Heat transfer jp holman 10th edition

7/21/2019 Heat transfer jp holman 10th edition

1/10

SolutionsManual

to accomparry

Heat Transfertenth edition

J. P. Holmansouthern Methodist Jniversity


2/10

Chapter I1-1

LT-(39oox9-022

= 625oc(0.2x0.6)l-2q _(0 .035X85) ,A 0.13

- 22f85 llt 'm2

= g2,3g6iln.^Z1-3

, ^ c l l f uc - _l...?1._-drc nrT

r- a x + b ; x = 0 ; r - 0 . 0 3 7 5

dT dT - k w '

x-0 .3 , r-A.0625r = 0.083 x+ 0.0375

| tuc- @+lq3-s4o)J rc(0.0g33x*AOW=-- l_ f

\"=or_ (zo4)(_447)rce.og33)o.og,n*+ .o3j )*=o= -

?q-2238 W

1{q _(0 .28X375-95 ) F^^A ffi-1s08 wl^z1-5

A- w2 q - -k4nrz r -4nk(ra r)d ; q-Y

ri rg

(r=4n(2xl0u)(zl + 196)1_ _ l .6 l7wm -0rt85

mass vaporated . 1'.u11= e.gl3 l0-s kg/s199'ooo- s.7az kglday


3/10

Chapter

l:7q 4 -T*

T-m=Zrlk ' httdo

l-8

30+20

(39) , r-

11.89W/m*arffi+er6rI

Like many kinds of homespun dvice, his s bad advice. Alltypes of heat ransfer;conductio4 convection, and radiation vary directly with area. The surface area of thehead s much ess han hat of the other portion of the body and hus will lose ess heat.This may be shown experimentally y comparing expozure n cold weather wearingheavy clothing and no hat, o that wearing a heavy hat and only undergarments


4/10

t-gq _(0.161X200-100)3ZZwl^,A 0.051-10

1-15il. q - (5.669 tO-t)t7n)4 -Q7r4J - t.9t 4x104 wl*'

b. q -(5.669x lo-r)ten)4 - (Til4I= (5 669x o-tlrtro )4 Q:R\41

T P = 6 4 1 K

q-8474-3 wl^ tReduced y 44.37o

=70.9 [^2

=188ym-

- (70)oJ 704.8W

KAATA r - : -

q

_ (roxlo-3 )tsoo)

= 0.0125m - 1.25 m

l-l I

z.Ls8 (2.7)(o"l(*)1o.to gz r\ 12) '

LT = 0.632"C

l-12

q--(5.669 x lo-*)t(ro:n)4 6214lA

l-13g =(5.669 lo-t)t(t 3n)4 - (698)41A

l-14

q = (s.669 10-r)(+ox0.32t(300)4

3 '


5/10

ChaPter

l-16

q-hA(Tw-Tpuia)

From Table -z wh-3soo;zJcq=(3500)rTd.L(40)=(3500)zr(0.025X3X40)=32987wq = mcoLTpaid32,987 [ - (0-5 kg/sX4180 /kg'oC)Af

LT=15 .78oC

l-17

hfs = 2257 kJ/kg

q = tuhfr= (3-78 g/hr (2257J/kg) 853

*= 2'37

FromTable_z h_Tsoo $-q=hA(T*-Tpu ia )

2370\{ - (7500x0.3)2 T*- 100)

T* =965oC

1-18

q= LALT

3x to4 Btu - hQ3z-zrz)"Fffi-h-1s00 =85 17

whr ' f t ' ^2 '

oC

1-19

q =oa{t( T)a - (Tz)41

2000W - (5.669 l0-sX0.85X0.006X3)t(n)4(298)41

\= t233K

1-20q - dr4= (5 66gx 0-*Xt000 273)a 1'489

lOs gm'

A

l-2rq -dt4Aro^

06 = (5 -66gx o-8 74T - 5556

kJ =2-37 wS


6/10

Chapter 1

r-22

q = oeAl(T1oTro\= 5.669 to-8;1o.oXazrX0.0aftl4 -2934) 29.t9w

r-23AT

q=lAi= M(Ta-T-)(1.4X315 4L)- h(41 38)

4.025

h-s |14 ym ' . o c

l-.251 0 0 T

q_(t .6) '#_ ro(r*z_ro)T*, - 35.7oC

q -10(357 -10) -2s7 gm2

r-26

From Table l-2 lt - 4.5 y for LT: 30oCm t . o c

q - uALT= (4.5X0,3)2(rO)12.15 ,'Conduqlion

q-kA+Arfr or air = 0.03 -

Wm - o C

, , , (o.o3xo.3)2(30) 3.24W -0^025

1-21

^r_(lsoo)(H)

- ,2soF25

7" .100 -1 .25 -98 .75oF

1t8700 (l r)(T*- 30)Tw = 93'6oC


7/10

Chapter 1

r-29

Q = Q c o n v * Q r a dqconv = hA(Tr-T*)

FromTable -2 ft= 180 +-m ' . o c

econv (1S0)z(0.05)0X200 30) = 4807y lengthm

Qrad oeAl(Tta Tza)

= (5.669 to-8xo.z)z(0.05X1)4:a4 -2834)

=272I t"ngthm

Qtotat 48O7 272 =5079 YmMost heat ransfer s by convection.

1-30

Q = f l c o n v * Q r a d

4conv = hA(T*-T*)

From Table J h= 4-5 -m' .oC

{convf;ll]il;r"so

- 20) (2sides)

erado4,(Tta -Tza)= (5.669 to-8X0.gX0.32X323a zgla) (2 sides)

=28.7W4total 24'3W +28'7 W = 53 WConvection nd adiation are about he same magnitude.

r_3I

e = econv erad= 0 (insulated)

4conv hA(T*-T*)

From Table -2 h:12 J-m ' ' o C

QradoeAl(Tl - Tzo'), = l'0, Tz = 35oC= 308 K

o = h\(Tr - T*)+ oArgt4 rz4)o = (12)(4 273)+ s.669 tO-8;1t.0X44 3084)Solution by iteration:T t - - 7 . = 2 8 5 K = 1 2 " C


8/10

Chapter 1

r-32(100X353 T,r)= (5.669 l0-8X& ,o

- T*rn)= l5(T*, 293)

ts(Twz 293)-(5.669

tO-8 [(rg7-

O.LsTw)a(T.r)a =

0=

f(T*r)

T*, f (Tnr)320 158.41350 907.223r0 -77.O3

313.3 0.058T*r=397 (0.15X313.3)350 K

1-36

wh=4'5 fr (Plate)w

h=6.5 + (cylinder)m- . "uT*= 2O"C 293KhA(T -T*) - otA(74 -T*4)Plate

9.5)Q zg3)=(5.668 to-8xr4 zgza)T= no realistic alue T =247 K, heat gained)Cylinder

(6.t(r - 293)= 5.668 to-8xr4 - 2%\T =320K= 47"Ct-37

The woman is probably correct. Her perceived comfort is based on both radiationand convection exchange with the surroundings. Even though a fan does not blowcool air on her from the refrigerator, her body will radiate o the cold interior andthereby contribute to her feeling of "coolness."

r-38This rs an old story. All things being equal, hot water does not freeze aster than

cold water. The only explanation or the observed aster cooling is that therefrigerator might be a non-self defrost model which accumulated an ice layer onthe freezing coils. Then, when the hot water tray was placed on the ice layer, itmelted and reduced he thermal insulation between he cooling coil and the icetray.


9/10

ChaPter

1-394

As in problem -36, it must be observed hat a person's omfortdepends n totalheat eichange with the surroundings y both radiation and convection. n thewinter he w-alls f the room will presumably e cooler han he room air andincrease he heat oss rom the bodies. n the sumrnr he walls are probably hotterthan he room air temperature nd hereby ncrease he heat gain or reduce heheat oss rom the people n the room.

l-40

Q = Q c o n v * Q n dqconv hA(T* - 7*) = Q)n(l')(6X78 68) 377 Btu/hrFot T2= 45oF= 505oR

erad=ocA1(T1aTzo)

= (0.r714x o-8xo.g)zr(lX6X53g4 so54)= 544 Btu/hr

Qwtat=377 5M = 921 Btu/hrFot T2= 80oF= 540oR

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

Qtotat 377 36'4 = 340'6 Btu/hrConitusion: Radiation lays a very mportant ole n "thermal comfort."

t-41

Ti=O"C=273K qss = 0.95

A - (12>(40) 480 m2 Ts = 25"C= 298 K Ta=

22"Cerad oeAl(Tsa \4 )= (5.668 I 0-8 X0.95X480X298

- 2t 3a') 60262w

{conv M(To-4)=(10X480X22-0) = 105,600

Q,rlvrr60,262+ 05,600 165,862W

For ce 1&= 80 calf =3.348 x tOs ft


10/10

t-42

The price of fuel and electric energy varies widely with time of year and ocationthroughout he world, so ndividual answers an differ substantially or this problem.

t-43

lelasswool0.038 thickness: 0.15 m

A:144 + (4X5Xl2):384m2

T (inside uildingsurface): -10 + 30:20oC

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

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

= (38a)(30/[0. 5l0.038 + r/r3] : 2862W

Energy saving by installing nsulation: 146,897W

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

t-44

This problem s quite open-ended nd he arurwers ill strongly depend on theassumptions ot/bunk materials etc.

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Heat transfer jp holman 10th edition