Q_dot_nofin=h*A_surface*(T_b-T_infinity) epsilon_fin=Q_dot_total_fin/Q_dot_nofin

εfin nfin

1.5 51.72 1.75 77.59 2 103.4 2.25 129.3 2.5 155.2 2.75 181 3 206.9 3.25 232.8 3.5 258.6 3.75 284.5 4 310.3 4.25 336.2 4.5 362.1 4.75 387.9 5 413.8 1.5 2 2.5 3 3.5 4 4.5 5

50

100

150

200

250

300

350

400

450

ε fin

n fin


3-136

3-189 An agitated vessel is used for heating an aqueous solution by saturated steam condensing in the jacket outside the vessel. The temperature of the outlet stream is to be determined.


Properties The thermal conductivity of steel is given to be k = 43 W/m⋅K.

Analysis (a) A heat balance on the system gives

))(115m 12(C)15(C)J/kg kg/s)(4180 60/500(

)()(2

steamin

TUT

TTUATTcm p

−=°−°⋅

−=−&

where

C W/m1585

C W/m000,101

C W/m43m 015.0

C W/m55001

111

1 2

22

°⋅=

°⋅++

°⋅+

°⋅

=++

=

oi hkL

h

U

S gubstitutin ,

bar. The rate of heat

sfer is two-dimensional (no change in the axial direction). 3 e temperature as the LNG.

⋅°C. The density and the specific heat of LNG are given to be 425 kg/m3 and 3.475 kJ/kg⋅°C, respectively,


C50.3°=−°⋅=°−°⋅

TTT ))(115m 12(C) W/m1585(C)15(C)J/kg kg/s)(4180 60/500( 22

3-190 A cylindrical tank containing liquefied natural gas (LNG) is placed at the center of a square solidtransfer to the tank and the LNG temperature at the end of a one-month period are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat tranThermal conductivity of the bar is constant. 4 The tank surface is at the sam

Properties The thermal conductivity of the bar is given to be k = 0.0002 W/m

-160°C

L = 1.9 m

D = 0.6 m

12°C

1.4 mm 92.12

m 4.108.1=

⎞⎛=

⎞⎛=

wS

m 6.008.1ln

)m 9.1(2

ln

2

⎟⎠

⎜⎝

⎟⎠

⎜⎝

ππ

D

L

f heat transfer to the tank becomes Then the steady rate o

[ ] W0.4444=°−−°=−= C)160(12)C W/m.0002.0)(m 92.12()( 21 TTSkQ&

The mass of LNG is

kg 07.486m) 6.0( 33D )kg/m 425(

63 === πρπρm V

perature of LNG at the end of the month becomes

=

The amount heat transfer to the tank for a one-month period is

J 10152.1s) 360024 W)(304444.0( 6×=××=∆= tQQ &

Then the tem

[ ]C153.1°−=

°−−°=×

−=

2

26

21

C)160()CJ/kg. kg)(3475 (48.07J 10152.1

)(

TT

TTmcQ p


3-137

3-191 A typical section of a building wall is considered. The temperature on the interior brick surface is to be determined.

Assumptions 1 Steady operating conditions exist.

Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K.

Analysis We consider 1 m2 of wall area. The thermal resistances are

C/Wm 1.0C) W/m0.1(

m 1.0

C/Wm 1032.10.0C)(0.6 W/m50( +°⋅ 05)

m 005.0m) 08.0(

)(

C/Wm 645.20.005)C)(0.6 W/m03.0(

m 6.0m) 08.0(

)(

C/Wm 02.0C) W/m5.0(

m 01.0

2

34

3434

25

23b

b2323

2

23a2323

212

°⋅=°⋅

==

°⋅×==

+=

°⋅=+°⋅

=

+=

°⋅=°⋅

==

−

kt

R

LLkL

tR

LLktR

kt

R

bab

baa

The total thermal resistance and the rate of heat transfer are

12

a

12

L

C/Wm 120.01.01032.1645.2

)1032.1)(645.2(645.202.0 2

5

5

342323

232312total

°⋅=+⎟⎟⎠

⎞⎜⎜⎝

⎛

×+

×+=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

−

−

RRR

RRRR

ba

ba

22

total

14 W/m125C/Wm 0.120C)2035(

=⋅

°−=

−=

RTT

q&

The temperature on the interior brick surface is

C22.5°=⎯→⎯⋅

°−=⎯→⎯

−= 32

32

34

34

C/Wm 0.1C)35(

W/m125 TT

RTT

q&


3-138

3-192 Ten rectangular aluminum fins are placed on the outside surface of an electronic device. The rate of heat loss from the electronic device to the surrounding air and the fin effectiveness are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 The heat transfer coefficient is constant and uniform over the entire fin surface. 4 The thermal properties of the fins are constant. 5 The heat transfer coefficient accounts for the effect of radiation from the fins.

Properties The thermal conductivity of the aluminum fin is given to be k = 203 W/m⋅K.

Analysis The fin efficiency is to be determined using Fig. 3-43 in the text.

93.0244.0)004.0)(203(

100)2/004.0020.0()/()2/()/( fin2/3 =⎯→⎯=+=+== ηξ kthtLkAhL

The rate of heat loss can be determined as follows

pc

W178=+=+= fintotal QQQ &&

=−=−=

=⎯→⎯−

=⎯→⎯−

==

=×=

=×+××=

∞

∞

6.16161

W6.16)2072)(004.0)(80()(

W161)2072)(0416.0)(80(

93.0)(

m 004.0)004.0100.0(10

m 0416.0)020.0004.0100.0020.0(102

base

basebase

finfin

fin

fin

max fin,

finfin

2base

2fin

TThAQ

QQ

TThAQ

QQ

A

A

b

b

&

&

&&&

&

&η

he fin eff tiveness is T ec

5.35=−×

=−

==∞ )2072)(100.0080.0)(80(

178)(fin no base,

fin

fin no

finfin TThA

QQQ

b

&

&

&ε


3-139

3-193 One wall of a refrigerated warehouse is made of three layers. The rates of heat transfer across the warehouse without and with the metal bolts, and the percent change in the rate of heat transfer across the wall due to metal bolts are to be determined.


Properties The thermal conductivities are given to be kAl = 200 W/m⋅K, kfiberglass = 0.038 W/m⋅K, kgypsum = 0.48 W/m⋅K, and kbolts = 43 W/m⋅K.

Analysis (a) The rate of heat transfer through the warehouse is

C W/m451.0

C W/m401

C W/m48.0m 03.0

C W/m038.0m 08.0

C W/m200m 01.0

C W/m401

1

111

2

22

1

°⋅=

°⋅+

°⋅+

°⋅+

°⋅+

°⋅

=

++++

=

ogy

gy

fg

fg

Al

Al

i hkL

kL

kL

h

U

[ ] W676=°−−×°⋅=−= C)10(20)m 105)(C W/m451.0()( 2211

&io TTAUQ

(b) The rate of heat transfer with the consideration of metal bolts is

[ ][ ] W8.674)10(20)02.0(25.0400510)451.0()( 2111 =−−×−×=−= πio TTAUQ&

C W/m94.18

C W/m401

C W/m43m 12.0

C W/m401

111

1 2

22

2 °⋅=

°⋅+

°⋅+

°⋅

=++

=

obolts

bolts

i hkL

h

U

[ ] W4.71C)10(20]m )02.0(25.0400)[C W/m94.18()( 222222 °−−×°⋅=−= πio TTAUQ& =

(c) The percent change in the rate of heat transfer across the wall due to metal bolts is

W746=+=+= 4.718.67421 QQQ &&&

10.3%==−

= 103.0676

676746change %


3-140

3-194 A spherical tank containing iced water is buried underground. The rate of heat transfer to the tank is to be determined for the insulated and uninsulated ground surface cases.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant. 4 The tank surface is assumed to be at the same temperature as the iced water because of negligible resistance through the steel.

Properties The thermal conductivity of the soil is given to be k = 0.55 W/m⋅°C.


m 93.17

m 4.2m 2.225.0125.01 −−

zD

)m 2.2(22==

ππDS

Then the steady rate of heat transfer from the tank becomes

C)018)(C W/m.55.0)(21

the grou surface is insulated,

T1 =18°C

T2 = 0°C z = 2.4 m

D = 2.2 m

=

=−= m 93.17()( TTSkQ& W178=°−°

If nd

m 25.11

m 4.2m 2.225.01

)m 2.2(2

25.01

2=

+=

+=

ππ

zD

DS

W111=°−°=−= C)018)(C W/m.55.0)(m 25.11()( 21 TTSkQ&


3-141

3-195 A square cross-section bar consists of a copper layer and an epoxy layer. The rates of heat transfer in different directions are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional.

Properties The thermal conductivities of copper and epoxy are given to be 380 and 0.4 W/m⋅K, respectively.

Analysis (a) Noting that the resistances in this case are in parallel, the heat transfer from front to back is

K/W 577.1

m 12.0m )01.002.0)(K W/m4.0(m )01.002.0)(K W/m380( 22

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛ ×⋅

+⎟⎟⎞

⎜⎜⎛ ×⋅

= m 12.0

1

1

=

⎥⎦⎢⎣ ⎠⎝⎠⎝

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

−

−

EpCu LkA

LkAR

12 cm

1 cm 1 cm

2 cm

copper epoxy

W31.7==∆

=K/W 1.577RK 50TQ&

(b) Noting that the resistances in this case are in series, the heat transfer from left to right is

K/W 43.10m )12.002.0)(K W/m4.0(

m 01.0m )12.002.0)(K W/m380(

m 01.022 =⎟⎟⎠

⎞⎜⎜⎝

⎛

×⋅+⎟

⎟⎠

⎞⎜⎜⎝

⎛

×⋅=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=+=

EpCuEpCu kA

LkALRRR

W4.8==∆

=K/W 10.43K 50

RTQ&

(c) Noting that the resistances in this case are in parallel, the heat transfer from top to bottom is

K/W 04381.0m 02.0

m )12.001.0)(K W/m4.0(m 02.0

m )12.001.0)(K W/m380(122

1

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ ×⋅+⎟

⎟⎠

⎞⎜⎜⎝

⎛ ×⋅=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

−

−

EpCu LkA

LkAR

W1141==∆

=K/W0.04381

K 50RTQ&


3-142

3-196 A spherical vessel is used to store a fluid. The thermal resistances, the rate of heat transfer, and the temperature difference across the insulation layer are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional.

Properties The thermal conductivity of the insulation is given to be 0.20 W/m⋅K.

Analysis (a) The thermal resistances are

K/W 13.3122 m) 10.3()K W/m10(

×=⋅

==πoo

o AhR 0

K/W 108.56

K/W 108.84 4

3

3

21

12

22

11)K W/m2.0(m) m)(1.55 (1.54

m)5.155.1(4

m) 3()Km11

−

−

−

×=⋅

−=

−=

×=⋅

==

ππ

π

ins krrrr

R

) The rate of heat transfer is

R W/40(ii

i Ah

(b

W1725=×+×+×

−=

++∆

=K/W )1031.3108.5610(8.84

K )022(3-3-4-

oinsi RRRTQ&

(c) The temperature difference across the insulation layer is

K 14.8=∆⎯→⎯×

∆=⎯→⎯

∆= ins

ins

ins

ins TT

RT

QK/W 108.56

W17253-

&


3-143

3-197 Using Table 3-4, the efficiency, heat transfer rate, and effectiveness of a straight triangular fin are to be determined.



Analysis From Table 3-3, for straight triangular fins, we have

4.0)m 055.0()m 004.0)(C W/m236(

)C W/m25(22 2=

°⋅°⋅

== LkthmL

22222 fin m 01211.0)2/m 004.0()m 055.0()m 110.0(2)2/(2 =+=+= tLwA

)2()2(1 1 mLI

=η 0

fin mLImL

rom Table 3-4, the modified Bessel functions

08.02 =−− ImL or

F are

)2(0 = emLIe 5241.0)8.0( 166.1)8.0(0 =I

1945.0)8.0()2( 18.0

12 == −− IemLIe mL or 4329.0)8.0(1 =I

Hence, the fin efficiency is

0.928=⎟⎠⎞

⎜⎝⎛==

4.01)2(1 1 mLIη

166.1329

4.0)2(0fin mLImL

a single fin is

The fin effectiveness is

The heat transfer rate for

W77.3=°−°⋅=−= ∞ C )25300)(m 01211.0)(C W/m25)(928.0()( 22finfinfin TThAQ bη&

25.5=°−

°⋅−− ∞∞ 0)(C W/m25())(()( 2fin TTtwhTThA bbb

===C )25300)(m 11.0)(m 004.

ε

iscussion The fin efficiency can also be determined using the EES with the following line:

eta_fin=1/0.4*Bessel_I1(0.8)/Bessel_I0(0.8)

W3.77finfin QQ &&

D


3-144

3-198 Aluminum pin fins of parabolic profile with blunt tips are attached to a plane surrface. The heat transfer rate from a single fin and the increase in the heat transfer as a result of attaching fins are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. Properties The thermal conductivity of the fin is given as 230 W/m · °C.

D = 4 mm k=230 W/m⋅°C

h = 45 W/m2⋅°C

L = 25 mm Analysis From Table 3-3, for pin fins of parabolic profile (blunt tip), we have

3497.0)m 025.0()m 004.0)(C W/m

230(kD

)C W/m45(44 2=

°⋅°⋅

== LhmL

24 m 10099.2 ×=

2/32

2

42/324

11m 004.0m 025.016

)m 025.0(96)m 004.0(

−

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=

⎭

⎪⎫

⎩

⎪⎧ ⎤⎡ ⎞⎛ ππ LD

2fin 111696 ⎪

⎬⎪⎨ −

⎥⎥⎦⎢

⎢⎣

+⎟⎠

⎜⎝

=DL

A

[ ][ ]3/)3497.0(4

3/)3497.0(4)

34.0(2)3/4(2 0 mLImL 973)3/4(3

0

11

IImLI

==η

opy the following line and paste on a blank EES screen to solve the above equation: n=3/(2*0.3497)*Bessel_I1(4*0.3497/3)/Bessel_I0(4*0.3497/3)

fficiency is =

fin

C eta_fiSolving by EES software, the fin e fin 9738.0η

C )25200)(m 10099.2)( W/m45)(9738.0()( 242TThAQ η&

The surface area of the unfinned portion is The heat transfer from the unfinned portion is The total heat transfer from the surface is

If there was no fin at the surface, The increase in heat transfer as a result of attaching fins is then

Discussion The values for the Bessel functions may also be approximated using Table 3-4.

The heat transfer rate for a single fin is C W1.610=°−×°⋅=−= −

∞finfinfin b

Heat transfer from 100 fins is

W161 W)610.1)(100(totalfin, ==Q&

2222unfin m 9987.04/)m 004.0(1001)4/(100m )11( =−=−×= ππDA

W7865C )25200)(C W/m45)(m 9987.0()( 22unfinunfin =°−°⋅=−= ∞TThAQ b

&

W80267865161unfintotalfin,total =+=+= QQQ &&&

W7875C )25200)(C W/m45)(m 1()( 22unfinnofin =°−°⋅=−= ∞TThAQ b

&

W151=−=−= 78758026nofintotalincrease QQQ &&&


3-145

3-199 The heat transfer rates are to be determined and the temperature variations are to be plotted for infinitely long fin, adiabatic fin tip, fin tip with temperature of 250 °C, and convection from the fin tip.



Analysis For a circular fin with uniform cross section, the perimeter and cross section area are

= m 03142.0)m 01.0( == ππDp

and 2522

m 10854.74

)m 01.0(4

−×===ππDAc

Also, we have

125

2m 41.20

)m 10854.7)(C W/m240(m) 03142.0C)( W/m250( −

−=

×°⋅

°⋅==

ckAhpm

C W/3848.0)m 10854.7)(C W/m240)(m 03142.0)(C W/m250( 2 °⋅°⋅=chpkA 25 °=× −

) For an finitely long fin, the heat transfer rate can be calculated as

(a in

W125=°−°=−= W/3848.0()( TThpkAQ& ∞ C )25350)(Cfin long bc

he temperature variation along the fin is given as

T

mxeTTTxT −∞ =

−−)(

b ∞

(b) For an adiabatic fin tip, the heat transfer rate can be calculated as

[ ] W96.3=

°−°°=

−= ∞

)m 050.0)(m 41.20(tanh)C 25C 350C W/3848.0(

tanh)(1

tipadiabatic

-

bc mLTThpkAQ&

perature variation along the fin is given as

)(

The tem

mL

xLmTTTxT

cosh)(cosh)( −

=−− ∞

b ∞

) For fin with tip temperature of 250 °C, the heat transfer rate can be calculated as (c

W90.7=

°−°°= )7250.0)(C 25C 350)(C W/3848.0(

The temperature variation along the fin is as

−−−−= ∞∞

∞ sinh)/()(cosh

)( tempspecified mLTTTTmL

TThpkAQ bLbc

&

mLTTb sinh− ∞

xLmmxTTTTTxT bL )(sinhsinh)/()()( −+−−=

− ∞∞∞

) For fin with convection from the tip, the heat transfer rate can be calculated as

(d

W98.8=°−°°=++

−=

7901.02C 50C W/3848.0(s)shcosh( tipconv mL

mLmLThpkAQ bc&

The temperature variation along t give

∞ cosinh)T

))(C 53)(inh mL/( mkh

)/( mkh

he fin is n as


3-146

mLmkhL (+m

LmkLmTTTxT

b )/cosh()/(cosh)( −−

=−−

∞

∞

The values for the temperature va for p a) to ( tabula the fo ng table:

°C

x) msinhhx () +sinh

riations arts ( d) are ted in llowi

T(x),L, m

Part (a) b) c) d) Part ( Part ( Part (0 350 350 350 350 0.005 318 326 328 325 0.010 290 305 308 304 0.015 264 288 292 285 0.020 241 272 279 270 0.025 220 260 268 256 0.030 201 250 259 246 0.035 184 242 253 237 0.040 169 237 250 231 0.045 155 233 249 227 0.050 142 232 250 224

The temperature variations for parts (a) to (d) are plotted in the following figure:

x, m

0.00 0.01 0.02 0.03 0.04 0.05

T, °

C

100

150

200

250

300

350

Infinitely long finAdiabatic fin tipFin with tip temperature of 250 °CConvection from the fin tip

Discussion The differences in the temperature variations show that applying the proper boundary condition is very important in order to perform the analysis correctly.


3-147

3-200 A tube carrying hot steam is centered at a square cross-section concrete bar. The width of the square concrete bar and the rate of heat loss in (W/m) are to be determined for the temperature difference between the outer surface of the square concrete bar and the ambient air to be maintained at 5 °C.

Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 Heat conduction through the tube wall is negligible. 5 Thermal contact resistance between the tube and the concrete bar is negligible.

Properties The thermal conductivity of the concrete is given as 1.7 W/m · °C.

Analysis Using Table 3-7 (Case 6), the shape factor is given to be

)/08.1ln( Dw

2 LS π=

rom energy balance, we have

F

)()( 221 ∞−=− TThATTkS s

or )(4)()/08.1ln(

2 kL221 ∞−=− TThwLTT

Dwπ

earrange to get R

⎟⎠⎞

⎜⎝⎛

−−

=⎟⎠⎞

⎜⎝⎛

∞2 hk

TTTT

Dww

208.1 21 π ln

⎥⎦⎢⎣ °⋅°⎠⎝ )C W/m20(2C 5m 127.0 ⎥⎤

⎢⎡ °⋅°−

=⎟⎞

⎜⎛ )C W/m7.1(C )0120(08.1ln πww

2

opy the following line and paste on a blank EES screen to solve the above equation:

/5*(3.1416*1.7)/(2*20)

olving by ES software, the width of the square concrete bar is

Discussion If the width of the concrete bar were less than 1.324 m, then the temperature difference between the outer surface of the concrete bar and the ambient air would be greater than 5 °C. This would mean more heat loss to the ambient air.

C

w*ln(1.08*w/0.127)=120

S E

m 1.324=w

The heat loss to the ambient air is

W/m530=°°⋅=−= ∞ C) 5)(m 324.1)(C W/m20(4)(4/ 22 TThwLQ&


3-148

Fundamentals of Engineering (FE) Exam Problems

3-201 Heat is lost at a rate of 275 W per m2 area of a 15-cm-thick wall with a thermal conductivity of k=1.1 W/m⋅ºC. The temperature drop across the wall is

(a) 37.5ºC (b) 27.5ºC (c) 16.0ºC (d) 8.0ºC (e) 4.0ºC

Answer (a) 37.5ºC

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

L=0.15 [m] k=1.1 [W/m-C] q=275 [W/m^2] q=k*DELTAT/L

3-202 Consider a wall that consists of two layers, A and B, with the following values: kA = 1.2 W/m⋅ºC, LA = 8 cm, kB = 0.2 W/m⋅ºC, LB = 5 cm. If the temperature drop across the wall is 18ºC, the rate of heat transfer through the wall per unit area of the wall is

(a) 56.8 W/m2 (b) 72.1 W/m2 (c) 114 W/m2 (d) 201 W/m2 (e) 270 W/m2

Answer (a) 56.8 W/m2


k_A=1.2 [W/m-C] L_A=0.08 [m] k_B=0.2 [W/m-C] L_B=0.05 [m] DELTAT=18 [C] R_total=L_A/k_A+L_B/k_B q_dot=DELTAT/R_total "Some Wrong Solutions with Common Mistakes" W1_q_dot=DELTAT/(L_A/k_A) "Considering layer A only" W2_q_dot=DELTAT/(L_B/k_B) "Considering layer B only"


3-149

3-203 A plane furnace surface at 150°C covered with 1-cm-thick insulation is exposed to air at 30°C, and the combined heat transfer coefficient is 25 W/m2⋅°C. The thermal conductivity of insulation is 0.04 W/m⋅°C. The rate of heat loss from the surface per unit surface area is

(a) 35 W (b) 414 W (c) 300 W (d) 480 W (e) 128 W

Answer (b) 414 W


Ts=150 Tinf=30 h=25 L=0.01 K=0.04 Rconv=1/h Rins=L/K Rtotal=Rconv+Rins Q=(Ts-Tinf)/(Rconv+Rins) “Some Wrong Solutions with Common Mistakes:” W1_Q=(Ts-Tinf)/Rins "Disregarding convection" W2_Q=(Ts-Tinf)/Rconv"Disregarding insulation" W3_Q=(Ts-Tinf)*(Rconv+Rins)"Multiplying by resistances"

3-204 Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at 26ºC with a heat transfer coefficient of 7.5 W/m2⋅ºC. The ball is to be covered with a material of thermal conductivity 0.15 W/m⋅ºC. The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is

(a) 0.5 cm (b) 1.0 cm (c) 1.5 cm (d) 2.0 cm (e) 2.5 cm

Answer (e) 2.5 cm


D=0.03 [m] r=D/2 T_infinity=26 [C] h=7.5 [W/m^2-C] k=0.15 [W/m-C] r_cr=(2*k)/h r_cr=(2*k)/h "critical radius of insulation for a sphere" thickness=r_cr-r "Some Wrong Solutions with Common Mistakes" W_r_cr=k/h W1_thickness=W_r_cr-r "Using the equation for cylinder"


3-150

3-205 Consider a 1.5-m-high and 2-m-wide triple pane window. The thickness of each glass layer (k = 0.80 W/m.°C) is 0.5 cm, and the thickness of each air space (k = 0.025 W/m.°C ) is 1.2 cm. If the inner and outer surface temperatures of the window are 10°C and 0°C, respectively, the rate of heat loss through the window is

(a) 3.4 W (b) 10.2 W (c) 30.7 W (d) 61.7 W (e) 86.8 W

Answer: (c) 30.7 W


"Using the thermal resistances per unit area, Q can be expressed as Q=A*DeltaT/R_total” Lglass=0.005 {m} kglass=0.80 {W/mC} Rglass=Lglass/kglass Lair=0.012 {m} kair=0.025 {W/mC} Rair=Lair/kair A=1.5*2 T1=10 T2=0 Q=A*(T1-T2)/(3*Rglass+2*Rair) “Some Wrong Solutions with Common Mistakes:” W1_Q=(T1-T2)/(3*Rglass+2*Rair) “Not using area” W2_Q=A*(T1-Tinf)*(3*Rglass+2*Rair) “Multiplying resistance instead of dividing” W3_Q=A*(T1-T2)/(Rglass+Rair) “Using one layer only” W4_Q=(T1-T2)/(3*Rglass+2*Rair)/A “Dividing by area instead of multiplying”

3-206 Consider a furnace wall made of sheet metal at an average temperature of 800°C exposed to air at 40°C. The combined heat transfer coefficient is 200 W/m2⋅°C inside the furnace, and 80 W/m2⋅°C outside. If the thermal resistance of the furnace wall is negligible, the rate of heat loss from the furnace per unit surface area is

(a) 48.0 kW/m2 (b) 213 kW/m2 (c) 91.2 kW/m2 (d) 151 kW/m2 (e) 43.4 kW/m2

Answer (e) 43.4 kW/m2


Ti=800 To=40 hi=200 ho=80 Rconv1=1/hi Rconv2=1/ho Rtotal=Rconv1+ Rconv2 Q=(Ti-To)/Rtotal “Some Wrong Solutions with Common Mistakes:” W1_Q=(Ti+To)/Rtotal “Adding temperatures” W2_Q=(hi+ho)*(Ti-To) “Adding convection coefficients” W3_Q=(hi-ho)*(Ti-To) “Subtracting convection coefficients”


3-151

3-207 Consider a jacket made of 5 layers of 0.1-mm-thick cotton fabric (k = 0.060 W/m.°C) with a total of 4 layers of 1-mm-thick air space (k = 0.026 W/m.°C) in between. Assuming the inner surface temperature of the jacket is 25°C and the surface area normal to the direction of heat transfer is 1.1 m2, determine the rate of heat loss through the jacket when the temperature of the outdoors is 0°C and the heat transfer coefficient on the outer surface is 18 W/m2.°C.

(a) 6 W (b) 115 W (c) 126 W (d) 287 W (e) 170 W

Answer (c) 126 W


"Using the thermal resistance concept, Q can be expressed as Q=A*DeltaT/R_total” Lcotton=0.0001 {m} kcotton=0.06 {W/mC} Rcotton=Lcotton/kcotton Lair=0.001 {m} kair=0.026 {W/mC} Rair=Lair/kair A=1.1 h=18 Rconv=1/h T1=25 Tinf=0 Q=A*(T1-Tinf)/(5*Rcotton+4*Rair+Rconv) “Some Wrong Solutions with Common Mistakes:” W1_Q=(T1-Tinf)/(5*Rcotton+4*Rair+Rconv) “Not using area” W2_Q=A*(T1-Tinf)*(5*Rcotton+4*Rair+Rconv) “Multiplying resistance instead of dividing” W3_Q=A*(T1-Tinf)/(Rcotton+Rair+Rconv) “Using one layer only” W4_Q=A*(T1-Tinf)/(5*Rcotton+4*Rair) “Disregarding convection”

3-208 Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase?

(a) Cleaning the surfaces to make them shinier

(b) Pressing the plates against each other with a greater force

(c) Filling the gab with a conducting fluid

(d) Using softer metals

(e) Coating the contact surfaces with a thin layer of soft metal such as tin

Answer (a) Cleaning the surfaces to make them shinier


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3-209 A 10-m-long, 8-cm-outer-radius cylindrical steam pipe is covered with 3-cm thick cylindrical insulation with a thermal conductivity of 0.05 W/m.°C. If the rate of heat loss from the pipe is 1000 W, the temperature drop across the insulation is

(a) 58°C (b) 101°C (c) 143°C (d) 282°C (e) 600°C

Answer (b) 101°C


R1=0.08 S=0.03 R2=0.11 L=10 K=0.05 Q=1000 R=ln(r2/r1)/(2*pi*L*k) dT=Q*R “Some Wrong Solutions with Common Mistakes:” W1_T=Q/k "Wrong relation" RR1=ln(s/r1)/(2*pi*L*k) W2_T=Q*RR1"Wrong radius" RR2=s/k W3_T=Q*RR2"Wrong radius"


3-153

3-210 Steam at 200ºC flows in a cast iron pipe (k = 80 W/m⋅ºC) whose inner and outer diameters are D1 = 0.20 m and D2 = 0.22 m, respectively. The pipe is covered with 2-cm-thick glass wool insulation (k = 0.05 W/m⋅ºC). The heat transfer coefficient at the inner surface is 75 W/m2⋅ºC. If the temperature at the interface of the iron pipe and the insulation is 194ºC, the temperature at the outer surface of the insulation is

(a) 32ºC (b) 45ºC (c) 51ºC (d) 75ºC (e) 100ºC

Answer (b) 45ºC


T_steam=200 [C] k_pipe=80 [W/m-C] k_ins=0.05 [W/m-C] D1=0.20 [m]; r1=D1/2 D2=0.22 [m]; r2=D2/2 t_ins=0.02 [m] r3=r2+t_ins L=1 [m] "Consider a unit length of pipe" h1=75 [W/m^2-C] T2=194 [C] A1=2*pi*r1*L R_conv1=1/(h1*A1) R_1=ln(r2/r1)/(2*pi*k_pipe*L) R_2=ln(r3/r2)/(2*pi*k_ins*L) Q_dot=(T_steam-T2)/(R_conv1+R_1) Q_dot=(T2-T3)/R_2


3-154

3-211 A 5-m diameter spherical tank is filled with liquid oxygen (ρ = 1141 kg/m3, cp = 1.71 kJ/kg⋅ºC) at -184ºC. It is observed that the temperature of oxygen increases to -183ºC in a 144-hour period. The average rate of heat transfer to the tank is

(a) 124 W (b) 185 W (c) 246 W (d) 348 W (e) 421 W

Answer (c) 246 W


D=5 [m] rho=1141 [kg/m^3] c_p=1710 [J/kg-C] T1=-184 [C] T2=-183 [C] time=144*3600 [s] V=pi*D^3/6 m=rho*V Q=m*c_p*(T2-T1) Q_dot=Q/time "Some Wrong Solutions with Common Mistakes" W1_Q_dot=Q "Using amount of heat transfer as the answer" Q1=m*(T2-T1) W2_Q_dot=Q1/time "Not using specific heat in the equation"

3-212 A 2.5-m-high, 4-m-wide, and 20-cm-thick wall of a house has a thermal resistance of 0.025ºC/W. The thermal conductivity of the wall is

(a) 0.8 W/m⋅ºC (b) 1.2 W/m⋅ºC (c) 3.4 W/m⋅ºC (d) 5.2 W/m⋅ºC (e) 8.0 W/m⋅ºC

Answer (a) 0.8 W/m⋅ºC


Height=2.5 [m] Width=4 [m] L=0.20 [m] R_wall=0.025 [C/W] A=Height*Width R_wall=L/(k*A) "Some Wrong Solutions with Common Mistakes" R_wall=L/W1_k "Not using area in the equation"


3-155

3-213 Consider two walls, A and B, with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is kA/kB = 4 and the ratio of the wall thicknesses is LA/LB = 2. The ratio of heat transfer rates through the walls is BA QQ && /

(a) 0.5 (b) 1 (c) 2 (d) 4 (e) 8

Answer (c) 2


k_A\k_B=4 L_A\L_B=2 Q_dot_A\Q_dot_B=k_A\k_B*(1/L_A\L_B) "From Fourier's Law of Heat Conduction"

3-214 A hot plane surface at 100°C is exposed to air at 25°C with a combined heat transfer coefficient of 20 W/m2⋅°C. The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of 0.10 W/m⋅°C. Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is

(a) 0.1 cm (b) 0.5 cm (c) 1.0 cm (d) 2.0 cm (e) 5 cm

Answer (b) 0.5 cm


Ts=100 Tinf=25 h=20 k=0.1 Rconv=1/h Rins=L/k Rtotal=Rconv+Rins Q1=h*(Ts-Tinf) Q2=(Ts-Tinf)/(Rconv+Rins) Q2=Q1/2


3-156

3-215 Consider a 4.5-m-long, 3.0-m-high, and 0.22-m-thick wall made of concrete (k = 1.1 W/m·ºC). The design temperatures of the indoor and outdoor air are 24ºC and 3ºC, respectively, and the heat transfer coefficients on the inner and outer surfaces are 10 and 20 W/m2⋅ºC. If a polyurethane foam insulation (k = 0.03 W/m⋅ºC) is to be placed on the inner surface of the wall to increase the inner surface temperature of the wall to 22ºC, the required thickness of the insulation is


Answer (e) 2.1 cm


Length=4.5 [m] Height=3.0 [m] L=0.22 [m] T_infinity1=24 [C] T_infinity2=3 [C] h1=10 [W/m^2-C] h2=20 [W/m^2-C] k_wall=1.1 [W/m-C] k_ins=0.03 [W/m-C] T1=22 [C] A=Length*Height R_conv1=1/(h1*A) R_wall=L/(k_wall*A) R_conv2=1/(h2*A) R_ins=L_ins/(k_ins*A) Q_dot=(T_infinity1-T_infinity2)/(R_conv1+R_wall+R_ins+R_conv2) Q_dot=(T_infinity1-T1)/R_conv1


3-157

3-216 Steam at 200ºC flows in a cast iron pipe (k = 80 W/m⋅ºC) whose inner and outer diameters are D1 = 0.20 m and D2 = 0.22 m. The pipe is exposed to room air at 35ºC. The heat transfer coefficients at the inner and outer surfaces of the pipe are 90 and 20 W/m2⋅ºC, respectively. The pipe is to be covered with glass wool insulation (k = 0.05 W/m⋅ºC) to decrease the heat loss from the steam by 90 percent. The required thickness of the insulation is


Answer (d) 3.3 cm


T_steam=200 [C] T_infinity=35 [C] k_pipe=80 [W/m-C] D1=0.20 [m] r1=D1/2 D2=0.22 [m] r2=D2/2 h1=90 [W/m^2-C] h2=20 [W/m^2-C] k_ins=0.05 [W/m-C] f=0.90 L=1 [m] "Consider a unit length of pipe" A1=2*pi*r1*L R_conv1=1/(h1*A1) R_1=ln(r2/r1)/(2*pi*k_pipe*L) A2=2*pi*r2*L R_conv2=1/(h2*A2) Q_dot_old=(T_steam-T_infinity)/(R_conv1+R_1+R_conv2) r3=r2+t_ins R_2=ln(r3/r2)/(2*pi*k_ins*L) A3=2*pi*r3*L R_conv2_new=1/(h2*A3) Q_dot_new=(1-f)*Q_dot_old Q_dot_new=(T_steam-T_infinity)/(R_conv1+R_1+R_2+R_conv2_new) "Some Wrong Solutions with Common Mistakes" W1_t_ins=r3 "Using outer radius as the result"


3-158

3-217 A 50-cm-diameter spherical tank is filled with iced water at 0ºC. The tank is thin-shelled and its temperature can be taken to be the same as the ice temperature. The tank is exposed to ambient air at 20ºC with a heat transfer coefficient of 12 W/m2⋅ºC. The tank is to be covered with glass wool insulation (k = 0.05 W/m⋅ºC) to decrease the heat gain to the iced water by 90 percent. The required thickness of the insulation is


Answer (a) 4.6 cm


T_ice=0 [C] T_infinity=20 [C] D1=0.50 [m] r1=D1/2 h=12 [W/m^2-C] k_ins=0.05 [W/m-C] f=0.90 A=pi*D1^2 Q_dot_old=h*A*(T_infinity-T_ice) r2=r1+t_ins R_ins=(r2-r1)/(4*pi*r1*r2*k_ins) D2=2*r2 A2=pi*D2^2 R_conv=1/(h*A2) Q_dot_new=(1-f)*Q_dot_old Q_dot_new=(T_infinity-T_ice)/(R_ins+R_conv) "Some Wrong Solutions with Common Mistakes" W1_t_ins=r2 "Using outer radius as the result"


3-159

3-218 A room at 20°C air temperature is loosing heat to the outdoor air at 0°C at a rate of 1000 W through a 2.5-m-high and 4-m-long wall. Now the wall is insulated with 2-cm-thick insulation with a conductivity of 0.02 W/m.°C. Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation.

(a) 20 W (b) 561 W (c) 388 W (d) 167 W (e) 200 W

Answer (d) 167 W


Tin=20 Tout=0 Q=1000 A=2.5*4 L=0.02 k=0.02 Rins=L/(k*A) Q=(Tin-Tout)/R Qnew=(Tin-Tout)/(R+Rins) "Some Wrong Solutions with Common Mistakes:" W1_Q=(Tin-Tout)/Rins "Disregarding original resistance" W2_Q=(Tin-Tout)*(R+L/k) "Disregarding area" W3_Q=(Tin-Tout)*(R+Rins)"Multiplying by resistances"

3-219 A 1-cm-diameter, 30-cm long fin made of aluminum (k = 237 W/m⋅ºC) is attached to a surface at 80ºC. The surface is exposed to ambient air at 22ºC with a heat transfer coefficient of 18 W/m2⋅ºC. If the fin can be assumed to be very long, the rate of heat transfer from the fin is

(a) 2.0 W (b) 3.2 W (c) 4.4 W (d) 5.5 W (e) 6.0 W

Answer (e) 6.0 W


D=0.01 [m] L=0.30 [m] k=237 [W/m-C] T_b=80 [C] T_infinity=22 [C] h=18 [W/m^2-C] p=pi*D A_c=pi*D^2/4 Q_dot=sqrt(h*p*k*A_c)*(T_b-T_infinity) "Some Wrong Solutions with Common Mistakes" a=sqrt((h*p)/(k*A_c)) W1_Q_dot=sqrt(h*p*k*A_c)*(T_b-T_infinity)*tanh(a*L) "Using the relation for insulated fin tip"


3-160

3-220 A 1-cm-diameter, 30-cm-long fin made of aluminum (k = 237 W/m⋅ºC) is attached to a surface at 80ºC. The surface is exposed to ambient air at 22ºC with a heat transfer coefficient of 11 W/m2⋅ºC. If the fin can be assumed to be very long, its efficiency is

(a) 0.60 (b) 0.67 (c) 0.72 (d) 0.77 (e) 0.88

Answer (d) 0.77


D=0.01 [m] L=0.30 [m] k=237 [W/m-C] T_b=80 [C] T_infinity=22 [C] h=11 [W/m^2-C] p=pi*D A_c=pi*D^2/4 a=sqrt((h*p)/(k*A_c)) eta_fin=1/(a*L) "Some Wrong Solutions with Common Mistakes" W1_eta_fin=tanh(a*L)/(a*L) "Using the relation for insulated fin tip"

3-221 A hot surface at 80°C in air at 20°C is to be cooled by attaching 10-cm-long and 1-cm-diameter cylindrical fins. The combined heat transfer coefficient is 30 W/m2⋅°C, and heat transfer from the fin tip is negligible. If the fin efficiency is 0.75, the rate of heat loss from 100 fins is

(a) 325 W (b) 707 W (c) 566 W (d) 424 W (e) 754 W

Answer (d) 424 W


N=100 Ts=80 Tinf=20 L=0.1 D=0.01 h=30 Eff=0.75 A=N*pi*D*L Q=Eff*h*A*(Ts-Tinf) “Some Wrong Solutions with Common Mistakes:” W1_Q= h*A*(Ts-Tinf) "Using Qmax" W2_Q= h*A*(Ts-Tinf)/Eff "Dividing by fin efficiency" W3_Q= Eff*h*A*(Ts+Tinf) "Adding temperatures" W4_Q= Eff*h*(pi*D^2/4)*L*N*(Ts-Tinf) "Wrong area"


3-161

3-222 A cylindrical pin fin of diameter 1 cm and length 5 cm with negligible heat loss from the tip has an effectiveness of 15. If the fin base temperature is 280°C, the environment temperature is 20°C, and the heat transfer coefficient is 65 W/m2.°C, the rate of heat loss from this fin is

(a) 20 W (b) 48 W (c) 156 W (d) 398 W (e) 418 W

Answer (a) 20 W


"The relation between for heat transfer from a fin is Q = h*A_base*(Tb-Tinf)*Effectiveness" D=0.01 {m} L=0.05 {m} Tb=280 Tinf=20 h=65 Effect=15 Q=h*(pi*D^2/4)*(Tb-Tinf)*Effect "Some Wrong Solutions with Common Mistakes:" W1_Q= h*(pi*D*L)*(Tb-Tinf)*Effect "Using fin area " W2_Q= h*(pi*D^2/4)*(Tb-Tinf) "Not using effectiveness" W3_Q= Q+W1_Q "Using wrong relation"

3-223 A cylindrical pin fin of diameter 0.6 cm and length of 3 cm with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is

(a) 0.3 (b) 0.7 (c) 2 (d) 8 (e) 14

Answer (e) 14


"The relation between fin efficiency and fin effectiveness is effect = (A_fin/A_base)*Efficiency" D=0.6 {cm} L=3 {cm} Effici=0.7 Effect=(pi*D*L/(pi*D^2/4))*Effici "Some Wrong Solutions with Common Mistakes:" W1_Effect= Effici "Taking it equal to efficiency" W2_Effect= (D/L)*Effici "Using wrong ratio" W3_Effect= 1-Effici "Using wrong relation" W4_Effect= (pi*D*L/(pi*D))*Effici "Using area over perimeter"


3-162

3-224 A 3-cm-long, 2 mm × 2 mm rectangular cross-section aluminum fin (k = 237 W/m⋅ºC) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is

(a) 39 (b) 30 (c) 24 (d) 18 (e) 7

Answer (a) 39


L=0.03 [m] s=0.002 [m] k=237 [W/m-C] eta_fin=0.65 A_fin=4*s*L A_b=s*s epsilon_fin=A_fin/A_b*eta_fin

3-225 Aluminum square pin fins (k = 237 W/m⋅ºC) of 3-cm-long, 2 mm × 2 mm cross-section with a total number of 150 are attached to an 8-cm-long, 6-cm-wide surface. If the fin efficiency is 78 percent, the overall fin effectiveness for the surface is

(a) 3.4 (b) 4.2 (c) 5.5 (d) 6.7 (e) 8.4

Answer (d) 6.7


s=0.002 [m] L=0.03 [m] k=237 [W/m-C] n_fin=150 Length=0.08 [m] Width=0.06 [m] eta_fin=0.78 A_fin=n_fin*4*s*L A_nofin=Length*Width A_unfin=A_nofin-n_fin*s*s epsilon_fin_overall=(A_unfin+eta_fin*A_fin)/A_nofin "Some Wrong Solutions with Common Mistakes" W1_epsilon_fin_overall=(A_unfin+A_fin)/A_nofin "Ignoring fin efficiency" A_fin1=4*s*L A_nofin1=Length*Width A_unfin1=A_nofin1-s*s W2_epsilon_fin_overall=(A_unfin1+eta_fin*A_fin1)/A_nofin1 "Considering a single fin in calculations"


3-163

3-226 Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one?

(a) higher efficiency and higher effectiveness

(b) higher efficiency but lower effectiveness

(c) lower efficiency but higher effectiveness

(d) lower efficiency and lower effectiveness

(e) equal efficiency and equal effectiveness

Answer (d) lower efficiency and lower effectiveness

Solution The efficiency of long fin is given by LhpkAc //= , which is inversely proportional to convection coefficienh. Therefore, efficiency of first finned surface with higher h will be lower. This is also the case for effectiveness since effectiveness is proportiona

η t

l to efficiency, )/( basefin AAηε = .

3-227 A 20-cm-diameter hot sphere at 120°C is buried in the ground with a thermal conductivity of 1.2 W/m⋅°C. The distance between the center of the sphere and the ground surface is 0.8 m, and the ground surface temperature is 15°C. The rate of heat loss from the sphere is

(a) 169 W (b) 20 W (c) 217 W (d) 312 W (e) 1.8 W

Answer (a) 169 W


D=0.2 T1=120 T2=15 K=1.2 Z=0.8 S=2*pi*D/(1-0.25*D/z) Q

=S*k*(T1-T2)

“Some Wrong Solutions with Common Mistakes:” A=pi*D^2 W1_Q=2*pi*z/ln(4*z/D) "Using the relation for cylinder" W2_Q=k*A*(T1-T2)/z "Using wrong relation" W3_Q= S*k*(T1+T2) "Adding temperatures" W4_Q= S*k*A*(T1-T2) "Multiplying vy area also"


3-164

3-228 A 25-cm-diameter, 2.4-m-long vertical cylinder containing ice at 0ºC is buried right under the ground. The cylinder is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are 18ºC and 0.85 W/m⋅ºC, respectively. The rate of heat transfer to the cylinder is

(a) 37.2 W (b) 63.2 W (c) 158 W (d) 480 W (e) 1210 W

Answer (b) 63.2 W


D=0.25 [m] L=2.4 [m] T_ice=0 [C] T_ground=18 [C] k=0.85 [W/m-C] S=(2*pi*L)/ln((4*L)/D) Q_dot=S*k*(T_ground-T_ice)

3-229 Hot water (c = 4.179 kJ/kg⋅K) flows through a 80 m long PVC (k = 0.092 W/m⋅K) pipe whose inner diameter is 2 cm and outer diameter is 2.5 cm at a rate of 1 kg/s, entering at 40°C. If the entire interior surface of this pipe is maintained at 35oC and the entire exterior surface at 20oC, the outlet temperature of water is

(a) 35oC (b) 36oC (c) 37oC (d) 38oC (e) 39°C

Answer (e) 39oC


do=2.5 [cm] di=2.0 [cm] k=0.092 [W/m-C] T2=35 [C] T1=20 [C] Q=2*pi*k*l*(T2-T1)/LN(do/di) Tin=40 [C] c=4179 [J/kg-K] m=1 [kg/s] l=80 [m] Q=m*c*(Tin-Tout)


3-165

3-230 The walls of a food storage facility are made of a 2-cm-thick layer of wood (k = 0.1 W/m⋅K) in contact with a 5-cm-thick layer of polyurethane foam (k = 0.03 W/m⋅K). If the temperature of the surface of the wood is -10oC and the temperature of the surface of the polyurethane foam is 20oC, the temperature of the surface where the two layers are in contact is

(a) -7oC (b) -2oC (c) 3oC (d) 8oC (e) 11°C

Answer (a) -7oC


kw=0.1 [W/m-C] tkw=0.02 [m] Tw=-10 [C] kf=0.03 [W/m-C] tkf=0.05 [W/m-C] Tf=20 [C] T=((kw*Tw/tkw)+(kf*Tf/tkf))/((kw/tkw)+(kf/tkf))

3-231 Heat transfer rate through the wall of a circular tube with convection acting on the outer surface is given per unit of its

length by

hrk o

rrq

io 1)/ln(+

&

er as long as

) ro < k/h (b) ro = k/h (c) ro > k/h (d) ro > 2k/h (e) increasing ro will always reduce the heat transfer

Answer (c) ro > k/h

TTL oi )(2 −=

π where i refers to the inner tube surface and o the outer tube surface. Increasing ro will

reduce the heat transf

(a


3-166

3-232 A typical section of a building wall is shown in the figure. This section extends in and out of the page and is repeated in the vertical direction. The correct thermal resistance circuit for this wall is

(a)

(b)

(c)

(d) (e) None of them Answer (b)

3-233 The 700 m2 ceiling of a building has a thermal resistance of 0.52 m2⋅K/W. The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is -10oC and the interior is at 20oC is

(a) 23.1 kW (b) 40.4 kW (c) 55.6 kW (d) 68.1 kW (e) 88.6 kW

Answer (b) 40.4 kW


R=0.52 [m^2-C/W] A=700 [m^2] T_1=20 [C] T_2=-10 [C] Q=A*(T_2-T_1)/R


3-167

3-234 A 1 m-inner diameter liquid oxygen storage tank at a hospital keeps the liquid oxygen at 90 K. This tank consists of a 0.5-cm thick aluminum (k = 170 W/m⋅K) shell whose exterior is covered with a 10-cm-thick layer of insulation (k = 0.02 W/m⋅K). The insulation is exposed to the ambient air at 20oC and the heat transfer coefficient on the exterior side of the insulation is 5 W/m2⋅K. The rate at which the liquid oxygen gains heat is

(a) 141 W (b) 176 W (c) 181 W (d) 201 W (e) 221 W

Answer (b) 176 W


R1=0.5 [m] R2=0.55 [m] R3=0.65 [m] k1=170 [W/m-K] k2=0.02 [W/m-K] h=5[W/m^2-K] T2=293 [K] T1=90 [K] R12=(R2-R1)/(4*pi*k1*R1*R2) R23=(R3-R2)/(4*pi*k2*R2*R3) R45=1/(h*4*pi*R3^2) Re=R12+R23+R45 Q=(T2-T1)/Re

3-235 A 1-m-inner diameter liquid oxygen storage tank at a hospital keeps the liquid oxygen at 90 K. This tank consists of a 0.5-cm-thick aluminum (k = 170 W/m⋅K) shell whose exterior is covered with a 10-cm-thick layer of insulation (k = 0.02 W/m⋅K). The insulation is exposed to the ambient air at 20oC and the heat transfer coefficient on the exterior side of the insulation is 5 W/m2⋅K. The temperature of the exterior surface of the insulation is

(a) 13oC (b) 9oC (c) 2oC (d) -3oC (e) -12°C

Answer (a) 13oC


R1=0.5 [m] R2=0.55 [m] R3=0.65 [m] k1=170 [W/m-K] k2=0.02 [W/m-K] h=5[W/m^2-K] T2=293 [K] T1=90 [K] R12=(R2-R1)/(4*pi*k1*R1*R2) R23=(R3-R2)/(4*pi*k2*R2*R3) R45=1/(h*4*pi*R3^2) Re=R12+R23+R45 Q=(T2-T1)/Re Q=(T2-T3)/R45


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3-236 The fin efficiency is defined as the ratio of the actual heat transfer from the fin to

(a) The heat transfer from the same fin with an adiabatic tip

(b) The heat transfer from an equivalent fin which is infinitely long

(c) The heat transfer from the same fin if the temperature along the entire length of the fin is the same as the base temperature

(d) The heat transfer through the base area of the same fin

(e) None of the above

Answer: (c)

3-237 Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A 512 MB memory chip dissipates 5 W of heat to air at 25oC. If the temperature of this chip is not exceed 60oC, the overall heat transfer coefficient – area product of the finned metal mount must be at least

(a) 0.14 W/oC (b) 0.20 W/oC (c) 0.32 W/oC (d) 0.48 W/oC (e) 0.76 W/oC

Answer (a) 0.14 W/oC


T_1=60 [C] T_2=25 [C] Q=5 [W] Q=UA*(T_1-T_2)

3-238 In the United States, building insulation is specified by the R-value (thermal resistance in h⋅ft2⋅oF/Btu units). A home owner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total R-value is increased from 15 to 25, the home owner can expect the heat loss through the ceiling to be reduced by

(a) 25% (b) 40% (c) 50% (d) 60% (e) 75%

Answer (b) 40%


R_1=15 R_2=25 DeltaT=1 "Any value can be chosen" Q1=DeltaT/R_1 Q2=DeltaT/R_2 Reduction%=100*(Q1-Q2)/Q1


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3-239 Coffee houses frequently serve coffee in a paper cup that has a corrugated paper jacket surrounding the cup like that shown here. This corrugated jacket

(a) serves to keep the coffee hot

(b) increases the coffee-to-surrounding thermal resistance

(c) lowers the temperature where the hand clasps the cup

(d) all of the above

(e) none of the above

Answer (d) all of the above

3-240 A triangular shaped fin on a motorcycle engine is 0.5-cm thick at its base and 3-cm long (normal distance between the base and the tip of the triangle), and is made of aluminum (k = 150 W/m⋅K). This fin is exposed to air with a convective heat transfer coefficient of 30 W/m2⋅K acting on its surfaces. The efficiency of the fin is 75 percent. If the fin base temperature is 130oC and the air temperature is 25oC, the heat transfer from this fin per unit width is

(a) 32 W/m (b) 57 W/m (c) 102 W/m (d) 124 W/m (e) 142 W/m

Answer (e) 142 W/m


h=30 [W/m-K] b=0.005 [m] l=0.03 [m] eff=0.75 Ta=25 [C] Tb=130 [C] A=2*(l^2+(b/2)^2)^0.5 Qideal=h*A*(Tb-Ta) Q=eff*Qideal

3-241 A plane brick wall (k = 0.7 W/m⋅K) is 10 cm thick. The thermal resistance of this wall per unit of wall area is

(a) 0.143 m2⋅K/W (b) 0.250 m2⋅K/W (c) 0.327 m2⋅K/W (d) 0.448 m2⋅K/W (e) 0.524 m2⋅K/W

Answer (a) 0.143 m2⋅K/W


k=0.7 [W/m-K] t=0.1 [m] R=t/k


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3-242 ··· 3-248 Design and Essay Problems

Heat 4e SM Chap03

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