Solution manual heat transfer j p holman 10th edition (www elsolucionario org)

once an insulation material is selected for this problem a commercial vendor mustUe consutted to determine the cost, as no cost figures ary- $ven in the.. Cost figures will also have to

q-2238 W1{

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mQrad = oeAl(Tta -Tza)

o = h\(Tr - T*) + o€Argt4 - rz4)

o = (12)(4 -273)+ (s.669 x tO-8;1t.0X44 - 3084)

Solution by iteration:

T t - - 7 = 2 8 5 K = 1 2 " C

Q = Q c o n v * Q n d

qconv = hA(T* - 7*) = Q)n(l')(6X78 - 68) -377 Btu/hr

Fot T2 = 45oF = 505oR

erad=ocA1(T1a -Tzo)

= (0.r714x to-8xo.g)zr(lX6X53g4 - so54)

= 544 Btu/hr

Qwtat=377 +5M = 921 Btu/hr

Fot T2 = 80oF = 540oR

erad =(0.1714 x to-8)(o.g)a(1)(6X5384 - 5+04) = -36.4 Btu/hr

For ice 1& = 80 calf 9=3.348 x tOs ft<g

Mass rate melted - 165'86? - = 0.495 kg/sec

3.348 x l0'density of ice - 1000 kel^3

volume rate melted = 0.495= 4.95x loa *34

The price of fuel and electric energy varies widely with time of year and location

throughout the world, so individual answers can differ substantially for this problem

t-43

lelasswool = 0.038 thickness: 0.15 m

A : 1 4 4 + ( 4 X 5 X l 2 ) : 3 8 4 m 2

T (inside building surface): -10 + 30:20oC

q lost (without insulation) : hAAT: (13X384X30): 149,760W

q lost (with inzulation): AAT/[Ax/k + l/h]

= (38a)(30/[0 l5l0.038 + r/r3] : 2862 WEnergy saving by installing insulation: 146,897 W

This number must be combined with the energy costs obtained in Problem 142 to obtainthe cost saving per hour (or per day, etc

t-44

This problem is quite open-ended and the arurwers will strongly depend on the

assumptions cot/bunk materials etc

- - J": (ffKF(o,ffios2 - o.ots'x250 - 3sx0.3e) = s 0E w

To = T-" = 93 + 25]t + 31'6 = 150'3oC

(o

-**."*t#r-*l**

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A

7,,

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k - 2 5 l\ =,75 (left)

Tt* = 50oC

I'D - 50 (right)T2* = 30oCax2

T - _ T * c 1 x * c z

T = T t a t x = - 0 0 4 ; T = T zat x = +{-04

( 1 )(2)(3)( 1 )(2)(3)

Use solution from Prob.2-28

rn : r o - Z k " ' q t *Tt \rz- (soo x t-q6xo'oos)2 * 100 t 200 = 46z.5oc

2 6

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ri-ro=-#r^, -,iz)+"r"[f) tu)

Heat Transfer is:

1 2

? o

single glass plate: R = 6.41x l0-3 + 0.0833 + 0.02 = 0.1097

u = != e.l I

-j{-R m ' o c

? l

Q = IJAT= (0.371X150 - l0) = 52 +

Insideofcopper= 150oC

Chapter 22'64i

e-rrx dx, = hP0o le-*16 =

Chapbr 2

2-70

0 = cp-* * c2€w In case II the end of the fin is insulated

+l = Q the boundary conditions are 0 =0s at x = 0

4 = total efficiency A/ = surface area of all fins 4f = ftnefficiency

A = total heat transfer area including fins and exposed tube or other surface.7O = base temp L = environment temp

Qact = h(A- Ay)Qo - ?i) + rUAyhQs - T*)

Tube heat transfer = (l10X0.0719)(20O- 93) = 846.6 W

-!l- = 1o.aT(2)z(1 fin 10X o.o2542 - o.onsz )e00 - 93) = 3 1 46 *

Total fin heat transfer = (31.46)(105.3) = 3312 W

Total heat transfer =846.6+3312 = 4159 W

t :

-2)(

-rl _ l

)J

i ) = c

X

).1i3:

26

.006:

0.01,

l c m.723

h = 4 0 T^ - )5f|0('-1, /v \r' T* - gOoc

l 2

=2.388(170)(0.101X0.002)

total: (225)(0.1029) = 23.1'5 W

A t

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Total: (8X6.62): 53 W

2-9L

Surface area fromProb 2'90 = (8X0.304)(0'02 : 0.04864

tuea per circular fin = (2)r(0 03252 - 0'01251= 0.00565

Number of circular fins:0.04865/0.0565 :8.6 Round offto 9 fins

Chapter 22-95

4r =0.934

q = (0.93X2)n(0.02592 -o.otz72x56)(125 - 30) = 15.84 w

+5

4f =o'97

q = (2r(o.97x0.0082 + 0.0012 f/2 $s)(zw - 25) = 1v.2 Wm

$ o

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Chapter 22-123

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Chaptcr 22-125

For 6 mm length total surface area

f+

For very long rod c2=0

Acts like two long rods l00oC on each end

\k4' ) 4,r = o.g5

q = (2)(r 6lvt (O On 52 - O.O P ) Q)A7(0 95) = 0.69 LT

Conclusion: Several thin fins are better than a few thick fins More heat trangferfor thc same weight of fins

once an insulation material is selected for this problem a commercial vendor must

Ue consutted to determine the cost, as no cost figures ary- $ven in the Pioblgmrtutrn1"nt Cost figures will also have to Ue Aetermined forthe material with

;;A;ti"r coatinglryhen the reflective material is installed it may have an

rntisi"iiyofO.f] bugafter a period of time the surface may oxidize or becomeiltA with foreign mattef sdctr that its emissivity will increase In that case theeconomic benefiiof the coated material will be reduced

2-138

For this problem, the net heat generarcd in the tube will bt tqqlg^It^P*:.1"n

will be detivered to the fluid by convection The temperature gradiery at:n:

,utror of the tube will be zero The problem does not state whethef the llulc ls onthe outside or insiOe of the tube so b6th cases must be examined In eithpr case themaximum tube tempefatufe will occur at the insulated surface' To effect the

;ild;;; *igftt fi,rt assume a surface tgmperyture for the tube surface in contactwith the fluid This will then determine the surface area Suitable combinations of

t JU" f"ngtft and diameier may then -be-examined to equal the total surface area'fh; trea;t;"t"tion .quuiiori fot a hollow gylil*t may then be solved for theother tube surface t"tp"tututt if a tube wali thickness is assumed (i'e"

egt"Ufi.fting the other iiameter)' The resultant value of temperature must be

reasonable, i.r., torilnffi ounioosrv, there are many combinations which will

be satisfactorY

5 a

f - \ = (c5e-1'x * c6eb)[clcos(,1.y) + cs sin(,try)]

or cg = 0 If ca - 0 then have trivial soln then,

c 5 * c 6 = 0 a n d c 5 = - c 6 0 = 0 a t x = W

0 - (trr-Aw + ,6rlwxcs sin(,try)l

3-2

" ' c 7 = O.' either c5 * c6 = 0

.' x = 0 but it was stated that rt < 0

Error is (1.00 - 0.gZ) x l00%o = 8Vo

o - ( t r r - h * c 6 e b ) ( c t ) Q

s- 2n(r) - 2.129

cosh -t (t zlo.lz5)

q - frSAf =(2.128X1 .8X67- 15) = tgg.Z Wm length

b s

1 8

1 6 14

1 2 10 8 6 4 2 0

0.25 0.3

,-l o'w'- ^-l *yf b- - (r t " ' L ta + r^a1) = ti

Chapter 3

I R,]

Chapter 33-55

Chapter 3 The Solution:

Chapter S The Solution:

other nodal equations are the same as in prob 3-6l

Equations for nodes | , 2, 4, 5, 7 , 8, 10, I I remain the same as in prob 3-5S:

Chapter 3

I - (2tO)(4.418 + 3.068)(10-5)

(2X0.015) = Q.524ht

( 6 ) B i T * + ; ( Z W + 7 5 + 7 7 ) - @ i + 2 \ 7 6 : O

( 7 ) Z 6 + 1 0 0 - ( B i + z 1 t i = g

8 4

fz

T1

4

-140-40-1000_200-100

7 r r + 1 5 0 + 2 T g4

[= (rz - ryXl) Per m dePth

e.-nalyticat solution is obtained from temperature equatlon

4 = 0 a n d

t)t

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For this problem the overall thermal resistance will be the sum of the resistance of

*," pif" *Al plus the resistance resulting from the shape factor for a buried pipe'F;g;l; practical standpoint one might *-ant to use several small pipes operating inseries to minimize pt"is.re losses due to fluid friction, but that is not a

consideration in thii problem Properties of plastic pipe may be taken as that ofPVC from the appendix The ouetall temperature Afer.encg may be t+"n a9 themean water temperature minus the soil tehperature 1t d"pth To start the selectionfro."r, take the outside wall temperature as that of the water and choose someionu"ni"nt depth f* th" pipe ThLn select a standard diameter (1 inch,2 inch) anddetermine ttt"itt"p" f*tot p"t unit length Then determine the required surfacerr"u u"O pipe length Then iefine calcu-iation by including pipe wall thermalresistance Repeat the calculation for several pipe sizes fqboth steel and plastic'iilor*iUf", obtain some cost information for ihi Plne as 1 function of diameterand examine the influence on the total cost Something like copper tubing is,pr"U"UfV

"ot a viable alternative because

of the high cbst PVC pipe will probablyiepresent the lowest installed cost option'

q"

ChaPter 3

3-101

For this problem only consider conventional electric burners which have spiral

heating coils Halot* ;O radiant burners have different characteristics' To

simplify the study, ;;;gh*onsiderjust the aluminum layer for an initial

calculation fne maioi tptE"afu effeciis caused by thislayer' After the heat

,prl"Ar, ,t

" stainleJs steel acts iainly as a series

thermal resistance and theobjective here is to

"uuiu"t" the effeciiveness

of the aluminum in bringing aboutuniform heating Ail;"gh th" pun is circular, an initial study could consider justtr, ,pir"ai;t.:ff;;i"ipr"iiii*lv^.0 mm heater elements spaced 18 mm oncenters in intimate

"ontJ"i with an infinite aluminum

plate exposed to the givenconvection

"onOition on A" ,"p surface

The objective.is to determine theuniformity of the surface temperature in contaciwith the later' The analysis canthen be refined to include a contact conductance between the heater element andthe pan, but this rt o"ia *"inly be evidenced as a lowering of the effective burnertemperature Keep in mind that the given value of h=1500,fru could vary byt4}Toormore,soelaboratenumericalmodelsarenotnecessary'

3-103

The analysis of this problem requires.that the heat gained by the building from allsources be equal to fie coofing iupplied by qe reffrgeration system' The heatgained from the /;d pu,t.itf,t*gft twb thermal resistances: that of the

concrete slab and that resulting from the shape factor of the slab in contact withthe ground In add'i-ii,orr, of coolse, there is an internal convection resistance' Theheat gained rro* tt

" *tside environment

air passes through thgrmalresistances '

of the outside convection, concrete wall, inside convection, and whatever

resistance results it"*i"tOUation of ift" outtia" insulating material' Note that theheat gained tnroug'iiire fi;;;;hb *ill nor depend on the outside insulation, asil;;; the interiJr conditions are maintained constant'

01+

Chapter 4 +l

T * = T m + A * s i n o n

Energy balance: q = h.4(T -L) = F{ {\

\dc )l.,et K = +dong pV with initial condition T = To at c =0i

correct value is 1.0 Error = 7 g4Vo

u

q = oA(74 - T*4 ) + hAe - T*) = -cpv t

dc

q {www.elsolucionario.org

Chapter 4

v - lrrdt2

J = - l l l ' 3

-1)(- 1)

= 308.3

1 00.gg0.980.gg

w

; 7 " c

0.650.410.370.33

0.320.080.040

k - 2 3 0

k - 1 2 ghL

Ti - 400

eo = 0.29ei

e = 8.42x l0-5

T=5'o

l o 7

Chaper 44-38

Chapter 44-41

ei 250 - l0

+=0.75 n-

e: - = ( 0 7 X 0 9 ) = 0 6 3

Chapbr 44_ffi

r 0 = 7 5 c m L = 1 5 c m 4 = 2 5 " C L = O o C I O = 6 o C

h=rz ,w-= a=7xr0-7 mzfs k=r.37,}

fi o.s+ +=r.075 hrs e: ei z5_o = !;\=0.24=[+J t'+.]

aJtz o.224

1 00.9

0.650.252

t t 2.

959518160 - I 175

10, 2001320 = 31.88

15, l00l2l0 = 7I.920,000/100 = 20010,000/50 - 2wUse A,t = 0.5558 sec

Compute for 2,2A, nO time increments.

t=16'600

I I

t t +

Chapter 4

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Chapter 44-74

(100x0.

= s = 6 6 6 6 70.015

Chapter 44-92

Tl= -c I +(tglt tt/t9.7).((cl{ I )o.2F2rtfczC rffi

2 TJ2- 'c2+(!ql I v.t4.$r.((c t€)o l )+((c.{afo.lxt I oczyrrlll

lJ= {t+(Ecl3l ul 79.4).((clclyo 2t+2.1 (c-.,f:trm osxrrcs-ffrvo ?rr

t 0 T T F -Clol{tC3lt/E2Ajsl /f;(Ct4l0)r1o 4l"t{fcrg-ClO}rn Ol\+l/alif tovo rr+rrir rf rrnfri r\./rr^-.r^\rD (rr

T l l - { l l +(tCl nEn.rSl'((ClOC l l )O.3)+((C l rc l t fn.C>rnC t z"c t r tnewr r o-e i i vr enr

t 2 Tr2? { l2+(!C3tt/t9.D.(((Ct t{l2l(}.z)+f(Ct64 r2vn 2*{ir o.c r zyr rnr

l 3 Tl3- "C l3+t!C3lUt ?9.4)f(((C941 3lm 2)+2 Uell4rwo oq\+/ttfrLn r t\{r ?n

Chapter 4

ll,7 Igll,7 lg23,43923,43923,43923,439

=(o l€!:(9{!}+o-rg!:(

t 7 t4

1 7 t 480024A040006000600080024008002400

I " ?

23.2323.2324.2624.2624.2624.26

t o 43.757t) ln.6973t t24.1308 129.9938 taJ.2565 I tt.65t3 _I3.r€

32s.8n

'f45.017t AA")A<

97.str7l 92.50r0r urlm

403.79'29

42t.t224 42r-tss4l tw.97591

6 3 41t.97t4 412-fi7r ts3.7s27 rn.$9 12t.tw 360.6248 422.5v8) 36t.ttt I 429.2fi3 422-rO9 t6r.wn

7 4 520.OtT' ff)4.lr2t 411 t692 s32.%l tt5.97J 420.rt95 s35.yTz 5tt.7tgl &2.2916 535.7{t3 5t9.2399 qn.653/l

t 5 6t5.t?t 5t9.(b55 16i2.sffi 631.t5{tt60f.173t 473.5tt9 636.{05 60t.4612a76.546,2637.2459609.45{3 4n.u59

9 6 rc5.t037 667.9n6 5(B.2tt!) T2S.3rt5 616.1659 tin.gyr6 731.6t{t 692.W7 526,06,t5 733.9256 693.73% trr.2a2r r0 7 7t9.t93t 7il.3159 552.23t65 ttz.?131 ?62.rc' ffi.7599 un.wt4 769.%2r 5?t.7An 8n.1209Tn.mt s73.Sns

0 lM2

84.4280.4280.4278.00

t t 1www.elsolucionario.org

Chapter 4

Excel solution for Lr = 25 sec and 75 sec shown below

rz - 190oC occurs at six, 25 sec time increments.

A?rn r,

16.9317.1617.1616.8317.1617.1617.517.5

Chapter 4

t S +

Approximate as inf plate with 2L = 0.5

Center plane is insulated

1 5 0175.94

t 4 5

(13 x 10-7)@X60)

= 0 2 9 1

at )

-| *j,

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Chapter 44-127

Excel solution shown for Ar = 5 sec

I minute = 12time increments

The Equations

t {t4OFB2+gY4 {4OtnZ+mY4 <100+A2+El+D2y4 <82+e+F2y4 {2fi}+q2+F2Y4 {tfi}rDl+p2Y4

4 {l4O}83+C3Y4 {4(}rA3+D3V4 {l0OtA3+El+D3Y4 {83+C3+F3y{ {20(}FC3+F3y4 {tfl}FD3+E}y4

5 {l'l0}E}4{r4y4 {40rA4+IXy4 {l0Orl4+Bl+114Y4 =(84+Of+F4y4 <20IJ+U+F4y4 {l00rD4+F,y4

6 {l40FBt+c5y4 {O}A5+D5y{ {I0O+A5+E}IP6Y4 {85+CS+Fsy4 {200+C5+FSY4 {Ifi}+Dt}E Y4

7 {I4O+86}C6Y4 {{o}A6fD6y4 {100+A6}E6}D6Y4 <BGFCftF6Y4 {2{,0{C6}F6Y4 {IflFDGtE6y4

t <l&+g7+.g;TY4 {4OrA7+p;N4 fl(fr+N+Et+DU4 =(87+CZ+Fry4 azU}+Ct+Fry4 {IOO}D?+EN4

9 {140+Bt{CtY4 {4(}rAt+DtV4 {loorA8+rg+nnyl =(Bt+cg+Fgy4 {200rct+Fry4 {lUl+D8+88y4t0 {I4O}89+C9Y4 {,f0+A}FD9y4 {1fi}+A9}89}D9Y{ :(89+C9FF9Y4 {z{)O+CgrF9Y4 {lm+Pe}89Y4

l l {l4orBlOrCl0Y4 {'lo}AI0+Dl0y4 {lq}rru(}rEl0+Dlt {BlO+ClOhFl0Y4 {2OO+CIO+Fl0Y4 {100+Dl0rElOy4 t2

- {l4O+Bl l+Cl tY4 {40+At l+Dt ty4 {l(XFAl l+Ell+Dt {Bl l+ct l+Fl ty4 <200{Ct l+Fl lY4 {100+Dt l+Et ly4 t3 {140}Bl2+ct2Y4 {4(}}Al2+Dl2Y4 {tm+AlZ+Et2+Dl <Bl2+Cl2+p12y4<2mrcl2+Fr2V4 {l0O+Ol2+g12Y4 t4 <140+8t3+cl3y4 {4(}}Al3+Dl3Y4 {l(X}+Al3+Et3+Dl {Bt3+Cl3+Fl3y4 {200}ct3+Fl3y4 {lUl+Dl3+Et3y4

l 5 {l4O}Bl4+Cl4Y4 {0+Al4+Dl4y4 {l0o}Al4+Et4+Dt <Bl4+Cl4+P14Y4 {200rcr4+Fl{y{ <lU)+Ol4+914y4l6 {14(}}8t5+Cl5Y4 {O}Als+DliY4 {l(X}+Al5+915.rP1 <Bls+Cl5+Ft5Y4 <200+Cl5+Fl5Y4 {l0OtDls+El5y4t7 {l4OFBl6+Cl6Y4 {4$}Al6+Dl5y4 {l0or,tll6}El6}Dl <Bl6+Ct6+Fl6Y4 {20OrCl6}Fl6y4 {l0O+Ol6+El6Y4

I E {l4O}Bl7+ClTtl4 {,lorAlz+plnt4 {100}Al7+g11Pt {BtZ+C17+P17Y4 {20o}ct?+Ft7v4 {IOO}D17+F,M4

l 9 {l'00}Blt+Clty{ {4(}}Att+Dtty4 {l0O}Alt+Elt+Dt {Blt+clt+Flty4 {2mrclt+Flty4 {lfi}+Dlt+El8Y4

n {l4O}Blg}Cl9Y4 {{o}Al9*Dl9V4 {100}Al+rEllrnt {Bl9+Cl9+Fl9y4 {20o}ct9}Fl9y4 {l0o+ntg}El9Y4

2r<l4olB20+c20y4 {4o}A20+Df2OV4 {lm+^l2O+ElO}D2l =(B2o+C20fF20Y4 <20OlC2O+F20Y4 {1fi}rOZ0}E20Y4z2{l4O}B2l+ClY4 {4O}A2l+pr2rY4 {100+A2l+Elt+D2 <B2l+C?l+FzlV4 {2OO+CZl+F2lY4 {lflr+nrzl+Elly4

23 {l,t0rB22+e2Y4 <n+AZ}+D'aV4 {lmr122+Vn+DZ -(B?J}+Q:2+FZ2ll4 {200r€22+FTlY4 {Ifi}rD22+EUtY4

24 {l4O}823+g2'3Y4 <40+A23+p41'Y4 {l0O}A23+E23+Dz <Bn+e3+F23V4 {20O}C23+F23Y4 {1fl}+PZ3+Et3Y4

25 lltfr+Bl24+e4Y4 -(40+lzl+v24y4 {1fl}rA24+84+D2,

<B?41g4+F24Y4 {2fl}FC24+F24Y4 {1fi}rD24+H24Y4

26 {l4O}825+C2SV4 {o}A25+m5y4 {l0O+A25+El5+Dz =(825+C25+F25Y4 <200{C25+F25Y4 <lflFE)lt5+Elsy4

n {I4OFB26+A6Y4 <n+AJ:6+[JI6V4 {lfi}+A26+Hlerutr -(826+Q6+F26Y4 <2Co+(26+F26Y4 {I0O}D26+g26V4 2t {l40+827+AW4 a4O+A2?+DZW4 {t00+A27+A7+D2' <827+A7+F2ry4 <20o+e7+FZW4

{lfi}+D27+nN4

29 {140+828{C2tY4 {40+A2t+mt,4 {100+A2t+El8+D2l <828+et+F2tV4 {200+C28+F2tY4 {1fi}+D28+E28V4

| +'?