Solutions manual for heat and mass transfer 2nd edition by kurt rolle

SOLUTIONS MANUAL for Heat and Mass Transfer 2nd Edition by Kurt Rolle Full download: http://downloadlink.org/p/solutions-manual-for-heatand-mass-transfer-2nd-edition-by-kurt-rolle/ Heat and Mass Transfer Solutions Manual Second Edition This solutions manual sets down the answers and solutions for the Discussion Questions, Class Quiz Questions, and Practice Problems. There will likely be variations of answers to the discussion questions as well as the class quiz questions. For the practice problems there will likely be some divergence of solutions, depending on the interpretation of the processes, material behaviors, and rigor in the mathematics. It is the author’s responsibility to provide accurate and clear answers. If you find errors please let the author know of them at rolle@uwplatt.edu.

Chapter 2 DiscussionQuestions Section2-1 1. Describe the physical significance of thermal conductivity.

Thermal conductivity is a parameter or coefficient used to quantitatively describe the amount of conduction heat transfer occurring across a unit area of a bounding surface, driven by a temperature gradient.

2. Why is thermal conductivity affected by temperature?

Conduction heat transfer seems to be the mechanism of energy transfer between adjacent molecules or atoms and the effectiveness of these transfers is strongly dependent on the temperatures. Thus, to quantify conduction heat transfer with thermal conductivity means that thermal conductivity is strongly affected by temperature.

3. Why is thermal conductivity not affected to a significant extent by material density?

Thermal conductivity seems to not be strongly dependent on the material density since thermal conductivity is an index of heat or energy transfer between adjacent molecules and while the distance separating these molecules is dependent on density, it is not strongly so.

Section2-2 4. Why is heat of vaporization, heat of fusion, and heat of sublimation accounted as energy generation in the usual derivation of energy balance equations?

Heats of vaporization, fusion, and sublimation are energy measures accounting for phase changes and not directly to temperature or pressure changes. It is

convenient, therefore, to account these phase change energies as lumped terms, or energy generation.

Section2-3 5. Why are heat transfers and electrical conduction similar?

Heat transfer and electrical conduction both are viewed as exchanges of energy between adjacent moles or atoms, so that they are similar.

6. Describe the difference among thermal resistance, thermal conductivity, thermal resistivity and R-Values.

Thermal Resistance is the distance over which conduction heat transfer occurs times the inverse of the area across which conduction occurs and the thermal conductivity, and thermal resistivity is the distance over which conduction occurs times the inverse of the thermal conductivity. The R-Value is the same as thermal resistivity, with the stipulation that in countries using the English unit system, 1 R-Value is 1 hr∙ft2 ∙0F per Btu. Section2-4 7. Why do solutions for temperature distributions in heat conduction problems need to converge?

Converge is a mathematical term used to describe the situation where an answer approaches a unique, particular value.

8. Why is the conduction in a fin not able to be determined for the case where the base temperature is constant, as in Figure 2-9?

The fin is an extension of a surface and at the edges where the fin surface coincides with the base, it is possible that two different temperatures can be ascribed at the intersection, which means there is no way to determine precisely what that temperature is. Conduction heat transfer can then not be completely determined at the base.

9. What is meant by an isotherm?

An isotherm is a line or surface of constant or the same temperature.

10. What is meant by a heat flow line?

A heat flow line is a path of conduction heat transfer. Conduction cannot cross a heat flow line.

Section2-5 11. What is a shape factor?

The shape factor is an approximate, or exact, incorporating the area, heat flow paths, isotherms, and any geometric shapes that can be used to quantify conduction heat flow between two isothermal surfaces through a heat conducting media. The product of the shape factor, thermal conductivity, and temperature difference of the two surfaces predicts the heat flow.

12. Why should isotherms and heat flow lines be orthogonal or perpendicular to each other?

Heat flow occurs because of a temperature difference and isotherms have no temperature difference. Thus heat cannot flow along isotherms, but must be perpendicular or orthogonal to isotherms.

Section2-6 13. Can you identify a physical situation when the partial derivatives from the left and right are not the same?

Often at a boundary between two different conduction materials the left and the right gradients could be different. Another situation could be if radiation or convection heat transfer occurs at a boundary and then again the left and right gradients or derivatives could be different.

Section2-7 14. Can you explain when fins may not be advantageous in increasing the heat transfer at a surface?

Fins may not be a good solution to situations where a highly corrosive, extremely turbulent, or fluid having many suspended particles is in contact with the surface.

15. Why should thermal contact resistance be of concern to engineers?

Thermal contact resistance inhibits good heat transfer, can mean a significant change in temperature at a surface of conduction heat transfer, and can provide a surface for potential corrosion.

ClassQuizQuestions 1. What is the purpose of the negative sign in Fourier’s law of conduction heat transfer?

The negative sign provides for assigning a positive heat transfer for negative temperature gradients.

2. If a particular 8 inch thick material has a thermal conductivity of 10 Btu/ hr∙ft∙0F, what is its R-value?

The R-value is the thickness times the inverse thermal conductivity; 0

R Value thicks / 8in / (12 in/ ft)(10 Bu/ hr ft 0 F) 0.0833 hr ft F / Btu 3. What is the thermal resistance of a 10 m2 insulation board, 30 cm thick, and having thermal conductivity of 0.03 W/m∙K? The thermal resistance is

x / A 0.3m / 10m 2 0.03W / m K 1.0K / W 4. What is the difference between heat conduction in series and in parallel between two materials?

The thermal resistance, or thermal resistivity are additive for series. In parallel the thermal resistance needs to be determined with the relationship

Req

R R /R 1

2

R 1

2

5. Write the conduction equation for radial heat flow of heat through a tube that has inside diameter of Di and outside diameter of D0. i

i

T ln D0 Di

Q2 L

6. Write the Laplace equation for two-dimensional conduction heat transfer through a homogeneous, isotropic material that has constant thermal conductivity.

2

T (x, y) x2

2

T (x, y) 0 y2

7. Estimate the heat transfer from an object at 1000F to a surface at 400F through a heat conducting media having conductivity Q ST B uthermal h F F of 5BBtu/hr∙ft∙ u h 0F if the shape factor is 1.0 ft. i 8 Sketch ve sotherms and appropr ate heat ow nes or heat trans er per 600 1.0 ft 5 t / r ft 0 300 t / r through a 5 cm x 5 cm square where the heat ow s rom a h gh temperat and another sotherma as the s de o the square . fi i i fl li f f unit depth fl i f i ure corner i l i f .

9

the therma contact res stance between a c utch sur ace and a dr v ng su 0 0023 m2 0C/W est mate the temperature drop across the contact ng su un t area when 200 W/m2 o heat s des red to be d ss pated . If l idrop s l f i i rface is The temperature . , i i rfaces, per i f i i i i . i 10 Wou d you expect the w re temperature to be greater or ess or a numbe i w re as compared to a number 142 copper w2re both conduct ng the same 0 0 T R 200W / m 0.0023m C / W 0.46 C current?

TCR

Q

.

l i

i

l

20

i ,

f i

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r 18 copper electrical

i

.

,

i

li

,

li l

i l

i , in whole or in part.

A number 18 copper wire has a smaller diameter and a greater electrical resistance per unit length. Therefore the number 18 wire would be expected to have a higher temperature than the number 14 wire.

PracticeProblems Section2-1 1. Compare the value for thermal conductivity of Helium at 200C using Equation 2-3 and the value from Appendix Table B-4.

Solution Using Equation 2-3 for helium

0.8762x104

T 0.0015W / c m K 0.15W / m K

From Appendix Table B-4 0.152W / m K

2. Predict the thermal conductivity for neon gas at 2000F. Use a value of 3.9 Ǻ for the collision diameter for neon.

Solution Assuming neon behaves as an ideal gas, with MW of 20, converting 2000F to 367K, and using Equation 2-1

8.328x104

367K 18.05x10 4 W / c m K 20 3.9

T 8.328x104 MW

3. Show that thermal conductivity is proportional to temperature to the 1/6-th power for a liquid according to Bridgeman’s equation (2-6).

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Solution

3.865x1023 V x 2

s m ⁄ From Bridgeman’s equation Also, Vs (sonic velocity) ~ -2/3 -1/2 / ⁄ ~ρ so ~ρ the mean separation distance between molecules =

∼

2/ 31/ 2

1/ 6

∼ T 1/ 6

that

4. Predict a value for thermal conductivity of liquid ethyl alcohol at 300 K. Use the equation suggested by Bridgman’s equation (2-6).

Solution ⁄ , which for ethyl Bridgeman’s equation (2-6) uses the sonic velocity in the liquid, 5 alcohol at 300 K is nearly 1.14 x 10 cm/s from Table 2-2. The equation also uses the mean distance between molecules, assuming a uniform cubic arrangement of the ⁄ , mm being the mass of one molecule in grams, the molecules, which is molecular mass divided by Avogadro’s number. Using data from a chemistry handbook the value of xm is nearly 0.459 x 10-7 cm. Using Equation 2-6, 4 3.865x1023 V x 2 s 20.9x10 W / c m K 0.209W / m K m

5. Plot the value for thermal conductivity of copper as a function of temperature as given

by Equation 2-10. Plot the values over a range of temperatures from -400F to 1600F.

Solution Using Equation 2-10 and coefficients from Appendix Table B-8E TO

T T0

227

Btu 0.0061 hr ft 0 R

Btu hr ft 0 R 2

T 4920 R

This can be plotted on a spreadsheet or other modes.

6. Estimate the thermal conductivity of platinum at -1000C if its electrical conductivity is 6 x

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107 mhos/m, based on the Wiedemann-Franz law. Note: 1 mho = 1 amp/volt = 1 coulomb/volt-s, 1 W = 1 J/s = 1 volt-coulomb/s.

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Solution Using the Wiedemann-Franz law, Equation 2-9 gives

Lz T 2.43x108 V 2

K 2 6x107 amp V m 173K 252.2W / m K

7. Calculate the thermal conductivity of carbon bisulfide using Equation 2-6 and compare this result to the listed value in Table 2-2.

Solution Equation 2-6 uses the sonic velocity in the material. This is ⁄ = 1.18 10 = / , where Eb is the bulk modulus. The mean distance between adjacent molecules, assuming a uniform cubic arrangement, is also used. This is = This gives

/ where mm is the mass of one molecule; MW/Avogadro’s number. = 0.466 10 . then 0 23 Vs

3.865x10

Section2-2

m

x

2

0.0021W / cm C

8. Estimate the temperature distribution in a stainless steel rod, 1 inch in diameter that is 1 yard long with 3 inches of one end submerged in water at 400F and the other end held by a person. Assume the person’s skin temperature is 820F, the temperature in the rod is uniform at any point in the rod, and steady state conditions are present.

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Sol ut i on Assuming the heat flow to be axi al and not radi al and al so 400F f or the fi rst 3 i nches of the rod, the temperature di stri buti on between x = 3 i nches and out to x = 36i ncheswe can use Fourier’s law of conduct ion and then for 3 in ≤ x ≤ 36 inches , ident ify ing the . sketched graph s here i slope and x-intercept T (x) 1.2727 x 36.1818 The nciudl ed.

One cou d now pred ct the heat ow ax a y through the rod us ng Four er s aw and us ng a therma conduct l i v ty or sta fl n ess i llstee , i i ’ l i

9

l

i i f

i l

l.

Der ve the genera energy equat on or conduct on heat trans er through a sotrop c med a ni cyf ndr ca coord . homogeneous i l i nates Equat f on 2 19 ,i i i i li i l i , i - .

So ut on l i Re err ng to the cy ndr ca e ement sketch you can app y an energy ba ance Energy f i li i l l , l l , in – Energy Out = Energy Accumulated in the Element. Then, accounting the energies in and out as conduction heat transfer we can write

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i

q

rz

T

r

r

r i

q

rz

r

i

1T r

i

i

qz

r

r

r

T z

2

z

z

i

i

qr

T

r rz

r

rr i

q

rz

rr

y

1T r

y

i

qz z

r

r

zrc 2

r

T x

2

zz

r

r

zz

rgy

T p

t

f l i l . If i terms and the three out terms on the left side of the energy balance and the accumulated energy on the right, divide all terms by + ⁄2 !" ∙ !$ ∙ ! , and take the limits as Δr →0, Δz → 0, and Δθ→ 0 gives, using calculus, Equation 2-19

1

r

T

1

T

T 26

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c

T rr

r

r2

z

p

z

t

an n energy

an n energy

an n energy

an out energ

an out energ

an out ene

The rate o energy accumu ated n the e ement

you put the three energy n

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10. Derive the general energy equation for conduction heat transfer through a homogeneous, isotropic media in spherical coordinates, Equation 2-20.

Solution Referring to the sketch of an element for conduction heat transfer in spherical coordinates, you can balance the energy in – the energy out equal to the energy accumulated in the element. Using Fourier’s law of conduction i

Q

r r sin

r r

T r

r

an in term

i

1

Q

r r sin

T

r

an in term

i

Q

1 T

rr

r sin

an in term

i

Q

r r r r sin

T

rr

r

rr i

Q

1

r sin r r

rr

an out term

T an out term

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i

Q

r T

r

r sin an out term

T Vc p

T r sin r r

t

t

Wh ch s the accumu ated energy nsert ng the three n terms as pos t ve on the e t o the ba ancel nsert ng the i energy i . I three i out terms asi negat ve on the i i e t s de ol fthside baf ance nsert ng l the ,accumu i i ated term on the r ght s de and i d v d ng al f terms i by f te quant ! !" gl ves the o ow ng i l ty, i &"!' i i , i i i ll he %

i 2

r r sin

∙

∙

i

T

f ll 2

r sin

i

T

rr r r sin r T r sin

rr

r

r sin

T

r 2 sin r r T

rr T rsin

r sin 2

r sin r

cp

T t

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Taking the limits as Δr →0, Δθ →0, Δφ → 0 and reducing

1

r2

T

r2 r

1

sin

T

1

r 2 sin

r

T

T

c

p

r 2 sin 2

t which

is Equation 2-20, conservation of energy for conduction heat transfer in spherical coordinates.

11 Determ ne a re at onsh p or the vo ume e ement n spher ca coord nates .

So ut on

i

l i

i f

l

l

i

i l

i

.

l i Re err ng to the sketch or an e ement n spher ca coord nates and gu ded by the concept o a vo ume e ement g ves f i f l i i l i , i V f l l i ,

r sin

r

r

28 © 2016 Cengage Learn ng® May not be scanned cop ed or dup cated or posted to a pub c y access b e webs te n who e o

i

.

,

i

li

,

li l

i l

i ,i

l

r in part.

Section2-3 12. An ice-storage facility uses sawdust as an insulator. If the outside walls are 2 feet thick sawdust and the sideboard thermal conductivity is neglected, determine the R-Value of the walls. Then, if the inside temperature is 250F and the outside is 850F, estimate the heat gain of the storage facility per square foot of outside wall.

Solution Assuming steady state conditions and that the thermal conductivity is the value listed in Appendix Table B-2E,

R Value i

x T

q A

x

2ft 58.8hr ft 2 0 F / Btu 58.8R Value 0 0.034Btu / hr ft F 850 F 250 F T 1.02Btu / hr ft 2 58.8hr ft 2 0 F / Btu R Value

13. The combustion chamber of an internal combustion engine is at 8000C when fuel is burnt in the combustion chamber. If the engine is made of cast iron with an average thickness of 6.4 cm between the combustion chamber and the outside surface, estimate the heat transfer per unit area if the outside surface temperature is 500C and the outside air temperature is 300C. Solution Assuming steady state one-dimensional conduction and using a thermal conductivity that is assumed constant and has a value from Table B-2, i

T

q A

39 x

W mK

0 0 800W 50W K

450kW / m 2

0.065m

14. Triple pane window glass has been used in some building construction. Triple pane glass is a set of three glass panels, each separated by a sealed air gap. Estimate the R-Value for triple pane windows and compare this to the R-Value for single pane glass.

Solution

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Assume the air in the gaps do not move so that they are essentially conducting media.

Then the R-Value is

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R Value 3

x

2

x

air 3

0.002

2

0.006

0.026

1.4

glass

0.4658m2 K / W 2.647hr ft 2 0 F / Btu

The R-Value for a single pane window is

R Value

x

0.002m k

0.1429m 2 K / W 0.008hr ft 2 0 F / Btu 0.008R Value

1.4W / m K

glass

The ratio of the R-Value for the triple pane to the R-Value for a single pane is roughly 324.

15. For the outside wall shown in Figure 2-50, determine the R-Value, the heat transfer through the wall per unit area and the temperature distribution through the wall if the outside surface temperature is 360C and the inside surface temperature is 150C.

Solution The R-Value is the sum of the three materials; pine, plywood, and limestone, with thermal conductivity

x

R Value R

x

x

V pine

0.04 0.02 0.06 k

plywood

lim estone

0.15

0.12

0.462m2 K / W

2.15

values obtained from Appendix Table B-2.The conversion to English units is 0.176 m2K/W = 1 R-Value so that ( − *+,- = 2.62 . The heat transfer per unit area is

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i

q

A

T 36 15 45.45W / m 2 RV 0.462

The temperature distribution is determined by

noting that the heat flow is the same through each material. For the pine,

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i

q

A, pine

45.45 / W m

2

T

0

1

C T 15 1 15 RV . p ine 0.04 / 0.15 so that T at the sur ac f e between 1

the pine and the plywood, is 27.10C. Similarly, to determine the temperature between the plywood and the limestone, again noting that the heat flow is the same as before i

q

45.45

A

gure fi

T T T 2 1 RV p ywood , l

2

27.1 so that T2 s 34 60C Th s s sketch 0.02 / 0.12 i . . i i ed in the

.

16 Determ ne the heat trans er per oot o ength through a copper tube hav d ameter o 2 nches and an ns de d ameter o 1 5 nches The p pe conta . ammon ia and s surrounded f by 80f 0F a rf l ing an outside i f i i i i f . i . i ins 1800F So ut on i i i. l i Assuming steady state and only conduction heat transfer, for a tube cylindrical coordinates is the appropriate means of analysis. Then i

q

2T l

ln D0 Di

2 231.16Btu / hr ft 0 R 180 800 R ln 2in 1.5in

504, 870Btu / hr ft

17. A steam line is insulated with 15 cm of rock wool. The steam line is a 5 cm OD iron pipe with a 5 mm thick wall. Estimate the heat loss through the pipe per meter length if steam at 1200C is in the line and the surrounding temperature is 200C. Also determine the temperature distribution through the pipe and insulation.

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i on Sol ut Assume heat flow is one-dimensiona l radia l and steady state . The heat flow is then th e overall temperature difference divided by the sum o f the radia l therma l res istances . We have i

q

rll ove a l

lnD0 Di

D lnD0 p pe

120 20

T ln54

i

ln355

12.91 / m 2 W

woo

51 p pe and 2 woo 0.04 nsu at on the 2 the temperature 2d str but on lthrough 2 the i To determ ne rad a heat ow w be the same through the ron p pe and the woo nsu at on The i at the nter ace between i i the i ron p pe and the i nsu at onl is determ l i ned temperature i l fl ill i i li l i .

T p peOD i f i i i l i i i by i From th s the nter ace temperature T =119 99 p peOd 2 51 1D 20 i =Twoo The temperature n a homogeneous rad a sect on s = . + 0 & For th / l ln 5 4 0 f i i , . 0C10C ron p pe the two boundary cond t ons 1 T = 120 C @ r = 2 cm and 2 Ti = 119 991 i ve or T(r) and resu ti ng l n two i iseparate equat+ons. So v ne r = 2 5lI cm can be used to so ithese itwo, s mu taneous y g ves that i i T = 120 .) 0280C and C = -0 040 For.)the ron p. pe t @ 0 f the woo nsulati oni the two boundary cond i t .onsl 1i g . . = 120 928 − 0 040 & l For l and i 2 T = 200C @. r = 17 5 cm can. be .subst tuted i nto i thehen = 119 9910Ci @ lr = 2 5 cm l i equat l ions s mu taneous y or T0 and i i C g v.) T equat on to .so ve or .T(r) +So .v ng these two that T.0 = 167 07 and .C = -51 385.)For the woo nsu at. on . = 157 07i − 51i 385 & i l f . l i i i l f on i es The o ow ng sketch nd cates the character o the temperature d strl but . . + . . . . li l i f ll i i i f i i i .

q

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18. Evaporator tubes in a refrigerator are constructed of 1 inch OD aluminum tubing with 1/8 in thick walls. The air surrounding the tubing is at 250F and the refrigerant in the evaporator is at 150F. Estimate the heat transfer to the refrigerant over 1 foot of length.

Solution Assume steady state one-dimensional radial conduction heat transfer and using a thermal conductivity value from Appendix Table B-2E i

Q

2L T ln r r 0

0

2 136.38Btu / hr ft 0 F 1 ft

Ti

ln 0.5in 0.375in

25 15

0

F 29, 786Btu / hr

i

19. Teflon tubing or 4 cm OD and 2.7 cm ID conducts 1.9 W/m when the outside temperature is 800C. Estimate the inside temperature of the tubing. Also predict the thermal resistance per unit length.

Solution Assume steady state one-dimensional radial conduction heat transfer. Reading the thermal conductivity from Appendix Table B-2, applying the Fourier’s Law of conduction i

q for radial heat flow

l

2 T Ti 1.9 W/ m 0 ln r 0 r i

1.9W/m ln 2cm/ 1.35cm 2 0.35W / m 0 C

Ti T0

and solving for Ti

0 80.34 C

and the thermal resistance per unit of length is

RTL

ln r0 ri 2

ln 42.7 0.1787m K / W 2 0.35

20. A spherical flask, 4 m diameter with a 5 mm thick wall, is used to heat grape juice.

During the heating process the outside surface of the flask is 1000C and the inside surface is 800C. Estimate the thermal resistance of the flask, the heat transfer through the flask, if it is assumed that only the bottom half is heated, and the temperature distribution through the flask wall.

Solution

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Assume steady state one-dimensional, radial conduction heat transfer with constant properties. Since only the bottom half is heated you need to recall that a surface area of

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a hemisphere is 2πr2 rather than 4π r2. Then i

21.4W/m K100 800 C

2TT 0

Q

i

7056W

1 1 ri r0

1 1 1.905m

2m

The thermal resistance for the full flask would be

1

1

1 1

1

1

0.001417m K / W

R T

r

r0 4

1.905m

2m

4 1.4W / m K

i

For such a small thermal resistance, the temperature distribution will be nearly constant through the wall. Yet for the bottom half of the flask we can write i

Q

T (r) T i

1 1

2 r

7056W

800 C

1 1

2 1.4 W/ m0 C 1.905m

r

r

i

or

T (r) 80 0 C 401m 0 C 0.525 m 1

1 r

21. A Styrofoam spherical container having a 1 inch thick wall and 2 foot diameter holds dry ice (solid carbon dioxide) at -850F. If the outside temperature is 600F, estimate the heat gain in the container and establish the temperature distribution through the 1 inch wall.

Solution Assuming steady state one-dimensional radial conduction heat transfer and using the thermal conductivity value for Styrofoam from Appendix Table B-2E i

Q

40.017Btu/hrft0F

4 1

T0 Ti

1 ri

r0

12 1

1

1 11 ft

ft

60 85 F

0

340.7Btu / hr

The temperature distribution for T(r) is

T (r) 850 F

340.7Btu/hr

12

12

4 0.017Btu / hr ft 0 F 11 ft

r

850 F 1739.80 F

0 19138W F in

r

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where r is in inches.

22. Determine the overall thermal resistance per unit area for the wall shown in Figure 2-51.

Exclude the thermal resistance due to convection heat transfer in the analysis. Then, if the heat transfer is expected to be 190W/m2 and the exposed brick surface is 100C, estimate the temperature distribution through the wall.

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Solution The overall thermal resistance will be the sum of the thermal resistances of the three components,

0.3m 0.025m 0.1m 1.6W / m K 0.029W / m K

RV

1.192m 2 K / W 0.7W / m K

Since there is expected to be 190W/m2 of conduction heat transfer through each of the three components, the temperatures at the inside surface and the two interface surfaces are

Tinside 19.0W / m 2 1.192m 2 0 C/ W 100 C 32.6 0 C

which is the

inside surface temperature. The temperature between the concrete and the Styrofoam i

T

T

Q

32.6 0 C 19W / m 2 0.1875m 2 0 C/ W 29.0 0 C

R

inside concrete is c styr and the temperature between the Styrofoam and the brick facing is i

Tstyr brick T 0

QR

100 C

brick

19W / m

2

0.143m 2 0 C / W 12.7 0 C

23. Determine the thermal resistance per unit length of the tubing (nylon) shown in Figure 2-52. Then predict the heat transfer through the tubing if the inside ambient temperature is -100C and the outside is 200C.

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Solution The nylon tubing has properties of Teflon, the inside diameter is 40mm, and the outside diameter is 60 mm. Then

ln D0 Di

ln r0 ri

ln 60mm40mm

2

2

2 0.35W / m K

RTL

0.184m K / W

Assuming steady state one-dimensional radial conduction heat transfer, i

q

30 0 C 0.184m 0 C / W

T R

l

162.7W / m TL

24. Determine the heat transfer through the wall of Example 2-5 if the thermal conductivity

9.2 0.007T is affected by temperature through the relationship where T is in degrees Fahrenheit.

Btu inch hr ft

2 0

F

Solution In Example 2-5 the wall is 15 inches thick, has a temperature of 550F on one side and 1000F on the other. Assuming steady state one-dimensional conduction heat transfer i

dT

q A i

9.2 0.007T dx

dT dx separating variables and integrating

i

q

dx

39

Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.2 0.007T Btu dT in9.2 (551100 0 F) Btuin 0.007 5522 100 2 0 F

15in

q

A

hr ft 2 0 F

2

hr ft 2 0 F 2

A

40 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

and then solving for the heat transfer per unit area gives i

1

Btu in Btuin Btu 414 24.4 29.2 15in hr ft 2 hr ft 2 hr ft 2

q A

0.001

25. Determine the temperature distribution through a slab if aT where T is in Kelvin degrees and a is constant. Then compare this to the case where κ = a.

Solution i

dT

q

aT 0.001 dx

A i

q

i A

dx

q

A

dT

dx If the variables are now separated and integrating a x a T 0.001dT T 1.001 C 1.001 defining a

boundary condition of T = T0 @ x = 0 allows the constant C to be defined as

a

C

T 1.001 1.001

0

the temperature distribution is then i

T (x)

1.001

T

1.001 0

i

q

x A

1.001 a

For κ = a and T = T0 @ x = 0

x

T T 0

q

A

a

Section2-4 26. Show that . => ? =@ >

, 6 = * %&7 + 8 9 7

-

:;

+ <- :; satisfies Laplace’s equation

=> ?

+ =;> = 0.

Solution

T ap cos px bp sin px ce py de py Taking first and second derivatives x 2

T 2 2 py de py 2 ap sin px bp cos px ce and x

taking the first and second

41 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

partial derivative with respect to y give, for the second derivative that 2

T 2 py de py 2 a sin px b cos px p ce y

summing these last two equations gives

Laplace’s equation.

42 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

27. For the wall of Example 2-11, determine the heat transfer in the y-direction at 3 feet above the base.

Solution

x T ( x, y) 50 0 F e y L sin L

The solution to the wall temperature of Example 2-11 is The heat transfer in the y-direction can be determined, i

Q

T

A y

y

L

W

y

T

50 0 F e y L sin

dx W

x

y

0

L

L

For W = 1 ft, L =

3 ft, and y = 3 ft this equation can then be finalized i

3 ft 0

Q

50 F y

x

0

e sin dx 50 F e 3 ft

3 ft

3 ft

3 ft

0

cos

3 ft

x

500 F e 2 3 ft 0

For a thermal conductivity of 0.925Btu/hr∙ft 0F from Appendix Table B-2E, the heat transfer is about 4.00 Btu/hr. The temperature distribution at y = 3 ft for 0 ≤ x ≤ 3ft is

T ( x, y 3 ft ) 2.15 sin

x 3 ft

28. Write the governing equation and the necessary boundary conditions for the problem of a tapered wall as shown in Figure 2-53.

43 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution For steady state conduction in two-dimensions the governing equation will be => ? =@ >

=> ?

+ =;> = 0. Calling Tg the ground temperature the following four (4) boundary

conditions may be used: B.C. 1 .

, 0 = .A

B9 0 <

≤E

B.C. 2 . E, 6 = .A

B9 0 ≤ 6 ≤ F

B.C. 3 .

, F = ./

B9 E − G ≤

B.C. 4 .

, 6 = ./

B9 0 ≤ 6 ≤ F

≤E *&< 6 = F⁄E − G

29. Write the governing equation and the necessary boundary conditions for the problem of a heat exchanger tube as shown in Figure 2-54.

Solution A heat exchanger tube with convection heat transfer at the inside and the outside surfaces can be analyzed for steady state one-dimensional radial heat transfer with the

1 d dT r 0 dr equation r dr

and with, as a possibility, the following two boundary

conditions B.C. 1 HJ = 2K L ℎL .L − . @ r = ri B.C. 2 HJ = 2K / ℎ/ . − ./ @ r = r0

30. Write the governing equation and the necessary boundary conditions for the problem of a spherical concrete shell as sketched in Figure 2-55.

44 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution For steady state one-dimensional radial conduction heat transfer in spherical coordinates the governing equation for analyzing this and two suggested boundary

1 d 2 dT r 0 2 dr conditions are r dr B.C. 1 T = T0 @ r = r0 B.C. 2 T = Ti @ r = ri

31. Determine the Fourier coefficient, An , for the problem resulting in a temperature OP;/Q distribution of . , 6 = ∑R %& &K /F involving a boundary OS/ GO temperature distribution given by .

, 0 = 9 K /F for 0 ≤ x ≤ L.

Solution The Fourier coefficient is defined as

A

L

2

n

T ( x, 0) sin L

nx

dx L

L

2

cos L

0

A

x

nx

sin L

dx L

0

L

2 1

nx x

sin

sin

and using an identity

nx x

L

dx

1

sin

nx x

sin

nx x

dx

n

L02

L

L

L

L

L0

L

L

L

For n = 0 the Fourier coefficient, A0 becomes

A0

1

L

sin L

O L

1

A

sin

x sin 0 L 2x

x

dx

0 L

L

For n = 1 the Fourier coefficient becomes

cos

2x

L

0

1

40 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

L

L0 A 2

is

n =4

2L

L L

1

x3

sin L A4

0

2 2 5 3

L

x sin

L

0

For n = 2, the Fourier coefficient

x3

1 dx L

cos 3

L

x1 cos

2 2 L

3

For

For any even integer of n, such as 6, 8, 10, etc. the Fourier

41 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2

An

2 n1

n1

coefficient is

By reviewing the first coefficient, A1 it turns out that for all odd integers of n, such as 3, 5, 7, 9, 11, etc, the Fourier coefficient is zero, 0.

32. Determine the Fourier coefficient An for the problem involving a boundary temperature @

distribution given by .

, 0 = ./ T1 − Q U and where the solution to the temperature

, 6 = ∑R OS/ GO -

field is .

OP;/Q

%& &K /F .

Solution

A

L

2

n

nx

2

T (x, 0) sin L

L

dx L

1

T 0

L

0

x

sin

nx

L

dx

L By inspection A0 = 0 for n =

0

0. For n = 1

A

2 T0

1

L

L

sin

x

dx

L2

L

0

2 T0

L

x sin

x

dx

L

0

Using integral tables in Appendix Table A-4 L

L

2T A1

L

0

2T L2

x 0

cos L

2

L

L

2

x sin L

L

4T

x 0

cos

0

0

2T

2T

0

L 0

For n

even, such as 2, 4, 6, 8, . . . .

An

An

2T0 n

and for n odd, such as 3, 5, 7, 9, . . . .

2T0 n

which is the same as for n even

33. Plot the Bessel’s function of the first kind of zero and first order, J0 and J1, for arguments from 0 to 10.

42 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution Appendix Table A-10-1 tabulates the Bessel’s Function of arguments from 0 to 10. The plot is shown.

43 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

34. Plot the Bessel’s Function of the second k ind o f zero and first order , Y0 and Y1 for arguments from 0 to 10. i on Sol ut The Bessel’s Functions o f the second k ind o f zeroth and first order are tabu lated in Appendix Table A-10-1, plotted in Appendix Figure A-10-2, and here shown.

44 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

35. A sili con rod 20 cm i n di ameter and 30 cm l ong i s exposed to a hi gh temperature at one end so that the end i s at 4000C whereas the remai ni ng surf aces are at 600C. Estimate the centerline temperature distr ibution through the rod.

Sol ut i on The ratio of the length to radius, L/R is 3.0 so, using Figure 2 -22 the fo llow ing values can be read: x/L

(T – T0)/(Tf – T0)

T(x) 0Celsius

0.0

0.000

60.00

0.2

0.002

60.68

0.4

0.020

67.20

0.6

0.086

90.96

0.8

0.360

189.6

1.0

1.000

400.0

45 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The values for T(x) are computed from the equation

(x, 0)

TT T0

400 600 C 60 0 C

T T f

0

36. A Teflon rod 6 inches in diameter and 2 feet long is at 2300F. It is then exposed at one end to cool air so that that end is at 800F whereas the cylindrical surface cools to 1500F. The other end remains at 2300F. Determine the expected temperature distribution.

Solution To determine the centerline temperature distribution you can use Figure 2-22b. Since the L/R value is 2ft/3/12ft = 8 we need to extrapolate on the graph for approximate values. Also, the centerline temperature will not change significantly for values of z/L less than about 0.6. In addition, a principle of superposition will provide the rigorous solution. Yet, since the axial lengths are such that the distance from the 2300F end will be the total length minus the length from the 800F end, $ / = F − $V/ . Since the temperature of the center of the rod, axially, does not change significantly from the 1500F (T0) for z/L ≤ 0.6, we can just consider each end separately. For the model of a rod at 1500F with one end at 800F we have, say at z/L of 0.8, from Figure 2-22b that (T – T0)/(800F – T0) = (T – 150)/(80 – 150) = (T – 80)/(-70) has a value of about 0.15.

Therefore, at z = 0.4 ft = 4.8 in (corresponding to z = 1.6 ft from the 1500F end) from the 800F end the centerline temperature is . 4.8%&, 0 = 0.15 80 − 150 + 150 = 139. 5/ W ) . At say z = 0.7 ft = 8.4 in (1.3 ft from the 1500F end), z/L = 1.3/2 = 0.65, and from Figure 2-22b, (T – T0)/(800F – T0) ≈ 0.03 and then the centerline temperature at 8.4 in from the 800F end is . 8.4%&, 0 = 0.03 80 − 150 + 150 = 147. 9/ . Similarly, for the end at 2300F with the rod at 1500F, at z/l = 0.8, corresponding to 4.8 in from the 2300F end, the centerline temperature is . 4.8%&, 0 = 0.15 230 − 150 + 150 = 0 162/ W). At z/L = 0.65 (corresponding to 8.4 in from the 230 F end) the centerline temperature is . 8.4%&, 0 = 0.03 230 − 150 + 150 = 152. 4/ W

Section2-5 37. A water line of 2 inch diameter is buried horizontally 4 feet deep in earth. Estimate the heat loss per foot from the water line if water at 500F flows through the line and the outside temperature of the line is assumed to be 500F. The surface temperature of the earth is -200F.

46 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution Using the shape factor from Table 2-3, item 8, where L » r

2L

S

2L

2L

cosh 1

Y r

cosh 1

4ft 1 12 ft

4.564 The thermal conductivity of earth is about 0.3

W/m∙K from Appendix Table B-2 so that the heat transfer per unit length is i

q

ST

l

1 2 L

0.301

Btu

70 0 F 29

Btu

28

hr ft 0 F

4.564

W

hr ft

m

38. A chimney is constructed of square concrete blocks with a round flue as shown in Figure 2-56. Estimate the heat loss through the cement blocks per meter of chimney if the outer surface temperature is -100C and the inner surface temperature is 1500C.

Solution Assuming steady state conduction and using the shape factor from Table 2-3, item 4, the heat loss can be estimated. From Appendix Table B-2 the thermal conductivity of

S

2 L

concrete may be taken as 1.4 W/m∙K so that The heat loss can then be calculated from

ln 0.54

2 W

4.725 ln 0.54 7

r

i

q

S l

T 4.725 1.4 L

W

0

150 10 C 1.058 mK

kW m

47 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

39. Nuclear waste is placed in drums 50 cm in diameter by 100 cm long and buried in sand.

Water lines are buried adjacent to the drums to keep them cool. The suggested typical arrangement is shown in Figure 2-57. Estimate the heat transfer between a drum and the water line.

Solution Assume steady state, infinite media, and all heat transfer occurs between the 100 cm long drum and an adjacent 10 cm long water line. Using, item 11 from Table 2-3 with r = r1/r2 = 25/5 = 5, and L = 50 cm/5 cm = 10, gives

2

S

2 2

cosh

1

L1r

2

2.336 cosh 1 7.4

2r

Assuming dry sand with a thermal

conductivity from Appendix Table B-2 of 0.3 W/m∙K, the heat transfer is i

Q

S LT 2.336 0.3

W

1m 135K 94.6W mK

40. Steel pins are driven into asphalt pavement as shown in Figure 2-58. Estimate the heat transfer between a pin when it is at 600F and the surface when it is at 1100F.

48 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution Assume steady state conduction. Using item 6 of Table B-3, with a value of 0.036 Btu/hr∙ft∙ F for the thermal conductivity of asphalt from Appendix Table B-2E, a uniform pin temperature of 600F and the asphalt surface is 1100F, i

Q

ST

2L

T

21.5 ft

0.036

2L ln

R

Btu

110 600 F 3.967Btu / hr hr ft 0 F

2 1.5ft ln

0.5 12 ft

41. A heat treat furnace sketched in Figure 2-59 has an inside surface temperature of 12000C and an outside surface temperature of 600C. If the walls are assumed to be homogeneous with thermal properties the same as asbestos, estimate the heat transfer from the walls, excluding the door.

49 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution The heat transfer between the inside and the outside is X = XYZ: + X X

[\]

+ 2X

+ 4X

L^_

L^

^A_

+ 2X

[\]_^A_

ZYYZ

+

+ 2X`:JLAaY_^A_ + 4X\ZJO_J . All of these

can be modeled with shape factors from Table 2-3. The first four terms are just onedimensional conduction through a sheet, or plate. The next three are edges and the last i

is a corner. Combining all this

T

Q

T

30 x65cm2 30x65 30x40 2x40x65 x

4 0.559 65 2 0.559 30 2 0.559 40 4 0.15 15

substituting the thermal conductivity, the thickness Δx , and the temperature difference i

Q 1634.8W 42. A small refrigerator freezer, 16 in x 16 in x 18 in outer dimensions, has an inside surface temperature of 100F and an outside surface temperature of 800F. If the walls are uniformly 3 inches thick, homogeneous, and with thermal properties the same as Styrofoam, estimate the heat transfer through the walls and door of the refrigerator.

Solution Using shape factor methods we can list i

i

i

QQ

Q door

i

i

Q Q back

top

i

i

i

i

i

2Q

4Q

4Q

4Q

8Q

bottom

side

edge

upedge

back / frontedge

0

The thermal conductivity of Styrofoam is 0.017 Btu/hr∙ft∙ F, the temperature difference is 700F. The first five terms are just heat transfer through a flat plate, the next three are edges, and the last is a corner. Using items 1, 17, and 18 from Table 2-3 we get i

Q

T

10 x10in 2 10x10 10x12 10 x12 2 x10x12 x T 0.559 4x12 4 x10 4 x10 T 8x0.5x3 i

Q 35.2

The total heat transfer is then

Btu hr

43. Using graphical methods, estimate the temperatue distribution and the heat transfer per meter depth between the two surfaces at the corner shown in Figure 2-60.

50 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

corner

i on Sol ut The sketch shown shows that there are 11 heat flow paths , M = 11 , and 4 temperature steps, N = 4. Thus, the shape factor is rough ly M/N = 2 .75 and the heat trans fer is i

Q

ST

2.75W 0.029 m K / K

110W m8.7725

/

44 Us ng graph ca methods est mate the temperature d str but on through the . d sk i surround i ng l the s con , ch i p sketched n F gure 2 61 i iThen i est mate the phenolic i i ili i i i . i heat trans er per un t depth f i .

49 Š 2016 Cengage Learn ingŽ . May not be scanned , cop ied or dup licated , or posted to a pub lic ly access ib le webs ite , in whole or in part.

Solution Here we have that the shape factor is the heat flow paths, M, divided by the temperature steps, N, so that S = M/N. From the sketch shown there are about 25 heat flow paths and 4 temperature steps. Us ing a t hermal conductivity of 0.35 W/m∙K for nylon as an approximation for pheno lic from Appendix Table B-2, we have i

q

ST l

25 4

Bu 0.202 h

t F r ft

180 800 F 126.25

0

Btu hr ft

45 Us ng graph ca techn ques est mate the temp around the e ectr ca power ne shown n F g . peri un t ength i l between i the ne i and the grou erature distribution through the earth l i l li i i ure 2-62. Then estimate the heat transfer i l li nd surface.

So ut on l iThe temperature d str but on and the heat tra o the heat ow nes and sotherms These tw i i i nsfer can be approximated with a sketch f fl li i . o sets of lines need to be orthogonal or

50 © 2016 Cengage Learn ing® . May not be scanned , cop ied or dup licated , or posted to a publicly accessible website, in whole or in part.

perpendicular at all times

and the spacing between adjacent isotherms and heat flow lines need to approximate a square. The Shape factor, S, will be the ratio of the heat flow paths, M, to the isotherms, N. The sketch shows a possible approximate solution where the temperature steps or isotherms is seven (7) and the number of heat flow paths is twenty seven (27). Then i

q

l

ST

M

T

27 W W 0.52 20K 40.1 N 7 mK

m

Notice that the shape factor is 27/7 = 3.86, which is a value in close agreement with item 8 of Table B-3 for a buried line,

S

2

2 4.77 4 1 Y cosh 1 cosh 2 R

46. Using graphical techniques estimate the temperature distribution through the cast iron engine block and head shown in Figure 2-63.

51 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution

52 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Referring to the sketch of the piston-cylinder and assuming symmetry, there are five (5) isothermal steps so N = 5. Also there are estimated to be twenty-two (22) heat flow paths for one half the cylinder for heat exchange between the cylinder at 6000F and the surroundings at 1600F. From Appendix Table B-2E the thermal conductivity for cast iron may be taken as 22.54 Btu/hr∙ft∙0F. Then the heat transfer is i

Btu

M

q

T 22.54 r

N

22

Btu 600 160 43, 637

0

hr ft

F 5

hr ft radians

and if we assume an effective radius of 0.3 ft and rotate the 22 heat flow paths through one revolution, 2πr, then the heat transfer will be i

Q

i

2 reffective

Btu

Btu

q

82, 255

2 0.3 ft 43, 637 hr ft

r

hr

47. A Bunsen burner is used to heat a block of steel. The surfaces of the steel may be taken as 500C except on the bottom, where the burner is heating the block. Figure 2-64 shows the overall configuration of the heating process. Using graphical techniques, estimate the temperature profile through the block and the heat transfer through the block.

53 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution

Using graphical techniques requires that a web of approximately square elements are formed between adjacent heat flow lines and isotherms. An approximate solution is shown, noting that the 8500C is assumed to be in the block. The number of heat flow paths for one-half the block is nine (9) and the number of isotherms is five (5). Assuming a carbon steel the thermal conductivity is taken as 60.5 W/m∙K from Appendix Table B2. Since the block is 25 cm square i

Q

L

M

T 60.5 N

W

0.25m mK

9

850 50K 21.78kW 5

Section2-6 48. Estimate the heat transfer from the fin shown in Figure 2-65. Write the necessary node equations and then solve for the temperatures. Assume the fin is aluminum.

54 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sol ut i on Referring to the sketch, assuming symmetry so that only 9 nodes need to be identified and using

node neighborhoods o f 1 inch squares (Δx = Δy = 1 inch), and assuming the temperature of node 5 is 3250F the node equat ions can be wr it ten for steady state conduction two0 dimensional heat transfer. The thermal conduct iv ity of aluminum is 136.38 Btu/hr∙ft∙ F, rounded to 136.4 Btu/hr∙ft∙0F from Appendix Table B-2E. For node 1:

y T5T1

x T2T1

2

x

y

h 2

y

1

2

equat on

Then th

and 9 i o 137.44T ow 1 68.2T2 22253.5 f ll

T

T

substituting into this node

e equations for nodes 2, 3, 4, 6, 7, 8,

274.88T2 68.2T1 68.2T3 136.4T6 177.08

274.88T3 68.2T2 68.2T4 136.4T7 177.08 232.7T4 68.2T3 136.4T8 25.4T9 227.6 272.8T6 136.4T2 58.2T7 22165

0

55

© 2016 Cengage 272.8T Learn7 ng 136.4T ® May not be scanned cop ed 3 68.2T 8 68.2T 6 or0dup cated or poste

i

.

,

i

li

,

d to a publicly accessible website, in whole or in part.

272.8T8 136.4T4 68.2T7 68.2T9 0 T 95.3T9 68.2

8

T 25.4

4

139.08 Then or the , f 8 x 8 matr x

i

49 Wr te the node equat ons or the mode o heat trans er through the compressor . i i f l f f hous ng sect on shown n F gure 2 66 Then so ve or the node temperatures by us ng i i i i - . l f i EES Mathcad or MATLAB , , .

56 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution Referring to Figure 2-66, which is a scale 1 to 6, the inner radius is assumed to be 21 cm and the outer radi usi s then 30 cm. The housing is cast iron so that the thermal conductivi ty i s 39 W/m∙ Kf rom Appendi x Table B-2 and assuming that the slots have qui escent fl ui d at wi th a thermal conducti iv yt of 1 W/m∙K, the node equations may be written out. Ref erri ng to the f oll owi ng sketches some of the nodes are identified, others need to be to be inferred, and node 1 is shown in some detail.

57 © 2016 Cengage Learn ing® . May not be scanned , cop ied or dup licated , o r posted to a publicly accessible website, in whole or in part.

Node 1 neighborhood. The angular displacement between nodes is 11 .25 or 0 .196 rad ians. For node 1 0

r

1

T T

0.196rad 0.285m T2T1

5 1 2 0.196 2 ve h Subst tut ng0.3m the therma conduct v ty convect

i i l i i , change node 1 9.95 T5 T1 36.3 T2 T1 5.292 35 T1 0

i

0.1960.3m

h r

0

350 C T 1

2

eat transfer coefficient, and radius which is the equation for

An energy ba ance or node 2 g ves l

r or

f

i

0.196 0.285m T1 T2 0.196 0.255m T3T2 T6T2 0 0.196 0.27m 2 r 22.1T6 T2 36.309 T1 T2 32.487 T3 T2 0

2

r

which is the equation for

node 2. An energy balance of the heat flows to each of the nodes can be made and the following equations result

24.872 T7 T3 32.487 T2 T3 28.665 T4 T3 0

t which is the equation for equation for node 4. node 3, which is the After applying energy balances to all of the 20 nodes the following set of equations result

14.21T8 T4 28.665 T3 T4 7.82 300 T4 0

51.542.b − 36.3. − 9.95. = 185.22 90.896. − 36.3.b − 22.1.c − 32.487. = 0 86.024. − 32.487. − 28.665.d − 24.872. = 0 50.695.d − 28.665. − 14.21.V = 2346 103.084. − 9.95.b − 72.6.c − 9.95.e = 370.44

58 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

0

142.077.c − 22.1. − 72.6. − 36.32. − 11.057.b/ = 0

59 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

130.96. − 24.87. − 36.32.c − 57.33.V − 12.44.bb = 0 101.39.V − 14.21.d − 57.33. − 14.21.b = 4692 103.084.e − 9.95. − 72.6.b/ − 9.95.b = 370.44 94.757.b/ − 11.057.c − 72.6.e − 0.043.bb − 11.057.bd = 0 82.253.bb − 12.44. − 0.043.b/ − 57.33.b − 12.44.b = 0 101.39.b − 14.21.V − 57.33.bb − 14.21.bc = 4692 103.084.b − 9.95.e − 72.6.bd − 9.95.b = 370.44 22.114.bd − 11.057.b/ − 72.6.b − 0.043.b − 11.057.bV = 0 82.253.b − 12.44.bb − 0.043.bd − 57.33.bc − 12.44.be = 0 101.39.bc − 14.21.b − 57.33.b − 14.21. / = 4692 51.542.b − 9.95.b − 36.3.bV = 185.22 47.3785.bV − 11.057.bd − 36.3.b − 0.0213.be = 0 41.126.be − 12.44.b − 0.0213.bV − 28.665. / = 0 50.695. / − 14.21.bc − 28.665.be = 2346 With this set of equations the temperatures can be determined. Using Mathcad, noting that the results are tabulated in the final column with node 1 being listed as 0, node 2 as 1, and so on.

60 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

;

60 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

50. Wri te the node equati ons for describing heat transfer through the buried waste shown in Fi gure 2- 67.

61 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution For doing a finite difference analysis the following grid may be used. Then the heat from the waste mass per unit depth (1 m) is A = 30 fg⁄ 0.2 = 1.2 fg. Estimate that the power or heat to node 1 is 0.6 kW and 0.3 kW to node 2. Using a thermal conductivity of 0.52 W/m∙K for earth or soil from Appendix Table B-2, and utilizing symmetry in the x-direction, one-half of the neighborhood for node 1 will be 0.05 m

x T3T1 wide and

0.52

T3T1

2

0.26 T 2

y T2T1

600W 0 2

y

x

or, for node 1

T 600W 0 2

1

Similarly, for the remaining nodes,

0.26 T1 T2 0.52 T5 T2 300W 0

which is for node 2,

62 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

0.26 T1 T3 0.26 T6 T3 0.52 T4 T3 0 f or node 3

3T 6

1

i i e node equation wh ch s th

4T4 T3 T2 T5 T7 0

T T 2 3

60 C 7

4T7 T 6T T4 1280 C 4T8 T 7T T5 1290 C 2T9 T 8120 C

51. Write the node equations for determining the temperature distribution through the cast iron lathe slide shown in Figure 2-68.

63 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution A proposed node layout is shown

The node ne ghborhoods are ! = !6 = 1 & ℎ 2assume the ho e has a r at 900F w th a 0 convect ve heat trans er coe c ent o 6 Btu/hr t F and the therma conduct v ty ior i % , l i 0 cast roni may be takenf as 22ffi 5 Btu/hr i f t F From∙f Append ∙ , x Tab e B 2E l App y ngi an i f i

.

∙f ∙

i

l

- .

l i

64 Cengage Learn ng® May not be scanned cop ed or dup cated or posted to a pub c y access b e webs te n who e or n part © 2016

i

.

,

i

li

,

li l

i l

i ,i

l

i

.

energy balance to node neighborhood 1, the following equation results

T2 T1

T6 T1

x

1000Btu / hr2 ft

0

Substituting for thermal conductivity and node neighborhood size, 2.b − . − .c = 3. 7/ W For nodes 2 through 6 3. − . − . = 3. 7/ W 1.044. − .d = 7. 7/ W 2.c − .b − . − .b/ = 0 Nodes 7, 8, 11, and 12 require some adjust ing . Re fe rring to the sketch for node 7

The energy balance can be approximated by

y

T T

y 6

7

T T2

y 7

T8 T7

x

T11 T7

whxch becomes x

3 017. − .2x c − . − 0 5.

S mi ar y or node 8

3.017.V − . − .e − 0.5. − 0.5. = 1.57 V bb

And 11 and 12 . 3 017.bb − .b/ − .b . − i il orl nodes ,f f

h

2y

x

900 F T7 0

2 2

− 0.5.b = 1.57

3 017. . b − .bc − .b − 0 0.5.b − 0.5. = 1.57

The rema n ng node equat ons are stra ght orward . .5.bb − 0.5.V = 1.57 For node 9 i i

33i 022.e − .di − f.V − .b energy balances and are,

65 © 2016 Cengage Learn ng® May not be scanned cop ed or dup cated or posted i . , i li , to a publicly accessible website, in whole or in part.

= 1.956/ W

.

i

.

,

i

li

,

to a publicly accessible website, in whole or in part.

For node 10

3.b/ − .c − .bb − .bd = 0

For node 13

33.022.b − .e − .b − .b = 1.956/W

For node 14

3.bd − .b/ − .b − .

For node 15

4.b − .bb − .bd − .bc − .

For node 16

4.bc − .b − .b − .b − .

For node 17

3.511.b − .b − .bc − . − 0.5.bV = 0.978/W

For node 18

2.022.bV − .b − .be − .

For node 19

1.533.be − 0.5.bV − .

For node 20

2. / − .bd − .

For node 21

3. b − .b − . / − .

=0

For node 22

3. − .bc − . b − .

=0

For node 23

33. − .b − . − .

For node 24

3. d − .bV − . − .

For node 25

2.022. − .be − .

b

/

=0

d

b

=0 =0

= 1.956/W

= 1.934/W

=0

d

d

=0

=0 = 2/ W

52. A concrete chimney flue is surrounded by a Styrofoam insulator as shown in Figure 2-69.

Construct an appropriate grid model and then write the node equations to determine the temperature distribution.

66 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

l i f i

,

i

l

i

ds to be

f

x 130 T1 2

f

i

ii

con

l

y

y

T2T1

con

x T5T1 con

x

2

0

y

Which can be reduced to

2.b − 0.5. − . = 65/ 0

Fore node 2

4. − .b − .c − . = 130/ 0

For node 3

4. − . − .d − . = 130/0

For node 4

2. 25.d − . − .V = 32. 5/ 0

For nodes 5 through 9 the Styrofoam impacts the energy balance so

x T1T5 con

con

2

y

T6T5

x 30 T sty

x

y

2 y 2

0 Also, since the

boundary temperature between the Styrofoam and the concrete is not yet known we i

Q

x TsT5 30C node5

con

x 30 Ts sty

2 y 2

2 y 2 Solving this equation for Ts

write and substituting back into the node equation gives

67 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

So ut on

h1.5 + i

ijkl mno p ijkl

q . − 0.5.b − .c =

ijkl imno pijkl

30/

Assume symmetry or the ch mney so that on y one quarter o the sect on nee cons dered as shown n the sketch

Wr t ng the energy ba ance or node 1

68 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

For node 6

h3 +

For node 7

h3 +

For node 8

h3 +

For node 9

h1 +

ijkl q .c imno p ijkl

ijkl imno p ijkl

ijkl imno p ijkl

ijkl imno p ijkl

−. −. − . =

ijkl imno pijkl

ijkl

q . − .V − .c − . =

imno pijkl

q .V − . − .d − .e =

imno pijkl

q .e − .V =

ijkl imno pijkl

ijkl

30/

30/

30/

30/

53. Consider the chimney flue of Figure 2-69. If the Styrofoam is removed and the outer boundary condition is the same, write the necessary node equations and solve for the node temperatures. What is the heat transfer through the chimney flue?

Solution Using the same node arrangement as for Problem 2-52 and referring to the sketch, the node equation for node 1 is 2.b − 0.5. − . = 65/ 0 For node 2

4. − .b − . − .c = 130/ 0

For node 3

4. − . − .d − . = 130/ 0

69 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

For node 4

2.25.d − . − .V = 32. 5/ 0

For node 5

2.5. − 0.5.b − .c = 15/ 0

For node 6

5.c − . − . − . = 60/ 0

For node 7

5. − . − .c − .V = 60/ 0

70 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

For node 8

5.V − .d − . − .e = 60/ 0

For node 9

3.e − .V = 60/ 0

The heat transfer can be approximated by the equat ion 8rXb

+X

c

+X

+ Xd

Vs

which can be wri tten

XYZY[t = 8u v0.5.b + . + . + .d − 0.5. − .c − . − .V w

71 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

70 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

54. Wri te the node equati ons for the nodes 1 and 2 of the model of the oak beam sketched in Fi gure 2- 32.

71 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution The model of the round beam is such that axial symmetry is assumed so that a hemispherical section will suffice for nodes. In Figure 2-32 the node numbering scheme follows the pattern of number 1 is in the center, 2, 3, and 4 are radially outward to the outside surface. Then on a 22.50 rotation numbers 5, 6, and 7 occur. On the next 22.50 rotation numbers 8, 9, and 10 occur. Continuing in this pattern there are three nodes at every 22.50 rotation for the first 28 nodes. On the next hemisphere axially parallel to the first hemisphere node number 29 will be on the center position with numbers 30. 31.

72 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

And 32 outward. Again at a 22 .50 rotat ion numbers 33, 34, and 35 occur. Continuing, the pattern is such that nodes on the hem isphe re parallel to the succeeding hemisphere will have a number of the previous node plus 28 . Thus, node 2 will be adjacent to nodes 29, axially, and also to nodes 3 , 5, and 1 . This model is sketched. Node 1 has nine (9) adjacent radial nodes; 2, 5, 8 , 11 , 14 , 17 , 20 , 23, and 26. Node 1 also has an adjacent node on the axis, number 29 . This mode l is s ketched.

For node 2 there four adjacent nodes w ith the thermal resistances of ,

( ?J b

=

cd iPxy

,

( ?J

,

=

, iPxy

(?y

xy

, /

=

i

PT

z

|}> UhJ > q

z{

(?•

=

T~xJU P⁄V

=

cd iPxy

, and

~

P

so that the node equ ation for node 2 can be formed.

Not ng that X = !.€( the node equat on becomes ? i i ,

xy⁄

xJ

= bcxy

z 64

1

2

32

3

2

T3 T r T T T T 16z T T 30 2 5 64 3

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2

0

i

.

,

i

li

,

publicly accessible website, in whole or in part.

The thermal resistances for conduction between node 1 and 3, 5, 8, 11, 14, 17, 20, 23, and 26 are

64 z

RTr ,21 R Tr ,261

RTr ,51 R Tr ,81 R

29 to 1

z

RTz ,291

T2 T26 2T1 64

Tr ,111

8z r2

and

R

Tr ,141

R

R

Tr ,171

R

Tr ,201

32 Tr ,231

z

and for node

and the node equation or energy balance for node 1 is

z

T5 T8 T11 T14 T17 T20 T23 7T1 32

r2

T29 T1 0

8z

55. Figure 2-70 shows a section of a large surface plate used for precision measurements. A person touches the surface and thereby induces heat transfer through the plate.

Neglecting radiation involved, write the node equations for nodes 1, 5, and 12.

Solution The sketch of the granite surface plate is shown.

74 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, l For node 1 the0energy ba ance becomes

y 65WFT

2

1

x

T T

x

6

yT T

1

2

y

2

wh ch reduces to

3.035T 0.5T T 99.79 F

1

x

h

xT

T 0 1

0

1 2 6 i n a s m ar ash on the node equat ons are

I

i il

f

i

,

3.035T5 0.5T4 T10 99.79 F

i 0

or node 5 and or node 12 f , f

4T12 T7 T13 T17 T11 0

56 Wr te the comp ete set o node equat ons or the gran te sur ace p ate shown n F gure . i l f i f i f l i i 2 70 and est mate the temperature through the p ate i l .

75 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution Referring to the sketch for the nodes of the surface plate, shown in the solution to Problem 2-55, the twenty node equations become 3.035.b − 0.5. − .c = 99.79/W 3.035. − 0.5.b − 0.5. − . = 99.79/W 3.035. − 0.5. − 0.5.d − .V = 99.79/W 3.035.d − 0.5.

− 0.5. − .e = 67.29/W

3.035. − 0.5.d − .b/ = 99.79/ W 4.c − .b − . − .bb = 65/ W 4. − .c − . − .V − .b = 0 4.V − . − . − .e − .b = 0 4.e − .V − .d − .b/ − .bd = 0 4.b/ − .e − . − .b = 65/ W 4.bb − .b − .bc − .c = 65/ W 4.b − .bb − . − .b − .b = 0 4.b − .b − .V − .bd − .bV = 0

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4.bd − .b − .e − .b − .be = 0 4.b − .bd − .b/ − .

/

= 65/ W

2.bc − 0.5.bb − .b = 32. 5/ W 2.b − .b − 0.5.bc − 0.5.bV = 0 2.bV − .b − 0.5.b − 0.5.be = 0 2.be − .bd − 0.5.bV − 0.5. / = 0 2. / − 0.5.be − .b = 32. 5/ W This set of 20 x 20 matrix ca be solved with Mathcad

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57. A pl utoni um nucl ear f uel rod shown i n Fi gure 2- 71 has energy generati on i n the amount of 3000 Btu/s∙f t3. For the gri d model shown, wri te the node equati ons and sol vef or the temperatures. Assume κ = 10 W/m∙K.

78 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sol ut i on From Figure 2-71, it can be assumed that the heat flow is rad ia lly outward and axially and angularly and axially symmetrical. The node mode l is sketc hed

Then for node 1 the adjacent nodes are 4 and 2 p lus a convect ive heat transfer.

Referring to the sketch for node 1

The node equation is r 2

z T2T1 2

2

2

r r

4

T4T1 z

2

r h

4

T T1

3000 x3600

Btu

z r

ft 3 hr

2

•Y`

Us ng the o ow ng va ues ℎR = 400 aJ‚Y > ƒ .R = 600 /W 5 779 i .

•Y` ! aJ‚Yƒ f ll

,

= i

b b

„ B† *&< !$ = 1 B† the o ow ng node e l , , , u = 10 = …

.

, f ll i 0 56586. b − 0 5. − 0 0009766.d =

quation results

79 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2

4

0

.

.

.

953

80 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A similar analysis for node 2, noting that it has three adjacent nodes, 1, 3, and 5, plus a convective heat transfer and energy generation, yielding 1.79417. − 0.5.b − 9.75. − 0.0347. = 7624.42

For node 3, the energy balance reduces to 17.727. − 91.5. − 0.00684.c = 16119.6 For node 4, 1.001954.d − 0.000977.b − 0.000977. − . = 1825 Node 5 is a bit more complicated. Re ferr ing to the sketch the node equation becomes

4.015625. − .d − 3.c − 0 .0078125. − 0.0078125.V = 7300 Node 6 has three adjacent nodes p lus convec tion and energy generation so its node equation is 20. 3.c − 3. − 0. 00684. − 0.00684.e = 18110.65 Node 7 energy balance s im ilar to node 4 , beco mes 1.001954. − 0.000977.d − 0.000977.b/ − .V = 1825 For the node 8 node equation , s im ilar to node 5 4.015625.V − . − 3.e − 0 .0078 125. − 0.0078125.bb = 7300

80 © 2016 Cengage Learn ng® May not be scanned cop ed or dup cated or i . , i li , posted to a publicly accessible website, in whole or in part.

For node 9, similar to node 6 20.3.e − 3.V − 0.00684.c − 0.00684.b = 18110.65 The energy balance for node 10 is similar to node 7 except it is only one-half as long as node 7 and there is no lower surface heat transfer.

0.500977.b/ − 0.000977. − 0.5.bb = 912.5 Node 11 equation is 1.328.bb − 0.5.b/ − 0.75.b − 0.0078125.V = 7300 And Node 12 is 9. 409.b − 0.75.bb − 0.00684.e = 11578.8 Using Mathcad for the predict ion o f the 12 node temperatures , the resu lts are

From the estimated inputs and the set of equat ions

81 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a pub licly access ib le webs ite , in who le or in part .

Secti on 2-7 58. Determine the heat transfer and fin efficiency for a copper fin shown in Figure 2 -72 . The fin can be assumed to be very long and its base temperature taken as 2000F .

82 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution For very long fins the fin efficiency is

1 A fin

L

hP where L = 4 in = 0.333..ft κ = 136.4 Btu/hrftF From Appendix Table B-2E A = 2 ft x (1/96 ft) = 0.020833 ft2 h = 70 Btu/hr∙ft2 ∙F P = perimeter = 4.020833 ft 0. 30 = 30%

Then

The heat transfer of the fin is i

i

0.30 hA 200 800 F 0.30 70 120 3393.5Btu / fin

Q

fin

Q

0

s

59. A square bronze fin, 30 cm wide. 1 cm thick, and 5 cm long is surrounded by air at 270C having a convective heat transfer coefficient of 300 W/m2 ∙K. Determine the fin tip temperature, the fin heat transfer, and the fin efficiency.

83 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution For a finite length fin the temperature distribution is given by the equation

h

cosh m L x x T x T0

sinh m L x

m 0

cosh mL

h

sinh mL

m

h = 300 W/m2K

For this fin

κ = 114 W/m∙K fin thickness, Y = 0.01 m,

fin width W = 0.3 m

fin length L = 0.05 m perimeter, P = 2W + 2Y = 0.62m ,

Area, A = WY = 0.003 m2

60. A square aluminum fin having base temperature of 1000C, 5 mm width, and 5 cm length is surrounded by water at 400C. Using h of 400 W/m2∙K, compare the heat transfer of the fin predicted by the three conditions: a) very long fin, b) adiabatic tip, and c) uniform convection heat transfer over the fin, including the tip. Assume a width of 1 m.

Solution From the Appendix Table B.2, κalum = 236 W/m∙K Also, Θ0 = 100 -40 = 60 K,

T∞ = 400C, h = 400 W/m2∙K, t = 0.005 m, L = 0.05 m, W = 1 m

P = 2t + 2W + 2W= 2.01 m, A = tW = 0.005 m2, and ‰ℎ = ˆ = 26.1 uG

For the very long fin, a) For a fin with an adiabatic tip, For a finite length fin,

i

Q fin

0

b

X‚LO = "/ √ℎ‰uG = 1848 g X‚LO = "/ √ℎ‰uGtanh

h sinh mL mL hP A

F = 1595 g

coshmL 1624W 84

© 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

cosh mL

h sinh mL mL

85 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

61. Show that the fin heat transfer for a square fin having an adiabatic tip is i

Q fin 0 hP A tanh mL Solution For a square fin with an adiabatic tip the temperature distribution is

x TxT

cosh m Lx cosh mL

0

The heat transfer is

i

AT

Q

A

fin

x

x x0

and

x0

m0sinh m Lx cosh mL

x

At x = 0 this is

m0sinh mL

tanh mL

m cosh mL

x x0

0

The fin hat transfer is

then i

Q

fin

m

A m0 tanh mL

but

Ph A

so that

i

Q

0

Ph A tanh mL

fin

62. Show that the heat transfer for a fin that is square and has fin tip convective heat transfer coefficient hL can be written

sinh mL

i

Q

cosh mL

m

hP A fin

hL

0

cosh mL

L

h

sinh mL

m Solution

86 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

For square fin with convective heat transfer coefficient hL at the tip, the temperature distribution is

87 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

hL

cosh m L x

sinh m L x

m

x TxT

0

cosh mL

hL m

sinh mL

The fin heat transfer is i

Q

A

T

A x

fin

x x0

m sinh

Also

x0

mhL

mL x

mL x

cosh

m x

cosh mL

sinh mL

i

Q

hL

Since

•a

= •i‘

cosh mL

m

hP A fin

hL sinh mL m

0

cosh mL

L

h

sinh mL

m

63. Derive an expression for the heat transfer from a tapered fin having base of Y thickness, L length, κ thermal conductivity, h0 convective coefficient, and T0 base temperature. The surrounding fluid temperature is T∞ .

Solution Referring to the sketch,

88 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6 = ’ T1 − U F From a heat balance through the fin

d2 h P 0 dx 2

A

"/ = ./ − .R

where " = . − .R ‰ = 26 + 2g ≈ 2g

1d

2

or y « W f .

h 2WL

2h L

0

dx 2

Then

0

WY L x

YLx

Using X = L – x and C =

2h9L/κY

1 d2 C dx 2 X

with two boundary conditions: B.C. 1, θ = θ0 @ X = L ”

B.C. 2, θ = 0 @ = 0

Now, assuming a series solution so that

c X c X 2 c X 3 c X 4 c X 5 ... c X n ...

c 0

1

2

3

4

5

n

From B.C. 2, c0 = 0 and

then

c1 X c2 X

2

c3 X

3

c4 X

4

c5 X

5

... cn X

n

...

for the second derivative

89 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2 dO

dx 2

2

3

2c2 6c3 X 12c4 X

4

20c5 X

30c6 X

^ >•

2c2 6c3 X 12c4 X

3

42c7 X

” " •

Using the differential equation − ^@ > = 2

4

20c5 X

42c7 X

= −0

, 6

d

20

= −0

7

72c9 X

8

...

90c10 X

we get

C ......

X

2

c1 X c2 X

”

9

b

= −0

, 12

6

56c8 X

5

30c6 X

Comparing coefficients, 2

5

=− ”>

9

= −0

9 9

d

d

=

4

c4 X

5

c5 X

b

=b ”

=−

”~

3

c3 X

VV/ b

b

bdd b

and so on…

For C less than or equal to 1.0, using the first four terms is suitable as higher terms will be significantly smaller. Then,

C

cX

cX

1

0

0

C

1

X

cX

1

2

cL

2

C

2

1

12

2

144

2

C

c L2 1

12

C

2

12

1

..

1

144

2

and using B.C.1

c L4 1

Solving this for c1 and substituting

3

X 3 C X 4 ...

144

C 2 C2 3 C L L L 12

c X4

C3

c L3

C2 X

2

C3

3

3

L4 ...

144

64. Show that the fin effectiveness is related to the fin efficiency by the equation fin

A 1fin AT

Afin fin

AT 90

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...

Solution For a fin and a base area between succeeding fins, the fin effectiveness is

91 Š 2016 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

i

i

Q Q fin

fin

i

Q

base

where

0

i

Q

hA

hAbase 0 hA

fin 0

0

G ? = G‚LO + G

Where

T

0

[ _

Also, i

Q

fin

fin

hA fin0

And

i

Q

hAbase0

base

fin

Substituting into the effectiveness equation

hA fin0hAbase0

fin

hAfin0h AT A fin0

fin

hAT 0

hAT0

Cancelling the h’s, θ0’s,

and rearranging,

fin

A fin Afin

1 fin

AT

AT

A 1fin

A fin fin

AT

AT

65. A circumferential steel fin is 8 cm long, 3 mm thick, and is on a 20 cm diameter rod. The surrounding air temperature is 200C and h = 35 W/m2K, while the surface temperature of the rod is 3000C. Determine a) Fin Efficiency and b) Heat transfer from the fin.

Solution Referring to Figure 2-41 L = 8 cm = 0.08 m, r1 = 0.1 m, y = 3 mm = 0.003 m, r2 = L + r1, LC = L + y/2 = 0.0815 m, r2C = r1 + LC = 0.1815 m, and Am = y(r2C - r1) = 0.0002445 m2 Using a thermal conductivity of 43 W/mK for steel from Appendix Table B-2

L3/C 2

h Am

1.342 and

r2C 1.815 r1 . Then, from Figure 2-41, 92

© 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

a)

η fin ≈ 44 %

93 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

b) i

Q

in

W

hA in f

in f

2

0.44 35

0

f

r2 r1

m2 K

2

2 r2 y

300 20K

W 318 fin

66 A bronze rod 1 cm n d ameter 0 band 30 cm long protrudes from a bronze surface at . 150 C The2rod us surrounded i i o 10 W/m K Determ ne the hey air at 100C with a convective heat transfer coefficient .

So ut fon

.

i

at transfer through the rod.

l i

Assume the bronze has the sam Append x Tab e B 2 Some o th e thermal conductivity as brass, 114 W/mK from 1500C i l - . f e other parameters are: h = 10 W/m2 K, T∞ = 100C, T0 = Θ = T, - T = 1400C P = πD = 0

0

∞

0.0314159 m, A = πr2 = 0.00007854 m2, and

,

hP 5.923m1 A

m

sinh mL

i

Q

and u sing the case III fin equation, the finite length fin,

0 fin

hP A

h

m h cosh mL m

cosh mL 7.024W / rod sinh mL

90 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

67. A circumferential cast iron fin attached to a compressor housing is 1 inch thick, 3 in long, 3 in diameter, and the convective heat transfer coefficient is 16 Btu/hr∙ft2∙ 0F. If the base temperature is 1600F and the surrounding air is 800F, determine the fin efficiency and the heat transfer through the fin.

Solution Referring to Figure 2-41, the following parameters are: 0.375 ft, L = 0.125 ft, y = 0.0833..ft,

r1 = 1.5 in = 1.25 ft, r2 =

LC = L + y/2 = 0.1666…ft, r2C = r1 + LC = 0.291666…ft, Am = y(r2C - r1) = 0.01388 ft2 ,

L3/C 2 r2C/r1 = 2.333, and

h Am

0.486 .

ηfin ≈ 82%.

The heat transfer is

i

Q

t f in

f

hAin

f

From Figure 2-41

Tin T

2

0.82 B 16u r ft 2 0 h F

0

r2 y

2

r1

0

2r2 F

160 80 Bu

t 618.26

68. A handle on a cooking pot can be mode led as a rod fin w ith an ad iabat ic t ip at the farthest section from the attachment po ints. For the hand le shown in the sketch , determine the temperature distribution and the heat trans fer through the hand le if the pot surface is 1900F, the surround ing a ir temperature is 900F , and the convect ive heat transfer coefficient is 160 Btu/hr∙ft2 ∙0F .

Sol ut i on

Treating this handle as a fin with an adiabatic tip, the important parameters are: Thermal conductivity of 22.5 Btu/hr∙ ft2 ∙ 0F from Appendix Tab le B -2E , L = πr/2 = π(3/24) ft = 0.3927 ft, P = π(1/12) ft = 0.2618 ft, A = π(1/12)2 (1/4) = 0.005454149 ft2 ,

m and

hP A

160 0.2618 22.54 0.005454

1 18.475 ft

For an adiabatic tipped fin,

91 © 2016 Cengage Learn ng® May not be scanned cop ed or dup cated or posted to a pub c y access b e webs te n who e or n part i . , i li , li l i l i ,i l i .

h

r

0

mL x cosh F cosh mL

100 0

Lcosh x 18.475 cosh 7.255

0.14128 cosh18.475 Lx

At the extreme outer po nt o the hand e i f l ,

F 0.141280

or

T = 90 141280F .

The heat trans er through the n s f fi i

Q i

in f

hP A 0

mL tanh

mL 135.956 tanh

Bu h 135.956 t / r

S nce the hand e has two ns so to speak i l fi , ,

Q i

handle

Bu h 271.912 t / r

69 An a um num n s attached at both ends n a compact heat exchanger as shown For . the sl tuat i on shown fi i determ ne the temperature i d str but on and the heat trans .er i i , i i i i f through the n fi .

So ut on l i

For the n fi 2 dO m2 2 dx

with boundary conditions, B.C. 1 " = "b = .b − .R = 180/W @ = 0 " = " = . − .R =

160/ W

@

=F

BC 2 . .

From th s equat on and the boundary cond t ons Equat on 2 114 s i i ii i i

92 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1 x

2 mL

e

e

A

mx

mL

e

2

1000.01m 236 2 0.005

hP

m

2

1

e 2 mL 1

mL

mx

e

1

e where L = 0.2 m,

13m 1

And then mL = 2.6 so that

1

x T x 100

180e5.25.2160e 2.6 e 13 x 160e 2.6 180e13 x e 1

The

maximum or minimum temperature occurs at the location predicted by Equation 2-115,

1

x

ln

m

e 2 mL e mL 2

1

2m

2

e

0.1079m

mL

m imum

1 Using x = 0.1079 m in the above equation The n heat trans er s the sum o the two ad abat c stems

for the temperature distribution, T Qin n Q186.08 Q 0 C n fihP A n i

i

m

f

i

m

f

hP A

m i

i

i

180 tanh (0.1079 ) 160 tanh 0.2 0.1079m 70.63W fi 1 2 70 For fithe tapered n fishown determ ne the temperature d str but on the n e c ency and the heat trans er through the n So ut on .

fi Re err ng to the gure f

,

i fi .

i

i

i

,

fi

ffi i

,

l i f

i

fi

,

The o ow ng parameters are known L = LC = 0 06 m 0 0006 m2 h = 1000 W/m2 K and

Κ = 236 W/mK f ll .

Y = 0 02 m Am = LY/2 =

i ,

: From F gure 2 40 ,

∙ ,

. , —‚LO ≈ 62%

. , and the heat trans er s

93

© 2016 Cengage Learn ng® May not be scanned cop ed or dup cated or posted to a pub c y access b e webs te n who e or n part

L3/C 2

h Am

1.235 i

-

,

f

i

i

.

,

i

li

,

li l

i l

i ,i

l

i

.

i

i

Q

fin

in

Q f

hA

in0

f

0.62 1000 0.073 50 W

0

2263

71 Determ ne the expected temperature drop at the contact between two 304 sta n ess . stee parts i i l the overa temperature drop across the two parts s 1000C l if ll i .

So ut on l i

From Tab e 2 12 us ng a va ue or therma contact l - , i l f l i

q

A

TTL T1T2 R A

T1T2 x

R

R

A

res stance o 304 sta n ess at 200C 0 assum ng t w be unchanged at 100 C 0 000528 m2 0C/W then 304 ss i f i l , i i ill , . ∙ , TC

i

q

A

V

2

TC

TT 100 W 0.031 2 0.0048137 m 2 then 2 0.000528 14

TTC

0.000528 1000 C 10.97 0 C 110 C 0.0048137

72 A m d stee we dment s bo ted to another m d stee sur ace The contact pressure s est mated at 20 atm and the expected heat trans er between the two parts s 300 2 l . il l i l il l f . i i f i Btu/hr∙in . Estimate the temperature drop at the contact due to thermal contact resistance.

94 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Sol ut i on

The temperature drop across the contact surface is i

TTC

q

A

RTL A 300

Btu hr in

2

RTL A The thermal contact resistance, from

Tab e 2 12 s l - ,i

RTC A 0.0022

ft 2 hr 0 F Btu

so that

TTC 950 F

73 For Examp e Prob em 2 26 est mate the temperature dr . heat trans ler s reduced l - to, 3 Btu/hr i op at the contact surface if the t2 f i ∙f .

So ut on l i The therma contact res stance o the concrete b ock/St i heat trans f er s reducedl to 3 Byrofoam for Example 2-26 is 2 152 hr t2 l 0F/Btu the . If f i tu/hr∙ft2, the temperature drop w. be ∙f ∙ ill , i

TTC

qR

A TC

A 3.1340 F

95 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

74. A guarded hot plate test results in the fo llow ing data :

Estimate the thermal conduct ivity o f the test mater ia l.

Sol ut i on The arithmetic averages are Amps = 0.05133 , volts = 8.6, thermocouple 1 = 2 .6677 mv , thermocoup le 2 = 2 .7753 mv.

The average power i s = amps∙ vol ts = 0. 44147 W. The average milli vol t diff erence between 1 and 2 i s 0. 10756 mv. For a 220C /mv setti ng, the average temperature difference will be 2.366 0C. From Four ier’s law i

Q

A

T 0.44147W x

l i f ( . ) f For a samp e th ckness o 2 cm 0 02 m and a test area o

0.01 m2 i

Qx0.373 AT

W mK

75. A stem line has an outer surface diameter of 3 cm and temperature of 1600C. If the line is surrounded by air ate 250C and the convective heat transfer coefficient is 3.0 W/m2∙K, determine the heat transfer per meter of line. Then determine the thickness of asbestos insulation needed to provide insulating qualities to the steam line.

Solution The heat transfer is by convection so i

q

l

h D Ts T 3

2

W 0.03m 160 25K 38.17W / m m K

96 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The critical radius of insulation needed to make the convection equal to the conduction through the line is

roc

0.156W m K 5.2cm h0 3W m 2 K

76. Electric power lines require convective cooling from the surrounding air to prevent excessive temperatures in the wire. If a 1 inch diameter line is wrapped with nylon to increase heat transfer with the surroundings, how much nylon can be wrapped around the wire before it begins to act as an insulator? The convective heat transfer coefficient is 5 Btu/hr∙ft2 ∙0F. Solution The critical thickness determines how much insulation wrapped around a cylinder decrease heat transfer. Using properties of Teflon from Appendix Table B-2E,

roc

0.2023 Btu hr ft 0 F 0.04 ft 0.48in h0 5 Btu hr ft2 0 F

77. Estimate the temperature distribution through a bare 16 gauge copper wire conducting 1.5 amperes of electric current if the surrounding air is at 100C and the convective heat transfer coefficient is 65 W/m2∙K. Solution Equation 2-123 will predict the temperature distribution through the wire. i

Tr T

e

gen

Here .R = 10/ 0

1

r0 2h0

4

r2 r2 0

ℎ/ = 65 g⁄

2. Then, from Appendix Table B-7,

˜, /

= 25.41

G/ = 2,583 % . (_ = 4.016 9ℎ

u = 400 g⁄ ˜ from Appendix Table B%+ = 0.0006454

%+ = 16.664 10

⁄1000B† = 13.1756 10 9ℎ

/

The energy generation is

97 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

i

2

1.5amps 13.1756x103 m

I 2R

1.779x10 4 W m3

e

e

gen

16.664 x10 7 m 2

A0

The temperature

distribution is

T r 10 0 C 17, 790

W

1

0.0006454m

4 400W m K

K

m3 2 65W m 2

0.0006454 2 m 2 r 2

and 0

0

2

2

T r 10 C 0.0883 C 11.11875

r0 r

At the center, where r = 0

T(r) = 10.0883050C

And at the outer surface, here r = r0

T(r) = 10.08830C

where r0 = 0.0006454 m

78. Aluminum wire has resistivity of 0.286 x 10-7 ohm-m where resistivity is defined as ohmarea/length. Determine the temperature distribution through an aluminum wire of ¼ inch diameter carrying 200 amperes of current if it is surrounded by air at 800F and with a convective heat transfer coefficient of 200 Btu/hr∙ft2∙0F. Solution Equation 2-123 predicts the wire temperature distribution i

r0

Tr T

e ℎ B†

gen

2h0

i

e

r2 r2

4

0

Here ,

/W

r0 = 1/8 in = 0.0104 ft,

Re

1

κ = 136.4 Btu/hr∙ft∙0F,

7 0.9383x10W ft 2 0.00034 ft

R A0

I 2R A0

ℎ/ = 200 ™†,/

I = 200 amps, A0 = 0.00034 ft2

2.7597 x10 4 / ft and

2

200amps 2.7597 x104 / ft

e gen

.R = 80/ W,

W 32, 467.1

0.00034 ft 2

ft 3

Btu 110, 844

hr ft3

then

98 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

0

T r 80 F 110, 844

0.0104ft

Btu

1

hr ft 3

0.0104 2 ft 2 r 2

4 136.4 Btu hr ft 0 F 2 200 Btu hr ft

2 0

F

99 © 2016 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

or

T r 82.9020 F 203.16r 2 T(r) = 82.9020F at the center, r 0 T(r) = 82.880F at the surface, r = r0

79. Determine the temperature distribution through a uranium slab shown. Assume energy 0 generation of 4,500 Btu/min∙ft3 and the slab is surrounded by water at 190 F with a 2 0 convective heat transfer coe fficient o f 450 Btu/hr∙ft ∙ F. Use a value of 21.96 Btu/hr∙ft∙0F for thermal conductivity of uranium. i on Sol ut Using the figure shown and the governing equation for one-dimensional conduction heat transfer with energy generat ion

i 2 dO T

dx

2

e

gen

0 with two boundary conditions: B.C.1

99 © 2016 Cengage Learn ng® May not be scanned cop ed or dup cate i . , i li d, or posted to a publicly accessible website, in whole or in part.

i

dT

4, 500 2.5in

e

dx

gen

2

28,125

12in / ft

Btu hr ft3

h T 1900 F 0

at x =0 ^?

And

^@

=0

@

= F/2

Separating variable once gives, i

e

dT dx

gen

xC

1

and then again

i

Tx

e

gen

x 2 C x C1 2

2

From B.C. 1

3 28,125Btu/hrftO 0 0 2 0 190 F 252.5 F 450Btu / hr ft F

T

at x = 0. This means that C2 = 252.50F

From B.C. 2 i

C1

e

gen

2

L so that the temperature distribution becomes i

Tx

e

gen

i

x2 2

e

gen

L 252.50 F 6147.5x 2 1280.5x 252.5 2

At the center of the slab, where x = 1.25 in = 0.104 ft, T = 319.180F

80. Plutonium plates of 6 cm thickness generate 60 kW/m3 of energy. It is exposed on one side to pressurized water which cannot be more than 2800C. The other surface is well insulated. What must the convective heat transfer coefficient be at the exposed surface?

Solution Using the governing energy balance equation 2

dOT

dx 2

i

e

gen

0

With B. C. 1,

^?

@ ^@

=0 =0

B. C. 2 -A_O F = â&#x201E;&#x17D;/ . â&#x2C6;&#x2019; .R

@x=L

Separating variables and integrat ing i

dT e gen xC dx

1

And separating variable once more , integrat ing gives, i

Tx

e

gen

x 2 C x C1 2

2

From B.C. 1 C1 = 0

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