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Overall Thermodynamic Balance
First Law of Thermodynamics Summary

Thermodynamics Heat Transfer and Fluid Flow Volume 1 of 3
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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







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