Solutions
Manual
to accomparry
Heat
Transfer
tenth
edition
J. P.
Holman
southern
Methodist
tJniversity
Chapter
I
1-1
LT- (39oox9-022
= 625oc
(0.2×0.6)
l-2
1-3
,^cll fuc
q – _l…?1._
– drc nrT
dT dT
– -kw’
1{
1-5
A- w2 q
– -k4nrz
ar -4nk(ra
-r)
d; q-Y
ri rg
Chapter
I
l:7
The surface
area
of the
heavy
t-g
A 0.05
1-10
1-15
il. q
– (5.669
x
tO-t)t7n)4 -Q7r4J – t.9t
4×104
wl*’
b. q
-(5.669
x
lo-r)ten)4 – (Til4I
=
(5
.669x
lo-tlrtro
)4
– Q:R\41
TP=641
K
q-8474-3
wl^t
Reduced by 44.37o
– (70)oJ
KAAT
Ar-: –
q
_
(roxlo-3
#)tsoo)
=
0.0125 m
– 1.25
cm
l-l I
– (2.7)(o”l(*)1o.to
l-12
l-13
l-14
3′
ChaPter
1
l-16
q-hA(Tw-Tpuia)
h-3soo
l-17
hfs
= 2257
kJ/kg
q
=
tuhfr=
(3-78
kg/hr
)(2257
kJ/kg)
=
853
1
* = 2’37
1-18
hr’ft’ ^2 ‘
1-19
q
=oa{t(
T)a
– (Tz)41
x l0-sX0.85X0.006X3)t(n)4
– (298)41
1-20
– 1’489
g
l-2r
q -dt4
kJ
=2-37
kw
S
Chapter 1
r-22
r-23
AT
q=lAi= M(Ta-T-)
l-.25
100
-T
q_(t.6)’#_ ro(r*z_ro)
r-26
1-21
Chapter 1
r-29
Q=Qconv*Qrad
qconv
= hA(Tr-T*)
1-30
Q=flconv*Qrad
4conv = hA(T* -T*)
r_3I
e = econv
* erad=
0 (insulated)
4conv
= hA(T*-T*)
Chapter
1
r-32
(100X353
– T,r)= (5.669
x l0-8X&
– T*rn)
= l5(T*,
– 293)
– 293)-
– O.LsTw)a
– (T.r)a
1-36
w
h=4’5
fr (Plate)
t-37
The woman is probably correct. Her perceived
comfort is based
on both radiation
r-38
This rs an old story.
All things being equal,
hot water
does
not freeze
faster
than
ChaPter
I
1-39 4
comfort
depends
on
total
convection.
In the
l-40
Q=Qconv*Qnd
qconv
= hA(T*
– 7*) = Q)n(l’)(6X78
– 68)
-377 Btu/hr
Conitusion:
Radiation
t-41
Ti
=O”C=273K qss
=
0.95
A – (12>(40)
= 480
m2 Ts
= 25″C
= 298
K Ta
= 22″C
= oeAl(Tsa
– \4 )= (5.668 ; I 0-8
X0.95X480X298
a
– 2t 3a’)
= 60262
w
t-42
t-43
lelasswool
= 0.038 thickness: 0.15
m
A:144 +
(4X5Xl2):384
m2
t-44