Thermodynamics FIRST LAW OF THERMODYNAMICSExample 6: Secondary Side of Heat ExchangerSteam flows through a condenser at 2.0 x 10^{6} 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 = 125.8kJ/kg). Cooling water is available at 18°C (h = 75.6 kJ/kg). Environmental requirementslimit the exit temperature to 25°C (h = 104.9 kJ/kg). Determine the required coolingwater 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 10^{6} kg/hr (125.8 - 2574 kj/kg)/(104.9 - 75.6 kj/kg)m_{cw}1.67 x 10^{8}kg/hrIn this example, we calculated the flow rate using the equation since a phase changeQmDhoccurred when the steam was condensed to liquid water. would not have workedQmc_{pD}Tsince DT=0 for a phase change. Had we attempted to solve the problem using , weQmc_{pD}Twould have discovered that an error occurs since the DT = 10^{o}C is the DT needed to subcool theliquid from saturation at 40^{o}C to a subcooled value of 30^{o}C. Therefore, the heat transfer processto condense the steam to a saturated liquid has not been taken into account.Rev. 0 Page 67 HT-01

Integrated Publishing, Inc.

6230 Stone Rd, Unit QPort Richey, FL34668

Phone For Parts Inquiries: (727) 493-0744
Google +