Chapter 1: One-Dimensional, Steady-State Conduction Section 1.1: What is Conduction? 1.1-1 Momentum transfer occurs in a fluid due to interactions between molecules that results in a

transfer of momentum. This process is characterized by viscosity, which relates the shear stress to a velocity gradient in the same way Fourier’s Law relates heat flux to a temperature gradient. It is not surprising, then, that the viscosity and thermal conductivity of an ideal gas are analogous transport properties. a.) Using reasoning similar to that provided in Section 1.1.2 for thermal conductivity, show that

the viscosity of an ideal gas can be estimated according to 2/T MWμ σ∝ .

1.1-2 (1-1 in text) Section 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at ideal gas conditions. Test the validity of this approximation by comparing the conductivity estimated using Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., low pressure argon) provided by the internal function in EES. Note that the molecular radius, σ, is provided in EES by the Lennard-Jones potential using the function sigma_LJ. a.) What is the value and units of the proportionality constant required to make Eq. (1-18) an

equality? b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and

100 MPa on a semi-log plot with pressure on the log scale. At what pressure does the approximation given in Eq. (1-18) begin to fail?

1.1-3 Equation (1-18) cannot be used to understand the thermal conductivity of a polyatomic ideal gas, such as low pressure oxygen, because the ideal gas thermal conductivity is the sum of two terms corresponding to translational and internal contributions.

trans intk k k= + Equation (1-18) only considers the translatonal contribution. Because thermal conductivity and

viscosity are analagous transport properties, the translation term for the thermal conductivity of a dilute gas can be estimated as a function of the viscosity (μ) of the gas according to:

154

univtrans

RkMW

μ=

where Runiv is the universal gas constant and MW is the molar mass of the of the gas. The

internal contribution for a polyatomic molecule results from the transfer of energy associated with rotational and vibrational degrees of freedom. An estimate of the internal contribution is provided by the Eucken1 correlation

52univ

int pRk c

MWμ ⎡ ⎤≈ −⎢ ⎥⎣ ⎦

where the viscosity is in units of Pa-s and the constant pressure specific heat and gas constant are

in units of J/kmol-K. The internal contribution is zero for a monotonic gas. Choose a gas and use the EES viscosity function to determine its viscosity as a function of

pressure and temperature. Then calculate and plot the thermal conductivity as a function of pressure at several temperatures. Compare the values you obtain from the dilute gas theory described above with the values provided at the same conditions obtained from the EES conductivity function. Use your program to answer the following questions. a.) The thermal conductivity of an ideal gas should only depend on temperature. At what

pressure does this requirement fail for the temperature and gas you have selected? b.) How does thermal conductivity vary with temperature? What causes this behavior? c.) How does thermal conductivity vary with the choice of gas. Is there are relationship between

the thermal conductivity and the number of atoms per molecule?

1 Hirschfelder, J.L., Curtiss, C.F, and Bird, R,B., “Molecular Theory of Gases and Liquids”, John Wiley and Sons, 1967

Section 1.2: Steady-State 1-D Conduction without Generation 1.2-1 A plane wall is a composite of a low conductivity material (with thickness L1 and conductivity

k1) and a high conductivity material (with thickness L2 = L1 and conductivity k2). The edge of the wall at x = 0 is at temperature T1 and the edge at x = L1 + L2 has temperature T2, as shown in Figure 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 in x.

T1 T2

k1 k2

L1 L2

xk1 k2

x

k1 k2

x

x x

Tq′′

L10 L1+L2 L10 L1+L2

T1

T2

(a) (b)

Figure P1.2-1: (a) Composite wall with k1 < k2, and (b) sketch of heat flux and temperature.

a.) Sketch the heat flux ( q′′ ) and temperature (T) as a function of position within the wall on the axes in Fig. 1.2-1(b). Make sure that your sketch reflects the fact that (1) the wall is at steady state, and (2) k1 < k2.

1.2-2 The temperature distribution for the shape shown in Figure P1.2-2 can be assumed to be 1-D in the coordinate s. The problem is at steady state and the area available for conduction changes with s according to an arbitrary function, A(s). The temperatures of the two ends of the shape are specified; TH at s1 and TC at s2.

s1

s2

THTCadiabatic

TH

TC

s

s

T

Figure P1.2-2: Conduction through a shape in which the cross-sectional area varies according to A(s).

a.) Sketch the temperature distribution through the shape on the axes below the figure. b.) Derive the governing differential equation for the problem; the governing differential

equation should include only temperature T and its derivatives with respect to s as well as the area and its derivatives with respect to s.

1.2-3 (1-2 in text) Figure P1.2-3 illustrates a plane wall made of a thin (thw = 0.001 m) and conductive

(k = 100 W/m-K) material that separates two fluids, A and B. Fluid A is at TA = 100°C and the heat transfer coefficient between the fluid and the wall is Ah = 10 W/m2-K while fluid B is at TB = 0°C with Bh = 100 W/m2-K.

thw = 0.001 m

k = 100 W/m-K

2

100 C

10 W/m -KA

A

T

h

= °

= 2

0 C100 W/m -K

B

B

Th

= °

=

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). b.) If you wanted to predict the heat transfer rate from fluid A to fluid B very 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.

1.2-4 Figure P1.2-4 illustrates a plane wall that is composed of two materials, A and B. The interface between the materials is characterized by a contact resistance. The left surface of material A is held at TH and the right surface of material B radiates to surroundings at TC and is also exposed to convection to a fluid at TC.

TH

material A

convection andradiation to TC

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 B Rconv = 1.0 K/W, resistance to convection Rrad = 10.0 K/W, resistance to radiation

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. Show where the temperatures TH and TC appear on your network.

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

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

1.2-5 Figure P1.2-5 illustrates a wafer that is being developed in an optical lithography process.

2 Wq =

wafer

chuck baseDw = 4 inch

thp = 0.5 cm

kch = 25 W/m-K

ε = 0.7

thch = 0.5 cm

20 CbT = °

-4 2chuck posts

5x10 K-m /W0.1

cRf′′ ==

220 C

15 W/m -KTh∞ = °=

Figure P1.2-5: Wafer being developed in an optical lithography process.

The energy required to develop the resist is deposited at a rate of q = 2 W near the center of the upper side of the wafer. The wafer has diameter Dw = 4 inch and is made of a conductive material; therefore, you may assume that the wafer is isothermal. The wafer is cooled by convection and radiation to the surroundings at T∞ as well as conduction to the chuck. The surrounding air is at T∞ = 20ºC and the heat transfer coefficient is h = 15 W/m2-K. The emissivity of the wafer surface is ε = 0.7. The chuck is made out of a single piece of material with conductivity kch = 25 W/m-K and consists of a base that is thb = 1.5 cm thick and an array of posts that are thp = 0.5 cm tall. The area of the base of the chuck is the same as the area of the wafer. The posts occupy f = 10% of the chuck area and the wafer rests on the top of the posts. There is an area specific contact resistance of cR′′ = 5x10-4 K-m2/W between the bottom of the wafer and the top of the posts. The bottom surface of the chuck base is maintained at Tb = 20 ºC. a.) What is the temperature of the wafer at steady-state? b.) Prepare a plot showing the wafer temperature as a function of the applied power, q . c.) What are the dominant heat transfer mechanisms for this problem? What aspects of the

problem are least important? d.) Radiation between the underside of the wafer and the top of the chuck base was ignored in

the analysis; is this an important mechanism for heat transfer? Assume that the chuck surface is black and justify your answer.

e.) In an effort to maintain the wafer temperature at Tw= 20ºC, you decide to try to reduce and control the chuck base temperature, Tb. What temperature do you need to reduce Tb to in order that Tw= 20ºC? If you can only control Tb to within ±0.5 K then how well can you control Tw?

f.) Perform the same analysis you carried out in (e), but this time evaluate the merit of controlling the surrounding temperature, T∞, rather than the chuck temperature. What are the advantages and disadvantages associated with controlling T∞?

1.2-6 You have designed a wall for a freezer. A cross-section of your freezer wall is shown in Figure P1.2-6. The wall separates the freezer air at Tf = -10°C from air within the room at Tr = 20°C. The heat transfer coefficient between the freezer air and the inner wall of the freezer is fh = 10 W/m2-K and the heat transfer coefficient between the room air and the outer wall of the freezer is

rh = 10 W/m2-K. The wall is composed of a thb = 1.0 cm thick layer of fiberglass blanket sandwiched between two thw = 5.0 mm sheets of stainless steel. The thermal conductivity of fiberglass and stainless steel are kb = 0.06 W/m-K and kw = 15 W/m-K, respectively. Assume that the cross-sectional area of the wall is Ac = 1 m2. Neglect radiation from either the inner or outer walls.

thw = 5 mm thw = 5 mmthb = 1 cm

220 C10 W/m -K

r

r

Th= °= 2

10 C10 W/m -K

f

f

Th

= − °=

stainless steel,kw = 15 W/m-K fiberglass blanket,

kb = 0.06 W/m-K Figure P1.2-6: Freezer wall.

a.) Draw a resistance network to illustrate this problem. Be sure to label the resistances in your

network so that it is clear what each resistance is meant to represent. b.) Enter all of the inputs in the problem into an EES program. Convert each input into the

corresponding base SI unit (i.e., m, kg, K, W, N, etc.) and set the unit for each variable using the Variable Information window. Using comments, indicate what each variable means. Make sure that you set and check units of each variable that you use in the remainder of the solution process.

c.) Calculate the net heat transfer to the freezer (W). d.) Your boss wants to make a more energy efficient freezer by reducing the rate of heat transfer

to the freezer. He suggests that you increase the thickness of the stainless steel wall panels in order to accomplish this. Is this a good idea? Justify your answer briefly.

e.) Prepare a plot showing the heat transfer to the freezer as a function of the thickness of the stainless steel walls. Prepare a second plot showing the heat transfer to the freezer as a function of the thickness of the fiberglass. Make sure that your plots are clear (axes are labeled, etc.)

f.) What design change to your wall would you suggest in order to improve the energy efficiency of the freezer.

g.) One of your design requirements is that no condensation must form on the external surface of your freezer wall, even if the relative humidity in the room reaches 75%. This implies that the temperature of the external surface of the freezer wall must be greater than 15°C. Does your freezer wall satisfy this requirement? Calculate the external surface temperature (°C).

h.) In order to prevent condensation, you suggest placing a heater between the outer stainless steel wall and the fiberglass. How much heat would be required to keep condensation from forming? Assume that the heater is very thin and conductive.

i.) Prepare a plot showing the amount of heat required by the heater as a function of the freezer air temperature.

1.2-7 You have designed the experimental apparatus shown in Figure P1.2-7 to measure contact resistance. Four thermocouples (labeled TC1 through TC4) are embedded in two sample blocks at precise locations. The thermocouples are placed L1 = 0.25 inch from the edges of the sample blocks and L2 = 1.0 inch apart, as shown. Heat is applied to the top of the apparatus and removed from the bottom using a flow of coolant. The sides of the sample blocks are insulated. The sample blocks are fabricated from an alloy with a precisely-known and nearly constant thermal conductivity, ks = 2.5 W/m-K. The apparatus is activated and allowed to reach steady state. The temperatures recorded by the thermocouples are TC1 = 53.3°C, TC2 = 43.1°C, TC3 = 22.6°C, and TC4 = 12.3°C. The contact resistance of interest is the interface between the sample blocks.

heat

sample block,ks = 2.5 W/m-Kδks = 0.4 W/m-K

insulation

cooled block

interface

TC1

TC2

TC3

TC4

L1 = 0.25 inchδL = 0.01 inch

L1 = 0.25 inchδL = 0.01 inch

L2 = 1 inchδL = 0.01 inch

Figure P1.2-7: Experimental device to measure contact resistance.

a.) Use the data provided above to compute the measured heat flux in the upper and lower

sample blocks. b.) Use the data to compute the temperature on the hot and cold sides of the interface. c.) Use the data to compute the measured contact resistance.

It is important to estimate the uncertainty in your measurement. The uncertainty in the distance measurements is δL= 0.01 inch, the uncertainty in the conductivity of the sample blocks is δks = 0.4 W/m-K, and the uncertainty in the temperature measurements is δT = 0.5 K. d.) Estimate the uncertainty in the measurement of the heat flux in the upper sample block, the

answer for (a), manually; that is carry out the uncertainty propagation calculations explicitly. e.) Verify that EES' uncertainty propagation function provides the same answer obtained in (d). f.) Use EES' uncertainty propagation function to determine the uncertainty in the measured

value of the contact resistance. What is the % uncertainty in your measurement? g.) Which of the fundamental measurements that are required by your test facility should be

improved in order to improve your measurement of the contact resistance? That is, would you focus your attention on reducing δks, δL, or δT? Justify your answer.

1.2-8 (1-3 in text) You have a problem with your house. Every spring at some point the snow immediately adjacent to your roof melts and runs along the roof line until it reaches the gutter. The water in the gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutter and causing water to run into your attic. The situation is shown in Figure P1.2-8.

snow melts at this surface2, 15 W/m -Kout outT h =

snow, ks = 0.08 W/m-K

222 C, 10 W/m -Kin inT h= ° =Lins = 3 inch

insulation, kins = 0.05 W/m-K

plywood,0.5 inch, 0.2 W/m-Kp pL k= =

Ls = 2.5 inch

Figure P1.2-8: Roof of your house.

The air in the attic is at Tin = 22°C and the heat transfer coefficient between the inside air and the inner surface of the roof is inh = 10 W/m2-K. The roof is composed of a Lins = 3.0 inch thick piece of insulation with conductivity kins = 0.05 W/m-K that is sandwiched between two Lp = 0.5 inch thick pieces of plywood with conductivity kp = 0.2 W/m-K. There is an Ls = 2.5 inch thick layer of snow on the roof with conductivity ks = 0.08 W/m-K. The heat transfer coefficient between the outside air at temperature Tout and the surface of the snow is outh = 15 W/m2-K. Neglect radiation and contact resistances for part (a) of this problem. a.) What is the range of outdoor air temperatures where you should be concerned that your

gutters will become blocked by ice? b.) Would your answer change much if you considered radiation from the outside surface of the

snow to surroundings at Tout? Assume that the emissivity of snow is εs = 0.82.

1.2-9 Computer chips tend to work better if they are kept cold. You are examining the feasibility of maintaining the processor of a personal computer at the sub-ambient temperature of Tchip = 0°F. Assume that the operation of the computer chip itself generates chipq =10 W of power. Model the processor unit as a box that is a = 2 inch x b = 6 inch x c = 4 inch. Assume that all six sides of the box is exposed to air at Tair = 70°F with a convection heat transfer coefficient of h =10 W/m2-K. The box experiences a radiation heat transfer with surroundings that are at Tsur = 70°F. The emissivity of the processor surface is ε =0.7 and all six sides experience the radiation heat transfer. You are asked to size the refrigeration system required to maintain the temperature of the processor. a.) What is the refrigeration load that your refrigeration system must be able to remove to

maintain the processor at a steady-state temperature (W)? b.) If the coefficient of performance (COP) of the refrigeration system is nominally 3.5, then

how much heat must be rejected to the ambient air (W)? Recall that COP is the ratio of the amount of refrigeration provided to the amount of input power required.

c.) If electricity costs 12¢/kW-hr, how much does it cost to run the refrigeration system for a year, assuming that the computer is never shut off.

1.2-10 You have been contracted by ASHRAE (the American Society of Heating, Refrigeration, and Air-Conditioning Engineers) to measure the thermal conductivity of various, new materials for insulating pipes. Your contract specifies that you will measure the thermal conductivity to within 10%. Your initial design for the test setup is shown in Figure P1.2-10. The test facility consists of a pipe (with conductivity kpipe = 120 W/m-K) with inner diameter, Di,pipe = 6.0 inch and thickness thpipe = 0.5 inch that carries a flow of chilled water, Twater = 10°C. The heat transfer coefficient between the water and the internal surface of the pipe is waterh = 300 W/m2-K. The pipe is covered by a thins = 2.0 inch thick layer of the insulation (with conductivity kins) that is being tested. Two thermocouples are embedded in the insulation, one connected to the outer surface (Tins,out) and the other to the inner surface (Tins,in). The insulation material is surrounded by a thm = 3.0 inch thick layer of a material with a well-known thermal conductivity, km = 2.0 W/m-K. Two thermocouples are embedded in the material at its inner and outer surface (Tm,in and Tm,out, respectively). Finally, a band heater is wrapped around the outer surface of the material. Assume that the thickness of the band heater is negligibly small. The band heater provides bandq = 3 kW/m. The outer surface of the band heater is exposed to ambient air at Tair = 20°C and has a heat transfer coefficient, airh = 10 W/m2-K and emissivity ε = 0.5. A contact resistance of cR′′ =1x10-4 m2-K/W is present at all 3 interfaces in the problem (i.e., between the pipe and the insulation, the insulation and the material, and the material and the band heater).

3 kWbandq =2

20 C10 W/m -K

0.5

air

air

Thε

= °=

=

Tm,outTm,inTins,outTins,in

210 C300 W/m -K

water

water

Th

= °=

band heater

pipe, kpipe = 120 W/m-K

thins = 2 inch

thm = 3 inchinsulation being tested, kins

material with known conductivity,km = 2.0 W/m-K

Di,pipe = 6 inch

thpipe = 0.5 inch

-4 21x10 K-m /WcR′′ =

Figure P1.2-10: Test facility for measuring pipe insulation

You may assume that the problem is 1-D (i.e., there are no variations along the length or circumference of the pipe) and do the problem on a per unit length of pipe (L=1 m) basis. a.) Draw a resistance network that represents the test facility. Clearly label each resistance and

indicate what it represents. Be sure to indicate where in the network the heat input from the band heater will be applied and also the location of the thermocouples mentioned in the problem statement.

b.) If the insulating material under test has a conductivity of kins = 1.0 W/m-K, then predict the heater temperature, Thtr, and determine how much of the heater power is transferred through the insulation ( insq ).

When the test facility is operated, the heater power is not measured nor are the contact resistances, emissivity, or heat transfer coefficients known with any precision. Also, the insulation thermal conductivity is not known, but rather must be calculated from the measured temperatures. The heat transferred through the material with known thermal conductivity is the same as the heat transferred through the insulation that is being measured. Therefore:

, , , ,

, ,

m out m in ins out ins inins

cond m cond ins

T T T Tq

R R− −

= = (P1.2-10-1)

and so the resistance of the insulation can be calculated based on the ratio of the temperature differences:

( )( )

, ,, ,

, ,

ins out ins incond ins cond m

m out m in

T TR R

T T−

=−

(P1.2-10-2)

Equation (P1.2-10-2) indicates that the test facility relies on accurately measuring the temperature differences across the insulation and the temperature difference across the material.

c.) Using your model, predict the temperature difference across the insulation ( , ,ins ins out ins inT T TΔ = − ) and the material ( , ,m m out m inT T TΔ = − ).

d.) Prepare a plot showing ΔTm as a function of the material thickness (thm) for thicknesses ranging from 5.0 mm to 50 cm. Explain the shape of your plot.

e.) Based on your plot from part (d), what is a reasonable value for thm? Remember that you need to measure the temperature difference and therefore you would like it to be large.

1.2-11 (1-4 in text) Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with kA = 1 W/m-K and B with kB = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A; the heating element provides 25000 W/mq′′ = of heat. The surface of the wall at x = 2L is exposed to fluid at Tf,in = 300 K with heat transfer coefficient inh = 100 W/m2-K.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ =

insulated

material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

Figure P1.2-11(a): Composite wall with a heater.

You may neglect radiation and contact resistance for parts (a) through (c) of this problem. a.) Draw a resistance network to represent this problem; clearly indicate what each resistance

represents and calculate the value of each resistance. b.) Use your resistance network from (a) to determine the temperature of the heating element. c.) Sketch the temperature distribution through the wall. Make sure that the sketch is consistent

with your solution from (b).

Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ =

insulated

material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

L = 1 cm

material CkC = 2 W/m-K

Figure P1.2-11(b): Composite wall with Material C.

Neglect radiation and contact resistance for parts (d) through (f) of this problem. d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly

indicate what each resistance represents and calculate the value of each resistance. e.) Use your resistance network from (d) to determine the temperature of the heating element. f.) Sketch the temperature distribution through the wall. Make sure that the sketch is consistent

with your solution from (e).

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a contact resistance between materials A and B, 20.01 K-m /WcR′′ = , and the surface of the wall at x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient outh = 10 W/m2-K.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ = material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

L = 1 cm

material CkC = 2 W/m-K

20.01 K-m /WcR′′ =

,2

400 K10 W/m -K

f out

out

Th

==

Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.

Neglect radiation for parts (g) through (i) of this problem. g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly

indicate what each resistance represents and calculate the value of each resistance. h.) Use your resistance network from (g) to determine the temperature of the heating element. i.) Sketch the temperature distribution through the wall.

1.2-12 (1-5) You have decided to install a strip heater under the linoleum in your bathroom in order to keep your feet warm on cold winter mornings. Figure P1.2-12 illustrates a cross-section of the bathroom floor. The bathroom is located on the first story of your house and is W = 2.5 m wide x L = 2.5 m long. The linoleum thickness is thL = 5.0 mm and has conductivity kL = 0.05 W/m-K. The strip heater under the linoleum is negligibly thin. Beneath the heater is a piece of plywood with thickness thP = 5 mm and conductivity kP = 0.4 W/m-K. The plywood is supported by ths = 6.0 cm thick studs that are Ws = 4.0 cm wide with thermal conductivity ks = 0.4 W/m-K. The center-to-center distance between studs is ps = 25.0 cm. Between each stud are pockets of air that can be considered to be stagnant with conductivity ka = 0.025 W/m-K. A sheet of drywall is nailed to the bottom of the studs. The thickness of the drywall is thd = 9.0 mm and the conductivity of drywall is kd = 0.1 W/m-K. The air above in the bathroom is at Tair,1 = 15°C while the air in the basement is at Tair,2 = 5°C. The heat transfer coefficient on both sides of the floor is h = 15 W/m2-K. You may neglect radiation and contact resistance for this problem.

strip heater

linoleum, kL = 0.05 W/m-K

thL = 5 mmthp = 5 mm

plywood, kp = 0.4 W/m-K

ths = 6 cm

Ws = 4 cm

studs, ks = 0.4 W/m-K

ps = 25 cm

thd = 9 mm

drywall, kd = 0.1 W/m-Kair, ka = 0.025 W/m-K

2,1 15 C, 15 W/m -KairT h= ° =

2,2 5 C, 15 W/m -KairT h= ° =

Figure P1.2-12: Bathroom floor with heater.

a.) Draw a thermal resistance network that can be used to represent this situation. Be sure to

label the temperatures of the air above and below the floor (Tair,1 and Tair,2), the temperature at the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat input to the strip heater ( hq ) on your diagram.

b.) Compute the value of each of the resistances from part (a). c.) How much heat must be added by the heater to raise the temperature of the floor to a

comfortable 20°C? d.) What physical quantities are most important to your analysis? What physical quantities are

unimportant to your analysis? e.) Discuss at least one technique that could be used to substantially reduce the amount of heater

power required while still maintaining the floor at 20°C. Note that you have no control over Tair,1 or h .

1.2-13 An electric burner for a stove is formed by taking a cylindrical piece of metal that is D = 0.32 inch in diameter and L = 36 inch long and winding it into a spiral shape. The burner consumes electrical power at a rate of q = 900 W. The burner surface has an emissivity of ε = 0.80. The heat transfer coefficient between the burner and the surrounding air ( h ) depends on the surface temperature of the burner (Ts) according to:

2 2 2

W W10.7 0.0048m -K m -K sh T⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

where h is in [W/m2-K] and Ts is in [K]. The surroundings and the surrounding air temperature are at Tsur = 20°C. a.) Determine the steady state surface temperature of the burner. b.) Prepare a plot showing the surface temperature as a function of the burner input power.

1.2-14 Ice for an ice skating rink is formed by running refrigerant at Tr = -30°C through a series of cast iron pipes that are embedded in concrete, as shown in Figure P1.2-14. The cast iron pipes have an outer diameter of Do,p = 4 cm and an inner diameter of Di,p = 3 cm. The pipes are spaced Lptp = 8.0 cm apart. The heat transfer coefficient between the refrigerant and the pipe surface is rh = 100 W/m2-K. The concrete slab is Lc = 8 cm thick and the pipes are in the center of the slab. The bottom of the slab is insulated (assume perfectly). The thermal conductivity of concrete and iron are kc = 4.5 W/m-K and kiron = 51 W/m-K, respectively.

An Lfill = 1 cm thick layer of water is placed on the top of the concrete slab. The refrigerant

cools the top of the slab and the water turns to ice slowly. Assume that the water is stagnant and can be treated as a solid. The conductivity of ice and water are kice = 2.2 W/m-K and kw = 0.6 W/m-K, respectively. The heat transfer coefficient between the top of the water layer and the surrounding air at Ta = 15°C is ah = 10.0 W/m2-K. The top of the water surface has an emissivity of ε = 0.90 and radiates to surroundings at Tsur = 15°C.

Lfill = 1 cmLice

ε = 0.9

215 C10 W/m -K

a

a

Th= °=

15 CsurT = °

Lc = 8 cm

water, kw = 0.6 W/m-Kice, kice = 2.2 W/m-K

concrete, kc = 4.5 W/m-K

230 C

10 W/m -Kr

r

Th= − °=

cast iron tubeskiron = 51 W/m-KDo,p = 4 cmDi,p = 3 cm

Lp = 4 cm

Lptp = 8 cm

insulation

Figure P1.2-14: Schematic of ice rink

a.) Draw a network that represents this situation using 1-D resistances. (Some of the resistances

must be approximate since it is not possible to exactly calculate a 1-D resistance to the conduction heat flow in the concrete). Include an energy term related to the energy that is added to the system by the generation of ice. Clearly label the resistors.

b.) Estimate the magnitude of each of the resistances in your network when the ice is 0.5 cm thick (i.e., Lice = 0.5 cm).

c.) Calculate the rate of change in the thickness of the ice when the ice thickness is 0.5 cm.

1.2-15 (1-6 in text) You are a fan of ice fishing but don't enjoy the process of augering out your fishing hole in the ice. Therefore, you want to build a device, the super ice-auger, that melts a hole in the ice. The device is shown in Figure P1.2-15.

D = 10 inchthins = 0.5 inch

thp = 0.75 inch

insulation,kins = 2.2 W/m-K

plate,kp = 10 W/m-K

ε = 0.9

250 W/m -K5 C

hT∞

== °

thice = 5 inchρice = 920 kg/m3

Δifus = 333.6 kJ/kg

heater, activated withV = 12 V and I = 150 A

Figure P1.2-15: The super ice-auger.

A heater is attached to the back of a D = 10 inch plate and electrically activated by your truck

battery, which is capable of providing V = 12 V and I = 150 A. The plate is thp = 0.75 inch thick and has conductivity kp = 10 W/m-K. The back of the heater is insulated; the thickness of the insulation is thins = 0.5 inch and the insulation has conductivity kins = 2.2 W/m-K. The surface of the insulation experiences convection with surrounding air at T∞ = 5°C and radiation with surroundings also at T∞ = 5°C. The emissivity of the surface of the insulation is ε = 0.9 and the heat transfer coefficient between the surface and the air is h = 50 W/m2-K. The super ice-auger is placed on the ice and activated, resulting in a heat transfer to the plate-ice interface that melts the ice. Assume that the water under the ice is at Tice = 0°C so that no heat is conducted away from the plate-ice interface; all of the energy transferred to the plate-ice interface goes into melting the ice. The thickness of the ice is thice = 5 inch and the ice has density ρice = 920 kg/m3. The latent heat of fusion for the ice is Δifus = 333.6 kJ/kg. a.) Determine the heat transfer rate to the plate-ice interface. b.) How long will it take to melt a hole in the ice? c.) What is the efficiency of the melting process? d.) If your battery is rated at 100 amp-hr at 12 V then what fraction of the battery's charge is

depleted by running the super ice-auger.

1.2-16 The temperature distribution across a L = 0.3 m thick wall at a certain instant of time is given by 2900 900 500T x x= − − where T is the temperature in the wall in °C at position x in m. The

density, specific heat capacity and thermal conductivity of the wall are ρ = 2050 kg/m3, c = 0.96 kJ/kg-K and k = 1.13 W/m-K, respectively. a) Calculate the rate of change of the average wall temperature. b) The left surface of the wall (at x = 0) is exposed to air at T∞ = 1000°C. Determine the

average convection coefficient between the air and the wall surface at x = 0 m.

1.2-17 Figure P1.2-17 illustrates the temperature distribution in a plane wall at a particular instant of time.

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

positive x direction), • 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.

1.2-18 Urea formaldehyde foam with conductivity kfoam = 0.020 Btu/hr-ft-F is commonly used as an insulation material in building walls. The major advantage of foam insulation is that it can be installed in existing walls by injection through a small hole. In a particular case, foam is to be installed in a wall consisting of a thplaster = 5/8 inch thick sheet of plaster board with conductivity kplaster = 0.028 Btu/hr-ft-F, a thas = 3.5 inch air space and a thbrick = 2.5 inch layer of brick with conductivity kbrick = 0.038 Btu/hr-ft-F. Before the air gap is filled with foam, there is natural convection associated with buoyancy induced air motion. The equivalent thermal resistance of the air gap on a per unit area basis due to the natural convection is agR′′ = 0.95 hr-ft2-F/Btu. The

convection coefficients for the inner and outer surface of the wall are ih = 1.5 Btu/hr-ft2-F and

oh = 3.5 Btu/hr-ft2-F, respectively. a) The R-value of a wall is the resistance of the wall on a per unit area basis, expressed in units

ft2-F-hr/Btu. Calculate the R-value of the wall before the foam insulation is applied. b) Calculate the R-value of the wall after insulation is applied. Assume that the insulating foam

completely fills the air gap. c) Foam insulation ordinarily shrinks after it is installed by an amount dependent upon

conditions such as the outdoor temperature. Calculate the R-value of the wall assuming that the foam shrinks by 3%. Assume that the air in the gap between the foam and the plaster and the foam and the brick is stagnant.

1.2-19 Figure P1.2-19 illustrates a cross-section of a thermal protection suit that is being designed for an astronaut.

37 CbT = °

tissue

thins = 4 cm

thext = 1 mmthliner = 1 cm

kliner = 0.06 W/m-Kheater

kins = 0.06 W/m-K

kext = 14.5 W/m-K

Tspace = 4 K

ε = 0.25

Figure P1.2-19: Cross-section of thermal protection suit.

The suit consists of a liner that is immediately adjacent to the skin. The skin temperature is maintained at Tb = 37ºC by the flow of blood in the tissue. The liner is thliner = 1 cm thick and has conductivity kliner = 0.06 W/m-K. A thin heater is installed at the outer surface of the liner. Outside of the heater is a layer of insulation that is thins = 4 cm with conductivity kins = 0.06 W/m-K. Finally, the outer layer of the suit is thext = 1 mm thick with conductivity kext = 14.5 W/m-K. The outer surface of the external layer has emissivity ε = 0.25 and is exposed by radiation only to outer space at Tspace = 4 K. a.) You want to design the heater so that it completely eliminates any heat loss from the skin.

What is the heat transfer per unit area required? b.) In order, rate the importance of the following design parameters to your result from (a): kins,

kext, and ε. c.) Plot the heat transfer per unit area required to eliminate heat loss as a function of the

emissivity, ε. While the average emissivity of the suit's external surface is ε = 0.25, you have found that this value can change substantially based on how dirty or polished the suit is. You are worried about these local variations causing the astronaut discomfort due to local hot and cold spots. d.) Assume that the heater power is kept at the value calculated in (a). Plot the rate of heat

transfer from the skin as a function of the fractional change in the emissivity of the suit surface.

1.2-20 Figure P1.2-20 illustrates a temperature sensor that is mounted in a pipe and used to measure the temperature of a flow of air.

0.1 Wg =

sensorAs = 0.001 m2

230 W/m -K50 C

hT∞

== °

L = 0.01 m

supportAc = 1x10-6 m2

k = 10 W/m-K

20 CwT = °

Figure P1.2-20: Temperature sensor.

The operation of the sensor leads to the dissipation of g = 0.1 W of electrical power. This power is either convected to the air at T∞ = 50°C or conducted along the support to the wall at Tw = 20°C. Treat the support as conduction through a plane wall (i.e., neglect convection from the edges of the support). The heat transfer coefficient between the air and the sensor is h = 30 W/m2-K. The surface area of the sensor is As = 0.001 m2. The support has cross-sectional area Ac = 1x10-6 m2, length L = 0.01 m, and conductivity k = 10 W/m-K. a.) What is the temperature of the temperature sensor? b.) What is the error associated with the sensor measurement (i.e., what is the difference

between the sensor and the air temperature)? Is the error primarily due to self-heating of the sensor associated with g or due to the thermal communication between the sensor and with the wall? Justify your answer.

c.) Radiation has been neglected for this problem. If the emissivity of the sensor surface is ε = 0.02, then assess whether radiation is truly negligible.

1.2-21 Figure P1.2-21 illustrates a plane wall made of a material with a temperature-dependent conductivity. The conductivity of the material is given by: k bT= where b = 1 W/m2-K2 and T is the temperature in K.

k bT=

TH

TC

L

x

Figure P1.2-21: Plane wall with temperature-dependent conductivity.

The thickness of the wall is L = 1 m. The left side of the wall (at x= 0) is maintained at TH = 500 K and the right side (at x= L) is kept at TC = 50 K. The problem is steady-state and 1-D. a.) Sketch the temperature distribution in the wall (i.e., sketch T as a function of x). Make sure

that you get the qualitative features of your sketch right. b.) Derive the ordinary differential equation that governs this problem. c.) What are the boundary conditions for this problem? d.) Solve the governing differential equation from (b) - you should end up with a solution that

involves two unknown constants of integration. e.) Use the boundary conditions from (c) with the solution from (d) in order to obtain two

equations in the two unknown constants. f.) Type the inputs for the problem and the equations from (e) into EES in order to evaluate the

undetermined constants. g.) Prepare a plot of the temperature as a function of position in the wall using EES.

1.2-22 You are designing a cubical case that contains electronic components that drive remotely located instruments. You have been asked to estimate the maximum and minimum operating temperature limits that should be used to specify the components within the case. The case is W = 8 inch on a side. The emissivity of the paint used on the case is ε = 0.85. The operation of the electronic components within the case generates between q = 5 and q = 10 W due to ohmic heating, depending on the intensity of the operation. The top surface of the case is exposed to a solar flux q′′ . All of the surfaces of the case convect (with average heat transfer coefficient h ) and radiate to surroundings at T∞. The case will be deployed in a variety of climates, ranging from very hot (T∞,max = 110°F) to very cold (T∞,max = -40°F), very sunny ( maxq′′ = 850 W/m2) to night ( minq′′ = 0 W/m2), and very windy ( maxh = 100 W/m2-K) to still ( minh = 5 W/m2-K). For the following questions, assume that the case is at a single, uniform temperature and at steady state. a.) Come up with an estimate for the maximum operating temperature limit. b.) Plot the maximum operating temperature as a function of the case size, W. Explain the shape

of your plot (why does the temperature go up or down with W? if there is an asymptotic limit, explain why it exists).

c.) Come up with an estimate for the minimum operating temperature limit (with W = 8 inch). d.) Do you feel that the emissivity of the case surface is very important for determining the

minimum operating temperature? Justify your answer.

1.2-23 Figure P1.2-23 illustrates a cross-sectional view of a water heater.

Din = 0.75 inch

Dout = 0.875 inch

ktube = 12 W/m-K2

20 C15 W/m -Kout

Th∞ = °

=

2

50 C18 psi250 W/m -K

f

f

in

Tph

= °==

ε = 0.5kins = 0.5 W/m-K

thins = 0.5 inch

2heater

10,000 W/mq′′ =

-4 21x10 K-m /WcR′′ =

Figure P1.2-23: Water heater.

The water flows through a tube with inner diameter Din= 0.75 inch and outer diameter Dout = 0.875 inch. The conductivity of the tube material is ktube = 12 W/m-K. The water in the tube is at mean temperature Tf = 50°C and pressure pf = 18 psi. The heat transfer coefficient between the water and the internal surface of the tube is inh = 250 W/m2-K. A very thin heater is wrapped around the outer surface of the tube. The heater provides a heat transfer rate of q′′= 10,000 W/m2. Insulation is wrapped around the heater. The thickness of the insulation is thins = 0.5 inch and the conductivity is kins= 0.5 W/m-K. There is a contact resistance between the heater and the tube and the heater and the insulation. The area specific contact resistance for both interfaces is

cR′′= 1x10-4 K-m2/W. The outer surface of the insulation radiates and convects to surroundings at T∞ = 20°C. The heat transfer coefficient between the surface of the insulation and the air is

outh = 15 W/m2-K. The emissivity of the outer surface of the insulation is ε = 0.5. a.) Draw a resistance network that represents this problem. Label each resistance and clearly

indicate what it represents. Show where the heater power enters your network. b.) Using EES, determine the temperature of the heater and the rate of heat transfer to the water. c.) What is the efficiency of the heater (the ratio of the power provided to the water to the power

provided to the heater)? d.) The efficiency of the heater is less than 100% due to heat lost to the atmosphere. Rank the

following parameters in terms of their relative importance with respect to limiting heat loss to the atmosphere: ε, cR′′ , kins, outh . Justify your answers using your resistance network and a discussion of the magnitude of the relevant resistances.

e.) Plot the efficiency as a function of the insulation thickness for 0 inch < thins < 1.5 inch. Explain the shape of your plot.

f.) The temperature on the internal surface of the tube must remain below the saturation temperature of the water in order to prevent any local boiling of the water. Based on this criteria, determine the maximum possible heat flux that can be applied to the heater (for thins = 0.5 inch).

Section 1.3: Steady-State 1-D Conduction with Generation 1.3-1 A plane wall is composed of two materials, A and B, with the same thickness, L, as shown in

Figure P1.3-1. The same, spatially uniform, volumetric rate of generation is present in both materials ( A Bg g g′′′ ′′′ ′′′= = ) and the wall is at steady state. The conductivity of material B is twice that of material A (kB = 2 kA). The left side of the wall is adiabatic and the right side is maintained at a temperature Tx=2L = To.

L L

material A material B

kA kB = 2kA

x

ToAg′′′ B Ag g′′′ ′′′=

Figure P1.3-1: Plane wall composed of materials A and B.

a.) Sketch the rate of heat transfer as a function of position within the wall ranging from x = 0

(the left face of material A) to x = 2L (the right face of material B). Note that the sketch should be qualitatively correct, but cannot be quantitative as you have not been given any numbers for the problem.

b.) Sketch the temperature as a function of position within the wall. Again, be sure that your sketch has the correct qualitative features.

1.3-2 A plane wall is composed of two materials, A and B, with the same conductivity k and thickness L, as shown in Figure P1.3-2. The left side of material A is adiabatic (i.e., well insulated) and the right side of material B is held at a temperature TL. There is no volumetric generation in material A but material B experiences a uniform rate of volumetric generation of thermal energy, g′′′ .

L L

material A material B

kA kB = kA

x

TL0Ag′′′ = Bg g′′′ ′′′=

Figure P1.3-2: Plane wall composed of materials A and B.

a.) Sketch the rate of heat transfer ( q ) as a function of position within the wall. Note that the

sketch should be qualitatively correct but cannot be quantitative as you have not been given any numbers for the problem.

b.) Sketch the temperature as a function of position within the wall. Again, be sure that your sketch has the correct qualitative features.

1.3-3 (1-7 in text) One of the engineers that you supervise has been asked to simulate the heat transfer problem shown in Figure P1.3-3(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 < x < L) has conductivity kA and experiences a uniform rate of volumetric thermal energy generation, g′′′ . The left side of material A (at x = 0) is completely insulated. Material B (from L < x < 2L) has lower conductivity, kB < kA. The right side of material B (at x= 2L) experiences convection with fluid at room temperature (20°C). Based on the facts above, critically examine the solution that has been provided to you by the engineer and is shown in Figure P1.3-3(b). There should be a few characteristics of the solution that do not agree with your knowledge of heat transfer; list as many of these characteristics as you can identify and provide a clear reason why you think the engineer’s solution must be wrong.

L L

material A material B

kA kB < kA

xAg g′′′ ′′′= 0Bg′′′ =

, 20 Cfh T = °

-100

-50

0

50

100

150

200

250

Position (m)

Tem

pera

ture

(°C

)

0 L 2L

Material A Material B

(a) (b)

Figure P1.3-3: (a) Heat transfer problem and (b) "solution" provided by the engineer.

1.3-4 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 for temperature (T) is:

( )2

1 2ln4

g rT C r Ck′′′

= − + +

where r is the radial location and C1 and C2 are 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 is L. 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

rinrout

Tout

,k g ′′′L

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

1.3-5 Figure P1.3-5(a) illustrates a motor that is constructed using windings that surround laminated iron poles. You have been asked to estimate the maximum temperature that will occur within the windings. The windings and poles are both approximated as being cylindrical, as shown in Figure P1.3-5(b).

pole

windings

L = 2 cm

r

rin = 1 cmrout = 2 cm

6 3

windings1 W/m-K

1x10 W/mkg=′′′ =

225 C

25 W/m -KairT

h= °

=

50 CpoleT = °

(a) (b) Figure P1.3-5: (a) Concentrated winding for a permanent magnet motor, (b) cylindrical model of the windings

The windings are a complicated composite formed from copper conductor, insulation and air that fills the gaps between adjacent wires. However, in most models, the windings are represented by a solid with equivalent properties that account for this underlying structure. You can therefore consider the windings in Figure P1.3-5(b) to be a solid. The electrical current in the windings causes an ohmic dissipation that can be modeled as a uniform volumetric generation rate of g′′′=1x106 W/m3. The conductivity of the windings is k = 1.0 W/m-K. The inner radius of the windings is rin = 1.0 cm and the outer radius is rout = 2.0 cm. The windings are L = 2.0 cm long and the upper and lower surfaces may be assumed to be insulated so that the temperature in the windings varies only in the radial direction. The stator pole is conductive and cooled externally; therefore, you can assume that the stator tooth has a uniform temperature of Tpole = 50°C. Neglect any contact resistance between the inner radius of the winding and the pole; therefore, the temperature of the windings at r = rin is Tpole. The outer radius of the windings is exposed to air at Tair = 20°C with a heat transfer coefficient of h = 25 W/m2-K. a.) Derive the governing differential equation for the temperature within the windings (i.e., the

differential equation that is valid from rin < r < rout). You should end up with an ordinary differential equation for T in terms of the symbols provided in the problem statement. Clearly show your steps, which should include:

1. define a differentially small control volume, 2. do an energy balance on your control volume, 3. expand the r + dr terms in your energy balance and take the limit as dr → 0, 4. substitute rate equations into your energy balance. b.) Specify the boundary conditions for your differential equation. This should be easy at the

inner radius, where the temperature is specified, but you will need to carry out an interface energy balance at r = rout.

c.) Solve the governing differential equation that you derived in part (a) by integrating twice. You should end up with a solution that involves two constants of integration, C1 and C2.

d.) Substitute your answer from part (c) into the boundary conditions specified in part (b) to obtain two equations for your two unknown constants of integration, C1 and C2.

e.) Implement your results from (c) and (d) in EES and prepare a plot of the temperature in the stator as a function of radius.

1.3-6 You need to transport water through a pipe from one building to another in an arctic environment, as shown in Figure P1.3-6. The water leaves the building very close to freezing, at Tw = 5°C, and is exposed to a high velocity, very cold wind. The temperature of the surrounding air is Ta = -35°C and the heat transfer coefficient between the outer surface of the pipe and the air is ah = 50 W/m2-K. The pipe has an inner radius of rh,in = 2 inch and an outer radius of rh,out = 4 inch and is made of a material with a conductivity kh = 5 W/m-K. The heat transfer coefficient between the water and the inside surface of the pipe is very large and therefore the inside surface of the pipe can be assumed to be at the water temperature. Neglect radiation from the external surface of the pipe.

water5 CwT = °

rh,in = 2 inchrh,out = 4 inch

235 C

50 W/m -Ka

a

Th= − °=

kh = 5 W/m-K Figure P1.3-6: Heated pipe transporting near freezing water in an arctic environment.

a.) Determine the rate of heat lost from the water to the air for a unit length, L=1 m, of pipe. b.) Plot the rate of heat loss from the water as a function of the pipe outer radius from 0.06 m to

0.3 m (keep the same inner radius for this study). Explain the shape of your plot.

In order to reduce the heat loss from the water and therefore prevent freezing, you run current through the pipe material so that it generates thermal energy at with a uniform volumetric rate, g′′′= 5x105 W/m3. c.) Develop an analytical model capable of predicting the temperature distribution within the

pipe. Implement your model in EES. d.) Prepare a plot showing the temperature as a function of position within the pipe. e.) Calculate the heat transfer rate from the water when you are heating the pipe. f.) Determine the volumetric generation rate that is required so that there is no heat transferred

from the water.

1.3-7 Figure P1.3-7 illustrates a spherical, nuclear fuel element which consists of a sphere of fissionable material (fuel) with radius rfuel = 5 cm and kfuel = 1 W/m-K that is surrounded by a spherical shell of metal cladding with outer radius rclad = 7 cm and kclad = 300 W/m-K. The outer surface of the cladding is exposed to helium gas that is being heated by the reactor. The convection coefficient between the gas and the cladding surface is gash = 100 W/m2-K and the temperature of the gas is Tgas = 500ºC. Neglect radiation heat transfer from the surface.

Inside the fuel element, fission fragments are produced which have high velocities. The products

collide with the atoms of the material and provide the thermal energy for the reactor. This process can be modeled as a volumetric source of heat generation in the material that is not uniform throughout the fuel. The volumetric generation ( g′′′ ) can be approximated by the function:

b

efuel

rg gr

⎛ ⎞′′′ ′′′= ⎜ ⎟⎜ ⎟

⎝ ⎠

where eg′′′= 5x105 W/m3 is the volumetric rate of heat generation at the edge of the sphere and b = 1.0; note that the parameter b is a dimensionless positive constant that characterizes how quickly the generation rate increases in the radial direction.

fissionable materialkfuel = 1 W/m-K

rfuel = 5 cm

rclad = 7 cm

claddingkclad = 300 W/m-K

g′′′2100W/m -K

500 Cgas

gas

hT

== °

Figure P1.3-7: Spherical fuel element surrounded by cladding

a.) Enter the problem inputs into EES; be sure to set the units appropriately. b.) Determine the governing differential equation that applies within the sphere (i.e., your

differential equation should be valid for 0 < r < rfuel). The differential equation should include only those symbols given in the problem statement. Clearly show your steps.

c.) Enter the governing differential equation into Maple and use Maple to obtain a solution that includes two constants of integration.

d.) The boundary condition at the center of the sphere is that the temperature must remain finite; this should eliminate one of the constants of integration in your Maple solution. Which constant must be zero?

e.) Determine a symbolic equation for the remaining boundary condition (the one at r = rfuel) in terms of the temperature and temperature gradient evaluated at r=rfuel.

f.) Use the expressions from Maple to determine the required constant of integration within EES. Copy the solution for the temperature in the cladding from Maple and paste it into

EES; modify the expression as necessary for compatibility (remember to eliminate the C1 term) and use it to generate plot of temperature vs radius within the cladding.

1.3-8 (1-8 in text) Freshly cut hay is not really dead; chemical reactions continue in the plant cells and therefore a small amount of heat is released within the hay bale. This is an example of the conversion of chemical to thermal energy. The amount of thermal energy generation within a hay bale depends on the moisture content of the hay when it is baled. Baled hay can become a fire hazard if the rate of volumetric generation is sufficiently high and the hay bale sufficiently large so that the interior temperature of the bale reaches 170°F, the temperature at which self-ignition can occur. Here, we will model a round hay bale that is wrapped in plastic to protect it from the rain. You may assume that the bale is at steady state and is sufficiently long that it can be treated as a one-dimensional, radial conduction problem. The radius of the hay bale is Rbale = 5 ft and the bale is wrapped in plastic that is tp = 0.045 inch thick with conductivity kp = 0.15 W/m-K. The bale is surrounded by air at T∞ = 20°C with h = 10 W/m2-K. You may neglect radiation. The conductivity of the hay is k = 0.04 W/m-K. a.) If the volumetric rate of thermal energy generation is constant and equal to g′′′ = 2 W/m3

then determine the maximum temperature in the hay bale. b.) Prepare a plot showing the maximum temperature in the hay bale as a function of the hay

bale radius. How large can the hay bale be before there is a problem with self-ignition? Prepare a model that can consider temperature-dependent volumetric generation. Increasing temperature tends to increase the rate of chemical reaction and therefore increases the rate of generation of thermal energy according to: g a bT′′′ = + where a = -1 W/m3 and b = 0.01 W/m3-K and T is in K. c.) Enter the governing equation into Maple and obtain the general solution (i.e., a solution that

includes two constants). d.) Use the boundary conditions to obtain values for the two constants in your general solution.

(hint: one of the two constants must be zero in order to keep the temperature at the center of the hay bale finite). You should obtain a symbolic expression for the boundary condition in Maple that can be evaluated in EES.

e.) Overlay on your plot from part (b) a plot of the maximum temperature in the hay bale as a function of bale radius when the volumetric generation is a function of temperature.

1.3-9 (1-9 in text) Figure P1.3-9 illustrates a simple mass flow meter for use in an industrial refinery.

rin = 0.75 inch

rout = 1 inch

L = 3 inch thins = 0.25

7 3test section

1x10 W/m10 W/m-K

gk

′′′ ==

insulationkins = 1.5 W/m-K2

20 C20 W/m -Kout

Th

∞ = °=

0.75kg/s18 Cf

mT

== °

Figure P1.3-9: A simple mass flow meter.

A flow of liquid passes through a test section consisting of an L = 3 inch section of pipe with inner and outer radii, rin = 0.75 inch and rout = 1.0 inch, respectively. The test section is uniformly heated by electrical dissipation at a rate 7 31x10 W/mg′′′ = and has conductivity k = 10 W/m-K. The pipe is surrounded with insulation that is thins = 0.25 inch thick and has conductivity kins = 1.5 W/m-K. The external surface of the insulation experiences convection with air at T∞ = 20°C. The heat transfer coefficient on the external surface is outh = 20 W/m2-K. A thermocouple is embedded at the center of the pipe wall. By measuring the temperature of the thermocouple, it is possible to infer the mass flow rate of fluid because the heat transfer coefficient on the inner surface of the pipe ( inh ) is strongly related to mass flow rate ( m ). Testing has shown that the heat transfer coefficient and mass flow rate are related according to:

[ ]

0.8

1 kg/sinmh C

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

where C= 2500 W/m2-K. Under nominal conditions, the mass flow rate through the meter is m = 0.75 kg/s and the fluid temperature is Tf = 18°C. Assume that the ends of the test section are insulated so that the problem is 1-D. Neglect radiation and assume that the problem is steady-state. a.) Develop an analytical model in EES that can predict the temperature distribution in the test

section. Plot the temperature as a function of radial position for the nominal conditions. b.) Using your model, develop a calibration curve for the meter; that is, prepare a plot of the

mass flow rate as a function of the measured temperature at the mid-point of the pipe. The range of the instrument is 0.2 kg/s to 2.0 kg/s.

The meter must be robust to changes in the fluid temperature. That is, the calibration curve developed in (b) must continue to be valid even as the fluid temperature changes by as much as 10°C. c.) Overlay on your plot from (b) the mass flow rate as a function of the measured temperature

for Tf = 8°C and Tf = 28°C. Is your meter robust to changes in Tf? In order to improve the meters ability to operate over a range of fluid temperature, a temperature sensor is installed in the fluid in order to measure Tf during operation.

d.) Using your model, develop a calibration curve for the meter in terms of the mass flow rate as a function of ΔT, the difference between the measured temperatures at the mid-point of the pipe wall and the fluid.

e.) Overlay on your plot from (d) the mass flow rate as a function of the difference between the measured temperatures at the mid-point of the pipe wall and the fluid if the fluid temperature is Tf = 8°C and Tf = 28°C. Is the meter robust to changes in Tf?

f.) If you can measure the temperature difference to within δΔT = 1 K then what is the uncertainty in the mass flow rate measurement? (Use your plot from part (d) to answer this question.)

You can use the built-in uncertainty propagation feature in EES to assess uncertainty automatically. g.) Set the temperature difference to the value you calculated at the nominal conditions and

allow EES to calculate the associated mass flow rate. Now, select Uncertainty Propagation from the Calculate menu and specify that the mass flow rate is the calculated variable while the temperature difference is the measured variable. Set the uncertainty in the temperature difference to 1 K and verify that EES obtains an answer that is approximately consistent with part (f).

h.) The nice thing about using EES to determine the uncertainty is that it becomes easy to assess the impact of multiple sources of uncertainty. In addition to the uncertainty δΔT, the constant C has an uncertainty of δC = 5% and the conductivity of the material is only known to within δk = 3%. Use EES' built-in uncertainty propagation to assess the resulting uncertainty in the mass flow rate measurement. Which source of uncertainty is the most important?

i.) The meter must be used in areas where the ambient temperature and heat transfer coefficient may vary substantially. Prepare a plot showing the mass flow rate predicted by your model for ΔT = 50 K as a function of T∞ for various values of outh . If the operating range of your meter must include -5°C < T∞ < 35°C then use your plot to determine the range of outh that can be tolerated without substantial loss of accuracy.

1.3-10 A current of 100 amps passes through a bare stainless-steel wire of D = 1.0 mm diameter. The thermal conductivity and electrical resistance per unit length of the wire are k = 15 W/m-K and

eR′ = 0.14 ohm/m, respectively. The wire is submerged in an oil that is maintained at T∞ = 30°C. The steady-state temperature at the center of the wire is measured to be 180°C, independent of axial position within the oil bath. a) What is the temperature at the outer surface of the wire? b) Estimate the convection coefficient between the submerged wire and the oil. c) A plastic material (kp = 0.05 W/m-K) can be applied to the outer surface of the wire. Can the

insulation result in a reduction of the center temperature? If so, what insulation thickness should be applied?

1.3-11 Figure P1.3-11 illustrates a spherical, nuclear fuel element which consists of a sphere of fissionable material (fuel) with radius rfuel = 5 cm and kfuel = 2 W/m-K that is surrounded by a spherical shell of metal cladding with outer radius rclad = 7 cm and kclad = 0.25 W/m-K. The outer surface of the cladding is exposed to fluid that is being heated by the reactor. The convection coefficient between the fluid and the cladding surface is h = 50 W/m2-K and the temperature of the fluid is T∞ = 500ºC. Neglect radiation heat transfer from the surface.

Inside the fuel element, thermal energy is being generated for the reactor. This process can be modeled as a volumetric source of heat generation in the material that is not uniform throughout the fuel. The volumetric generation ( g′′′ ) can be approximated by the function:

grβ′′′ =

where β = 5x103 W/m2. fissionable materialkfuel = 2 W/m-K

rfuel = 5 cm

rclad = 7 cm

claddingkclad = 0.25 W/m-K

g ′′′ 250 W/m -K500 C

hT∞

== °

Figure P1.3-11: Spherical fuel element surrounded by cladding

a.) Determine an analytical solution for the temperature distribution within the fuel element.

Implement your solution in EES and plot the temperature as a function of radius for 0 < r < rfuel.

b.) The maximum allowable temperature in the fuel element is Tmax = 1100ºC. What is the maximum value of β that can be used? What is the associated total rate that heat is transferred to the gas?

c.) You are designing the fuel elements. You can vary rfuel and β. The cladding must always be 2 cm thick (that is rclad = rfuel + 2 cm). The constraint is that the fuel temperature cannot exceed Tmax = 1100ºC and the design target (the figure of merit to be maximized) is the rate of heat transfer per unit volume of material (fuel and cladding). What values of rfuel and β are optimal?

1.3-12 Figure P1.3-12 illustrates a plane wall. The temperature distribution in the wall is 1-D and the problem is steady state.

LxLq′′ ,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′′′ = where a is a constant and x is 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 L and the conductivity of the wall material, k, is constant. a.) Derive the ordinary differential equation that governs this problem. Clearly show your steps. b.) Solve the differential equation that you obtained in (a). Your solution should include two

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

Section 1.4: Numerical Solutions to Steady-State 1-D Conduction Problems using EES 1.4-1 Figure P1.4.1(a) illustrates a plane wall with thickness L and cross-sectional area A that has a

specified temperature TH on the left side (at x = 0) and a specified temperature TC on the right side (at x = 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).

THTC

L

k = b+cT

T1 T2 T3Δx

(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: Δx = 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: TH, TC, Δx, A, b, and c.

1.4-2 (1-10 in text) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). The conductivity of the material that is used to make the test section is not actually constant, as was assumed in Problem 1-9, but rather depends on temperature according to:

[ ]( )2

W W10 0.035 300 Km-K m-K

k T⎡ ⎤= + −⎢ ⎥⎣ ⎦

a.) Develop a numerical model of the mass flow meter using EES. Plot the temperature as a

function of radial position for the conditions shown in Figure P1.3-9 (Figure P1-9 in the text) with the temperature-dependent conductivity.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in the text) in the limit that the conductivity is constant.

c.) What effect does the temperature dependent conductivity have on the calibration curve that you generated in part (d) of Problem 1.3-9 (1-9)?

1.4-3 Reconsider Problem 1.3-7 using a numerical model developed in EES. a.) Plot the temperature as a function of position within the fuel. b.) Verify that your answer agrees with the analytical solution obtained in Problem 1.3-7. c.) Plot some aspect of the solution as a function of the number of nodes used in the numerical

model and determine the number of nodes required for an accurate solution.

1.4-4 If you bale hay without allowing it to dry sufficiently then the hay bales will contain a lot of water. Besides making the bales heavy and therefore difficult to put in the barn, the water in the hay bails causes an exothermic chemical reaction to occur within the bale (i.e., the hay is rotting). The chemical reaction proceeds at a rate that is related to temperature and the bales may be thermally isolated (they are placed in a barn and surrounded by other hay bales); as a result, the hay can become very hot and even start a barn fire. Figure P1.4-4 illustrates a cross-section of a barn wall with hay stacked against it.

L = 5 mthw = 1 cm

x

,2

,

20 C15 W/m -K

a in

a in

Th

= °=

,2

,

5 C45 W/m -K

a out

a out

Th

= − °=

haykh = 0.05 W/m-K

barn wallkw = 0.11 W/m-K

g′′′

Figure P1.4-4: Barn wall with hay.

The air within the barn is maintained at Ta,in = 20°C and the heat transfer coefficient between the air and the inner surface of the hay is ,a inh = 15 W/m2-K. The outside air is at Ta,out = -5°C with

,a outh = 45 W/m2-K. Neglect radiation from the surfaces in this problem. The barn wall is composed of wood (kw = 0.11 W/m-K) and is thw = 1 cm thick. The hay has been stacked L = 5 m thick against the wall. Hay is a composite structure composed of plant fiber and air. However, hay can be modeled as a single material with an effective conductivity kh = 0.05 W/m-K. The volumetric generation of the hay due to the chemical reaction is given by:

[ ]

0.5

3

W1.5 expm 320

TgK

⎡ ⎤⎛ ⎞⎡ ⎤′′′ = ⎢ ⎥⎜ ⎟⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎢ ⎥⎝ ⎠⎣ ⎦

where T is temperature in K. a.) Develop a numerical model that can predict the temperature distribution within the hay. b.) Prepare a plot that shows the temperature distribution as a function of position in the hay. c.) Prepare a plot that shows that you are using a sufficient number of nodes in your numerical

solution. d.) Verify that your solution is correct by comparing it with an analytical solution in an

appropriate limit. Prepare a plot that overlays your numerical solution and the analytical solution in this limit.

e.) What is the maximum allowable thickness of hay (Lmax) based on keeping the maximum temperature below Tfire?

f.) If L = Lmax from (e) then how much of the hay will remain usable (what percent of the hay is lost to heat degradation)?

1.4-5 Solve the problem stated in EXAMPLE 1.3-2 numerically rather than analytically. a.) Develop a numerical model that can predict the temperature distribution within the lens.

Prepare a plot of the temperature as a function of position. b.) Plot some characteristic of your solution as a function of the number of nodes to show that

you are using a sufficient number of nodes. c.) Think of a sanity check that you can use to gain confidence in your model; that is, can you

change some input parameter and show that the solution behaves as you would expect. Support your answer with a plot.

d.) Plot the maximum lens temperature as a function of the heat transfer coefficient, h .

1.4-6 A current lead must be designed to carry current to a cryogenic superconducting magnet, as shown in Figure P1.4-6.

TH = 300 K

TC = 100 K

D = 1 cm

L = 20 cmIc = 1000 amp

current lead

Figure P1.4-6: Current lead.

The current lead carries Ic = 1000 amp and therefore experiences substantial generation of thermal energy due to ohmic dissipation. The electrical resistivity of the lead material depends on temperature according to:

[ ] [ ]( )9 11 ohm-m17x10 ohm-m 300 K 5x10Ke Tρ − − ⎡ ⎤= + − ⎢ ⎥⎣ ⎦

(1)

The length of the current lead is L = 20 cm and the diameter is D = 1 cm. The hot end of the lead (at x = 0) is maintained at Tx=0 = TH = 300 K and the cold end (at x = L) is maintained at Tx=L = TC = 100 K. The conductivity of the lead material is k = 400 W/m-K. The lead is installed in a vacuum chamber and therefore you may assume that the external surfaces of the lead (the outer surface of the cylinder) are adiabatic. a.) Develop a numerical model in EES that can predict the temperature distribution within the

current lead. Plot the temperature as a function of position. b.) Determine the rate of energy transfer into the superconducting magnet at the cold end of the

current lead. This parasitic must be removed in order to keep the magnet cold and therefore must be minimized in the design of the current lead.

c.) Prepare a plot showing the rate of energy transfer into the magnet as a function of the number of nodes used in your model.

d.) Plot the rate of heat transfer to the cold end as a function of the diameter of the lead. You should see a minimum value and therefore an optimal diameter - explain why this occurs.

e.) Prepare a plot showing the optimal diameter and minimized rate of heat transfer to the cold end as a function of the current that must be carried by the lead. You may want to use the Min/Max Table selection from the Calculate menu to accomplish this.

1.4-7 Figure P1.4-7 illustrates a plane wall. The temperature distribution in the wall is 1-D and the problem is steady state.

Lx

Lq′′

,h T∞

,g a x k′′′ =

node 1node 2 node 3

Figure P1.4-7: Three-node model of a 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′′′ = (2) where a is a constant and x is 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 L and the conductivity of the wall material, k, is constant. You are going to develop a numerical model with 3 nodes, as shown in Figure P1.4-7. The nodes are distributed uniformly throughout the domain. Derive the three equations that must be solved in order to implement the numerical model. Do not solve these equations.

Section 1.5: Numerical Solutions to Steady-State 1-D Conduction Problems using MATLAB 1.5-1 (1-11 in text) Reconsider Problem 1.3-8, but obtain a solution numerically using MATLAB. The

description of the hay bale is provided in Problem 1.3-8. Prepare a model that can consider the effect of temperature on the volumetric generation. Increasing temperature tends to increase the rate of reaction and therefore increase the rate of generation of thermal energy; the volumetric rate of generation can be approximated by: g a bT′′′ = + where a = -1 W/m3 and b = 0.01 W/m3-K. Note that at T = 300 K, the generation is 2 W/m3 but that the generation increases with temperature. a.) Prepare a numerical model of the hay bale using EES. Plot the temperature as a function of

position within the hay bale. b.) Show that your model has numerically converged; that is, show some aspect of your solution

as a function of the number of nodes and discuss an appropriate number of nodes to use. c.) Verify your numerical model by comparing your answer to an analytical solution in some,

appropriate limit. The result of this step should be a plot that shows the temperature as a function of radius predicted by both your numerical solution and the analytical solution and demonstrates that they agree.

d.) Prepare a numerical model of the hay bale using MATLAB. Plot the temperature as a function of position within the hay bale.

1.5-12 (1-12 in text) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). Assume that the conductivity of the material that is used to make the test section is not actually constant as was assumed in Problem 1.3-9 (1-9 in text) but rather depends on temperature according to:

[ ]( )2

W W10 0.035 300 Km-K m-K

k T⎡ ⎤= + −⎢ ⎥⎣ ⎦

a.) Develop a numerical model of the mass flow meter using MATLAB. Plot the temperature as

a function of radial position for the conditions shown in Figure P1.3-9 with the temperature-dependent conductivity.

b.) Verify that your numerical solution limits to the analytical solution from P1.3-9 in the limit that the conductivity is constant.

Section 1.6: Analytical Solutions for Constant Cross-Section Extended Surfaces 1.6-1 (1-13 in text) A resistance temperature detector (RTD) utilizes a material that has a resistivity that

is a strong function of temperature. The temperature of the RTD is inferred by measuring its electrical resistance. Figure P1.6-1 shows an RTD that is mounted at the end of a metal rod and inserted into a pipe in order to measure the temperature of a flowing liquid. The RTD is monitored by passing a known current through it and measuring the voltage across it. This process results in a constant amount of ohmic heating that may tend to cause the RTD temperature to rise relative to the temperature of the surrounding liquid; this effect is referred to as a self-heating error. Also, conduction from the wall of the pipe to the temperature sensor through the metal rod can result in a temperature difference between the RTD and the liquid; this effect is referred to as a mounting error.

pipe

RTD

L = 5.0 cm 20°CwT =

xD = 0.5 mm

k = 10 W/m-K2.5 mWshq =5.0 CT∞ = °

2150 W/m -Kh =

Figure P1.6-1: Temperature sensor mounted in a flowing liquid.

The thermal energy generation associated with ohmic heating is shq = 2.5 mW. All of this ohmic heating is assumed to be transferred from the RTD into the end of the rod at x = L. The rod has a thermal conductivity k = 10 W/m-K, diameter D = 0.5 mm, and length L = 5 cm. The end of the rod that is connected to the pipe wall (at x = 0) is maintained at a temperature of Tw = 20°C. The liquid is at a uniform temperature, T∞ = 5°C and the heat transfer coefficient between the liquid and the rod is h = 150 W/m2-K. a.) Is it appropriate to treat the rod as an extended surface (i.e., can we assume that the

temperature in the rod is a function only of x)? Justify your answer. b.) Develop an analytical model of the rod that will predict the temperature distribution in the

rod and therefore the error in the temperature measurement; this error is the difference between the temperature at the tip of the rod and the liquid.

c.) Prepare a plot of the temperature as a function of position and compute the temperature error. d.) Investigate the effect of thermal conductivity on the temperature measurement error. Identify

the optimal thermal conductivity and explain why an optimal thermal conductivity exists.

1.6-2 (1-14 in text) Your company has developed a micro-end milling process that allows you to easily fabricate an array of very small fins in order to make heat sinks for various types of electrical equipment. The end milling process removes material in order to generate the array of fins. Your initial design is the array of pin fins shown in Figure P1.6-2. You have been asked to optimize the design of the fin array for a particular application where the base temperature is Tbase = 120°C and the air temperature is Tair = 20°C. The heat sink is square; the size of the heat sink is W = 10 cm. The conductivity of the material is k = 70 W/m-K. The distance between the edges of two adjacent fins is a, the diameter of a fin is D, and the length of each fin is L.

D

L

W = 10 cm

array of finsk = 70 W/m-K

120 CbaseT = °

20 C,airT h= °

a

Figure P1.6-2: Pin fin array

Air is forced to flow through the heat sink by a fan. The heat transfer coefficient between the air and the surface of the fins as well as the unfinned region of the base, h , has been measured for the particular fan that you plan to use and can be calculated according to:

[ ] [ ]

0.4 0.3

2

W40 m K 0.005 m 0.01 m

a Dh−

⎛ ⎞ ⎛ ⎞⎡ ⎤= ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠

Mass is not a concern for this heat sink; you are only interested in maximizing the heat transfer rate from the heat sink to the air given the operating temperatures. Therefore, you will want to make the fins as long as possible. However, in order to use the micro-end milling process you cannot allow the fins to be longer than 10x the distance between two adjacent fins. That is, the length of the fins may be computed according to: 10L a= . You must choose the most optimal value of a and D for this application. a.) Prepare a model using EES that can predict the heat transfer coefficient for a given value of a

and D. Use this model to predict the heat transfer rate from the heat sink for a = 0.5 cm and D = 0.75 cm.

b.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the distance between adjacent fins, a, for a fixed value of D = 0.75 cm. Be sure that the fin length is calculated using 10L a= . Your plot should exhibit a maximum value, indicating that there is an optimal value of a.

c.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the diameter of the fins, D, for a fixed value of a = 0.5 cm. Be sure that the fin length is

calculated using 10L a= . Your plot should exhibit a maximum value, indicating that there is an optimal value of D.

d.) Determine the optimal values of a and D using EES' built-in optimization capability.

1.6-3 A water heater consists of a copper tube that carries water through hot gas in a furnace, as shown in Fig. P1.6-3(a). The copper tube has an outer radius, ro,tube = 0.25 inch and a tube wall thickness of th = 0.033 inch. The conductivity of the copper is ktube = 300 W/m-K. Water flows through the pipe at a temperature of Tw = 30ºC. The heat transfer coefficient between the water and the internal surface of the pipe is wh = 500 W/m2-K. The external surface of the tube is exposed to hot gas at Tg = 500ºC. The heat transfer coefficient between the gas and the outer surface of the pipe is gh = 25 W/m2-K. Neglect radiation from the tube surface.

230 C500 W/m -K

w

w

Th

= °=

2500 C25 W/m -K

g

g

Th

= °=

th = 0.033 inch

ro,tube = 0.25 inch

ktube = 300 W/m-K

Figure P1.6-3(a): Copper tube in a water heater.

a.) At what rate is heat is added to the water for a unit length of tube, L = 1 m, for this

configuration (W/m)? b.) What is the dominant resistance to heat transfer in your water heater?

In order to increase the capacity of the water heater, you decide to slide washers over the tube, as shown in Fig. P1.6-3(b). The washers are w = 0.06 inch thick with an outer radius of ro,washer = 0.625 inch and have a thermal conductivity of kwasher = 45 W/m-K. The contact resistance between the washer and the tube is cR′′ = 5x10-4 m2-K/W. The distance between two adjacent washers is b = 0.25 inch.

-4 25x10 m -K/WcR′′ =

kwasher = 45 W/m-Kb = 0.25 inch

w = 0.06 inch

ro,washer = 0.625 inch

Figure P1.6-3(b): Water heater with washers installed.

c.) Can the brass washers be treated as extended surfaces (i.e. can the temperature in the washers

be considered to be only a function of radius)? Justify your answer with a calculation.

Assume that your answer to (c) showed that the washers can be treated as extended surfaces and therefore modeled as a fin with the appropriate fin resistance. d.) Draw a thermal resistance network that can be used to represent this situation. Be sure to

draw and label resistances associated with convection through the water (Rconv,w), conduction

through the copper tube (Rcond), heat transfer through contact resistance (Rcontact), heat transfer through the washers (Rwashers), and convection to gas from unfinned outer surface (Rconv,g,unfinned).

e.) How much heat is added to the water with the washers installed on the tube for a 1 m length of tube?

1.6-4 A wire is subjected to ohmic heating (i.e., a current runs through it) while it is convectively cooled. The ends of the wire have fixed temperatures. Describe how you would determine whether the extended surface approximation (i.e., the approximation in which you treat the temperature of the wire as being one-dimensional along its length and uniform radially) is appropriate when solving this problem.

1.6-5 A cylindrical bracket that is L = 4 cm long with diameter D = 5 mm extends between a wall at TH = 100°C (at x= 0) and a wall at TC = 20°C (at x= L). The conductivity of the bracket is k = 25 W/m2-K. The cylinder is surrounded by gas at T∞ = 200°C and the heat transfer coefficient is h = 250 W/m2-K. a.) Is an extended surface approximation appropriate for this problem? Justify your answer. b.) Assume that your answer to (a) was yes. Develop an analytical model in EES. Plot the

temperature as a function of position within the bracket. c.) Overlay on your plot from (b) the temperature as a function of position with h = 2.5, 25 and

h = 2500 W/m2-K. Explain the shape of your plots. d.) Plot the heat transfer from the wall at TH into the bracket (i.e., the heat transfer into the

bracket at x = 0) as a function of h . Explain the shape of your plot.

1.6-6 Your company has developed a technique for forming very small fins on a plastic substrate. The diameter of the fins at their base is D = 1 mm. The ratio of the length of the fin to the base diameter is the aspect ratio, AR = 10. The fins are arranged in a hexagonal close packed pattern. The ratio of the distance between fin centers and to the base diameter is the pitch ratio, PR = 2. The conductivity of the plastic material is k = 2.8 W/m-K. The heat transfer coefficient between the surface of the plastic and the surrounding gas is h = 35 W/m2-K. The base temperature is Tb = 60°C and the gas temperature is T∞ = 35°C. a.) Determine the number of fins per unit area and the thermal resistance of the unfinned region

of the base. b.) You have been asked to evaluate whether triangular, parabolic concave, or parabolic convex

pin fins will provide the best performance. Plot the heat transfer per unit area of base surface for each of these fin shapes as a function of aspect ratio. Note that the performance of these fins can be obtained from the EES functions eta_fin_spine_parabolic_ND, eta_fin_spine_parabolic2_ND, and eta_fin_spine_triangular_ND. Explain the shape of your plot.

c.) Assume that part (b) indicated that parabolic convex fins are the best. You have been asked whether it is most useful to spend time working on techniques to improve (increase) the aspect ratio or improve (reduce) the pitch ratio. Answer this question using a contour plot that shows contours of the heat transfer per unit area in the parameter space of AR and PR.

Section 1.7: Analytical Solutions for Advanced Constant Cross-Section Extended Surfaces 1.7-1 You are designing a manipulator for use within a furnace. The arm must penetrate the side of the

furnace, as shown in Figure P1.7-1. The arm has a diameter of D = 0.8 cm and protrudes Li = 0.5 m into the furnace, terminating in the actuator that can be assumed to be adiabatic. The portion of the arm in the manipulator is exposed to flame and hot gas; these effects can be represented by a heat flux of 4 21x10 W/mq′′ = and convection to gas at Tf = 500°C with heat transfer coefficient

fh = 50 W/m2-K. The conductivity of the arm material is k = 150 W/m-K. The arm outside of the furnace has the same diameter and conductivity, but is exposed to air at Ta = 20°C with heat transfer coefficient ah = 30 W/m2-K. The length of the arm outside of the furnace is Lo = 0.75 m and this portion of the arm terminates in the motor system which can also be considered to be adiabatic.

x

4 21x10 W/mradq′′ =

D = 0.8 cm

Lo = 0.75 m

Li = 0.5 m230 W/m -K

20 Ca

a

hT

== °

250 W/m -K500 C

f

f

hT

== °

k = 150 W/m-K

Figure P1.7-1: Manipulator arm for a furnace.

a.) Is an extended surface model appropriate for this problem? Justify your answer. b.) Develop an analytical model of the manipulator arm; implement your model in EES. Plot the

temperature as a function of axial position x (see Figure P1.7-1) for -Lo < x < Li. c.) Prepare a plot showing the maximum temperature at the end of the arm (within the furnace)

as a function of the internal length of the arm (Li) for various values of the diameter (D).

1.7-2 Figure P1.7-2 illustrates a metal ring of radius R = 5.0 cm that is rotating with an angular velocity ω = 0.1 rad/s. During each rotation, the ring material passes from compartment #1 containing hot fluid at Tf,1 = 200ºC to compartment #2 containing fluid at Tf,2 = 20ºC. The heat transfer coefficient between the ring surface and the fluid in compartments #1 and #2 are 1h = 10 W/m2-K and 2h = 20 W/m2-K, respectively. The ring has a circular cross-section with diameter d = 1.0 mm. For this problem you can assume that d/R<<1. The metal has conductivity k = 100 W/m-K, density ρ = 2700 kg/m3, and specific heat capacity c = 900 J/kg-K.

,12

1

compartment #1200 C

10W/m -KfT

h= °

=

,22

2

compartment #220 C

20 W/m -KfT

h= °

=

ω = 0.1 rad/sR = 5 cm

d = 1 mm

ρ= 2700 kg/m3

c = 900 J/kg-Kk = 100 W/m-K

Figure P1.7-2: Metal ring rotates between two compartments.

a.) Is it appropriate to treat the ring as an extended surface? b.) Develop an analytical model that can predict the temperature distribution in the ring. c.) Plot the temperature distribution in the ring as a function of position along the circumference

of the ring for various values of the ring rotation rate. d.) Compute the total rate of energy transferred between the two compartments. e.) Determine the dimensionless parameter that characterizes the relative importance of energy

transfer due to ring rotation and energy transfer by conduction. Plot the heat transfer rate between the two compartments as a function of this parameter (varied by changing the ring rotation rate).

1.7-3 (1-15 in text) Figure P1.7-3 illustrates a material processing system.

x

0.75 m/s300 Kin

uT==

D = 5 cm

gap filled with gasth = 0.6 mmkg = 0.03 W/m-K

extruded materialk = 40 W/m-Kα = 0.001 m2/s

oven wall temperature varies with x

Figure P1.7-3: Material processing system.

Material is extruded and enters the oven at Tin = 300 K with velocity u = 0.75 m/s. The material has diameter D = 5 cm. The conductivity of the material is k = 40 W/m-K and the thermal diffusivity is α = 0.001 m2/s. In order to precisely control the temperature of the material, the oven wall is placed very close to the outer diameter of the extruded material and the oven wall temperature distribution is carefully controlled. The gap between the oven wall and the material is th = 0.6 mm and the oven-to-material gap is filled with gas that has conductivity kg = 0.03 W/m-K. Radiation can be neglected in favor of convection through the gas from the oven wall to the material. For this situation, the heat flux experienced by the material surface can be approximately modeled according to:

( )gconv w

kq T T

th′′ ≈ −

where Tw and T are the oven wall and material temperatures at that position, respectively. The oven wall temperature varies with position x according to:

( ),0 expw f f wc

xT T T TL

⎛ ⎞= − − −⎜ ⎟

⎝ ⎠

where Tw,0 is the temperature of the wall at the inlet (at x = 0), Tf = 1000 K is the temperature of the wall far from the inlet, and Lc is a characteristic length that dictates how quickly the oven wall temperature approaches Tf. Initially, assume that Tw,0 = 500 K, Tf = 1000 K, and Lc = 1 m. Assume that the oven can be approximated as being infinitely long. a.) Is an extended surface model appropriate for this problem? b.) Assume that your answer to (a) was yes. Develop an analytical solution that can be used to

predict the temperature of the material as a function of x. c.) Plot the temperature of the material and the temperature of the wall as a function of position

for 0 < x < 20 m. Plot the temperature gradient experienced by the material as a function of position for 0 < x < 20 m.

The parameter Lc can be controlled in order to control the maximum temperature gradient and therefore the thermal stress experienced by the material as it moves through the oven. d.) Prepare a plot showing the maximum temperature gradient as a function of Lc. Overlay on

your plot the distance required to heat the material to Tp = 800 K (Lp). If the maximum temperature gradient that is allowed is 60 K/m, then what is the appropriate value of Lc and the corresponding value of Lp?

1.7-4 (1-16 in text) The receiver tube of a concentrating solar collector is shown in Figure P1.7-4.

r = 5 cmth = 2.5 mm

280 C100 W/m -K

w

w

Th

= °=

sq′′

φ

225 C25 W/m -K

a

a

Th= °=

k = 10 W/m-K

Figure P1.7-4: A solar collector

The receiver tube is exposed to solar radiation that has been reflected from a concentrating mirror. The heat flux received by the tube is related to the position of the sun and the geometry and efficiency of the concentrating mirrors. For this problem, you may assume that all of the radiation heat flux is absorbed by the collector and neglect the radiation emitted by the collector to its surroundings. The flux received at the collector surface ( sq′′ ) is not circumferentially uniform but rather varies with angular position; the flux is uniform along the top of the collector, π < φ < 2π rad, and varies sinusoidally along the bottom, 0 < φ < π rad, with a peak at φ = π/2 rad.

( ) ( ) ( )+ sin for 0

for 2t p t

s

t

q q qq

q

φ φ πφ

π φ π

⎧ ′′ ′′ ′′− < <⎪′′ = ⎨′′ < <⎪⎩

where tq′′ = 1000 W/m2 is the uniform heat flux along the top of the collector tube and pq′′ = 5000 W/m2 is the peak heat flux along the bottom. The receiver tube has an inner radius of r = 5.0 cm and thickness of th = 2.5 mm (because th/r << 1 it is possible to ignore the small difference in convection area on the inner and outer surfaces of the tube). The thermal conductivity of the tube material is k = 10 W/m-K. The solar collector is used to heat water, which is at Tw = 80°C at the axial position of interest. The average heat transfer coefficient between the water and the internal surface of the collector is wh = 100 W/m2-K. The external surface of the collector is exposed to air at Ta = 25°C. The average heat transfer coefficient between the air and the external surface of the collector is ah = 25 W/m2-K. a.) Can the collector be treated as an extended surface for this problem (i.e., can the temperature

gradients in the radial direction in the collector material be neglected)? b.) Develop an analytical model that will allow the temperature distribution in the collector wall

to be determined as a function of circumferential position.

1.7-5 A flux meter is illustrated in Figure P1.7-5.

2900 W/mq′′ = evacuatedspace

L = 2 cm

-4 21x10 K-m /WR′′ =

th = 1 mm

k = 10 W/m-K

20 CT∞ = °

x

20 CcT = °

Figure P1.7-5: Flux meter.

A thin plate is clamped on either end to a casing that is maintained at Tc = 20ºC. There is a

contact resistance between the plate and the casing, R′′ = 1x10-4 K-m2/W. The thickness of the plate is th = 1 mm and the half-width of the plate is L = 2 cm. The plate conductivity is k = 10 W/m-K. The back of the plate is insulated and the front of the plate is mounted within an evacuated enclosure in order to eliminate convection. You may model radiation loss from the plate surface using an effective "radiation" heat transfer coefficient, calculated according to:

34radh Tε σ≈ , where T is the average absolute temperature of the plate and the surroundings (take T = 300 K for this problem). The plate radiates to surroundings at T∞ = 20ºC. The nominal flux on the plate is q′′ = 900 W/m2. You may assume that the plate temperature distribution is 1-D in the x-direction. You may also assume that the emissivity of the plate, ε, is one and therefore all of the flux on the plate is absorbed. The flux meter operates by correlating the difference between the temperature at the center of the plate and the casing with the applied heat flux. a.) Derive the governing differential equation that governs the temperature within the plate.

Clearly show your steps. b.) What are the boundary conditions for the differential equation? c.) Determine the solution to the differential equation from (a) subject to the boundary

conditions from (b) without using Maple. d.) Plot the temperature in the plate as a function of position. e.) Use Maple to solve the differential equation and obtain symbolic expressions for the

boundary conditions. Implement the Maple expressions into EES and show that your answer is identical to the one determined in (c).

f.) Prepare a calibration curve for the flux meter - plot the heat flux as a function of the difference between the temperature at the center of the plate and the casing.

g.) If the uncertainty in the measurement of the temperature difference is δΔT = 0.5 K then what is the uncertainty in the measurement of the heat flux?

Section 1.8: Analytical Solutions for Non-Constant Cross-Section Extended Surfaces 1.8-1 (1-17 in text) Figure P1.8-1 illustrates a disk brake for a rotating machine. The temperature

distribution within the brake can be assumed to be a function of radius only. The brake is divided into two regions. In the outer region, from Rp = 3.0 cm to Rd = 4.0 cm, the stationary brake pads create frictional heating and the disk is not exposed to convection. The clamping pressure applied to the pads is P = 1.0 MPa and the coefficient of friction between the pad and the disk is μ = 0.15. You may assume that the pads are not conductive and therefore all of the frictional heating is conducted into the disk. The disk rotates at N = 3600 rev/min and is b = 5.0 mm thick. The conductivity of the disk is k = 75 W/m-K and you may assume that the outer rim of the disk is adiabatic.

center line

clamping pressureP = 1 MPa

Rp = 3 cmRd = 4 cm

b = 5 mm 30 C,aT h= °

k = 75 W/m-K

coefficient of friction, μ = 0.15stationarybrake pads

disk, rotates at N = 3600 rev/min Figure P1.8-1: Disk brake.

The inner region of the disk, from 0 to Rp, is exposed to air at Ta = 30°C. The heat transfer coefficient between the air and disk surface depends on the angular velocity of the disk, ω, according to:

[ ]

1.25

2 2

W W20 1500m -K m -K 100 rad/s

h ω⎛ ⎞⎡ ⎤ ⎡ ⎤= + ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠

a.) Develop an analytical model of the temperature distribution in the disk brake; prepare a plot

of the temperature as a function of radius for r = 0 to r = Rd. b.) If the disk material can withstand a maximum safe operating temperature of 750°C then what

is the maximum allowable clamping pressure that can be applied? Plot the temperature distribution in the disk at this clamping pressure. What is the braking torque that results?

c.) Assume that you can control the clamping pressure so that as the machine slows down the maximum temperature is always kept at the maximum allowable temperature, 750°C. Plot the torque as a function of rotational speed for 100 rev/min to 3600 rev/min.

1.8-2 Figure P1.8-2 illustrates a thin, disk-shaped window that is used to provide optical access to a combustion chamber. The thickness of the window is b and the outer radius of the window is Ro. The window is composed of material with conductivity k and absorption coefficient α. The combustion chamber side of the window is exposed to convection with hot gas at Tg and heat transfer coefficient h. Convection with the air outside of the chamber can be neglected. There is a radiation heat flux, radq′′ , that is incident on the combustion chamber side of the glass. The amount of this radiation that is absorbed by the glass is, approximately, radq bα′′ . The remainder of this radiation, ( )1radq bα′′ − , exits the opposite surface of the glass. The outer edge (at r = Ro) of the glass is held at temperature Tedge.

combustion chamber

window, k and α,fT h

radiant flux, radq′′

( )unabsorbed radiant flux, 1 radb qα ′′−

Tedge Tedge

x

rRo

b

outside of chamber Figure P1.8-2: Disk-shaped window.

You are to develop a 1-D, steady state analytical model that can predict the temperature

distribution in the glass as a function of radial position, r. a.) How would you justify using a 1-D model of the glass? What number would you calculate in

order to verify that the temperature does not vary substantially in the x direction? b.) Derive the ordinary differential equation in r that must be solved. Make sure that your

differential equation includes the effect of conduction, convection with the gas within the chamber, and generation of thermal energy due to absorption.

c.) What are the boundary conditions for the ordinary differential equation that you derived in part (b)?

d.) Solve the ordinary differential equation in order to obtain an expression for the temperature as a function of radius.

1.8-3 It is often necessary to provide a fluid outlet port that will allow very cold gas to escape from a cryogenic facility; for example, the boil-off of liquid nitrogen or liquid helium from within a vacuum vessel must be allowed to vent to the atmosphere. The seal between the base of the flange that contains the fluid passage and the surrounding vessel is usually made with an o-ring; these seals are convenient in that they are hermetic and easily demountable. However, most convenient o-ring materials do not retain their ductility at temperatures much below 0°C and therefore it is important that the o-ring be kept at or above this temperature so that it continues to provide a good seal; if the o-ring “freezes” then the cryogenic facility will lose its vacuum. The o-ring temperature is maintained at an appropriate level using a thermal standoff, as shown in Figure P1.8-3

The fluid passage is attached to the flange via a separate, slightly larger tube made of a stainless

steel with thermal conductivity k = 15 W/m-K. This thermal standoff has an outer radius, rts = 1 cm, thickness tts = 1 mm, and length Lts = 5 cm. The cryogenic gas is at 77 K and you may assume that the point x = 0 in Figure P1.8-3 is at Tcold = 77 K. The inside of the tube is exposed to a vacuum and you may assume that it experiences negligible radiation heat transfer and no convective heat transfer. The outside of the tube is exposed to air at Tair = 20°C with heat transfer coefficient, h = 7 W/m2-K. The bottom of the tube (x = Lts) is welded to the flange. The flange has an outer radius rfl = 8 cm and a thickness tfl = 1.5 mm. The inside of the flange is exposed to vacuum and therefore, for the purposes of this problem, adiabatic. The outside of the flange is exposed to the same 20°C air with the same 7 W/m2-K heat transfer coefficient.

thermal stand off

o-ring seal

flange

cryogenic gas

Lts = 5 cm

tts = 1 mm

Tcold = 77 K

x

rts = 1 cm

rfl = 8 cm

tfl = 1.5 mm

220 C

7 W/m -KairT

h= °

=

vacuum

k = 15 W/m-K

Figure P1.8-3: Cryogenic thermal standoff.

a.) Is it appropriate to treat the thermal standoff and the flange as extended surfaces? That is,

can the temperature within the stand-off be treated as being only a function of x and the temperature in the flange only a function of r?

b.) Develop an analytical model of the problem that can predict the temperature distribution as a function of x in the thermal stand-off and as a function of r in the flange.

1.8-4 Figure P1.8-4 shows a typical fin design that is fabricated by attaching a thin washer to the outer radius of a tube. The inner and outer radii of the fin are rin and rout, respectively. The thickness of the fin is th and the fin material has conductivity, k. The fin is surrounded by fluid at T∞ and the average heat transfer coefficient is h . The base of the fin is maintained at Tb and the tip is adiabatic.

th

rin

rout

k

,h T∞

Tb

Figure P1.8-4: Circular fin.

Determine an analytical solution for the temperature distribution in the fin and the fin efficiency.

1.8-5 (1-18 in text) Figure P1.8-5 illustrates a fin that is to be used in the evaporator of a space conditioning system for a space-craft.

L = 2 cm

th = 1 mm

Wb = 1 cm

k = 50 W/m-Kρ= 3000 kg/m3

10 CbT = °

2120 W/m -K20 C

hT∞

== °

thb = 2 mm thg = 2 mm

ρb = 8000 kg/m3

x

Figure P1.8-5: Fin on an evaporator.

The fin is a plate with a triangular shape. The thickness of the plate is th = 1 mm and the width

of the fin at the base is Wb = 1 cm. The length of the fin is L = 2 cm. The fin material has conductivity k = 50 W/m-K. The average heat transfer coefficient between the fin surface and the air in the space-craft is h = 120 W/m2-K. The air is at T∞ = 20°C and the base of the fin is at Tb = 10°C. Assume that the temperature distribution in the fin is 1-D in x. Neglect convection from the edges of the fin. a.) Obtain an analytical solution for the temperature distribution in the fin. Plot the temperature

as a function of position. b.) Calculate the rate of heat transfer to the fin. c.) Determine the fin efficiency.

The fin has density ρ = 3000 kg/m3 and the fin is installed on a base material with thickness thb = 2 mm and density ρb = 8000 kg/m3. The half-width of the gap between adjacent fins is thg = 2 mm. Therefore, the volume of the base material associated with each fin is thb Wb (th + 2 thg). d.) Determine the ratio of the absolute value of the rate of heat transfer to the fin to the total

mass of material (fin and base material associated with the fin). e.) Prepare a contour plot that shows the ratio of the heat transfer to the fin to the total mass of

material as a function of the length of the fin (L) and the fin thickness (th). f.) What is the optimal value of L and th that maximizes the absolute value of the fin heat

transfer rate to the mass of material?

1.8-6 As an alternative to surgery, cancer tumors may be destroyed by placing cylindrically-shaped cryoprobes into the body, as shown in Figure P1.8-6. The probe surface is cooled causing the temperature of the surrounding tissue to drop to a lethal level, killing the tumor.

rp = 5 mm

tissuek = 0.6 W/m-Kβ = 40000 W/m3-K

230000 W/mpq′′ =

37 CbT = °

Figure 1.8-6: Cryosurgical probe.

The probe radius is rp = 5 mm and the heat flux at the surface of the probe (leaving the tissue) is

pq′′ = 30000 W/m2. The tissue has conductivity k = 0.6 W/m-K. The blood flow through the tissue results in a volumetric heating effect ( g′′′ ) that is proportional to the difference between the local temperature and the blood temperature, Tb = 37ºC: ( )bg T Tβ′′′ = − where β = 40000 W/m3-K. The temperature of the tissue far from the probe is Tb. Assume that the temperature distribution is 1-D and steady-state. a.) Develop an analytical model that can be used to predict the temperature distribution in the

tissue. Implement your solution in EES and prepare a plot of the temperature distribution as a function of radius.

b.) The lethal temperature for cell death is Tlethal = -30ºC. Plot the radius of the cryolesion (i.e., the kill radius - all tissue inside of this radius is dead) as a function of the heat flux provided by the cryoprobe.

1.8-7 A disk-shaped bracket connects a cylindrical heater to an outer shell, as shown in Figure P1.8-7(a).

r

center line

heaterbracket

shell

qrb

rt

th

kTt

,h T∞

,h T∞

Figure P1.8-7(a): Disk-shaped bracket.

The thickness of the bracket is th and it is made of material with conductivity k. The bracket extends radially from rb at the heater to rt at the outer shell. The temperature of the bracket location where it intersects the shell (at r = rt) is Tt. The heater provides a rate of heat transfer to the bracket, q , at r = rb. Both the upper and lower surfaces of the bracket are exposed to fluid at T∞ with average 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 temperature is only a function of r); provide an expression in terms of the symbols in the problem statement.

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

r

center line

qrb

rt

Tt,h T∞

,h T∞

rb rtr

T

Tt

T∞

Figure P1.8-7(b): Qualitative sketch of the temperature distribution expected if h → 0 and h → ∞.

b.) On the axes in Figure P1.8-7(b), sketch the temperature distribution that you would expect if the heat transfer coefficient h is very low ( h → 0). Note that the qualitative values of Tt and T∞ are indicated in the plot - your sketch should be consistent with these values.

c.) On the axes in Figure P1.8-7(c), sketch the temperature distribution that you would expect if the heat transfer coefficient h is very high ( h → ∞). Note that the qualitative values of Tt and T∞ are indicated in the plot - your sketch should be consistent with these values.

d.) What dimensionless number would you calculate in order to determine whether the actual temperature distribution is closer to your sketch from (b) or (c)? Provide an expression in terms of the symbols in the problem statement.

e.) Derive the governing ordinary differential equation for the bracket. f.) What are the boundary conditions for the ordinary differential equation from (e)? g.) Obtain a solution to your ordinary differential equation that includes two undetermined

constants. h.) Write the two algebraic equation that could be solved to provide the undetermined constants

(don't solve these equations).

1.8-8 Figure P1.8-8 illustrates a triangular fin with a circular cross-section.

Figure P1.8-5: Fin on an evaporator.

The fin is surrounded by fluid at T∞ with heat transfer coefficient h . The base of the fin is at Tb

and the fin conductivity is k. a.) Derive the governing differential equation and the boundary conditions for the problem. b.) Normalize the governing differential equation and the boundary conditions. This process

should lead to the identification of a dimensionless fin parameter that governs the solution. Identify the physical significance of this parameter.

c.) Solve the differential equation subject to the boundary conditions. d.) Prepare a plot of dimensionless temperature as a function of dimensionless position for

various values of the remaining dimensionless parameter, identified in (b). e.) The fin efficiency is defined as the ratio of the heat transfer into the base of the fin to the heat

transfer that would occur if the entire fin were isothermal and at the base temperature (i.e., if the fin material were infinitely conductive). Develop an equation for the fin efficiency and plot the fin efficiency as a function of the dimensionless fin parameter identified in (b).

1.8-9 One side of a thin circular membrane is subjected to a flux of energy that varies according to: 2q a r′′ = Both sides of the membrane are exposed to fluid at T∞ with heat transfer coefficient h . The outer edge of the membrane is held at T∞. The radius of the membrane is ro and the thickness is th. The conductivity of the membrane material is k. Assume that the temperature distribution in the membrane is only a function of radius. a.) Derive the governing differential equation for the temperature in the membrane and the

boundary conditions. b.) Define a dimensionless temperature difference and radius. Use them to non-dimensionalize

the governing differential equation and boundary conditions from (a). This process should lead to the identification of another dimensionless parameter. Explain the significance of this parameter.

c.) Solve the normalized problem from (b). Prepare a plot of the dimensionless temperature as a function of the dimensionless radius for various values of the dimensionless parameter identified in (b).

Section 1.9: Numerical Solution of Extended Surface Problems 1.9-1 A fuse is a long, thin piece of metal that will heat up when current is passed through it. If a large

amount of current is passed through the fuse, then the material will melt and this protects the electrical components downstream of the fuse. Figure P1.9-1 illustrates a fuse that is composed of a piece of metal with a square cross-section (a x a) that has length L. The conductivity of the fuse material is k. The fuse surface experiences convection with air at temperature Ta with heat transfer coefficient h . Radiation from the surface can be neglected for this problem. The ohmic heating associated with the current passing through the fuse results in a uniform rate of volumetric thermal energy generation, g′′′ . The two ends of the fuse (at x = 0 and x = L) are held at temperature Tb. You have been asked to generate a numerical model of the fuse. Figure P1.9-1 also shows a simple numerical model that includes only four nodes which are positioned uniformly along the length of the fuse.

xL

,k g ′′′ A

A

,aT h

section A-A

a

a

node 1node 2

node 3node 4

Figure P1.9-1: Fuse and a four node numerical model of the fuse.

a.) How would you determine if the extended surface approximation was appropriate for this

problem? For the remainder of this problem, assume that you can use the extended surface approximation. b.) Derive a system of algebraic equations that can be solved in order to predict the temperatures

at each of the four nodes in Figure P1.9-1 (T[1], T[2], T[3], and T[4]). Your equations should include only those symbols defined in the problem statement. Do not solve these equations.

c.) How would you determine the amount of heat transferred from the fuse to the wall at x = 0 using your solution from (b)?

d.) Derive the differential equation and boundary conditions that you would need in order to solve this problem analytically. Show your steps clearly.

1.9-2 Reconsider the disk-shaped bracket that was discussed in Problem P1.8-7. You have decided to generate a numerical model of the bracket that has three nodes, positioned as shown in Figure P1.9-2.

r

center line

rb

rt

node 1 node 2

node 3

Figure P1.9-2: A 3-node numerical model of the disk-shaped bracket.

a.) 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 three nodes (r1, r2, and r3). Do not solve these equations.

1.9-3 (1-19 in text) A fiber optic bundle (FOB) is shown in Figure P1.9-3 and used to transmit the light for a building application.

5 21x10 W/mq′′ =

25 W/m -K20 C

hT∞

== °

x

rout = 2 cm

fiber optic bundle Figure P1.9-3: Fiber optic bundle used to transmit light.

The fiber optic bundle is composed of several, small diameter fibers that are each coated with a thin layer of polymer cladding and packed in approximately a hexagonal close-packed array. The porosity of the FOB is the ratio of the open area of the FOB face to its total area. The porosity of the FOB face is an important characteristic because any radiation that does not fall directly upon the fibers will not be transmitted and instead contributes to a thermal load on the FOB. The fibers are designed so that any radiation that strikes the face of a fiber is “trapped” by total internal reflection. However, radiation that strikes the interstitial areas between the fibers will instead be absorbed in the cladding very close to the FOB face. The volumetric generation of thermal energy associated with this radiation can be represented by:

expch ch

q xgL Lφ ⎛ ⎞′′

′′′= −⎜ ⎟⎝ ⎠

where q′′= 1x105 W/m2 is the energy flux incident on the face, φ = 0.05 is the porosity of the FOB, x is the distance from the face, and Lch = 0.025 m is the characteristic length for absorption of the energy. The outer radius of the FOB is rout = 2 cm. The face of the FOB as well as its outer surface are exposed to air at T∞ = 20°C with heat transfer coefficient h = 5 W/m2-K. The FOB is a composite structure and therefore conduction through the FOB is a complicated problem involving conduction through several different media. Section 2.9 discusses methods for computing the effective thermal conductivity for a composite. The effective thermal conductivity of the FOB in the radial direction is keff,r = 2.7 W/m-K. In order to control the temperature of the FOB near the face, where the volumetric generation of thermal energy is largest, it has been suggested that high conductivity filler material be inserted in the interstitial regions between the fibers. The result of the filler material is that the effective conductivity of the FOB in the axial direction varies with position according to:

, , , , expeff x eff x eff xk

xk k kL∞

⎛ ⎞= + Δ −⎜ ⎟

⎝ ⎠

where keff,x,∞ = 2.0 W/m-K is the effective conductivity of the FOB in the x-direction without filler material, Δkeff,x = 28 W/m-K is the augmentation of the conductivity near the face, and Lk = 0.05 m is the characteristic length over which the effect of the filler material decays. The length of the FOB is effectively infinite. a.) Is it appropriate to use a 1-D model of the FOB? b.) Assume that your answer to (a) was yes. Develop a numerical model of the FOB.

c.) Overlay on a single plot the temperature distribution within the FOB for the case where the filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0).

1.9-4 (1-20 in text) An expensive power electronics module normally receives only a moderate current. However, under certain conditions it might experience currents in excess of 100 amps. The module cannot survive such a high current and therefore, you have been asked to design a fuse that will protect the module by limiting the current that it can experience, as shown in Figure P1.9-4.

L = 2.5 cm

D = 0.9 mm

25 C

5 W/m -KTh∞ = °=

ε = 0.9

20 CendT = ° 20 CendT = °

k = 150 W/m-Kρr = 1x10-7 ohm-mI = 100 amp

Figure 1.9-4: A fuse that protects a power electronics module from high current.

The space available for the fuse allows a wire that is L = 2.5 cm long to be placed between the module and the surrounding structure. The surface of the fuse wire is exposed to air at T∞ = 20°C. The heat transfer coefficient between the surface of the fuse and the air is h = 5.0 W/m2-K. The fuse surface has an emissivity of ε = 0.90. The fuse is made of an aluminum alloy with conductivity k = 150 W/m-K. The electrical resistivity of the aluminum alloy is ρe = 1x10-7 ohm-m and the alloy melts at approximately 500°C. Assume that the properties of the alloy do not depend on temperature. The ends of the fuse (i.e., at x=0 and x=L) are maintained at Tend =20°C by contact with the surrounding structure and the module. The current passing through the fuse, I, results in a uniform volumetric generation within the fuse material. If the fuse operates properly, then it will melt (i.e., at some location within the fuse, the temperature will exceed 500°C) when the current reaches 100 amp. Your job will be to select the fuse diameter; to get your model started, you may assume a diameter of D = 0.9 mm. Assume that the volumetric rate of thermal energy generation due to ohmic dissipation is uniform throughout the fuse volume. a.) Prepare a numerical model of the fuse that can predict the steady state temperature

distribution within the fuse material. Plot the temperature as a function of position within the wire when the current is 100 amp and the diameter is 0.9 mm.

b.) Verify that your model has numerically converged by plotting the maximum temperature in the wire as a function of the number of nodes in your model.

c.) Prepare a plot of the maximum temperature in the wire as a function of the diameter of the wire for I=100 amp. Use your plot to select an appropriate fuse diameter.

1.9-5 A triangular fin is shown in Figure P1.9-5.

Figure P1.9-5: Wedge fin

The fin infinitely long (in the z-direction) and can be treated as an extended surface. The

thickness of the fin base is th = 1 cm and the length is L = 10 cm. The conductivity of the material is k = 24 W/m-K. The base temperature is Tb = 140°C and the ambient temperature is T∞ = 25°C. The heat transfer coefficient is h = 15 W/m2-K. The width of the fin, W, is much larger than its length. a.) Develop a numerical model of the fin. b.) Plot the temperature distribution within the fin. c.) Determine the fin efficiency. Compare your answer with the fin efficiency obtained from the

EES function eta_fin_straight_triangular. d.) Plot the fin efficiency as a function of the number of nodes used in the solution.

1.9-6 Your company manufacturers heater wire. Heater wire is applied to surfaces that need to be heated and then current is passed through the wire in order to develop ohmic dissipation. A key issue with your product is failures that occur when a length of the wire becomes detached from the surface and therefore the wire is not well connected thermally to the surface. The wire in the detached region tends to get very hot and melt.

The wire diameter is D = 0.4 mm and the current passing through the wire is current = 10 amp.

the detached wire is exposed to surroundings at T∞ = 20ºC through convection and radiation. The average convection heat transfer coefficient is h = 30 W/m2-K and the emissivity of the wire surface is ε = 0.5. The length of the wire that is detached from the surface is L = 1 cm. The ends of the wire at x = 0 and x = L are held at Tend = 50ºC. The wire conductivity is k = 10 W/m-K and the electrical resistivity is ρe = 1x10-7 ohm-m. a.) Develop a numerical model of the wire and use the model to plot the temperature distribution

within the wire. b.) Plot the maximum temperature in the wire as a function of the number of nodes in the

numerical model. c.) Plot the maximum temperature in the wire as a function of the length of the detached region.

If the maximum temperature of the wire before failure is Tmax = 400ºC, then what is the maximum allowable length of detached wire?

You are looking at methods to alleviate this problem and have identified an alternative material, material D, in which the electrical resistivity is ρe = 1x10-7 ohm-m but the conductivity increases with temperature according to:

[ ]( )2

W W10 0.05 300 Km K m K

k T⎡ ⎤ ⎡ ⎤

= + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

d.) Modify your numerical model in order to simulate material D. Overlay on your plot from (c) the maximum temperature as a function of the length of the detached wire for material D.

You have identified another alternative material, material E, in which the thermal conductivity is k = 10 W/m-K but the electrical resistivity decreases with temperature according to:

[ ] [ ]( )7 -10 ohm m1x10 ohm m 1x10 300 KK

Tρ − ⎡ ⎤= − −⎢ ⎥⎣ ⎦

e.) Modify your numerical model in order to simulate material E. Overlay on your plot from (c) the maximum temperature as a function of the length of the detached wire for material E.

1.9-7 Figure P1.9-7 illustrates a flat plate solar collector.

L = 8 cm

th = 1.5 mm

2900 W/msq′′ =

ε = 1k = 75 W/m2-K

25 W/m -K10 C

hT∞

== °

50 CwT = °

Figure 1.9-7: Flat plate solar collector.

The collector consists of a flat plate that is th = 1.5 mm thick with conductivity k = 75 W/m-K. The plate is insulated on its back side and experiences a solar flux of sq′′= 900 W/m2 which is all absorbed. The surface is exposed to convection and radiation to the surroundings. The emissivity of the surface is ε = 1. The heat transfer coefficient is h = 5 W/m2-K and the surrounding temperature is T∞ = 10ºC. The temperature of the water is Tw = 50ºC. The center-to-center distance between adjacent tubes is 2 L where L = 8 cm. a.) Develop a numerical model that can predict the rate of energy transfer to the water per unit

length of collector. b.) Determine the efficiency of the collector; efficiency is defined as the ratio of the energy

delivered to the water to the solar energy incident on the collector. c.) Plot the efficiency as a function of the number of nodes used in the solution. d.) Plot the efficiency as a function of Tw - T∞. Explain your plot.

1.1-2 (1-1 in text) Section 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at ideal gas conditions. Test the validity of this approximation by comparing the conductivity estimated using Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., low pressure argon) provided by the internal function in EES. Note that the molecular radius, σ, is provided in EES by the Lennard-Jones potential using the function sigma_LJ. a.) What is the value and units of the proportionality constant required to make Eq. (1-18) an

equality? b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and

100 MPa on a semi-log plot with pressure on the log scale. At what pressure does the approximation given in Eq. (1-18) begin to fail?

1.2-3 (1-2 in text) Figure P1.2-3 illustrates a plane wall made of a thin (thw = 0.001 m) and conductive

(k = 100 W/m-K) material that separates two fluids, A and B. Fluid A is at TA = 100°C and the heat transfer coefficient between the fluid and the wall is Ah = 10 W/m2-K while fluid B is at TB = 0°C with Bh = 100 W/m2-K.

thw = 0.001 m

k = 100 W/m-K

2

100 C

10 W/m -KA

A

T

h

= °

= 2

0 C100 W/m -K

B

B

Th

= °

=

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). b.) If you wanted to predict the heat transfer rate from fluid A to fluid B very 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.

1.9-4 (1-20 in text) An expensive power electronics module normally receives only a moderate current. However, under certain conditions it might experience currents in excess of 100 amps. The module cannot survive such a high current and therefore, you have been asked to design a fuse that will protect the module by limiting the current that it can experience, as shown in Figure P1.9-4.

L = 2.5 cm

D = 0.9 mm

25 C

5 W/m -KTh∞ = °=

ε = 0.9

20 CendT = ° 20 CendT = °

k = 150 W/m-Kρr = 1x10-7 ohm-mI = 100 amp

Figure 1.9-4: A fuse that protects a power electronics module from high current.

The space available for the fuse allows a wire that is L = 2.5 cm long to be placed between the module and the surrounding structure. The surface of the fuse wire is exposed to air at T∞ = 20°C. The heat transfer coefficient between the surface of the fuse and the air is h = 5.0 W/m2-K. The fuse surface has an emissivity of ε = 0.90. The fuse is made of an aluminum alloy with conductivity k = 150 W/m-K. The electrical resistivity of the aluminum alloy is ρe = 1x10-7 ohm-m and the alloy melts at approximately 500°C. Assume that the properties of the alloy do not depend on temperature. The ends of the fuse (i.e., at x=0 and x=L) are maintained at Tend =20°C by contact with the surrounding structure and the module. The current passing through the fuse, I, results in a uniform volumetric generation within the fuse material. If the fuse operates properly, then it will melt (i.e., at some location within the fuse, the temperature will exceed 500°C) when the current reaches 100 amp. Your job will be to select the fuse diameter; to get your model started, you may assume a diameter of D = 0.9 mm. Assume that the volumetric rate of thermal energy generation due to ohmic dissipation is uniform throughout the fuse volume. a.) Prepare a numerical model of the fuse that can predict the steady state temperature

distribution within the fuse material. Plot the temperature as a function of position within the wire when the current is 100 amp and the diameter is 0.9 mm.

b.) Verify that your model has numerically converged by plotting the maximum temperature in the wire as a function of the number of nodes in your model.

c.) Prepare a plot of the maximum temperature in the wire as a function of the diameter of the wire for I=100 amp. Use your plot to select an appropriate fuse diameter.

1.2-8 (1-3 in text) You have a problem with your house. Every spring at some point the snow immediately adjacent to your roof melts and runs along the roof line until it reaches the gutter. The water in the gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutter and causing water to run into your attic. The situation is shown in Figure P1.2-8.

snow melts at this surface2, 15 W/m -Kout outT h =

snow, ks = 0.08 W/m-K

222 C, 10 W/m -Kin inT h= ° =Lins = 3 inch

insulation, kins = 0.05 W/m-K

plywood,0.5 inch, 0.2 W/m-Kp pL k= =

Ls = 2.5 inch

Figure P1.2-8: Roof of your house.

The air in the attic is at Tin = 22°C and the heat transfer coefficient between the inside air and the inner surface of the roof is inh = 10 W/m2-K. The roof is composed of a Lins = 3.0 inch thick piece of insulation with conductivity kins = 0.05 W/m-K that is sandwiched between two Lp = 0.5 inch thick pieces of plywood with conductivity kp = 0.2 W/m-K. There is an Ls = 2.5 inch thick layer of snow on the roof with conductivity ks = 0.08 W/m-K. The heat transfer coefficient between the outside air at temperature Tout and the surface of the snow is outh = 15 W/m2-K. Neglect radiation and contact resistances for part (a) of this problem. a.) What is the range of outdoor air temperatures where you should be concerned that your

gutters will become blocked by ice? b.) Would your answer change much if you considered radiation from the outside surface of the

snow to surroundings at Tout? Assume that the emissivity of snow is εs = 0.82.

1.2-11 (1-4 in text) Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with kA = 1 W/m-K and B with kB = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A; the heating element provides 25000 W/mq′′ = of heat. The surface of the wall at x = 2L is exposed to fluid at Tf,in = 300 K with heat transfer coefficient inh = 100 W/m2-K.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ =

insulated

material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

Figure P1.2-11(a): Composite wall with a heater.

You may neglect radiation and contact resistance for parts (a) through (c) of this problem. a.) Draw a resistance network to represent this problem; clearly indicate what each resistance

represents and calculate the value of each resistance. b.) Use your resistance network from (a) to determine the temperature of the heating element. c.) Sketch the temperature distribution through the wall. Make sure that the sketch is consistent

with your solution from (b).

Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ =

insulated

material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

L = 1 cm

material CkC = 2 W/m-K

Figure P1.2-11(b): Composite wall with Material C.

Neglect radiation and contact resistance for parts (d) through (f) of this problem. d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly

indicate what each resistance represents and calculate the value of each resistance. e.) Use your resistance network from (d) to determine the temperature of the heating element. f.) Sketch the temperature distribution through the wall. Make sure that the sketch is consistent

with your solution from (e).

Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a contact resistance between materials A and B, 20.01 K-m /WcR′′ = , and the surface of the wall at x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient outh = 10 W/m2-K.

L = 1 cm

L = 1 cm

x

25000 W/mq′′ = material AkA = 1 W/m-K

material BkB = 5 W/m-K

,2

300 K100 W/m -K

f in

in

Th

==

L = 1 cm

material CkC = 2 W/m-K

20.01 K-m /WcR′′ =

,2

400 K10 W/m -K

f out

out

Th

==

Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.

Neglect radiation for parts (g) through (i) of this problem. g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly

indicate what each resistance represents and calculate the value of each resistance. h.) Use your resistance network from (g) to determine the temperature of the heating element. i.) Sketch the temperature distribution through the wall.

1.2-12 (1-5) You have decided to install a strip heater under the linoleum in your bathroom in order to keep your feet warm on cold winter mornings. Figure P1.2-12 illustrates a cross-section of the bathroom floor. The bathroom is located on the first story of your house and is W = 2.5 m wide x L = 2.5 m long. The linoleum thickness is thL = 5.0 mm and has conductivity kL = 0.05 W/m-K. The strip heater under the linoleum is negligibly thin. Beneath the heater is a piece of plywood with thickness thP = 5 mm and conductivity kP = 0.4 W/m-K. The plywood is supported by ths = 6.0 cm thick studs that are Ws = 4.0 cm wide with thermal conductivity ks = 0.4 W/m-K. The center-to-center distance between studs is ps = 25.0 cm. Between each stud are pockets of air that can be considered to be stagnant with conductivity ka = 0.025 W/m-K. A sheet of drywall is nailed to the bottom of the studs. The thickness of the drywall is thd = 9.0 mm and the conductivity of drywall is kd = 0.1 W/m-K. The air above in the bathroom is at Tair,1 = 15°C while the air in the basement is at Tair,2 = 5°C. The heat transfer coefficient on both sides of the floor is h = 15 W/m2-K. You may neglect radiation and contact resistance for this problem.

strip heater

linoleum, kL = 0.05 W/m-K

thL = 5 mmthp = 5 mm

plywood, kp = 0.4 W/m-K

ths = 6 cm

Ws = 4 cm

studs, ks = 0.4 W/m-K

ps = 25 cm

thd = 9 mm

drywall, kd = 0.1 W/m-Kair, ka = 0.025 W/m-K

2,1 15 C, 15 W/m -KairT h= ° =

2,2 5 C, 15 W/m -KairT h= ° =

Figure P1.2-12: Bathroom floor with heater.

a.) Draw a thermal resistance network that can be used to represent this situation. Be sure to

label the temperatures of the air above and below the floor (Tair,1 and Tair,2), the temperature at the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat input to the strip heater ( hq ) on your diagram.

b.) Compute the value of each of the resistances from part (a). c.) How much heat must be added by the heater to raise the temperature of the floor to a

comfortable 20°C? d.) What physical quantities are most important to your analysis? What physical quantities are

unimportant to your analysis? e.) Discuss at least one technique that could be used to substantially reduce the amount of heater

power required while still maintaining the floor at 20°C. Note that you have no control over Tair,1 or h .

1.2-15 (1-6 in text) You are a fan of ice fishing but don't enjoy the process of augering out your fishing hole in the ice. Therefore, you want to build a device, the super ice-auger, that melts a hole in the ice. The device is shown in Figure P1.2-15.

D = 10 inchthins = 0.5 inch

thp = 0.75 inch

insulation,kins = 2.2 W/m-K

plate,kp = 10 W/m-K

ε = 0.9

250 W/m -K5 C

hT∞

== °

thice = 5 inchρice = 920 kg/m3

Δifus = 333.6 kJ/kg

heater, activated withV = 12 V and I = 150 A

Figure P1.2-15: The super ice-auger.

A heater is attached to the back of a D = 10 inch plate and electrically activated by your truck

battery, which is capable of providing V = 12 V and I = 150 A. The plate is thp = 0.75 inch thick and has conductivity kp = 10 W/m-K. The back of the heater is insulated; the thickness of the insulation is thins = 0.5 inch and the insulation has conductivity kins = 2.2 W/m-K. The surface of the insulation experiences convection with surrounding air at T∞ = 5°C and radiation with surroundings also at T∞ = 5°C. The emissivity of the surface of the insulation is ε = 0.9 and the heat transfer coefficient between the surface and the air is h = 50 W/m2-K. The super ice-auger is placed on the ice and activated, resulting in a heat transfer to the plate-ice interface that melts the ice. Assume that the water under the ice is at Tice = 0°C so that no heat is conducted away from the plate-ice interface; all of the energy transferred to the plate-ice interface goes into melting the ice. The thickness of the ice is thice = 5 inch and the ice has density ρice = 920 kg/m3. The latent heat of fusion for the ice is Δifus = 333.6 kJ/kg. a.) Determine the heat transfer rate to the plate-ice interface. b.) How long will it take to melt a hole in the ice? c.) What is the efficiency of the melting process? d.) If your battery is rated at 100 amp-hr at 12 V then what fraction of the battery's charge is

depleted by running the super ice-auger.

1.3-3 (1-7 in text) One of the engineers that you supervise has been asked to simulate the heat transfer problem shown in Figure P1.3-3(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 < x < L) has conductivity kA and experiences a uniform rate of volumetric thermal energy generation, g′′′ . The left side of material A (at x = 0) is completely insulated. Material B (from L < x < 2L) has lower conductivity, kB < kA. The right side of material B (at x= 2L) experiences convection with fluid at room temperature (20°C). Based on the facts above, critically examine the solution that has been provided to you by the engineer and is shown in Figure P1.3-3(b). There should be a few characteristics of the solution that do not agree with your knowledge of heat transfer; list as many of these characteristics as you can identify and provide a clear reason why you think the engineer’s solution must be wrong.

L L

material A material B

kA kB < kA

xAg g′′′ ′′′= 0Bg′′′ =

, 20 Cfh T = °

-100

-50

0

50

100

150

200

250

Position (m)

Tem

pera

ture

(°C

)

0 L 2L

Material A Material B

(a) (b)

Figure P1.3-3: (a) Heat transfer problem and (b) "solution" provided by the engineer.

1.3-8 (1-8 in text) Freshly cut hay is not really dead; chemical reactions continue in the plant cells and therefore a small amount of heat is released within the hay bale. This is an example of the conversion of chemical to thermal energy. The amount of thermal energy generation within a hay bale depends on the moisture content of the hay when it is baled. Baled hay can become a fire hazard if the rate of volumetric generation is sufficiently high and the hay bale sufficiently large so that the interior temperature of the bale reaches 170°F, the temperature at which self-ignition can occur. Here, we will model a round hay bale that is wrapped in plastic to protect it from the rain. You may assume that the bale is at steady state and is sufficiently long that it can be treated as a one-dimensional, radial conduction problem. The radius of the hay bale is Rbale = 5 ft and the bale is wrapped in plastic that is tp = 0.045 inch thick with conductivity kp = 0.15 W/m-K. The bale is surrounded by air at T∞ = 20°C with h = 10 W/m2-K. You may neglect radiation. The conductivity of the hay is k = 0.04 W/m-K. a.) If the volumetric rate of thermal energy generation is constant and equal to g′′′ = 2 W/m3

then determine the maximum temperature in the hay bale. b.) Prepare a plot showing the maximum temperature in the hay bale as a function of the hay

bale radius. How large can the hay bale be before there is a problem with self-ignition? Prepare a model that can consider temperature-dependent volumetric generation. Increasing temperature tends to increase the rate of chemical reaction and therefore increases the rate of generation of thermal energy according to: g a bT′′′ = + where a = -1 W/m3 and b = 0.01 W/m3-K and T is in K. c.) Enter the governing equation into Maple and obtain the general solution (i.e., a solution that

includes two constants). d.) Use the boundary conditions to obtain values for the two constants in your general solution.

(hint: one of the two constants must be zero in order to keep the temperature at the center of the hay bale finite). You should obtain a symbolic expression for the boundary condition in Maple that can be evaluated in EES.

e.) Overlay on your plot from part (b) a plot of the maximum temperature in the hay bale as a function of bale radius when the volumetric generation is a function of temperature.

1.3-9 (1-9 in text) Figure P1.3-9 illustrates a simple mass flow meter for use in an industrial refinery.

rin = 0.75 inch

rout = 1 inch

L = 3 inch thins = 0.25

7 3test section

1x10 W/m10 W/m-K

gk

′′′ ==

insulationkins = 1.5 W/m-K2

20 C20 W/m -Kout

Th

∞ = °=

0.75kg/s18 Cf

mT

== °

Figure P1.3-9: A simple mass flow meter.

A flow of liquid passes through a test section consisting of an L = 3 inch section of pipe with inner and outer radii, rin = 0.75 inch and rout = 1.0 inch, respectively. The test section is uniformly heated by electrical dissipation at a rate 7 31x10 W/mg′′′ = and has conductivity k = 10 W/m-K. The pipe is surrounded with insulation that is thins = 0.25 inch thick and has conductivity kins = 1.5 W/m-K. The external surface of the insulation experiences convection with air at T∞ = 20°C. The heat transfer coefficient on the external surface is outh = 20 W/m2-K. A thermocouple is embedded at the center of the pipe wall. By measuring the temperature of the thermocouple, it is possible to infer the mass flow rate of fluid because the heat transfer coefficient on the inner surface of the pipe ( inh ) is strongly related to mass flow rate ( m ). Testing has shown that the heat transfer coefficient and mass flow rate are related according to:

[ ]

0.8

1 kg/sinmh C

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

where C= 2500 W/m2-K. Under nominal conditions, the mass flow rate through the meter is m = 0.75 kg/s and the fluid temperature is Tf = 18°C. Assume that the ends of the test section are insulated so that the problem is 1-D. Neglect radiation and assume that the problem is steady-state. a.) Develop an analytical model in EES that can predict the temperature distribution in the test

section. Plot the temperature as a function of radial position for the nominal conditions. b.) Using your model, develop a calibration curve for the meter; that is, prepare a plot of the

mass flow rate as a function of the measured temperature at the mid-point of the pipe. The range of the instrument is 0.2 kg/s to 2.0 kg/s.

The meter must be robust to changes in the fluid temperature. That is, the calibration curve developed in (b) must continue to be valid even as the fluid temperature changes by as much as 10°C. c.) Overlay on your plot from (b) the mass flow rate as a function of the measured temperature

for Tf = 8°C and Tf = 28°C. Is your meter robust to changes in Tf? In order to improve the meters ability to operate over a range of fluid temperature, a temperature sensor is installed in the fluid in order to measure Tf during operation.

d.) Using your model, develop a calibration curve for the meter in terms of the mass flow rate as a function of ΔT, the difference between the measured temperatures at the mid-point of the pipe wall and the fluid.

e.) Overlay on your plot from (d) the mass flow rate as a function of the difference between the measured temperatures at the mid-point of the pipe wall and the fluid if the fluid temperature is Tf = 8°C and Tf = 28°C. Is the meter robust to changes in Tf?

f.) If you can measure the temperature difference to within δΔT = 1 K then what is the uncertainty in the mass flow rate measurement? (Use your plot from part (d) to answer this question.)

You can use the built-in uncertainty propagation feature in EES to assess uncertainty automatically. g.) Set the temperature difference to the value you calculated at the nominal conditions and

allow EES to calculate the associated mass flow rate. Now, select Uncertainty Propagation from the Calculate menu and specify that the mass flow rate is the calculated variable while the temperature difference is the measured variable. Set the uncertainty in the temperature difference to 1 K and verify that EES obtains an answer that is approximately consistent with part (f).

h.) The nice thing about using EES to determine the uncertainty is that it becomes easy to assess the impact of multiple sources of uncertainty. In addition to the uncertainty δΔT, the constant C has an uncertainty of δC = 5% and the conductivity of the material is only known to within δk = 3%. Use EES' built-in uncertainty propagation to assess the resulting uncertainty in the mass flow rate measurement. Which source of uncertainty is the most important?

i.) The meter must be used in areas where the ambient temperature and heat transfer coefficient may vary substantially. Prepare a plot showing the mass flow rate predicted by your model for ΔT = 50 K as a function of T∞ for various values of outh . If the operating range of your meter must include -5°C < T∞ < 35°C then use your plot to determine the range of outh that can be tolerated without substantial loss of accuracy.

1.4-2 (1-10 in text) Reconsider the mass flow meter that was investigated in Problem 1.3-9 (1-9 in text). The conductivity of the material that is used to make the test section is not actually constant, as was assumed in Problem 1-9, but rather depends on temperature according to:

[ ]( )2

W W10 0.035 300 Km-K m-K

k T⎡ ⎤= + −⎢ ⎥⎣ ⎦

a.) Develop a numerical model of the mass flow meter using EES. Plot the temperature as a

function of radial position for the conditions shown in Figure P1.3-9 (Figure P1-9 in the text) with the temperature-dependent conductivity.

b.) Verify that your numerical solution limits to the analytical solution from Problem 1.3-9 (1-9 in the text) in the limit that the conductivity is constant.

c.) What effect does the temperature dependent conductivity have on the calibration curve that you generated in part (d) of Problem 1.3-9 (1-9)?

Section 1.5: Numerical Solutions to Steady-State 1-D Conduction Problems using MATLAB 1.5-1 (1-11 in text) Reconsider Problem 1.3-8, but obtain a solution numerically using MATLAB. The

description of the hay bale is provided in Problem 1.3-8. Prepare a model that can consider the effect of temperature on the volumetric generation. Increasing temperature tends to increase the rate of reaction and therefore increase the rate of generation of thermal energy; the volumetric rate of generation can be approximated by: g a bT′′′ = + where a = -1 W/m3 and b = 0.01 W/m3-K. Note that at T = 300 K, the generation is 2 W/m3 but that the generation increases with temperature. a.) Prepare a numerical model of the hay bale using EES. Plot the temperature as a function of

position within the hay bale. b.) Show that your model has numerically converged; that is, show some aspect of your solution

as a function of the number of nodes and discuss an appropriate number of nodes to use. c.) Verify your numerical model by comparing your answer to an analytical solution in some,

appropriate limit. The result of this step should be a plot that shows the temperature as a function of radius predicted by both your numerical solution and the analytical solution and demonstrates that they agree.

d.) Prepare a numerical model of the hay bale using MATLAB. Plot the temperature as a function of position within the hay bale.

Section 1.6: Analytical Solutions for Constant Cross-Section Extended Surfaces 1.6-1 (1-13 in text) A resistance temperature detector (RTD) utilizes a material that has a resistivity that

is a strong function of temperature. The temperature of the RTD is inferred by measuring its electrical resistance. Figure P1.6-1 shows an RTD that is mounted at the end of a metal rod and inserted into a pipe in order to measure the temperature of a flowing liquid. The RTD is monitored by passing a known current through it and measuring the voltage across it. This process results in a constant amount of ohmic heating that may tend to cause the RTD temperature to rise relative to the temperature of the surrounding liquid; this effect is referred to as a self-heating error. Also, conduction from the wall of the pipe to the temperature sensor through the metal rod can result in a temperature difference between the RTD and the liquid; this effect is referred to as a mounting error.

pipe

RTD

L = 5.0 cm 20°CwT =

xD = 0.5 mm

k = 10 W/m-K2.5 mWshq =5.0 CT∞ = °

2150 W/m -Kh =

Figure P1.6-1: Temperature sensor mounted in a flowing liquid.

The thermal energy generation associated with ohmic heating is shq = 2.5 mW. All of this ohmic heating is assumed to be transferred from the RTD into the end of the rod at x = L. The rod has a thermal conductivity k = 10 W/m-K, diameter D = 0.5 mm, and length L = 5 cm. The end of the rod that is connected to the pipe wall (at x = 0) is maintained at a temperature of Tw = 20°C. The liquid is at a uniform temperature, T∞ = 5°C and the heat transfer coefficient between the liquid and the rod is h = 150 W/m2-K. a.) Is it appropriate to treat the rod as an extended surface (i.e., can we assume that the

temperature in the rod is a function only of x)? Justify your answer. b.) Develop an analytical model of the rod that will predict the temperature distribution in the

rod and therefore the error in the temperature measurement; this error is the difference between the temperature at the tip of the rod and the liquid.

c.) Prepare a plot of the temperature as a function of position and compute the temperature error. d.) Investigate the effect of thermal conductivity on the temperature measurement error. Identify

the optimal thermal conductivity and explain why an optimal thermal conductivity exists.

Section 1.6: Analytical Solutions for Constant Cross-Section Extended Surfaces 1.6-1 (1-13 in text) A resistance temperature detector (RTD) utilizes a material that has a resistivity that

is a strong function of temperature. The temperature of the RTD is inferred by measuring its electrical resistance. Figure P1.6-1 shows an RTD that is mounted at the end of a metal rod and inserted into a pipe in order to measure the temperature of a flowing liquid. The RTD is monitored by passing a known current through it and measuring the voltage across it. This process results in a constant amount of ohmic heating that may tend to cause the RTD temperature to rise relative to the temperature of the surrounding liquid; this effect is referred to as a self-heating error. Also, conduction from the wall of the pipe to the temperature sensor through the metal rod can also result in a temperature difference between the RTD and the liquid; this effect is referred to as a mounting error.

pipe

RTD

L = 5.0 cm 20°CwT =

xD = 0.5 mm

k = 10 W/m-K2.5 mWshq =5.0 CT∞ = °

2150 W/m -Kh =

Figure P1.6-1: Temperature sensor mounted in a flowing liquid.

The thermal energy generation associated with ohmic heating is shq = 2.5 mW. All of this ohmic heating is assumed to be transferred from the RTD into the end of the rod at x = L. The rod has a thermal conductivity k = 10 W/m-K, diameter D = 0.5 mm, and length L = 5 cm. The end of the rod that is connected to the pipe wall (at x = 0) is maintained at a temperature of Tw = 20°C. The liquid is at a uniform temperature, T∞ = 5°C and the heat transfer coefficient between the liquid and the rod is h = 150 W/m2-K. a.) Is it appropriate to treat the rod as an extended surface (i.e., can we assume that the

temperature in the rod is a function only of x)? Justify your answer. b.) Develop an analytical model of the rod that will predict the temperature distribution in the

rod and therefore the error in the temperature measurement; this error is the difference between the temperature at the tip of the rod and the liquid. You may find it easiest to use Maple for this process.

c.) Prepare a plot of the temperature as a function of position and compute the temperature error. d.) Investigate the effect of thermal conductivity on the temperature measurement error. Identify

the optimal thermal conductivity and explain why an optimal thermal conductivity exists.

1.6-2 (1-14 in text) Your company has developed a micro-end milling process that allows you to easily fabricate an array of very small fins in order to make heat sinks for various types of electrical equipment. The end milling process removes material in order to generate the array of fins. Your initial design is the array of pin fins shown in Figure P1.6-2. You have been asked to optimize the design of the fin array for a particular application where the base temperature is Tbase = 120°C and the air temperature is Tair = 20°C. The heat sink is square; the size of the heat sink is W = 10 cm. The conductivity of the material is k = 70 W/m-K. The distance between the edges of two adjacent fins is a, the diameter of a fin is D, and the length of each fin is L.

D

L

W = 10 cm

array of finsk = 70 W/m-K

120 CbaseT = °

20 C,airT h= °

a

Figure P1.6-2: Pin fin array

Air is forced to flow through the heat sink by a fan. The heat transfer coefficient between the air and the surface of the fins as well as the unfinned region of the base, h , has been measured for the particular fan that you plan to use and can be calculated according to:

[ ] [ ]

0.4 0.3

2

W40 m K 0.005 m 0.01 m

a Dh−

⎛ ⎞ ⎛ ⎞⎡ ⎤= ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠

Mass is not a concern for this heat sink; you are only interested in maximizing the heat transfer rate from the heat sink to the air given the operating temperatures. Therefore, you will want to make the fins as long as possible. However, in order to use the micro-end milling process you cannot allow the fins to be longer than 10x the distance between two adjacent fins. That is, the length of the fins may be computed according to: 10L a= . You must choose the most optimal value of a and D for this application. a.) Prepare a model using EES that can predict the heat transfer coefficient for a given value of a

and D. Use this model to predict the heat transfer rate from the heat sink for a = 0.5 cm and D = 0.75 cm.

b.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the distance between adjacent fins, a, for a fixed value of D = 0.75 cm. Be sure that the fin length is calculated using 10L a= . Your plot should exhibit a maximum value, indicating that there is an optimal value of a.

c.) Prepare a plot that shows the heat transfer rate from the heat sink as a function of the diameter of the fins, D, for a fixed value of a = 0.5 cm. Be sure that the fin length is

calculated using 10L a= . Your plot should exhibit a maximum value, indicating that there is an optimal value of D.

d.) Determine the optimal values of a and D using EES' built-in optimization capability.

1.7-3 (1-15 in text) Figure P1.7-3 illustrates a material processing system.

x

0.75 m/s300 Kin

uT==

D = 5 cm

gap filled with gasth = 0.6 mmkg = 0.03 W/m-K

extruded materialk = 40 W/m-Kα = 0.001 m2/s

oven wall temperature varies with x

Figure P1.7-3: Material processing system.

Material is extruded and enters the oven at Tin = 300 K with velocity u = 0.75 m/s. The material has diameter D = 5 cm. The conductivity of the material is k = 40 W/m-K and the thermal diffusivity is α = 0.001 m2/s. In order to precisely control the temperature of the material, the oven wall is placed very close to the outer diameter of the extruded material and the oven wall temperature distribution is carefully controlled. The gap between the oven wall and the material is th = 0.6 mm and the oven-to-material gap is filled with gas that has conductivity kg = 0.03 W/m-K. Radiation can be neglected in favor of convection through the gas from the oven wall to the material. For this situation, the heat flux experienced by the material surface can be approximately modeled according to:

( )gconv w

kq T T

th′′ ≈ −

where Tw and T are the oven wall and material temperatures at that position, respectively. The oven wall temperature varies with position x according to:

( ),0 expw f f wc

xT T T TL

⎛ ⎞= − − −⎜ ⎟

⎝ ⎠

where Tw,0 is the temperature of the wall at the inlet (at x = 0), Tf = 1000 K is the temperature of the wall far from the inlet, and Lc is a characteristic length that dictates how quickly the oven wall temperature approaches Tf. Initially, assume that Tw,0 = 500 K, Tf = 1000 K, and Lc = 1 m. Assume that the oven can be approximated as being infinitely long. a.) Is an extended surface model appropriate for this problem? b.) Assume that your answer to (a) was yes. Develop an analytical solution that can be used to

predict the temperature of the material as a function of x. c.) Plot the temperature of the material and the temperature of the wall as a function of position

for 0 < x < 20 m. Plot the temperature gradient experienced by the material as a function of position for 0 < x < 20 m.

The parameter Lc can be controlled in order to control the maximum temperature gradient and therefore the thermal stress experienced by the material as it moves through the oven. d.) Prepare a plot showing the maximum temperature gradient as a function of Lc. Overlay on

your plot the distance required to heat the material to Tp = 800 K (Lp). If the maximum temperature gradient that is allowed is 60 K/m, then what is the appropriate value of Lc and the corresponding value of Lp?

1.7-4 (1-16 in text) The receiver tube of a concentrating solar collector is shown in Figure P1.7-4.

r = 5 cmth = 2.5 mm

280 C100 W/m -K

w

w

Th

= °=

sq′′

φ

225 C25 W/m -K

a

a

Th= °=

k = 10 W/m-K

Figure P1.7-4: A solar collector

The receiver tube is exposed to solar radiation that has been reflected from a concentrating mirror. The heat flux received by the tube is related to the position of the sun and the geometry and efficiency of the concentrating mirrors. For this problem, you may assume that all of the radiation heat flux is absorbed by the collector and neglect the radiation emitted by the collector to its surroundings. The flux received at the collector surface ( sq′′ ) is not circumferentially uniform but rather varies with angular position; the flux is uniform along the top of the collector, π < φ < 2π rad, and varies sinusoidally along the bottom, 0 < φ < π rad, with a peak at φ = π/2 rad.

( ) ( ) ( )+ sin for 0

for 2t p t

s

t

q q qq

q

φ φ πφ

π φ π

⎧ ′′ ′′ ′′− < <⎪′′ = ⎨′′ < <⎪⎩

where tq′′ = 1000 W/m2 is the uniform heat flux along the top of the collector tube and pq′′ = 5000 W/m2 is the peak heat flux along the bottom. The receiver tube has an inner radius of r = 5.0 cm and thickness of th = 2.5 mm (because th/r << 1 it is possible to ignore the small difference in convection area on the inner and outer surfaces of the tube). The thermal conductivity of the tube material is k = 10 W/m-K. The solar collector is used to heat water, which is at Tw = 80°C at the axial position of interest. The average heat transfer coefficient between the water and the internal surface of the collector is wh = 100 W/m2-K. The external surface of the collector is exposed to air at Ta = 25°C. The average heat transfer coefficient between the air and the external surface of the collector is ah = 25 W/m2-K. a.) Can the collector be treated as an extended surface for this problem (i.e., can the temperature

gradients in the radial direction in the collector material be neglected)? b.) Develop an analytical model that will allow the temperature distribution in the collector wall

to be determined as a function of circumferential position.

Section 1.8: Analytical Solutions for Non-Constant Cross-Section Extended Surfaces 1.8-1 (1-17 in text) Figure P1.8-1 illustrates a disk brake for a rotating machine. The temperature

distribution within the brake can be assumed to be a function of radius only. The brake is divided into two regions. In the outer region, from Rp = 3.0 cm to Rd = 4.0 cm, the stationary brake pads create frictional heating and the disk is not exposed to convection. The clamping pressure applied to the pads is P = 1.0 MPa and the coefficient of friction between the pad and the disk is μ = 0.15. You may assume that the pads are not conductive and therefore all of the frictional heating is conducted into the disk. The disk rotates at N = 3600 rev/min and is b = 5.0 mm thick. The conductivity of the disk is k = 75 W/m-K and you may assume that the outer rim of the disk is adiabatic.

center line

clamping pressureP = 1 MPa

Rp = 3 cmRd = 4 cm

b = 5 mm 30 C,aT h= °

k = 75 W/m-K

coefficient of friction, μ = 0.15stationarybrake pads

disk, rotates at N = 3600 rev/min Figure P1.8-1: Disk brake.

The inner region of the disk, from 0 to Rp, is exposed to air at Ta = 30°C. The heat transfer coefficient between the air and disk surface depends on the angular velocity of the disk, ω, according to:

[ ]

1.25

2 2

W W20 1500m -K m -K 100 rad/s

h ω⎛ ⎞⎡ ⎤ ⎡ ⎤= + ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠

a.) Develop an analytical model of the temperature distribution in the disk brake; prepare a plot

of the temperature as a function of radius for r = 0 to r = Rd. b.) If the disk material can withstand a maximum safe operating temperature of 750°C then what

is the maximum allowable clamping pressure that can be applied? Plot the temperature distribution in the disk at this clamping pressure. What is the braking torque that results?

c.) Assume that you can control the clamping pressure so that as the machine slows down the maximum temperature is always kept at the maximum allowable temperature, 750°C. Plot the torque as a function of rotational speed for 100 rev/min to 3600 rev/min.

1.8-5 (1-18 in text) Figure P1.8-5 illustrates a fin that is to be used in the evaporator of a space conditioning system for a space-craft.

L = 2 cm

th = 1 mm

Wb = 1 cm

k = 50 W/m-Kρ= 3000 kg/m3

10 CbT = °

2120 W/m -K20 C

hT∞

== °

thb = 2 mm thg = 2 mm

ρb = 8000 kg/m3

x

Figure P1.8-5: Fin on an evaporator.

The fin is a plate with a triangular shape. The thickness of the plate is th = 1 mm and the width

of the fin at the base is Wb = 1 cm. The length of the fin is L = 2 cm. The fin material has conductivity k = 50 W/m-K. The average heat transfer coefficient between the fin surface and the air in the space-craft is h = 120 W/m2-K. The air is at T∞ = 20°C and the base of the fin is at Tb = 10°C. Assume that the temperature distribution in the fin is 1-D in x. Neglect convection from the edges of the fin. a.) Obtain an analytical solution for the temperature distribution in the fin. Plot the temperature

as a function of position. b.) Calculate the rate of heat transfer to the fin. c.) Determine the fin efficiency.

The fin has density ρ = 3000 kg/m3 and the fin is installed on a base material with thickness thb = 2 mm and density ρb = 8000 kg/m3. The half-width of the gap between adjacent fins is thg = 2 mm. Therefore, the volume of the base material associated with each fin is thb Wb (th + 2 thg). d.) Determine the ratio of the absolute value of the rate of heat transfer to the fin to the total

mass of material (fin and base material associated with the fin). e.) Prepare a contour plot that shows the ratio of the heat transfer to the fin to the total mass of

material as a function of the length of the fin (L) and the fin thickness (th). f.) What is the optimal value of L and th that maximizes the absolute value of the fin heat

transfer rate to the mass of material?

1.9-3 (1-19 in text) A fiber optic bundle (FOB) is shown in Figure P1.9-3 and used to transmit the light for a building application.

5 21x10 W/mq′′ =

25 W/m -K20 C

hT∞

== °

x

rout = 2 cm

fiber optic bundle Figure P1.9-3: Fiber optic bundle used to transmit light.

The fiber optic bundle is composed of several, small diameter fibers that are each coated with a thin layer of polymer cladding and packed in approximately a hexagonal close-packed array. The porosity of the FOB is the ratio of the open area of the FOB face to its total area. The porosity of the FOB face is an important characteristic because any radiation that does not fall directly upon the fibers will not be transmitted and instead contributes to a thermal load on the FOB. The fibers are designed so that any radiation that strikes the face of a fiber is “trapped” by total internal reflection. However, radiation that strikes the interstitial areas between the fibers will instead be absorbed in the cladding very close to the FOB face. The volumetric generation of thermal energy associated with this radiation can be represented by:

expch ch

q xgL Lφ ⎛ ⎞′′

′′′= −⎜ ⎟⎝ ⎠

where q′′= 1x105 W/m2 is the energy flux incident on the face, φ = 0.05 is the porosity of the FOB, x is the distance from the face, and Lch = 0.025 m is the characteristic length for absorption of the energy. The outer radius of the FOB is rout = 2 cm. The face of the FOB as well as its outer surface are exposed to air at T∞ = 20°C with heat transfer coefficient h = 5 W/m2-K. The FOB is a composite structure and therefore conduction through the FOB is a complicated problem involving conduction through several different media. Section 2.9 discusses methods for computing the effective thermal conductivity for a composite. The effective thermal conductivity of the FOB in the radial direction is keff,r = 2.7 W/m-K. In order to control the temperature of the FOB near the face, where the volumetric generation of thermal energy is largest, it has been suggested that high conductivity filler material be inserted in the interstitial regions between the fibers. The result of the filler material is that the effective conductivity of the FOB in the axial direction varies with position according to:

, , , , expeff x eff x eff xk

xk k kL∞

⎛ ⎞= + Δ −⎜ ⎟

⎝ ⎠

where keff,x,∞ = 2.0 W/m-K is the effective conductivity of the FOB in the x-direction without filler material, Δkeff,x = 28 W/m-K is the augmentation of the conductivity near the face, and Lk = 0.05 m is the characteristic length over which the effect of the filler material decays. The length of the FOB is effectively infinite. a.) Is it appropriate to use a 1-D model of the FOB? b.) Assume that your answer to (a) was yes. Develop a numerical model of the FOB.

c.) Overlay on a single plot the temperature distribution within the FOB for the case where the filler material is present (Δkeff,x = 28 W/m-K) and the case where no filler material is present (Δkeff,x = 0).