ch06
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ch06


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these conditions follows from an
energy balance.
in out
in conv evap
E E 0
q q q 0
- =
- - =
& &
where qin is the heat supplied to sustain the losses by convection and evaporation.
( )
in conv evap
in L s s fg
-4 3
in
q q q
q h A T T mh
q 454 W 9.243 10 kg/s 2414 10 J/kg
¥
= +
= - +
= + ´ ´ ´
&
( )inq 254 2231 W 2685 W.= + = <
COMMENTS: Note that the loss from the surface by evaporation is nearly 5 times that due to
convection.
PROBLEM 6.76
KNOWN: Thickness, temperature and evaporative flux of a water layer. Temperature of air flow
and surroundings.
FIND: (a) Convection mass transfer coefficient and time to completely evaporate the water, (b)
Convection heat transfer coefficient, (c) Heater power requirement per surface area, (d) Temperature
of dry surface if heater power is maintained.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Applicability of heat and mass transfer analogy with n = 1/3,
(3) Radiation exchange at surface of water may be approximated as exchange between a small surface
and large surroundings, (4) Air is dry (\u3c1A,\u221e = 0), (5) Negligible heat transfer from unwetted surface of
the plate.
PROPERTIES: Table A-6, Water (Tw = 340K): \u3c1f = 979 kg/m3, 1 3A,sat gv 0.174 kg / m ,\u3c1 \u2212= =
fgh 2342 kJ / kg.= Prescribed, Air: \u3c1 = 1.08 kg/m
3
, cp = 1008 J/kg\u22c5K, k = 0.028 W/m\u22c5K. Vapor/Air:
DAB = 0.29 × 10
-4
 m
2/s. Water: \u3b5w = 0.95. Plate: \u3b5p = 0.60.
ANALYSIS: (a) The convection mass transfer coefficient may be determined from the rate equation
( )A m A,s A,hn ,\u3c1 \u3c1 \u221e= \u2212\u2032\u2032 where ( )A,s A,sat w A,T and 0.\u3c1 \u3c1 \u3c1 \u221e= = Hence,
2
A
m 3A,sat
n 0.03kg / s mh 0.172m / s
0.174kg / m\u3c1
\u2032\u2032 \u22c5
= = = <
The time required to completely evaporate the water is obtained from a mass balance of the form
A fn d / dt,\u3c1 \u3b4\u2032\u2032\u2212 = in which case
i
0 t
f A 0
d n dt\u3b4\u3c1 \u3b4 \u2032\u2032= \u2212\u222b \u222b
( )3f i
2A
979 kg / m 0.002m
t 65.3s
n 0.03kg / s m
\u3c1 \u3b4
= = =
\u2032\u2032
\u22c5
<
(b) With n = 1/3 and Le = \u3b1/DAB = k/\u3c1cp DAB = 0.028 W/m\u22c5K/(1.08 kg/m3 × 1008 J/kg\u22c5K × 0.29 ×
10-4 m2/s) = 0.887, the heat and mass transfer analogy yields
( )
( )
2m
1/3 1/34 2
AB
0.028W / m K 0.172m / sk hh 173W / m K
D Le 0.29 10 m / s 0.887\u2212
\u22c5
= = = \u22c5
×
<
The electrical power requirement per unit area corresponds to the rate of heat loss from the water.
Hence,
Continued \u2026..
PROBLEM 6.76 (Cont.)
( ) ( )4 4elec evap conv rad A fg w w w surP q q q n h h T T T T\u3b5 \u3c3\u221e\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032= + + = + \u2212 + \u2212
( ) ( ) ( )2 6 2 8 2 4 4 4elecP 0.03 kg / s m 2.342 10 J / kg 173 W / m K 40K 0.95 5.67 10 W / m K 340 300\u2212\u2032\u2032 = \u22c5 × + \u22c5 + × × \u22c5 \u2212
2 2 2 2
elecP 70, 260 W / m 6920 W / m 284 W / m 77,464 W / m\u2032\u2032 = + + = <
(c) After complete evaporation, the steady-state temperature of the plate is determined from the
requirement that
( ) ( )4 4elec p p p surP h T T T T\u3b5 \u3c3\u221e\u2032\u2032 = \u2212 + \u2212
( ) ( )2 2 8 2 4 4 4p p77, 464 W / m 173W / m K T 300 0.60 5.67 10 W / m K T 300\u2212= \u22c5 \u2212 + × × \u22c5 \u2212
pT 702K 429 C= = ° <
COMMENTS: The evaporative heat flux is the dominant contributor to heat transfer from the water
layer, with convection of sensible energy being an order of magnitude smaller and radiation exchange
being negligible. Without evaporation (a dry surface), convection dominates and is approximately an
order of magnitude larger than radiation.
PROBLEM 6.77
KNOWN: Heater power required to maintain water film at prescribed temperature in dry ambient
air and evaporation rate.
FIND: (a) Average mass transfer convection coefficient mh , (b) Average heat transfer convection
coefficient h, (c) Whether values of mh and h satisfy the heat-mass analogy, and (d) Effect on
evaporation rate and disc temperature if relative humidity of the ambient air were increased from 0 to
0.5 but with heater power maintained at the same value.
SCHEMATIC:
ASSUMPTIONS: (1) Water film and disc are at same temperature; (2) Mass and heat transfer
coefficient are independent of ambient air relative humidity, (3) Constant properties.
PROPERTIES: Table A-6, Saturated water (305 K): vg = 29.74 m
3
/kg, hfg = 2426 ´ 10
3
 J/kg;
Table A-4, Air ( )T 300 K, 1 atm := k = 0.0263 W/m×K, a = 22.5 ´ 10-6 m2/s, Table A-8, Air-
water vapor (300 K, 1 atm): DAB = 0.26 ´ 10
-4
 m
2
/s.
ANALYSIS: (a) Using the mass transfer convection rate equation,
( ) ( )A m s A,s A, m s A,satn h A h A 1r r r f¥ ¥= - = -
and evaluating rA,s = rA,sat (305 K) = 1/vg (305 K) with f¥ ~ rA,¥ = 0, find
( )
A
m
s A,s A,
n
h
A r r ¥
=
-
( )
( )( ) ( )
4
3
m 2 3
2.55 10 kg/hr/ 3600s/hr
h 6.71 10 m/s.
0.020 m / 4 1/29.74 0 kg/mp
-
-´= = ´
-
<
(b) Perform an overall energy balance on the disc,
( )conv evap s s A fgq q q hA T T n h¥= + = - +
and substituting numerical values with hfg evaluated at Ts, find h:
( ) ( )23 -8 3200 10 W h 0.020 m / 4 305 295 K 7.083 10 kg/s 2426 10 J/kgp-´ = - + ´ ´ ´
2h 8.97 W/m K.= × <
Continued \u2026..
PROBLEM 6.77 (Cont.)
(c) The heat-mass transfer analogy, Eq. 6.67, requires that
1/3?
AB
m AB
h k D
.
h D a
æ ö= ç ÷è ø
Evaluating k and DAB at ( )sT T T / 2 300 K¥= + = and substituting numerical values,
1/32 4 2
-3 -4 2 -6 2
8.97 W/m K 0.0263 W/m K 0.26 10 m / s
1337 1061
6.71 10 m/s 0.26 10 m / s 22.5 10 m / s
-æ ö× × ´ç ÷= ¹ =ç ÷´ ´ ´è ø
Since the equality is not satisfied, we conclude that, for this situation, the analogy is only
approximately met (» 30%).
(d) If f¥ = 0.5 instead of 0.0 and q is unchanged, nA will decrease by nearly a factor of two, as will
nAhfg = qevap. Hence, since qconv must increase and h remains nearly constant, Ts - T¥ must
increase. Hence, Ts will increase.
COMMENTS: Note that in part (d), with an increase in Ts, hfg decreases, but only slightly, and
rA,sat increases. From a trial-and-error solution assuming constant values for mh and h, the disc
temperature is 315 K for f¥ = 0.5.
PROBLEM 6.78
KNOWN: Power-time history required to completely evaporate a droplet of fixed diameter
maintained at 37°C.
FIND: (a) Average mass transfer convection coefficient when droplet, heater and dry ambient air
are at 37°C and (b) Energy required to evaporate droplet if the dry ambient air temperature is 27°C.
SCHEMATIC:
ASSUMPTIONS: (1) Wetted surface area of droplet is of fixed diameter D, (2) Heat-mass
transfer analogy is applicable, (3) Heater controlled to operate at constant temperature, Ts = 37°C,
(4) Mass of droplet same for part (a) and (b), (5) Mass transfer coefficients for parts (a) and (b) are
the same.
PROPERTIES: Table A-6, Saturated water (37°C = 310 K): hfg = 2414 kJ/kg, rA,sat = 1/vg =
1/22.93 = 0.04361 kg/m
3
; Table A-8, Air-water vapor (Ts = 37°C = 310 K, 1 atm): DAB = 0.26
´ 10-6m2/s(310/289)3/2 = 0.276 ´ 10-6 m2/s; Table A-4, Air (T = (27 + 37)°C/2 = 305 K, 1
atm): r = 1.1448 kg/m3, cp = 1008 J/kg×K, n = 16.39 ´ 10
-6
 m
2
/s, Pr = 0.706.
ANALYSIS: (a) For the isothermal conditions (37°C), the electrical energy Q required to
evaporate the droplet during the interval of time Dt = te follows from the area under the P-t curve
above,
( ) ( )
te
0
3 -3Q Pdt 20 10 W 50 60 s 0.5 20 10 W 100 50 60s
Q 90 J.
-é ù= = ´ ´ ´ + ´ ´ - ´ê úë û
=
ò
From an overall energy balance during the interval of time Dt = te, the mass loss due to evaporation
is
fg fg
3 -5
Q Mh or M Q/h
M 90 J/2414 10 J/kg 3.728 10 kg.
= =
= ´ = ´
To obtain the average mass transfer coefficient, write the rate equation for an interval of time Dt = te,
( ) ( )e m s A,s A, e m s A,s eM m t h A t h A 1 tr r r f¥ ¥= × = - × = - ×&
Substituting numerical values with f¥ = 0, find
( )( ) ( )25 3m3.278 10 kg h 0.004 m / 4 0.04361 kg/m 100 60 sp-´ = ´ ´
Continued \u2026..
PROBLEM 6.78 (Cont.)
mh 0.0113 m/s.= <
(b) The energy required to evaporate