Chapter
11
11-l
1t-3
T*
=25oC u- =
1.5
m/sec .b=
30 cm
h-*to.urlt)Re
LttzPrtt3 hD= E [:rlt”
L’ L’J “Ir pfcp[6r)
Ig)’
t6r,/ ‘[ofJ -l’098 Prl/3-0-884
OF Tw
OC hfr
xl0{
cw Tf
OK
v
xl0{
kcp Pr Py
59 r5 2.47 0.0129 293 l4.gl 0.026 r005 0.69 1.2
4to
Chapter
1l
Tw qconv qnuss qtot
15 -7.993 22.99 15.01
35 7.93 69.97 76.9
tt4
hfr=3.77xt05[kg Tw=26″c=299K p=l3.3kr'{/*t
1l-5
hfs=z.4sxt06
/tg t4n=3m/sec ,=ffi=1.185
Ir
= 1.98
x l0-5 k
=0.026 Pr
=
0.7 Sc
= 0.6
1ryr
Chapbr tl
11-6
0.4593
psia T -25″C=
298
K pn,
=ry=0.2292 psia
.256
=
Z.S6xl0-5
^zfr
Pr, =
D=0
1l:7
11-8
,, =”!o =l2.5oC=285.5
K p=t.34 lt=1.&xl0-s
4z*
Chapter
ll
1l-9
Tw
=32″C= 305
K =
89.6oF Cn
=O.O342 hf, =2.42xt06 rytg
t1-10
u* = 10 mph
T*
=
I
l5oF
L =ll5oF It*=lO mph=52,800 ft/hr L=lft.
{sun
= 350(d”*)e1 €r
= 1.0
Neglect internal
heat
generation
= M(T* Tr) 4raalarm
-‘ eoAT*4 eevap
= hDA(Cw
C-)hfs
{rudlrun
+ {*nnl”i, = 4ool”r,,,
+ {“n”pl.r- qr”d[un
=350d’L= I16.5 Btu/hr
4zq
Chapter
11
1l-tt
h(T- -T,)= hDcwhfs+o(Tw+
-Tr4) T* =43″C 4 = lgoC
.L
=
30 cm W =12 m/sec Pr
= 0.71 Sc
ffi(#)”‘ =0.0226
tt-tz
4)lA
11-13
T* I lsoF ,6 l0 mph = 52,800 ft/hr
Q
genV
+ Qradlrun
* Q
conu
fui,
= Q
radl”* * e
ev
ap
l-rn
4z*
Chapter | 1
11-16
L=65oC l=0.3m u*=6m/sec Ir=38t=3llK
i 38+65
rt =
T=5l.5oC =324.5K py
=1.088 ltf =2.03×10-5
= 1008 kf =0.0281 Sc
4zr
tb2a
T- 35″C
95oF,
u 250 m/min
224 mr/day
Evap
rate
3 50 glnf-hr
: 32.52
gftt-hr
: 780
glft’-day
Relative
humidity
87
ll-21
kpa
= 0.0691
in Hg
Pw
= (0.4X0.0691)
= 0.0276
inHg
Eto
=
[0.37
+(0.0041×1
t
g)X0.0691
0.0
27610.sa
= 0.052
in.f
day
0.
13
t
3 cm/da!
= 0.0055 cm/hr
A
rEQ)2
I
2.57
ft}
=
l.1675
m2
Ero
_o.oo55 n A
A
-ffi=o.oo47r
;ft =47.r#”
For
E
=o E,!-
=20.4__L_
A mt.hr
4Zc.
Chapler
ll
lt-22
Eb
=
0.3
# =
1080
# =
0080X1’167
s)
= 126r
Elh
=
l’02
in/dav
= 1,/}l.6
0’004kr)(P’
P’)o’E
t1-23
I
\”,i/ ^)asadgna RH
=30Vo
Assume
breeze
at 3 mi/hr = 72
= l ‘033
= (0.3X0.5073)
=0’152
psia
= 0’31
in
Hg
Eso
= (0.8)[0.37
+ 0.004
1(72)
l(2’6 -0′ 1 3)0’88
tl-24
=in/day I m/s
= 53
.69
mi/daY
Pw
T* = 80oF
mi/daY
=fr
4n1