This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Gift from Professor Muller: to save yall 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...
View
Full
Document
This note was uploaded on 02/08/2011 for the course ME 650:462 taught by Professor Muller during the Fall '10 term at Rutgers.
 Fall '10
 Muller

Click to edit the document details