ch06
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ch06


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the droplet of mass M = 3.728 ´ 10-5 kg follows from an
overall energy balance,
( )fg s sQ Mh hA T T¥= + -
where h is obtained from the heat-mass transfer analogy, Eq. 6.67, using n = 1/3,
2/3
pnm AB
h k
 c Le
h D Le
r= =
where
6 2
-4 2AB
16.39 10 m /s
Sc 0.594
D 0.276 10 m / s
Sc 0.594
Le 0.841.
Pr 0.706
n -´= = =
´
= = =
Hence,
( )2/33 2h 0.0113 m/s 1.1448 kg/m 1008 J/kg K 0.841 11.62 W/m K.= ´ ´ × = ×
and the energy requirement is
( )( ) ( )2-5 2Q 3.728 10 kg 2414 kJ/kg 11.62 W/m K 0.004 m / 4 37 27 Cp= ´ ´ + × - o
( )Q 90.00 0.00145 J 90 J.= + = <
The energy required to meet the convection heat loss is very small compared to that required to
sustain the evaporative loss.
PROBLEM 6.79
KNOWN: Initial plate temperature Tp (0) and saturated air temperature (T¥) in a dishwasher at the
start of the dry cycle. Thermal mass per unit area of the plate Mc/As = 1600 J/m
2
×K.
FIND: (a) Differential equation to predict plate temperature as a function of time during the dry cycle
and (b) Rate of change in plate temperature at the start of the dry cycle assuming the average
convection heat transfer coefficient is 3.5 W/m2×K.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is spacewise isothermal, (2) Negligible thermal resistance of water film
on plate, (3) Heat-mass transfer analogy applies.
PROPERTIES: Table A-4, Air (T =(55 + 65)°C/2 = 333 K, 1 atm): r = 1.0516 kg/m
3
, cp = 1008
J/kg×K, Pr = 0.703, n = 19.24´ 10
-6
 m
2
/s; Table A-6, Saturated water vapor, (Ts = 65°C = 338 K): rA
= 1/vg = 0.1592 kg/m
3
, hfg = 2347 kJ/kg; (Ts = 55°C = 328 K): rA = 1/vg = 0.1029 kg/m
3
; Table A-8,
Air-water vapor (Ts = 65°C = 338 K, 1 atm): DAB = 0.26 ´ 10
-4
 m
2
/s (338/298)
3/2
 = 0.314 ´ 10-
4
m
2
/s.
ANALYSIS: (a) Perform an energy balance on a rate basis on the plate,
( )( )in out st conv evap s pE E E q q Mc/A dT /dt .¢¢ ¢¢- = - =& & &
Using the rate equations for the heat and mass transfer fluxes, find
( ) ( ) ( ) ( )( )p m A,s s A, fg sh T T t h T T h Mc/A dT/dt .r r¥ ¥ ¥é ù é ù- - - =ë ûë û <
(b) To evaluate the change in plate temperature at t = 0, the start of the drying process when Tp (0) =
65°C and T¥ = 55°C, evaluate mh from knowledge of 
2h 3.5 W/m K= × using the heat-mass
transfer analogy, Eq. 6.67, with n = 1/3,
2 /3 2/3
2/3 AB
p p p
m
h Sc / D
 c Le c c
h Pr Pr
nr r ræ ö æ ö= = =ç ÷ ç ÷è ø è ø
and evaluating thermophysical properties at their appropriate temperatures, find
2 / 32 -6 2 -4 2
3 3
m
m
3.5 W/m K 19.24 10 m /s/0.314 10 m /s
1.0516 kg/m 1008 J/kg K h 3.619 10 m/s.
h 0.703
-æ ö× ´ ´ç ÷= ´ × = ´ç ÷è ø
Substituting numerical values into the conservation expression of part (a), find
( ) ( ) ( )2 -3 3 3 2 p3.5 W/m K 55 65 C 3.619 10 m/s 0.1592 0.1029 kg/m 2347 10 J/kg 1600 J/m K dT /dt× - - ´ - ´ ´ = ×o
[ ] 2 2pdT /dt 35.0 478.2 W/m K/1600 J/m K 0.32 K/s.= - + × × = - <
COMMENTS: This rate of temperature change will not be sustained for long, since, as the plate
cools, the rate of evaporation (which dominates the cooling process) will diminish.
	Problems 6.1-6.25
	6.1
	6.2
	6.3
	6.4
	6.5
	6.6
	6.7
	6.8
	6.9
	6.10
	6.11
	6.12
	6.13
	6.14
	6.15
	6.16
	6.17
	6.18
	6.19
	6.20
	6.21
	6.22
	6.23
	6.24
	6.25
	Problems 6.26-6.50
	6.26
	6.27
	6.28
	6.29
	6.30
	6.31
	6.32
	6.33
	6.34
	6.35
	6.36
	6.37
	6.38
	6.39
	6.40
	6.41
	6.42
	6.43
	6.44
	6.45
	6.46
	6.47
	6.48
	6.49
	6.50
	Problems 6.51-6.79
	6.51
	6.52
	6.53
	6.54
	6.55
	6.56
	6.57
	6.58
	6.59
	6.60
	6.61
	6.62
	6.63
	6.64
	6.65
	6.66
	6.67
	6.68
	6.69
	6.70
	6.71
	6.72
	6.73
	6.74
	6.75
	6.76
	6.77
	6.78
	6.79