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10-116
10-123 Saturated steam condenses on a suspended silver sphere which is initially at 25C. The time needed for the
temperature of the sphere to rise to 50C and the amount of steam condenses are to be determined.
Assumptions 1 The temperature of the sphere changes uniformly and thus the lumped system analysis is applicable. 2 The
average condensation heat transfer coefficient evaluated for the average temperature can be used for the entire process. 3
C W/m429
CJ/kg 235
/sm10174
kg/m 500,10
26
3
=
=
=
=
−
l
p
k
c
:BallSilver
C W/m663.0
CJ/kg 4190
skg/m10404.0
kg/m 5.977
3
3
=
=
=
=
−
l
pl
l
l
k
c
: WaterLiquid
AnalysisThe modified latent heat of vaporization is
J/kg 102435=C)5.37C(100J/kg 41900.68+J/kg 102257
)(68.0
33
sat
*
−=
−+= splfgfg TTchh
Silver
sphere
Steam
100C
Ti = 25C
1.2 cm
10-117
10-118
10-125 There is film condensation on the outer surfaces of 8horizontal tubes arranged in a horizontal or vertical tier. The ratio
of the condensation rate for the cases of the tubes being arranged in a horizontal tier versus in a vertical tier is to be
determined.
Assumptions Steady operating conditions exist.
Analysis The heat transfer coefficients for the two cases are related
10-119
10-126E Saturated steam at a saturation pressure of 0.95 psia and thus at a saturation temperature of Tsat = 100F (Table A-
9E) condenses on the outer surfaces of 100 horizontal tubes which are maintained at 80F by circulating cooling water and
arranged in a 10 10 square array. The rate of heat transfer to the cooling water and the rate of condensation of steam are to
=+= 2/)( sat sf TTT
(100 + 80)/2 = 90F are (Table A-9E),
hlbm/ft 842.1slbm/ft 10117.5
lbm/ft 12.62
2
4
3
==
=
−
l
l
n = 100 tubes
80F
P = 0.95 psia
Cooling
water
10-120
10-121
10-128 Ammonia is liquefied in a horizontal condenser at 37C by a coolant at 20C. The average value of overall heat
transfer coefficient and the tube length are to be determined.
C W/m4602.0
CJ/kg 4867
skg/m10303.1
kg/m 0.589
4
3
=
=
=
=
−
l
pl
l
l
k
c
10-122
10-123
9.275)98.7()440,44(023.0PrRe023.0 4.08.04.08.0 ===Nu
C) W/m590.0( 2=
k
10-124
10-130 Saturated steam at 270.1 kPa pressure and thus at a saturation temperature of Tsat = 130C (Table A-9) condenses
inside a horizontal tube which is maintained at 110C. The average heat transfer coefficient and the rate of condensation of
C W/m.683.0
CJ/kg. 4244
/sm10246.0/
kg/m.s10232.0
kg/m 4.943
26
3
3
=
=
==
=
=
−
−
l
pl
lll
l
l
k
c
Steam
270.1 kPa
110C
Dtube = 2.5 cm
Ltube = 10 m
10-125
10-126
10-135 Saturated water vapor is condensing on a 0.5 m2 vertical flat plate in a continuous film with an average heat transfer
coefficient of 5 kW/m2K. The temperature of the water is 80oC (hfg = 2309 kJ/kg) and the temperature of the plate is 60oC.
The rate at which condensate is being formed is
A=0.5 [m^2]
10-136 Steam condenses at 50ºC on a 1.2-m-high and 2.4-m-wide vertical plate that is maintained at 30ºC. The condensation
heat transfer coefficient is
(a) 4260 W/m2K (b) 4780 W/m2K (c) 5510 W/m2K (d) 6260 W/m2K (e) 6940 W/m2K
w=2.4 [m]
h_fg=2383E3 [J/kg] "at 50 C from Table A-9"
g=9.81 [m/s^2]
h_fg_star=h_fg+0.68*c_p_l*(T_sat-T_s)
10-127
10-137 An air conditioner condenser in an automobile consists of 2 m2 of tubular heat exchange area whose surface
temperature is 30oC. Saturated refrigerant 134a vapor at 50oC (hfg = 152 kJ/kg) condenses on these tubes. What heat transfer
coefficient must exist between the tube surface and condensing vapor to produce 1.5 kg/min of condensate?
10-138 Saturated water vapor at 40C is to be condensed as it flows through a tube at a rate of 0.2 kg/s. The condensate
leaves the tube as a saturated liquid at 40C. The rate of heat transfer from the tube is
(a) 34 kJ/s (b) 268 kJ/s (c) 453 kJ/s (d) 481 kJ/s (e) 515 kJ/s
Answer (d) 481 kJ/s
10-128
10-129
10-140 Steam condenses at 50ºC on a tube bank consisting of 20 tubes arranged in a rectangular array of 4 tubes high and 5
tubes wide. Each tube has a diameter of 3 cm and a length of 5 m and the outer surfaces of the tubes are maintained at 30ºC.
The rate of condensation of steam is
D=0.03 [m]
L=5 [m]
N=4
g=9.81 [m/s^2]
h_fg_star=h_fg+0.68*c_p_l*(T_sat-T_s)
