ch06
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ch06


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ASSUMPTIONS: (1) Steady-state conditions, (2) Water insulated from ground, (3) Water
surface has e = 1, (4) Heat-mass transfer analogy applies, (5) Ambient air is dry.
PROPERTIES: Table A-4, Air (Tf » 2.5°C » 276K, 1 atm): r = 1.2734 kg/m
3
, cp = 1006
J/kg×K, a = 19.3 ´ 10-6m2/s; Table A-6, Water vapor (273.15K): hfg = 2502 kJ/kg, rg = 1/vg =
4.847 ´ 10-3kg/m3; Table A-8, Water vapor - air (298K): D m s.AB
2= ´ -0 26 10 4. /
ANALYSIS: (a) Neglecting evaporation and performing an energy balance,
( ) ( ) ( ) ( )
conv rad
4 4 4 4
s s s ssky sky
q q 0
h T T T T 0 or T =T / h T Tes es¥ ¥
¢¢ ¢¢- =
- - - = + -
( ) ( )
-8 2 4
4 4
2
1 5.667 10 W/m K
T 0 C 0 273 30 273 4.69 C.
25 W/m K
¥
´ ´ × é ù= + + - - + =ê úë û×
o o <
(b) Invoking the heat-mass transfer analogy in the form of Eq. 6.67 with n = 1/3,
2/3 2/3
p m p AB
m
h
 c Le or h h/ c Le where Le /D
h
r r a= = =
( ) ( )
2 / 3-6 2
2 3
m -4 2
19.3 10 m / s
h 25 W/m K /1.273 kg/m 1006 J/kg K 0.0238 m/s.
0.26 10 m / s
´
= × × =
´
é ù
ê ú
ê úë û
<
(c) Including evaporation effects and performing an energy balance gives conv rad evapq q q 0¢¢ ¢¢ ¢¢- - =
where ( )evap fg m A,s A, fgq m h h h ,r r ¥¢¢ ¢¢= = -& A,s gr r= and A, 0.r ¥ = Hence,
( ) ( ) ( )( )4 4s s m g fgsky
3 3 6
2
T T / h T T h / h 0 h
0.0238 m/s
T 4.69 C 4.847 10 kg/m 2.502 10 J/kg
25 W/m K
es r¥
-
¥
= + - + -
= + ´ ´ ´ ´
×
o
T 4.69 C 11.5 C 16.2 C.¥ = + =
o o o <
PROBLEM 6.73
KNOWN: Wet-bulb and dry-bulb temperature for water vapor-air mixture.
FIND: (a) Partial pressure, pA, and relative humidity, f, using Carrier\u2019s equation, (b) pA and f
using psychrometric chart, (c) Difference between air stream, T¥, and wet bulb temperatures based
upon evaporative cooling considerations.
SCHEMATIC:
ASSUMPTIONS: (1) Evaporative cooling occurs at interface, (2) Heat-mass transfer analogy
applies, (3) Species A and B are perfect gases.
PROPERTIES: Table A-6, Water vapor: pA,sat (21.1°C) = 0.02512 bar, pA,sat (37.8°C) =
0.06603 bar, hfg (21.1°C) = 2451 kJ/kg; Table A-4, Air (Tam = [TWB + TDB]/2 @ 300K, 1 atm):
a = 22.5 ´ 10-6m2/s, cp = 1007 J/kg×K; Table A-8, Air-water vapor (298K): DAB = 0.26 ´ 10
-
4
m
2
/s.
ANALYSIS: (a) Carrier\u2019s equation has the form
( ) ( )gw DB WB
v gw
WB
p p T T
p p
1810 T
- -
= -
-
where pv = partial pressure of vapor in air stream, bar
pgw = sat. pressure at TWB = 21.1°C, 0.02512 bar
p = total pressure of mixture, 1.033 bar
TDB = dry bulb temperature, 37.8°C
TWB = wet bulb temperature, 21.1°C.
Hence,
( ) ( )
( )v
1.013 0.02512 bar 37.8 21.1 C
p 0.02512 bar 0.0142 bar.
1810 21.1 273.1 K
- ´ -
= - =
- +
o
The relative humidity, f, is then
( )
vA
A,sat A
pp 0.0142 bar
0.214.
p 0.06603 barp 37.8 C
f º = = =
o
<
(b) Using a psychrometric chart
WB
DB
T 21.1 C 70 F 0.225
T 37.8 C 100 F
f
üï= = »ý
= = ïþ
o o
o o
<
v satp p 0.225 0.06603 bar 0.0149 bar.f= = ´ = <
Continued \u2026..
PROBLEM 6.73 (Cont.)
(c) An application of the heat-mass transfer analogy is the process of evaporative cooling which
occurs when air flows over water. The change in temperature is estimated by Eq. 6.73.
( ) ( ) ( )A B fg A,sat s A,s 2/3
p
M / M h p T p
T T
p pc Le
¥
¥
é ù
- = -ê ú
ê úë û
where cp and Le are evaluated at Tam = (T¥ + Ts)/2 and pA,¥ = pv, as determined in Part (a).
Substituting numerical values, using Le = a/DAB,
( )
( ) 3
s 2/3-6 2
-4 2
J
18 kg/kmol/29 kg/kmol 2451 10
0.02491 bar 0.0149 barkgT T 
1.013 bar 1.013 bar
22.5 10 m / s
1007 J/kg K 
0.26 10 m / s
¥
´ ´
é ù- = -ê úë ûé ù´× ê ú
ê ú´ë û
( )sT T 17.6 C.¥ - = o <
Note that cp and a are associated with the air.
COMMENTS: The following table compares results from the two calculation methods.
Carrier\u2019s Eq. Psychrometric Chart
pv (bar) 0.0142 0.0149
 f 0.214 0.225
Evaporative Cooling T¥ - Ts = 17.6°C
Observed Difference TDB - TWB = 16.7°C
% Difference: 
17.6 16.7
100 5.4%.
16.7
- ´ =
PROBLEM 6.74
KNOWN: Wet and dry bulb temperatures.
FIND: Relative humidity of air.
SCHEMATIC:
ASSUMPTIONS: (1) Perfect gas behavior for vapor, (2) Steady-state conditions, (3) Negligible
radiation, (4) Negligible conduction along thermometer.
PROPERTIES: Table A-4, Air (308K, 1 atm): r = 1.135 kg/m3, cp = 1007 J/kg×K, a = 23.7 ´
10
-6
m
2
/s; Table A-6, Saturated water vapor (298K): vg = 44.25 m
3
/kg, hfg = 2443 kJ/kg; (318K):
vg = 15.52 m
3
/kg; Table A-8, Air-vapor (1 atm, 298K): DAB = 0.26 ´ 10
-4
m
2
/s, DAB (308K) =
0.26 ´ 10-4m2/s ´ (308/298)3/2 = 0.27 ´ 10-4m2/s, Le = a/DAB = 0.88.
ANALYSIS: From an energy balance on the wick, Eq. 6.71 follows from Eq. 6.68. Dividing Eq.
6.71 by rA,sat(T¥),
( )
( )
( ) ( )
A,sat s A,s m
fg
A,sat A,sat A,sat
TT T h
h .
T h T T
r r
r r r
¥¥
¥ ¥ ¥
é ù- é ù= -ê úê úë û ê úë û
With ( )A, A,sat/ Tr r f¥ ¥ ¥é ù »ë û for a perfect gas and h/hm given by Eq. 6.67,
( )
( ) ( )
( )pA,sat s s2/3A,sat A,sat fg
 cT
T T .
T Le T h
rr
f
r r
¥ ¥
¥ ¥
= - -
Using the property values, evaluate
( )
( ) ( )
( ) ( )
gA,sat s
A,sat g s
13 3
A,sat
v TT 15.52
0.351
T v T 44.25
T 15.52 m /kg 0.064 kg/m .
r
r
r
¥
¥
-
¥
= = =
= =
Hence,
( )
( ) ( ) ( )
3
2/3 3 6
1.135 kg/m 1007 J/kg K
0.351 45 25 K
0.88 0.064 kg/m 2.443 10 J/kg
f¥
×
= - -
´
0.351 0.159 0.192.f¥ = - = <
COMMENTS: Note that latent heat must be evaluated at the surface temperature (evaporation
occurs at the surface).
PROBLEM 6.75
KNOWN: Heat transfer correlation for a contoured surface heated from below while experiencing
air flow across it. Flow conditions and steady-state temperature when surface experiences
evaporation from a thin water film.
FIND: (a) Heat transfer coefficient and convection heat rate, (b) Mass transfer coefficient and
evaporation rate (kg/h) of the water, (c) Rate at which heat must be supplied to surface for these
conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Heat-mass transfer analogy applies, (3)
Correlation requires properties evaluated at Tf = (Ts + T¥)/2.
PROPERTIES: Table A-4, Air (Tf = (Ts + T¥)/2 = (290 + 310)K/2 = 300 K, 1 atm): n =
15.89 ´ 10-6m2/s, k = 0.0263 W/m×K, Pr = 0.707; Table A-8, Air-water mixture (300 K, 1 atm):
DAB = 0.26 ´ 10
-4
 m
2
/s; Table A-6, Sat. water (Ts = 310 K): rA,sat = 1/vg = 1/22.93 m
3
/kg =
0.04361 kg/m
3
, hfg = 2414 kJ/kg.
ANALYSIS: (a) To characterize the flow, evaluate ReL at Tf
5
L -6 2
VL 10 m/s 1 m
Re 6.293 10
15.89 10 m / sn
´= = = ´
´
and substituting into the prescribed correlation for this surface, find
( ) ( )0.58 0.45LNu 0.43 6.293 10 0.707 864.1= ´ =
L 2
L
Nu k 864.1 0.0263 W/m K
h 22.7 W/m K.
L 1 m
× ´ ×= = = × <
Hence, the convection heat rate is
( )conv L s sq h A T T¥= -
( )2 2convq 22.7 W/m K 1 m 310 290 K 454 W= × ´ - = <
(b) Invoking the heat-mass transfer analogy
0.58 0.4mL L
AB
h L
Sh 0.43Re Sc
D
= =
where
6 2
-4 2AB
15.89 10 m / s
Sc 0.611
D 0.26 10 m / s
n -´= = =
´
and n is evaluated at Tf. Substituting numerical values, find
Continued \u2026..
PROBLEM 6.75 (Cont.)
( ) ( )0.58 0.45LSh 0.43 6.293 10 0.611 815.2= ´ =
4 2
L 2AB
m
Sh D 815.2 0.26 10 m / s
h 2.12 10 m/s.
L 1 m
-
-× ´ ´= = = ´ <
The evaporation rate, with ( )A,s A,sat sT , isr r=
( )
( )
m s A,s A,
-2 2 3
m h A
m 2.12 10 m/s 1 m 0.04361 0 kg/m
r r ¥= -
= ´ ´ -
&
&
-4m 9.243 10 kg/s 3.32 kg/h.= ´ =& <
(c) The rate at which heat must be supplied to the plate to maintain