Secondary Side of Heat Exchanger

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: Q_stm Q_cw m_stm(h_out h_in)_stm m_cw(h_out h_in)_cw m_cw m_stm(h_out h_in)_stm/(h_out h_in)_cw =^-2.0 x 10⁶ kg/hr (125.8 - 2574 kj/kg)/(104.9 - 75.6 kj/kg) m_cw 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 mc_pDT since DT=0 for a phase change. Had we attempted to solve the problem using , we Q mc_pDT would have discovered that an error occurs since the DT = 10^oC is the DT needed to subcool the liquid from saturation at 40^oC to a subcooled value of 30^oC. 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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