Exam 1 Review Problems

8/11/2019 Exam 1 Review Problems

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ME 364 Review Problems for Exam #1

Problem 1

A plane wall is a composite of a low conductivity material (with thickness L1and conductivity

k1) and a high conductivity material (with thickness L2= L1and conductivity k2). The edge of

the wall atx= 0 is at temperature T1and the edge atx=L1+L2has temperature T2, as shown inFigure P1.2-1(a). T1 is greater than T2. The wall is at steady-state and the temperature

distribution in the wall is one-dimensional inx.

T1 T2

k1 k2

L1 L2

x

k1 k2

x

k1 k2

x

x x

Tq

L1

0 L1

+L2

L1

0 L1

+L2

T1

T2

(a) (b)Figure P1.2-1: (a) Composite wall withk1


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Solving Eq. (2) for the temperature gradient leads to:

xqdT

dx k

(3)

The numerator of Eq. (3), the heat flux, is constant while the denominator changes depending on

whether you are in material 1 or material 2. In the low conductivity material 1, the temperature

gradient will be higher than in the high conductivity material 2. Within each material, thetemperature gradient must be constant (i.e., the temperature must be linear withx). The solution

is shown in Figure 3.

x

x

L1 L1+L2

q

x

x

L1 L1+L2

T

T1

T2

k1 k2 k1 k2

xq

(a) (b)

Figure 3: (a)Heat transfer rate and (b)temperature as a function of position within wall.

Problem 2

Figure P1.2-3 illustrates a plane wall made of a very thin ( thw= 0.001 m) and conductive (k=100 W/m-K) material that separates two fluids, A and fluid B. Fluid A is at TA= 100C and the

heat transfer coefficient between the fluid and the wall is Ah = 10 W/m2-K while fluid B is at TB

= 0C with Bh = 100 W/m2-K.

thw= 0.001 m

k= 100 W/m-K

2

100 C

10 W/m -K

A

A

T

h

2

0 C

100 W/m -K

B

B

T

h

Figure P1.2-3: Plane wall separating two fluids

a.) Draw a resistance network that represents this situation and calculate the value of each

resistor (assuming a unit area for the wall,A= 1 m2).


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Heat flowing from fluid A to fluid B must pass through a fluid A-to-wall convective resistance(Rconv,A), a resistance to conduction through the wall (Rcond), and a wall-to-fluid B convective

resistance (Rconv,B). These resistors are in series. The network and values of the resistors are

shown in Figure 2.

TA = 100C

,

1conv A

A

Rh A

wcond

tR

k A

K0.1

W

K0.0001

W

,

1cond B

B

Rh A

K0.01

W

TB = 0C

Figure 2: Thermal resistance network representing the wall.

b.) If you wanted to predict the heat transfer rate from fluid AtoBvery accurately, then which

parameter (e.g., thw, k, etc.) would you try to understand/measure very carefully and which

parameters are not very important? Justify your answer.

The largest resistance in a series network will control the heat transfer. For the wall above, the

largest resistance is Rconv,A. Therefore, I would focus on predicting this resistance accurately.

This would suggest that Ah is the most important parameter and the others do not matter much.

Problem 3

Figure P1.2-4 illustrates a plane wall that is composed of two materials, A and B. The interfacebetween the materials is characterized by a contact resistance. The left surface of material A is

held at THand the right surface of materialBradiates to surroundings at TCand is also exposed to

convection to a fluid at TC.

TH

material A

convection andradiation toT

C

material B

contact resistance Figure P1.2-4: Composite wall with contact resistance, convection and radiation

The resistance network that represents the situation in Figure P1.2-4 should include five thermal

resistors; their values are provided below:Rcond,A= 0.05 K/W, resistance to conduction through material A

Rcontact= 0.01 K/W, contact resistance

Rcond,B = 0.05 K/W, resistance to conduction through material BRconv= 1.0 K/W, resistance to convection

Rrad= 10.0 K/W, resistance to radiation


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a.) Draw a resistance network that represents the situation in Figure P1.2-4. Each resistance in

the network should be labeled according to Rcond,A, Rcontact, Rcond,B, Rconv, and Rrad. Showwhere the temperatures THand TCappear on your network.

Figure 2: Resistance network that represents Figure P1.2-4.

b.) What is the most important resistor in the network? That is, the heat transfer from THto TCismost sensitive to which of the five resistances?

The most important resistor in a series combination is the largest. The largest resistance is theparallel combination ofRconvandRrad. The most important resistance in a parallel combination is

the smallest; the smallest ofRconvandRradisRconv. Thus,Rconvis the most important resistance.

c.) What is the least important resistor in the network?

The least important resistance is the contact resistance; it is the smallest in a series of resistorsthat are themselves unimportant relative to convection and radiation.

Problem 4Figure P1.2-17 illustrates the temperature distribution in a plane wall at a particular instant oftime.

T

x Figure P1.2-17: Temperature distribution in a plane wall at a certain instant in time.

Select the correct statement from those listed below and justify your answer briefly.

The heat transfer at the left-hand face of the wall (i.e., at x= 0) is into the wall (in the

positivexdirection),

The heat transfer at the left-hand face of the wall is out of the wall (in the negative x

direction),

It is not possible to tell the direction of the heat transfer at the left-hand face of the wall.

Fouriers law states that conduction is proportional to the negative of the temperature gradient.

At the left-hand face of the wall the temperature gradient is positive; therefore, the heat transfer

must be in the negativex-direction or out of the wall.


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Problem 5

A cylinder with conductivity k experiences a uniform rate of volumetric generation g , as

shown in Figure P1.3-4. The cylinder experiences 1-D, steady state conduction heat transfer in

the radial direction and therefore the general solution to the ordinary differential equation fortemperature (T) is:

2

1 2ln4

g rT C r C

k

(4)

where ris the radial location and C1and C2are undetermined constants. At the inner radius of

the cylinder (r= rin), a heater applies a uniform rate of heat transfer, inq . At the outer radius of

the cylinder (r= rout), the temperature is fixed at Tout. The length of the cylinder isL. Write the

two algebraic equations that can be solved in order to obtain the constants C1 and C2. Your

equations must contain only the following symbols in the problem statement: inq

, Tout, k, rin, rout,L, g , C1, and C2. Do not solve these equations.

inq

rin

rout

Tout

,k gL

Figure P1.3-4: Cylinder with uniform volumetric generation.

At the outer surface, the temperature is specified and therefore the boundary condition is:

2

1 2ln4

outout out

g rT C r C

k

(5)

At the inner surface, the temperature is not specified and therefore it is necessary to do an energy

balance on this interface, as shown in Figure 2.


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Figure 2: Interface balance at r= rin.

The interface energy balance is:

@ inin r r q q (6)

Substituting Fourier's law for @ inr rq leads to:

2

in

in in

r r

dTq k r L

dr

(7)

Substituting the general solution, Eq. (4), into Eq. (7) leads to:

122

inin in

in

g r Cq k r L

k r

(8)

Problem 6

Figure P1.3-12 illustrates a plane wall. The temperature distribution in the wall is 1-D and the

problem is steady state.


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L

xLq ,h T

,g a x k

Figure P1.3-12: Plane wall.

There is generation of thermal energy in the wall. The generation per unit volume is not uniform

but rather depends on position according to:

g a x (9)

where ais a constant andxis position. The left side of the wall experiences a specified heat flux,

Lq . The right side of the wall experiences convection with heat transfer coefficient h to fluid at

temperature T. The thickness of the wall is Land the conductivity of the wall material, k, is

constant.

a.) Derive the ordinary differential equation that governs this problem. Clearly show your steps.

A differential control volume is shown in Figure 2 and leads to:

x x dxq g q (10)

x

dx

xq x dxq

g

Figure 2: Differential control volume with energy terms.

After expanding thex+ dx term:

x x

dqq g q dx

dx

(11)

The rate of thermal energy generation within the control volume is:

cg g A dx (12)


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where Ac is the cross-sectional area of the wall. The conduction term is expressed usingFouriers law:

c

dTq k A

dx

(13)

Substituting Eqs. (13) and (12) into Eq. (11) results in

c c

d dTg A dx k A dx

dx dx

(14)

which can be simplified:

d dT g

dx dx k

(15)

Substituting the position dependent generation into Eq. (15) leads to:

d dT a x

dx dx k

(16)

b.) Solve the differential equation that you obtained in (a). Your solution should include twoundetermined constants.

Equation (15) is separated and integrated:

dT a xd dx

dx k

(17)

which leads to:

2

12

dT aC

dx k (18)

where CB1Bis a constant of integration. Equation (18) is integrated again:

2

12

adT x C dx

k

(19)

which leads to:


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3 1 26

aT x C x C

k (20)

c.) Specify the boundary conditions for the differential equation that you derived in (a).

An interface energy balance atx= 0 leads to:

0

L

x

dTq k

dx (21)

An interface energy balance atx=Lleads to:

x Lx L

dTk h T T

dx

(22)

d.) Use the results of (b) and (c) to obtain two equations that can be solved for the twoundetermined constants.

Substituting Eq. (18) into Eq. (21) leads to:

1Lq k C (23)

Substituting Eqs. (18) and (20) into Eq. (22) leads to:

2 3

1 1 22 6

a a

k L C h L C L C T k k

(24)

Equations (23) and (24) can be solved for C1 and C2.

Problem 7

Figure P1.4.1(a) illustrates a plane wall with thickness L and cross-sectional area A that has a

specified temperature THon the left side (at x= 0) and a specified temperature TCon the rightside (atx= L). There is no volumetric generation in the wall. However, the conductivity of the

wall material is a function of temperature such that: k b cT where a and b are constants.

You would like to model the wall using a finite difference solution; a model with only 3 nodes is

shown in Figure P1.4-1(b).


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THTC

L

k = b+cT

T1 T2 T3

x

(a) (b)

Figure P1.4-1: (a) A plane wall and (b) a numerical model with 3 nodes.

The distance between adjacent nodes for the 3 node solution is: x=L/2.a.) Write down the system of equations that could be solved in order to obtain the temperatures

at the three nodes. Your equations should include the temperature of the nodes (T1, T2, and

T3) and the other parameters listed in the problem statement: TH, TC, x,A, b, and c.

The equations for T1and T3are easy, their temperatures are specified:

[1] HT T (25)

[3] CT T (26)

Figure 2 illustrates the control volume for the 2nd

node.

Figure 2: Control volume for node 2.

An energy balance for the control volume shown in Fig. 3 leads to:

[2] [2] 0RHS LHSq q (27)

The energy transfer rates must be approximated according to:


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3 2 3 2

[2]2

RHS

T T T Tq A b c

x

(28)

and

1 2 1 2[2]

2LHS

T T T Tq A b c

x

(29)

Notice that the temperature differences agree with the sign convention used in Figure 3 and that

the conductivity is evaluated at the temperature of the interface. Substituting Eqs. (28) and (29)into Eq. (27) leads to:

3 2 1 23 2 1 20

2 2

T T T T T T T T A b c A b c

x x

(30)

Equations (25), (26), and (30) together represent a system of three equations in the threeunknown temperatures.

Problem 8

Figure 1 illustrates a plane wall that is composed of material A (from 0


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L L

cg A xq

x

Figure 2: Control volume with material A.

The rate of conduction heat transfer within material A is given by:

cq g A x (31)

A control volume is defined that extends fromx= 0 toxwithin material B, as shown in Figure 3.

L L

q

x

cg A L

Figure 3: Control volume with material B.

The rate of conduction heat transfer within material B is given by:

cq g A L (32)

These characteristics are reflected in Figure 4.

Material A

2A Bk k

g

contact resistance cR

x

L L

Material B

Bk,h T

q

x0 L 2L

0

cg A L

Figure 4: Conduction heat transfer as a function ofx.


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b.) On the axes below, sketch the temperature as a function of position, x. This should be aqualitative sketch - the shape of the curve should be correct.

Fourier's law states that:

c

dT q

dx k A

(33)

Figure 5 is consistent with Eq. (33) - the temperature gradient is zero at x= 0 and becomes morenegative as the rate of heat transfer increases in material A. The temperature gradient is constant

in material B. The temperature gradient in material A at the interface is half that of material Bbecause it has twice the conductivity. Also, there are temperature drops associated with the

contact resistance and due to convection.

Material A

2A Bk k

g


x

L L

Material B

Bk,h T

x0 L 2L

T

T

Figure 5: Temperature as a function ofx.

You have decided to develop an analytical model of the temperature distribution in material A(i.e., the temperature from 0 < x


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An interface energy balance atx= 0 leads to:

0

0x

dT

dx (35)


1 0C (36)

An interface energy balance atx=Lleads to:

01

x LA c

cx

c B c c

T TdTk A

R Ldx

A k A h A

(37)


2

1 2

1

2

1A

A ccA

c B c c

gL C L C T

kgk A L C

R Lk

A k A h A

(38)

Equations (36) and (38) can be solved for C1and C2.

You have decided to develop a numerical model of the temperature distribution in material A,

shown in Figure 1. Your numerical model consists of three nodes, as shown in Figure 6.

L L

T1

T2 T3

Material A

, Ag k


Material B

Bk,h T

Figure 6: Numerical model of material A.

d.) Write the three equations that must be solved in order to determine the temperatures at each

of the nodes (T1, T2, and T3). Do not attempt to solve these equations.


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T1T2 T3

RHSq

Figure 7: Energy balance on the control volume for node 1.

An energy balance on the control volume for node 1 is shown in Figure 7 and leads to:

0RHSq g (39)

or

2 1 02

A cc

k A xT T g A

x

(40)

where

2

Lx (41)

T1T2 T3

RHSq

LHSq



0RHS LHSq q g (42)

or

3 2 1 2 0A c A c

c

k A k AT T T T g A x

x x

(43)

T1

T2 T3

LHSq inq



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0in LHS q q g (44)

or

3 2 3 01 2

A cc

c

c B c c

T T k A xT T g A

R L x

A k A h A

(45)

Equations (40), (43), and (45) together are 3 equations in the 3 unknown temperatures.

Figure 10 provides specific values of the parameters that govern the problem.

L= 0.1 m

Material A

2 W/m-KAk

g

Material B

1 W/m-KBk

-3 21x10 W-m /KcR

2100 W/m -Kh

T

L= 0.1 m

Figure 10 Parameters for the problem.

The conductivity of materials A and B are kA= 2 W/m-K and kB= 1 W/m-K, respectively. Thethickness of these materials is L= 0.1 m. The contact resistance is cR = 1x10-3

m2-K/W. The

heat transfer coefficient is h = 100 W/m2-K.

e.) Rank the following parameters in terms of their importance to the problem: kA, kB, cR , and

h . Note that the only things that you should need to calculate are the values of the thermal

resistances that govern the problem. Justify your answer using the values of these thermal

resistances.

The elevation of the temperature in the heater is due to several resistances in series, including the

thermal resistance to conduction through material B, contact resistance, and convectionresistance. In addition, there is the resistance to conduction through material A. The thermal

resistance to conduction through material B is:

, 2

0.1 m m-K0.1 K/W

1 W 1 mcond B

B c

LR

k A (46)


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where the resistance is computed on a unit area (Ac= 1 m2) basis. The thermal resistance

associated with the contact resistance is:

-3 2

2

1x10 K-m0.001 K/W

W 1 m

c

cc

RR

A

(47)

The thermal resistance associated with convection is:

2

2

1 m -K 0.01 K/W

100 W 1 mconv

c

Rh A

(48)

Material A cannot be represented exactly by a thermal resistance yet it is possible to

approximately determine its importance by calculating the resistance to conduction through A

according to:

, 2

0.1 m m-K0.025 K/W

2 2 2 W 1 mcond A

A c

LR

k A (49)

In descending order, the values of the resistances are ,cond BR , ,cond AR , convR , and cR . Therefore,

the ranking of the parameters are:kB, kA, h , and cR .

Problem 9Figure 1 illustrates a thin metal rod that has a heater installed on its tip.

L

Tb x

,h T

hq

Figure 1: Thin metal rod exposed to convection with a heater installed at its tip.

The rate of heat transfer provided by the heater, hq , is transferred by conduction into the tip of

the rod (atx=L), as shown. The base of the rod (atx= 0) is maintained at temperature, Tb. The

cross sectional area of the rod isAcand the perimeter of the rod isper. The rod is surrounded byfluid at temperature Twith heat transfer coefficient h . The conductivity of the rod material is

k. Assume that the extended surface approximation is appropriate (i.e., the temperature withinthe rod is a function only ofx). Neglect radiation for this problem.

a.) Sketch the temperature distribution in the rod. Justify your choice by listing some of the

characteristics that you expect to see.


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The correct solution is shown in Figure 2. Note that the solution satisfies the two boundary

conditions (the base temperature and the required temperature gradient corresponding to the heattransfer into the tip) and also remains always above the fluid temperature.

L

Tb x ,h T

hq

T

xL0

Tb

T

The ordinary differential equation that governs this problem is the same as the ODE that was

derived for a constant cross-sectional area fin:

22 2

2 where

c

er hd Tm T m T m

k Adx (50)

The general solution to this ODE is:

1 2exp expT C m x C m x T (51)

b.) Use the correct boundary conditions in order to obtain two equations that can be solvedsimultaneously in order to provide C1 and C2. IT IS NOT NECESSARY THAT YOU

SOLVE THESE EQUATIONS.

The boundary conditions are:

0x bT T (52)

c h

x L

dTk A qdx

(53)

Note that the heat transfer is INTO the tip of the fin and therefore in the negative x-direction,leading to the sign change in Eq. (53). Substituting Eq. (51) into Eqs. (52) and (53) leads to:

1 2 bC C T T (54)


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1 2exp expc hk A C m m L C m m L q (55)

Equations (54) and (55) can be solved to provide C1and C2.

You have decided to model the rod using a three node numerical solution, as shown in Figure 2.

L

Tb x

,h T

hqT

1

T2

T3

Figure 2: Three node numerical solution.

c.) Derive the three equations that must be solved simultaneously in order to provide thetemperatures of the three nodes (T1, T2, and T3). IT IS NOT NECESSARY TO SOLVE

THESE EQUATIONS.

The temperature of node 1 is specified:

1 bT T (56)

An energy balance on the control volume around node 2 leads to:

1 2 3 2 2 0c ck A k A

T T T T h per x T T x x

(57)

An energy balance on the control volume around node 3 leads to:

2 3 3 02

c

h

k A xT T h per T T q

x

(58)

The distance between adjacent nodes, xin Eqs. (57) and (58), is:

2

Lx (59)

Problem 10Figure 1 illustrates the wall of a freezer.


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freezer airTf= -10C

Rconv= 3 K/W

surroundingsTo= 20C

Rconv= 3 K/W

Rrad= 30 K/Wfreezer wall

Rcond= 10 K/W

q

inner surface outer surface

Figure 1: Outdoor wall in a building.

The inner surface of the wall experiences convection with the air in the freezer at Tf= -10C.

The outer surface of the wall experiences both convection and radiation with the surroundings atTo= 20C. The resistance to convection on the inside and outside of the wall isRconv= 3 K/W.

The resistance to radiation on the outer surface isRrad= 30 K/W, you may neglect radiation from

the inner surface. The resistance to conduction through the wall isRcond= 10 K/W.

a.) Draw a resistance diagram that represents the freezer wall.

Figure 2 illustrates the resistance diagram.

Tf T

o

Rrad= 30 K/W

Rconv= 3 K/W

Rconv= 3 K/W

Rcond= 10 K/W

Figure 2: Resistance diagram.

b.) Rank the following effects in terms of their importance (most important to least important):

convection, radiation, conduction. Justify your answer.

The radiation resistance is in parallel with a much smaller convection resistance it is not

important. The convection resistances are in series with a larger conduction resistance they are

less important (although still important).

The most important effect is conduction, followed by convection, with radiation being least

important.

c.) What is the rate of heat transfer to the freezer through the wall, q ?

The rate of heat transfer is:


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1 1

20 10 K 1.91 W

1 1 K1 13 10

3 30 W

o f

conv cond

conv rad

T Tq

R RR R

(60)

d.) What is the temperature of the outer surface of the wall?

The temperature of the outer surface of the wall is:

3 10 K 1.91 W

10 C 14.8 CW

s f conv condT T R R q

(61)

If the temperature of the outer surface of the wall is too low then condensation will form on the

freezer. This is undesirable and therefore a thin heater is installed on the outer surface in order to

maintain the temperature of the outer surface above the dew point temperature.

e.) What is the rate of heat transfer that must be provided by the heater in order to keep thetemperature of the outer surface of the wall at Ts = 18C?

The resistance diagram with the heater power added is shown in Figure 3.

Tf To

Rrad

Rconv

Rconv

Rcond

hq

Ts

Figure 3: Resistance network with heater power.

An energy balance on the heater leads to:

1 1

18 10 K 18 20 K 1.42 W

3 10 K/W 1 11 1K/W

3 30

s f s o

h

conv cond

conv rad

T T T Tq

R R

R R

(62)

Problem 11A disk-shaped bracket is used to hold a pipe in place, as shown in Figure 1.


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r

center line

pipe

disk shaped bracket

rb

rt

th

kTp

,h T

,h T

Figure 1: Disk-shaped bracket.

The thickness of the bracket is thand it is made of material with conductivity k. The bracket

extends radially from inner radius rbwhere it is connected to the pipe surface to outer radius rt.

The temperature of the bracket at the inner radius where it connects to the pipe is Tp. The outeredge of the bracket is adiabatic. Both the upper and lower surfaces of the bracket are exposed to

fluid at Twith heat transfer coefficient h .a.) What calculation would you do in order to justify treating the bracket as an extended surface

(i.e., justify the assumption that the temperature within the bracket is only a function ofradius, r); write your answer in terms of the symbols that are used in the problem statement.

The appropriate Biot number is the ratio of the resistance to conduction across the thickness of

the bracket to the resistance to convection from the bracket surface. The resistance to conduction

across the thickness of the bracket (i.e., in thexdirection) is:

, 2 22

cond x

t b

thR

k r r

(63)

The resistance to convection from the fin surface is:

2 21

conv

t b

Rh r r

(64)

The Biot number is therefore:

2 2,2 2 1 2

2

t bcond x

conv t b

h r rR th th hBi

R kk r r

(65)

Any answer within a factor of 2 of Eq. (65) is acceptable.

For the remainder of the problem, assume that the bracket can be treated as an extended surface.

b.) Derive the governing ordinary differential equation and the associated boundary conditions

that should be solved in order to obtain an analytical solution for the temperature in the


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bracket. Your differential equation and boundary conditions should only involve the

symbols provided in the problem statement. Do not solve the differential equation.

An energy balance on a differential segment of the bracket is shown in Figure 2.

r

center line

rb

rt

dr

rq r drq

convqT

p

Figure 2: Energy balance on a differential segment of the bracket.

The energy balance in Figure 2 leads to:

r r dr convq q q (66)

or

0 convdq

dr qdr

(67)

Substituting rate equations into Eq. (67) leads to:

0 2 4d dTk r th dr h r dr T T dr dr

(68)

or

2d dT h

r r T T dr dr k th

(69)

The boundary conditions are:

2 0

t

t

r r

dTk r th

dr

(70)

which can be written as:


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0

tr r

dT

dr (71)

and

br r pT T (72)

You have decided to generate a numerical model of the bracket that has three nodes, positionedas shown in Figure 3.

r

center line

rb

rt

node 1 node 2

node 3

Figure 3: A 3-node numerical model of the disk-shaped bracket.

c.) Derive a system of algebraic equations that can be solved in order to predict the temperatures

at each of the three nodes in Figure P1.9-2 (T1, T2, T3). Your equations should include only

those symbols defined in the problem statement as well as the radial locations of the threenodes (r1, r2, and r3). Do not solve these equations.

The temperature at node 1 is specified:

1 pT T (73)

An energy balance on node 2 is shown in Figure 4.

r

center line

node 1 node 2

node 3

LHSq RHSq

convq

Figure 4: Energy balance on node 2.


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and leads to:

LHS RHS convq q q (74)

or, with rate equations:

2 2

1 2 3 2 2 3 2 12

32

1 2

2 2 22 2

ln ln

T T T T r r r rk th k th h T T

rr

r r

(75)

Note that this could also be written as:

2 2

1 2 2 3 2 3 2 11 2 3 2 2

2 1 3 2

2 22

2 2 2 2

r r k th r r k th r r r rT T T T h T T

r r r r

(76)

An energy balance on node 3 is shown in Figure 5.

r

center line

node 1 node 2

node 3

convq

LHSq

Figure 3: Energy balance on node 3.

and leads to:

LHS convq q (77)

or

2

3 2 2 2 33 3

3

2

2 22

ln

T T r rk th h r T T

r

r

(78)

Equation (78) can also be written as:


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2

2 3 2 2 33 2 3 3

3 2

22

2 2

r r k th r rT T h r T T

r r

(79)

d.) On the axes in Figure 4, sketch the temperature distribution that you would expect if the

conductivity of the bracket, k, is very low (label your sketch k0). Note that the qualitativevalues of Tp and Tare indicated in the plot - your sketch should be consistent with these

values.

r

center line

rb

rt

Tp

,h T

,h T

rb rtr

T

Tp

T

k

0k

Figure 4: Qualitative sketch of the temperature distribution expected askbecomes very small andkbecomes

very large.

The sketch is shown in Figure 4 and has the following characteristics.

1. The slope at r= rtshould be zero because the tip is insulated.2. The temperature at r= rbmust be Tp.

3. If k is small then the resistance to radial conduction is large and all of the temperature

drop is related to conduction. Thus the temperature decreases rapidly from Tpat the baseto T.

e.) On the axes in Figure 4, sketch the temperature distribution that you would expect ifconductivity, k, is very high (label your sketch k).

The sketch is also shown in Figure 4 and has the following characteristics.

1. The slope at r= rtshould be zero because the tip is insulated.2. The temperature at r= rbmust be Tp.


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3. If k is large then the resistance to radial conduction is small and all of the temperature

drop is related to convection. Thus the temperature remains close to Tp throughout thebracket.

Problem 12

One of the engineers that you supervise has been asked to simulate the heat transfer problemshown in Figure 1(a). This is a 1-D, plane wall problem (i.e., the temperature varies only in the

x-direction and the area for conduction is constant with x). Material A (from 0

Exam 1 Review Problems

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