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Solutions

Manual

to accomparry

Heat Transfer

tenth

edition

J. P. Holman

southern

MethodisttJniversity

I

LT-

(39oox9-

= 625oc

(0.2x0.6)

l-

q

_ ( 0. 0 3 5 X 8 5 )

_ ,

A

22f85 llt'm

= g2,3g

iln.^Z

, ^ c l l f u c

q

• 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. 0 6 2 5

r = 0.

3x + 0.

tuc

@+lq3-s4o)

J rc(0.0g33x*AOW=-

• l _ f

\ " = o r _

( z o 4 ) ( _ 4 4 7 )

rce.og33)

[ o.og,n*+

o.o

lj )*=o

q-2238 W

q

_ ( 0. 2 8 X 3 7 5 - 9 5 )4 F ^ ^

A

ffi-1s

wl^z

A- w

q - -k4nrz

ar -4nk(ra

-r)

d ;

q - Y

ri rg

(r=4n(2xl0u)(zl

1 _

_ l. 6 l 7 w

m

-0rt

mass

evaporated=

. 1'.u11=

= e.gl3x l0-s kg/s

199'ooo

• s.7az kglday

t-g

q

_(0.161X200-100)

=3ZZwl^,

A 0.

il. q

(5.669x tO-t)t7n)4 -Q7r4J - t.9t 4x104 wl*'

b. q

x lo-r)ten)4 - (Til4I

= (5.669xlo-tlrtro

Q:R\

T P = 6 4 1K

q - 8 4 7 4 - 3w l ^ t

Reducedby 44.37o

g [

^

=

ry

m-

• (70)oJ

W

KAAT

A r - : -

q

_

(roxlo-

#)tsoo)

m

cm

l-l I

z.Ls

• (2.7)(o"l(*)1o.to

4gzLr

' \ 1 2

LT = 0.632"C

l-

q-

-(5.669x lo-*)t(r o:n)4-6214l

A

l-

g

x lo-t)t(t 3n)4 - (698)

A

l-

q

= (s.

x 10-r)(+ox0.

,2t(300)

ChaPter

1

l-

q-hA(Tw-Tpuia)

FromTablet-z

w

h-3soo

;zJc

q=(3500)rTd.L(40)=(3500)zr(0.025X3X40)=32987w

q = mc

oLTpaid

[

• (0-5kg/sX4180J/kg'oC)Af

L T = 1 5. 7 8 o C

l-

hfs = 2257

kJ/kg

q = tuhfr= (3-78kg/hr

kJ/kg)

FromTable

r_z

h_Tsoo

q = h A ( T * - T p u i a )

2370{

• (7500x0.3)

(T* - 100)

T* =965oC

1-

q=

LALT

3 x to

Btu -

hQ3z-zrz)"F

ffi-

h-1s00+=

17

w

h r ' f t '

^ 2

' o C

q =oa{t( T)a

• (Tz)

2000W

x l0-sX0.85X0.006X3)t(n)

\ = t 2 3 3 K

q

dr4= (5.66gxt0-*Xt

+273)a

x lOs

g

m'

A

l-2r

q

-dt

A'ro^

06 = (5 -66gx

1o-8)

T

K

kJ

=2-37kw

S

r-

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

qconv = hA(Tr-T*)

FromTablel-2 ft

m '. o c

econv

=(1S0)z(0.05)0X

y

length

m

Qrad

= oeAl(Tta

-Tza)

x to-8xo.z)z(0.05X1)

(4:a4 -2834)

=272I

t"ngth

m

Qtotat

= 48O

+ 272=5079 Y

m

Most heattransferis by convection.

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

4conv

= hA(T*

-T*)

FromTable

lJ h= 4-

m '

. o C

{conv

=

f;ll]il;r"so

• 1. (2sides)

erad

= o4,(Tta

-Tza)= (5.

x to-8X0.gX0.32X323a

-zgla) (2 sides)

=28.7W

4total

= 24'3W +28'7 W = 53

W

Convectionandradiationareaboutthe

same

magnitude.

r_3I

e

=

econv

erad=

0

(insulated)

4conv

= hA(T-T)

From Table l-

h: J-

m ' ' o C

Qrad

= oeAl(Tl

Tzo'),€ = l'0,

Tz = 35oC= 308 K

o = h(Tr - T*)

+ o€Argt

• rz4)

o = (12)(4-273)+(s.

x tO-8;1t.0X

Solutionby iteration:

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

r-

(100X

• T,r)= (5.

x l0-8X&

,o

• Trn)= l5(T,

ts(Twz

x t O-

)[(rg7 - O.LsTw)a

• (T.r)a] = 0 =

f

(T*r)

T*,

f

(Tnr)

3r

-77.O

T*r

=397- (0.15X313.3)

= 350K

w

h=4'

fr

(Plate)

w

h=6.5 +

(cylinder)

m -. " u

T* = 2O"C

= 293K

hA(T

-T*)

• otA(

-T*4)

Plate

9.5)Q

-zg3)=(5.

x to-8xr4-zgza)

T= no realisticvalue(T =247 K, heatgained)

Cylinder

(6.t(r - 293)=

x to-8xr4 - 2%\

T =320K= 47"C

t-

The woman is probably correct. Her perceivedcomfort is basedon both radiation

and convection exchangewith the surroundings.Even though a fan does

not blow

cool air on her from the refrigerator, her body will radiate

to the cold interior and

thereby

contribute to her feeling of "coolness."

r-

This rs an old story. All things being equal,

hot water does not freezefaster

than

cold water. The only explanation for the observedfaster

cooling is that the

refrigerator might be a non-self defrost model which accumulatedan ice layer on

the freezing coils. Then, when the hot water tray was placed

on the ice layer, it

melted and reducedthe thermal insulation betweenthe cooling coil and the ice

tray.

t-

The price of fuel and electricenergyvarieswidely with time of year and location

throughoutthe world, so individual answerscandiffer substantiallyfor this problem.

t-

lelasswool

= 0.038 thickness:0.15m

A:144 + (4X5Xl2):384m

T (insidebuildingsurface):

• 30:20oC

q lost (withoutinsulation)

: hAAT: (13X384X30):

149,760W

q lost (with inzulation): AAT/[Ax/k

• l/h]

= (38a)(30/[0.

l5l0.038+ r/r3] : 2862W

Energysavingby installinginsulation: 146, W

This numbermust be combinedwith the energycostsobtainedin Problem 142 to obtain

the cost savingper hour

(or per day,etc.

t-

This problemis quite open-ended andthe arurwerswill stronglydependon the

assumptionscot/bunkmaterialsetc.

2-l

T

• T o

i - F

Et**#),",

Ax = 0.

m

r? 0 0 _ 3 0

ffi

Assume Linear variation:k =

he+

pf

:=

-frlq_-

r,

:"!,:'^T^:J=

#1,

*,r,

-,r|

Q_=95oC,

4=62"C, Tt=35"C,

Ar=ti.OiS

,b,

• zs

f

2-g

l l

h = 0. 0 1 5 + 0. 0 0 2 = 0. 0 1 7

4Ai

(1500)a(0.03)

v'vv

ln(ro

l=r)_ ln(0.0120,015)

0.000 433

Znk Zn(46)

v'

l l

hh Q97)n(0.034)

v'' I

s t

o C. m

LR-0.

?

q. -? 2 3 * = 3 0 1 7

W m

L 0,

= 300x-

ox

^ t

d"T

'nn

I aT

; T = J O U = - _

h e a t i n g u p

dx' a dr

Dr

• = - 3 0 a t x = 0

\

ox

• = 6 0 a t x = 0. 3

l -

ox

=-(0.04X-30)

u, x = 0

A

\ , / \ - -. ,

^ 2

+=--(0.04X60)-

atr=0.

A \

A m '

2-Il

d:0.000025;

k:16;

L=O.gm; h=500;

T*=20oC;

To=Tl=200oC

m

(4)/(r6)(o.oooozs11tn

0'0:0'L =200{

lg

q = -2kA(ffi '

lfuh= -2kA(_m0,6)

Q)06>n(o.oaoozs,

6)(rBly

W

lz

ku=W-5.35x10-

RA, =

RFi

=

*=#=313#

R--

"c

ozx o'5778t

R f = -.. ,.

= 7. s 9

r

(o.o3gxo.577g)

' 'J-

&=#

(0.577gJ

Rr

=

*=0.

& =+=e.

r

q

_ 7 2

• Z O

A - E - - r ' r u

h ,. f t ,

u

• • = 0. 0 9 9 6

In

2-r

T

&veralt

U I2O

• 0 '. 0 0 2 5 = 0. 3

A%veralt

$veralt

A 4 r = ( 0. 3 X 6 0 ) - t g o c

Rr=+=

l b

2-

AT

o - _

t I n

& t u , , , = # - - &

= 9. 7 5 2 x r 0 - 3

arcQOa)

I

_ _ l

p.

• m4- -7.95g

' \ n s - @ =

& o n u -

• • 3
    • 1. 5 9 2

vv'!r'

hA (ZO)(4,n)(0.0,

C T

_

• Q A 1 \ I /

q -

l.S9Z

W

2-lg

di

in. do

3.50 in. k - 43

W

m. o C

&teet

ln(3.5f2.9)

Rin, =

(zrc)(43X1)

ln(5.5f3.5)

(2n)(0.06X1)

(10)zr(5.5X0.0254)

Ro,

I.

LT 250_

w

q -

R 1.

m

7-r

kA- 0. 166 kf :0.

3 t 5 - 4

m ' o c

-T34E -

-296.7oC

= 9.

xl0-

1.lggg

I

&onu

• -l-

v v r. ,

h 4

L 6

2

Qr

= -k4nr

dT

dr

n,f'o

+dr

• -k4ttf'

dr

ry.

r' J

(r

r)

a , l ; - -

l - - 4 r c k ( T o - T i )

\t

rt

q=-+ryn$o

r ,4)

[ v - a )

\ b

r i

I

(t-r)

&n_

\r r/

4rck

2a

r ; = Q. 5 m m - 5 x l O a m

k

r s

x 1 0 - 4+ 2 x l O a - J x l O a

k

• (7xlo-4xt2o)

= o.og

w

q(barewire)= x(a.ootxt

zuff4r,,li-

I 3s.

Wm

4(insulated)

Wm

q -

In(bfsxto-oF

ffiT-----

@+ffi

By iteration:

rc

= 135mm

thickness

1345mm

(1)(r2)

(30)rc(2.067)

ln(Z.37512.067)

2rc(27)

ln(3.

2n(0.An)

6.16x l0-

1. 1 8 8 x l 0 - 4

Rp

(1x12)

&=

• 1. 6 5 9 x 1 0 - l

2n(3.

I G

2-

Fiberglass

Urethane

Mineral Wool

K R

0.091 I 1.

R : 1

k

CalciumSilicate 0.058 I7.z

M

\

1000"c

rz:4oooC

Ti - 55"C

k m - 9 o

m w

h = t 5 Y

m. o c

r e L a '

^ 2. o c

kr = 42

mW

T* =40oC

m. o C

g

= h(r?- T*)

= (15X

• 40)-

g

A

• • \ - J ' @ /

\ ^ " / \ u "

' u ,

^ 2

q

_ t.

(1000- 400)

i = o ^ T

L x m - 0. 2 4 m

q

_k,

(4oo

A

' n r '

L x F

L x F : 0 - 0 6 4 4 m

r

Uniformly distributedheat sources

n

d ' T , q

• • • • 0

T = T t a t x - - L

d x "

k

T = T z. a t x = + L

. )

T -

q x -

|

= -

• c 1 x * c 2

2k

n

.? t

T

qL-

1 1= - - - C 1 L + c 2

' 2 k

rz: -

+*cl

L+c

L 2 k

T

+ U 3

• • Z y + T ) : T t

x * T t + \ '

2 k \

2 L 2

2-3r

r y = 2. 5 r y : 3. 5 r y =6. 5

R,_

ln(ry|ry)

-ln(3-512.5)

= a.

r

0.22(2n) 0.22(2rc)

Rr _ln(rylr) _1.

L

0.46(2n)

R*=

^

w

hA

(60)n(2X0.065)

R

r.

"c'm

w

LLT

/.' - .Of}? \I/

L

R r.

lcl