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Unformatted text preview: Gift from Professor Muller: to save y’all work, the enthalpy of the water entering the feedwater heater is 92.27 kJ/kg and the entropy is 0.3249 kJ/kgK Solution: a) 0.1 mPa = 1 bar 1 bar = 100 kPa 7 mPa, = 70 bar  saturation temperature = 285.9 °C …. superheated temp = 400 °C Enthalpy = 3158.1 kJ/kg; entropy = 6.4478 kJ/kgK b) After the pump, h = 92.27 kJ/kg, entropy = 0.3249 kJ/kgK For saturated liquid, s= 3.1216 h = 1267.4 60 * 92.27 + X * 3158.1 = (60 +X) * 1267.4 (3158.1 – 1267.4) X = 70,507.8 X = 37.29 kg/min Center for Advanced Energy Systems Michael R. Muller, Director C) entropy in = 60*0.3249 + 37.29* 6.4478 = 259.93 kJ/K Entropy 0ut = 97.29* 3.1216 = 303.70 Entropy production = 43.77 kJ/minK...
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 Fall '10
 Muller
 Thermodynamics, Feedwater heater, Michael R. Muller, Advanced Energy Systems, Professor Muller

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