Thermodynamics
FIRST LAW OF THERMODYNAMICS
Example 6: Secondary Side of Heat Exchanger
Steam flows through a condenser at 2.0 x 106 kg/hr, entering as saturated vapor at 40°C
(h = 2574 kj/kg), and leaving at the same pressure as subcooled liquid at 30°C (h = 125.8
kJ/kg). Cooling water is available at 18°C (h = 75.6 kJ/kg). Environmental requirements
limit the exit temperature to 25°C (h = 104.9 kJ/kg). Determine the required cooling
water flow rate.
Solution:
Thermal balance gives the following:
Qstm
Qcw
mstm(hout
hin)stm
mcw(hout
hin)cw
mcw
mstm(hout hin)stm/(hout hin)cw
=-2.0 x 106 kg/hr (125.8 - 2574 kj/kg)/(104.9 - 75.6 kj/kg)
mcw
1.67 x 108 kg/hr
In this example, we calculated the flow rate using the equation
since a phase change
Q
mDh
occurred when the steam was condensed to liquid water.
would not have worked
Q
mcpDT
since DT=0 for a phase change. Had we attempted to solve the problem using
, we
Q
mcpDT
would have discovered that an error occurs since the DT = 10oC is the DT needed to subcool the
liquid from saturation at 40oC to a subcooled value of 30oC. Therefore, the heat transfer process
to condense the steam to a saturated liquid has not been taken into account.
Rev. 0
Page 67
HT-01
