HT3eChap02_129_164

2-63

2-129 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity of the tank is given to be k = 18 W/m⋅°C. Also, hfg = 198 kJ/kg for nitrogen. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as

02 =⎟⎠⎞

⎜⎝⎛

drdTr

drd

and C196)( 11 °−== TrT

])([)(2

2∞−=− TrTh

drrdTk

(b) Integrating the differential equation once with respect to r gives

r1

r2

h T∞

r -196°C

N2

12 C

drdTr =

Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,

21

rC

drdT

= → 21)( C

rC

rT +−=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

r = r1: 121

11 )( TC

rC

rT =+−=

r = r2: ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=− ∞TC

rCh

rCk 2

2

12

2

1

Solving for C1 and C2 simultaneously gives

1

2

21

2

11

1

112

21

2

121

1 and

1

)(rr

hrk

rr

TTT

rC

TC

hrk

rr

TTrC

−−

−+=+=

−−

−= ∞∞

Substituting C1 and C2 into the general solution, the variation of temperature is determined to be

196)/1.205.1(8.549C)196(1.221.2

)m 1.2)(C W/m25(C W/m18

21.21

C)20196(

1

11)(

2

12

1

2

21

2

11

11

1

11

1

−−=°−+⎟⎠⎞

⎜⎝⎛ −

°⋅

°⋅−−

°−−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

−=+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=++−= ∞

rr

Trr

rr

hrk

rr

TTT

rrC

rC

Tr

CrT

(c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from

negative) since tank the(to W 261,200

)m 1.2)(C W/m25(C W/m18

21.21

C)20196(m) 1.2()C W/m18(4

1

)(44)4(

2

21

2

1212

12

−=

°⋅

°⋅−−

°−−°⋅−=

−−

−−=−=−=−= ∞

π

πππ

hrk

rr

TTrkkC

rC

rkdxdTkAQ&

kg/s 1.32===J/kg 000,198J/s 200,261

fghQm&

&

PROPRIETARY MATERIAL. © 2007 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.

2-64

2-130 A spherical liquid oxygen container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of oxygen are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity of the tank is given to be k = 18 W/m⋅°C. Also, hfg = 213 kJ/kg for oxygen. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as

02 =⎟⎠⎞

⎜⎝⎛

drdTr

drd

and C183)( 11 °−== TrT

])([)(2

2∞−=− TrTh

drrdTk

(b) Integrating the differential equation once with respect to r gives

r1

r2

h T∞

r -183°C

O2

12 C

drdTr =

Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating,

21

rC

drdT

= → 21)( C

rC

rT +−=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

r = r1: 121

11 )( TC

rC

rT =+−=

r = r2: ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=− ∞TC

rCh

rCk 2

2

12

2

1

Solving for C1 and C2 simultaneously gives

1

2

21

2

11

1

112

21

2

121

1 and

1

)(rr

hrk

rr

TTT

rC

TC

hrk

rr

TTrC

−−

−+=+=

−−

−= ∞∞

Substituting C1 and C2 into the general solution, the variation of temperature is determined to be

183)/1.205.1(7.516C)183(1.2

21.2

)m 1.2)(C W/m25(C W/m18

21.21

C)20183(

1

11)(

2

12

1

2

21

2

11

11

1

11

1

−−=°−+⎟⎠⎞

⎜⎝⎛ −

°⋅

°⋅−−

°−−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

−=+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=++−= ∞

rr

Trr

rr

hrk

rr

TTT

rrC

rC

Tr

CrT

(c) The rate of heat transfer through the wall and the rate of evaporation of oxygen are determined from

negative) since tank the(to

)m 1.2)(C W/m25(C W/m18

21.21

C)20183(m) 1.2()C W/m18(4

1

)(44)4(

2

21

2

1212

12

W245,450−=

°⋅

°⋅−−

°−−°⋅−=

−−

−−=−=−=−= ∞

π

πππ

hrk

rr

TTrkkC

rC

rkdxdTkAQ&

kg/s 1.15===J/kg 000,213J/s 450,245

fghQm&

&

PROPRIETARY MATERIAL. © 2007 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.

2-65

2-131 A large plane wall is subjected to convection, radiation, and specified temperature on the right surface and no conditions on the left surface. The mathematical formulation, the variation of temperature in the wall, and the left surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the wall. Properties The thermal conductivity and emissivity are given to be k = 8.4 W/m⋅°C and ε = 0.7. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, and the mathematical formulation of this problem can be expressed as

02

2=

dxTd

and ])273[(][])([])([)( 4surr

422

4surr

4 TTTThTLTTLThdx

LdTk −++−=−+−=− ∞∞ εσεσ

C45)( 2 °== TLT

45°Cε

x

h T∞

L

Tsurr(b) Integrating the differential equation twice with respect to x yields

1CdxdT

=

21)( CxCxT +=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

Convection at x = L kTTTThC

TTTThkC

/]})273[(][{

])273[(][ 4

surr4

221

4surr

4221

−+εσ+−−=→

−+εσ+−=−

∞

∞

Temperature at x = L: LCTCTCLCLT 122221 )( −=→=+×=

Substituting C1 and C2 into the general solution, the variation of temperature is determined to be

)4.0(23.4845

m )4.0(C W/m4.8

]K) 290()K 318)[(K W/m100.7(5.67+C)2545)(C W/m14(C45

)(])273[(][)()()(

444282

4surr

422

212121

x

x

xLk

TTTThTCxLTLCTxCxT

−+=

−°⋅

−⋅×°−°⋅+°=

−−+εσ+−

+=−−=−+=

−

∞

(c) The temperature at x = 0 (the left surface of the wall) is C64.3°=−+= )04.0(23.4845)0(T

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2-66

2-132 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. 4 Heat loss through the upper part of the iron is negligible. Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7. Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be

ε

x

h T∞

L

Tsurrq 2

24base

00 W/m667,66

m 10150 W1000

=×

==−A

Qq

&&

Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as

02

2=

dxTd

and 20 W/m667,66

)0(==− q

dxdT

k &

])273[(][])([])([)( 4surr

422

4surr

4 TTTThTLTTLThdx

LdTk −++−=−+−=− ∞∞ εσεσ

(b) Integrating the differential equation twice with respect to x yields

1CdxdT

=

21)( CxCxT +=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

x = 0: k

qCqkC 0

101 &

& −=→=−

x = L: ])273[(][ 4surr

4221 TTTThkC −++−=− ∞ εσ

Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2,

04

surr4

22 ])273[()( qTTTTh &=−++− ∞ εσ

(c) Substituting the known quantities into the implicit relation above gives

2442

4282

2 W/m667,66]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =−+⋅×+−°⋅ − TT

Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above to be T2 = 759°C

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2-67

2-133 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. 4 Heat loss through the upper part of the iron is negligible. Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7. Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be

ε

x

h T∞

L

Tsurrq 2

24base

00 W/m000,100

m 10150 W1500

=×

==−A

Qq

&&

Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as

02

2=

dxTd

and 20 W/m000,100

)0(==− q

dxdT

k &

])273[(][])([])([)( 4

surr4

224

surr4 TTTThTLTTLTh

dxLdT

k −++−=−+−=− ∞∞ εσεσ

(b) Integrating the differential equation twice with respect to x yields

1C

dxdT

=

21)( CxCxT +=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

x = 0: k

qCqkC 0

101 &

& −=→=−

x = L: ])273[(][ 4surr

4221 TTTThkC −++−=− ∞ εσ

Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2,

04

surr4

22 ])273[()( qTTTTh &=−++− ∞ εσ

(c) Substituting the known quantities into the implicit relation above gives

2442

4282

2 W/m000,100]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =−+⋅×+−°⋅ − TT

Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above to be T2 = 896°C

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2-68

2-134E The concrete slab roof of a house is subjected to specified temperature at the bottom surface and convection and radiation at the top surface. The temperature of the top surface of the roof and the rate of heat transfer are to be determined when steady operating conditions are reached. Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the roof area is large relative to its thickness, and the thermal conditions on both sides of the roof are uniform. 3 Thermal properties are constant. 4 There is no heat generation in the wall. Properties The thermal conductivity and emissivity are given to be k = 1.1 Btu/h⋅ft⋅°F and ε = 0.8. Analysis In steady operation, heat conduction through the roof must be equal to net heat transfer from the outer surface. Therefore, taking the outer surface temperature of the roof to be T2 (in °F),

])460[()( 4sky

422

21 TTATThAL

TTkA −++−=

−∞ σε x T∞

h

T1

L

Tsky

Canceling the area A and substituting the known quantities,

4442

428

222

R]310)460)[(RftBtu/h 101714.0(8.0

F)50)(FftBtu/h 2.3(ft 0.8

F)62()FftBtu/h 1.1(

−+⋅⋅×+

°−°⋅⋅=°−

°⋅⋅

− T

TT

Using an equation solver (or the trial and error method), the outer surface temperature is determined to be T2 = 38°F Then the rate of heat transfer through the roof becomes

Btu/h 28,875=°−

×°⋅⋅=−

=ft 0.8

F)3862()ft 3525)(FftBtu/h 1.1( 221

LTT

kAQ&

Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the house is losing heat as expected.

PROPRIETARY MATERIAL. © 2007 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.

2-69

2-135 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since this two-layer heat transfer problem possesses symmetry about the center line and involves no change in the axial direction, and thus T = T(r) . 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform. Properties It is given that C W/m18wire °⋅=k and C W/m8.1plastic °⋅=k .

Analysis Letting denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in the wire can be expressed as

IT

01 gen =+⎟⎠⎞

⎜⎝⎛

ke

drdTr

drd

r

&

r2r

r1

egen

T∞

h

with and ITrT =)( 1 0)0(=

drdT

Multiplying both sides of the differential equation by r, rearranging, and integrating give

rk

edrdTr

drd gen&

−=⎟⎠⎞

⎜⎝⎛ → 1

2gen

2Cr

ke

drdTr +−=

& (a)

Applying the boundary condition at the center (r = 0) gives

B.C. at r = 0: 0 02

)0(0 11

gen =→+×−=× CCk

edr

dT &

Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating,

rk

edrdT

2gen&

−= → 22gen

4)( Cr

ke

rT +−=&

(b)

Applying the other boundary condition at 1rr = ,

B. C. at : 1rr = 21

gen22

21

gen

4

4r

ke

TCCrk

eT II

&&+=→+−=

Substituting this relation into Eq. (b) and rearranging give 2C

)(4

)( 221

wire

genwire rr

ke

TrT I −+=&

(c)

Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as

0=⎟⎠⎞

⎜⎝⎛

drdTr

drd

with and ITrT =)( 1 ])([)(2

2∞−=− TrTh

drrdTk

The solution of the differential equation is determined by integration to be

1CdrdTr = →

rC

drdT 1= → 21 ln)( CrCrT +=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1: 112211 ln ln rCTCTCrC II −=→=+

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2-70

r = r2: ])ln[( 2212

1∞−+=− TCrCh

rCk →

21

21

lnhrk

rr

TTC I

+

−= ∞

Substituting C1 and C2 into the general solution, the variation of temperature in plastic is determined to be

1

2

plastic

1

2111plastic ln

lnlnln)(

rr

hrk

rr

TTTrCTrCrT I

II

+

−+=−+= ∞

We have already utilized the first interface condition by setting the wire and plastic layer temperatures equal to at the interface . The interface temperature is determined from the second interface condition that the heat flux in the wire and the plastic layer at

IT 1rr = IT

1rr = must be the same:

1

2

plastic

1

2plastic

1gen1plasticplastic

1wirewire

1

ln2

)()(

rhr

krr

TTk

redr

rdTk

drrdT

k I

+

−−=→−=− ∞

&

Solving for and substituting the given values, the interface temperature is determined to be IT

C97.1°°+⎟⎟⎠

⎞⎜⎜⎝

⎛

°⋅

°⋅+

°⋅×

=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∞

=C25m) C)(0.007 W/m(14

C W/m1.8m 0.003m 007.0ln

C) W/m2(1.8m) 003.0)( W/m105.1(

ln2

2

236

2

plastic

1

2

plastic

21gen T

hrk

rr

kre

TI

&

Knowing the interface temperature, the temperature at the center line (r = 0) is obtained by substituting the known quantities into Eq. (c),

C97.3°=°⋅×

×°=+=

C) W/m(184m) 003.0)( W/m105.1(+C1.97

4)0(

236

wire

21gen

wire kre

TT I&

Thus the temperature of the centerline will be slightly above the interface temperature.

PROPRIETARY MATERIAL. © 2007 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.

2-71

2-136 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the shell is to be determined.

r2

T2

r r1

T1k(T)

Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be

. )1()( 20 TkTk β+=

Analysis When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between is determined from

21 and TT

( ) ( )

( )⎥⎦⎤

⎢⎣⎡ +++=

−

⎥⎦⎤

⎢⎣⎡ −+−

=−

⎟⎠⎞

⎜⎝⎛ +

=−

+=

−=

∫∫

2121

220

12

31

32120

12

30

12

20

12avg

31

33)1()(2

1

2

1

2

1

TTTTk

TT

TTTTk

TT

TTk

TT

dTTk

TT

dTTkk

T

T

T

T

T

T

β

βββ

This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity equals the rate of heat transfer through the same medium with variable conductivity k(T).

avgk

Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be

( ))/ln(3

12)/ln(

212

212121

220

12

21avgcylinder rr

TTLTTTTk

rrTT

LkQ−

⎥⎦⎤

⎢⎣⎡ +++=

−=

βππ&

Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-77, and performed the indicated integration. 2-137 Heat is generated uniformly in a cylindrical uranium fuel rod. The temperature difference between the center and the surface of the fuel rod 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 change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity of uranium at room temperature is k = 27.6 W/m⋅°C (Table A-3). Analysis The temperature difference between the center and the surface of the fuel rods is determined from Ts

e D C92.8°=°

×==−

C) W/m.6.27(4m) 016.0)( W/m104(

4

2372gen

kre

TT oso

&

PROPRIETARY MATERIAL. © 2007 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.

2-72

2-138 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 0.77 W/m⋅°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the mathematical formulation of this problem can be expressed as

02

2=

dxTd

k

h1

T∞1

L

h2

T∞2

and

dx

dTkTTh

)0()]0([ 11 −=−∞

])([)(

22 ∞−=− TLThdx

LdTk

(b) Integrating the differential equation twice with respect to x yields

1CdxdT

=

21)( CxCxT +=

where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: 12111 )]0([ kCCCTh −=+×−∞

x = L: ])[( 22121 ∞−+=− TCLChkC

Substituting the given values, these equations can be written as 12 77.0)27(5 CC −=−

)82.0)(12(77.0 211 −+=− CCC

Solving these equations simultaneously give 20 45.45 21 =−= CC

Substituting into the general solution, the variation of temperature is determined to be 21 and CC

xxT 45.4520)( −=

(c) The temperatures at the inner and outer surfaces are

C10.9

C20°=×−=

°=×−=2.045.4520)(

045.4520)0(LT

T

PROPRIETARY MATERIAL. © 2007 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.

2-73

2-139 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in the pipe. The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the centerline. 2 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 14 W/m⋅°C. Analysis The rate of heat generation is determined from

[ ]3

2221

22

gen W/m750,264/)m 17(m) 3.0(m) 4.0(

W000,254/)(

=−

=−

==ππ LDD

WWe&&

&V

Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as

01 gen =+⎟⎠⎞

⎜⎝⎛

ke

drdTr

drd

r

&

r2

T2

r r1

T1egen

and C60)( 11 °== TrT C80)( 22 °== TrTRearranging the differential equation

0gen =−

=⎟⎠⎞

⎜⎝⎛

kre

drdTr

drd &

and then integrating once with respect to r,

1

2gen

2C

kre

drdTr +

−=

&

Rearranging the differential equation again

r

Ck

redrdT 1gen

2+

−=

&

and finally integrating again with respect to r, we obtain

21

2gen ln4

)( CrCk

rerT ++

−=

&

where C1 and C2 are arbitrary constants. Applying the boundary conditions give

r = r1: 211

21gen

1 ln4

)( CrCk

rerT ++

−=

&

r = r2: 221

22gen

2 ln4

)( CrCk

rerT ++

−=

&

Substituting the given values, these equations can be written as

21

2)15.0ln(

)14(4)15.0)(750,26(60 CC ++

−=

21

2)20.0ln(

)14(4)20.0)(750,26(80 CC ++

−=

Solving for simultaneously gives 21 and CC 8.257 58.98 21 == CCSubstituting into the general solution, the variation of temperature is determined to be 21 and CC

rrrrrT ln58.987.4778.2578.257ln58.98)14(4

750,26)( 22

+−=++−

=

The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the inner radius and outer radius. C71.3°=+−= )175.0ln(58.98)175.0(7.4778.257)( 2rT

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2-74

2-140 Heat is generated in a plane wall. Heat is supplied from one side which is insulated while the other side is subjected to convection with water. The convection coefficient, the variation of temperature in the wall, and the location and the value of the maximum temperature in the wall 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 the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Analysis (a) Noting that the heat flux and the heat generated will be transferred to the water, the heat transfer coefficient is determined from the Newton’s law of cooling to be

Ts

L

k

x

Heater

Insulation T∞ , h

sq&

gene&C W/m400 2 °⋅=°−

+=

−

+=

∞

C40)(90m) )(0.04 W/m(10) W/m(16,000 352

gen

TTLeq

hs

s &&

(b) The variation of temperature in the wall is in the form of T(x) = ax2+bx+c. First, the coefficient a is determined as follows

k

e

dTTdke

dxTdk gen

2

2

gen2

20

&& −=→=+

cbxxk

eTbx

ke

dxdT

++−=+−= 2gengen

2 and

&& →

235

gen C/m2500)C W/m20(2

W/m102

°−=°⋅

=−=k

ea

&

Applying the first boundary condition: x = 0, T(0) = Ts → c = Ts = 90ºC As the second boundary condition, we can use either

sLx

qdxdTk −=−

=

→ sqbk

Lek =⎟

⎟⎠

⎞⎜⎜⎝

⎛+− gen&

→ ( ) ( ) C/m100004.010160002011 5

gen °=×+=+= Leqk

b s &

or

)(0

∞=

−−=− TThdxdTk s

x

k(a×0+b) = h(Ts -T∞) → C/m1000)4090(20400

°=−=b

Substituting the coefficients, the variation of temperature becomes

9010002500)( 2 ++−= xxxT

(c) The x-coordinate of Tmax is xvertex= -b/(2a) = 1000/(2×2500) = 0.2 m = 20 cm. This is outside of the wall boundary, to the left, so Tmax is at the left surface of the wall. Its value is determined to be

C126°=++−=++−== 90)04.0(1000)04.0(25009010002500)( 22max LLLTT

The direction of qs(L) (in the negative x direction) indicates that at x = L the temperature increases in the positive x direction. If a is negative, the T plot is like in Fig. 1, which shows Tmax at x=L. If a is positive, the T plot could only be like in Fig. 2, which is incompatible with the direction of heat transfer at the surface in contact with the water. So, temperature distribution can only be like in Fig. 1, where Tmax is at x=L, and this was determined without using numerical values for a, b, or c.

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2-75

Fig. 1

qs(0)

qs(L) Slope Fig. 2 qs(L)

qs(0) Slope

Here, heat transfer and slope are incompatible

This part could also be answered to without any information about the nature of the T(x) function, using qualitative arguments only. At steady state, heat cannot go from right to left at any location. There is no way out through the left surface because of the adiabatic insulation, so it would accumulate somewhere, contradicting the steady state assumption. Therefore, the temperature must continually decrease from left to right, and Tmax is at x = L.

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2-76

2-141 Heat is generated in a plane wall. The temperature distribution in the wall is given. The surface temperature, the heat generation rate, the surface heat fluxes, and the relationship between these heat fluxes, the heat generation rate, and the geometry of the wall 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 the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Analysis (a) The variation of temperature is symmetric about x = 0. The surface temperature is

C72°=°×−°=−=−== 2242 )m 02.0)(C/m102(C80)()( bLaLTLTTs

The plot of temperatures across the wall thickness is given below.

-0.02 -0.01 0 0.01 0.0270

72

74

76

78

80

82

x [m]

T [C

]

T∞ h

x

k

gene&

72ºC

L

72ºC

-L

(b) The volumetric rate of heat generation is

35 W/m102×=°×°⋅=−−=⎯→⎯=+ )C/m102(C) W/m5(2)2(0 24gengen2

2bkee

dxTdk &&

(c) The heat fluxes at the two surfaces are

[ ] 2

2

W/m4000

W/m4000

−=°×°⋅−=−−−=−=−

=°×°⋅=−−=−=

m) 02.0)(C/m102(C) W/m5(2)(2()(

m) 02.0)(C/m102(C) W/m5(2)2()(

24

24

LbkdxdTkLq

bLkdxdTkLq

Ls

Ls

&

&

(d) The relationship between these fluxes, the heat generation rate and the geometry of the wall is

[ ]

[ ]LeLqLq

LWHeWHLqLq

eALqLq

EE

ss

ss

ss

gen

gen

gen

genout

2)()(

)2()()(

)()(

&&&

&&&

&&&

&&

=−+

=−+

=−+

=

V

Discussion Note that in this relation the absolute values of heat fluxes should be used. Substituting numerical values gives 8000 W/m2 on both sides of the equation, and thus verifying the relationship.

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2-77

2-142 Steady one-dimensional heat conduction takes place in a long slab. It is to be shown that the heat

flux in steady operation is given by ⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

wTTTT

Wkq

*0

**ln& . Also, the heat flux is to be calculated for a given

set of parameters. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. Analysis The derivation is given as follows

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+

−−=+

−=+

+

−=−=

∫∫

w

w

T

T

WT

T

TTTT

Wkq

kWq

TTTT

Wkq

TT

dxkq

TTdT

dxdT

TTk

dxdTkq

w

w

*0

**

*0

*

*

**

0**

*

*

ln

ln

)0()ln(

or

)(

0

0

&

&

&

&

&

The heat flux for the given values is

25 W/m101.42×−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−×

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

K)4001000(K)6001000(ln

m 0.2 W/m107ln

4

*0

**

wTTTT

Wkq&

2-143 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at the surface of the ball 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., and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k = 45 W/m⋅°C.

h T∞

gene&

D Analysis The temperatures at the center and at the surface of the ball are determined directly from

C86.7°=°

×+°=+= ∞

C). W/m1200(3m) 12.0)( W/m106.2(

C03 2

36gen

hre

TT os

&

C225°=°

×+°=+=

C) W/m.45(6m) 12.0)( W/m106.2(

C7.866

2362gen

0 kre

TT os

&

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2-78

Fundamentals of Engineering (FE) Exam Problems

2-144 The heat conduction equation in a medium is given in its simplest form as 01gen =+⎟

⎠⎞

⎜⎝⎛ e

drdTrk

drd

r&

Select the wrong statement below. (a) the medium is of cylindrical shape. (b) the thermal conductivity of the medium is constant. (c) heat transfer through the medium is steady. (d) there is heat generation within the medium. (e) heat conduction through the medium is one-dimensional. Answer (b) thermal conductivity of the medium is constant 2-145 Consider a medium in which the heat conduction equation is given in its simplest form as

tT

rTr

rr ∂∂

α=⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ 11 2

2

(a) Is heat transfer steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the thermal conductivity of the medium constant or variable? (e) Is the medium a plane wall, a cylinder, or a sphere? (f) Is this differential equation for heat conduction linear or nonlinear? Answers: (a) transient, (b) one-dimensional, (c) no, (d) constant, (e) sphere, (f) linear 2-146 An apple of radius R is losing heat steadily and uniformly from its outer surface to the ambient air at temperature T∞ with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute temperatures). Also, heat is generated within the apple uniformly at a rate of

per unit volume. If Tgene& s denotes the outer surface temperature, the boundary condition at the outer surface of the apple can be expressed as

(a) )()( 4surr

4 TTTThdrdTk ss

Rr−+−=− ∞

=

εσ (b) genssRr

eTTTThdrdTk &+−+−=− ∞

=

)()( 4surr

4εσ

(c) )()( 4surr

4 TTTThdrdTk ss

Rr−+−= ∞

=

εσ (d) genssRr

eR

RTTTThdrdTk &

2

34

surr4

43/4)()(

ππεσ +−+−= ∞

=

(e) None of them

Answer: )()( 4surr

4 TTTThdrdTk ss

Rr−+−=− ∞

=

εσ

Note: Heat generation in the medium has no effect on boundary conditions.

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2-79

2-147 A furnace of spherical shape is losing heat steadily and uniformly from its outer surface of radius R to the ambient air at temperature T∞ with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute temperatures). If To denotes the outer surface temperature, the boundary condition at the outer surface of the furnace can be expressed as

(a) )()( 4surr

4 TTTThdrdTk oo

Rr−+−=− ∞

=

εσ (b) )()( 4surr

4 TTTThdrdTk oo

Rr−−−=− ∞

=

εσ

(c) )()( 4surr

4 TTTThdrdTk oo

Rr−+−= ∞

=

εσ (d) )()( 4surr

4 TTTThdrdTk oo

Rr−−−= ∞

=

εσ

(e) )()()4( 4surr

42 TTTThdrdTRk oo

Rr−+−= ∞

=

εσπ

Answer (a) )()( 4surr

4 TTTThdrdTk oo

Rr−+−=− ∞

=

εσ

2-148 A plane wall of thickness L is subjected to convection at both surfaces with ambient temperature T∞1 and heat transfer coefficient h1 at inner surface, and corresponding T∞2 and h2 values at the outer surface. Taking the positive direction of x to be from the inner surface to the outer surface, the correct expression for the convection boundary condition is

(a) [ ))0()0(

11 ∞−= TThdx

dTk ] (b) [ ))(

)(22 ∞−= TLTh

dxLdT

k ]

(c) [ ))0(211 ∞∞ −=− TTh

dxdTk ] (d) [ ]))(

212 ∞∞ −=− TThdx

LdTk

(e) None of them

Answer (a) [ ]))0()0(11 ∞−= TTh

dxdTk

2-149 Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them Answer (a) plane wall

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2-80

2-150 Consider a large plane wall of thickness L, thermal conductivity k, and surface area A. The left surface of the wall is exposed to the ambient air at T∞ with a heat transfer coefficient of h while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is

(a) ∞−

= Tk

xLhxT )()( (b) ∞+= T

LxhkxT

)5.0()(

(c) ∞⎟⎠⎞

⎜⎝⎛ −= T

kxhxT 1)( (d) ∞−= TxLxT )()( (e) ∞= TxT )(

Answer (e) ∞= TxT )(

2-151 The variation of temperature in a plane wall is determined to be T(x)=65x+25 where x is in m and T is in °C. If the temperature at one surface is 38ºC, the thickness of the wall is (a) 2 m (b) 0.4 m (c) 0.2 m (d) 0.1 m (e) 0.05 m Answer (c) 0.2 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 38=65*L+25 2-152 The variation of temperature in a plane wall is determined to be T(x)=110-48x where x is in m and T is in °C. If the thickness of the wall is 0.75 m, the temperature difference between the inner and outer surfaces of the wall is (a) 110ºC (b) 74ºC (c) 55ºC (d) 36ºC (e) 18ºC Answer (d) 36ºC Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. T1=110 [C] L=0.75 T2=110-48*L DELTAT=T1-T2

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2-81

2-153 The temperatures at the inner and outer surfaces of a 15-cm-thick plane wall are measured to be 40ºC and 28ºC, respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) (b) 4028)( += xxT 2840)( +−= xxT

(c) (d) 2840)( += xxT 4080)( +−= xxT

(e) 8040)( −= xxT

Answer (d) 4080)( +−= xxT

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. T1=40 [C] T2=28 [C] L=0.15 [m] "T(x)=C1x+C2" C2=T1 T2=C1*L+T1 2-154 Heat is generated in a long 0.3-cm-diameter cylindrical electric heater at a rate of 150 W/cm3. The heat flux at the surface of the heater in steady operation is (a) 42.7 W/cm2 (b) 159 W/cm2 (c) 150 W/cm2 (d) 10.6 W/cm2 (e) 11.3 W/cm2 Answer (e) 11.3 W/cm2

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. "Consider a 1-cm long heater:" L=1 [cm] e=150 [W/cm^3] D=0.3 [cm] V=pi*(D^2/4)*L A=pi*D*L "[cm^2]” Egen=e*V "[W]" Qflux=Egen/A "[W/cm^2]" “Some Wrong Solutions with Common Mistakes:” W1=Egen "Ignoring area effect and using the total" W2=e/A "Threating g as total generation rate" W3=e “ignoring volume and area effects”

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2-82

2-155 Heat is generated in a 8-cm-diameter spherical radioactive material whose thermal conductivity is 25 W/m.°C uniformly at a rate of 15 W/cm3. If the surface temperature of the material is measured to be 120°C, the center temperature of the material during steady operation is (a) 160°C (b) 280°C (c) 212°C (d) 360°C (e) 600°C Answer (b) 280°C D=0.08 Ts=120 k=25 e_gen=15E+6 T=Ts+g*(D/2)^2/(6*k) “Some Wrong Solutions with Common Mistakes:” W1_T= e_gen*(D/2)^2/(6*k) "Not using Ts" W2_T= Ts+e_gen*(D/2)^2/(4*k) "Using the relation for cylinder" W3_T= Ts+e_gen*(D/2)^2/(2*k) "Using the relation for slab" 2-156 Heat is generated in a 3-cm-diameter spherical radioactive material uniformly at a rate of 15 W/cm3. Heat is dissipated to the surrounding medium at 25°C with a heat transfer coefficient of 120 W/m2⋅°C. The surface temperature of the material in steady operation is (a) 56°C (b) 84°C (c) 494°C (d) 650°C (e) 108°C Answer (d) 650°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. h=120 [W/m^2-C] e=15 [W/cm^3] Tinf=25 [C] D=3 [cm] V=pi*D^3/6 "[cm^3]" A=pi*D^2/10000 "[m^2]" Egen=e*V "[W]" Qgen=h*A*(Ts-Tinf)

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2-83

2-157 Heat is generated uniformly in a 4-cm-diameter, 16-cm-long solid bar (k = 2.4 W/m⋅ºC). The temperatures at the center and at the surface of the bar are measured to be 210ºC and 45ºC, respectively. The rate of heat generation within the bar is (a) 240 W (b) 796 W b) 1013 W (c) 79,620 W (d) 3.96×106 W Answer (b) 796 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. D=0.04 [m] L=0.16 [m] k=2.4 [W/m-C] T0=210 [C] T_s=45 [C] T0-T_s=(e*(D/2)^2)/(4*k) V=pi*D^2/4*L E_dot_gen=e*V "Some Wrong Solutions with Common Mistakes" W1_V=pi*D*L "Using surface area equation for volume" W1_E_dot_gen=e*W14_1 T0=(W2_e*(D/2)^2)/(4*k) "Using center temperature instead of temperature difference" W2_Q_dot_gen=W2_e*V W3_Q_dot_gen=e "Using heat generation per unit volume instead of total heat generation as the result" 2-158 A solar heat flux is incident on a sidewalk whose thermal conductivity is k, solar absorptivity is α

sq&

s and convective heat transfer coefficient is h. Taking the positive x direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is

(a) ss qdxdTk &α=− (b) )( ∞−=− TTh

dxdTk (c) ss qTTh

dxdTk &α−−=− ∞ )(

(d) ss qTTh &α=− ∞ )( (e) None of them

Answer (c) ss qTThdxdTk &α−−=− ∞ )(

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2-84

2-159 Hot water flows through a PVC (k = 0.092 W/m⋅K) pipe whose inner diameter is 2 cm and outer diameter is 2.5 cm. The temperature of the interior surface of this pipe is 35oC and the temperature of the exterior surface is 20oC. The rate of heat transfer per unit of pipe length is (a) 22.8 W/m (b) 38.9 W/m (c) 48.7 W/m (d) 63.6 W/m (e) 72.6 W/m

Answer (b) 38.9 W/m

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. do=2.5 [cm] di=2.0 [cm] k=0.092 [W/m-C] T2=35 [C] T1=20 [C] Q=2*pi*k*(T2-T1)/LN(do/di) 2-160 The thermal conductivity of a solid depends upon the solid’s temperature as k = aT + b where a and b are constants. The temperature in a planar layer of this solid as it conducts heat is given by (a) aT + b = x + C2 (b) aT + b = C1x2 + C2 (c) aT2 + bT = C1x + C2 (d) aT2 + bT = C1x2 + C2 (e) None of them Answer (c) aT2 + bT = C1x + C2

2-161 Harvested grains, like wheat, undergo a volumetric exothermic reaction while they are being stored. This heat generation causes these grains to spoil or even start fires if not controlled properly. Wheat (k = 0.5 W/m⋅K) is stored on the ground (effectively an adiabatic surface) in 5-m thick layers. Air at 20°C contacts the upper surface of this layer of wheat with h = 3 W/m2⋅K. The temperature distribution inside this layer is given by

2

01 ⎟

⎠⎞

⎜⎝⎛−=

−−

Lx

TTTT

s

s

where Ts is the upper surface temperature, T0 is the lower surface temperature, x is measured upwards from the ground, and L is the thickness of the layer. When the temperature of the upper surface is 24oC, what is the temperature of the wheat next to the ground? (a) 39oC (b) 51oC (c) 72oC (d) 84oC (e) 91°C

Answer (d) 84oC k=0.5 [W/m-K] h=3 [W/m2-K] L=5[m] Ts=24 [C] Ta=20 [C] To=(h*L/(2*k))*(Ts-Ta)+Ts

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2-85

2-162 The conduction equation boundary condition for an adiabatic surface with direction n being normal to the surface is (a) T = 0 (b) dT/dn = 0 (c) d2T/dn2 = 0 (d) d3T/dn3 = 0 (e) -kdT/dn = 1 Answer (b) dT/dn = 0 2-163 Which one of the followings is the correct expression for one-dimensional, steady-state, constant thermal conductivity heat conduction equation for a cylinder with heat generation?

(a) tTce

rTrk

rr ∂∂

=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂ ρgen

1& (b)

tT

ke

rTr

rr ∂∂

=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

α11 gen&

(c) tT

rTr

rr ∂∂

α=⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ 11 (d) 01 gen =+⎟

⎠⎞

⎜⎝⎛

ke

drdTr

drd

r

& (e) 0=⎟

⎠⎞

⎜⎝⎛

drdTr

drd

Answer (d) 01 gen =+⎟⎠⎞

⎜⎝⎛

ke

drdTr

drd

r

&

2-164 .... 2-167 Design and Essay Problems

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