(5a). (6p) A rigid tank with a volume of 0.25 m3 contains 3.5 kg of Refrigerant R-134a at 200 kPa. Please identify the phase (or phases), quality, temperature of this substance, as well as its total internal energy and total enthalpy. (5b). (6p) Please solve Problem 5a for 0.5 kg of H2O at 1.0 MPa in the same tank. Namely, a rigid tank with a volume 0.25 m3 contains 0.5 kg of H2O at 1.0 MPa. Please determine the phase, quality, temperature, total internal energy and total enthalpy of this substance. Solution For R134a properties, see http://www.irc.wisc.edu/properties/ Density = m/V = 3.5 / 0.25 = 14 kg/m^3 At 200 kPa and density = 14 kg/m^3, we get quality x = 0.713, Temperature T = -10.1 deg C, internal energy u = 171 kJ/kg, enthalpy h = 185 kJ/kg Total internal energy U = m*u = 3.5*171 = 598.5 kJ Total enthalpy H = m*h = 3.5*185 = 647.5 kJ 2) For H2O properties, see http://www.irc.wisc.edu/properties/ Density = m/V = 0.5 / 0.25 = 2 kg/m^3 At 1 MPa and density = 2 kg/m^3, we get quality x = superheated vapor, Temperature T = 812 deg C, internal energy u = 3680 kJ/kg, enthalpy h = 4180 kJ/kg Total internal energy U = m*u = 0.5*3680 = 1840 kJ Total enthalpy H = m*h = 0.5*4180 = 2090 kJ.

1. (5a). (6p) A rigid tank with a volume of 0.25 m3 contains 3.5 kg of Refrigerant R-134a at 200
kPa. Please identify the phase (or phases), quality, temperature of this substance, as well as its
total internal energy and total enthalpy.
(5b). (6p) Please solve Problem 5a for 0.5 kg of H2O at 1.0 MPa in the same tank. Namely, a
rigid tank with a volume 0.25 m3 contains 0.5 kg of H2O at 1.0 MPa. Please determine the
phase, quality, temperature, total internal energy and total enthalpy of this substance.
Solution
For R134a properties, see http://www.irc.wisc.edu/properties/
Density = m/V = 3.5 / 0.25 = 14 kg/m^3
At 200 kPa and density = 14 kg/m^3, we get quality x = 0.713, Temperature T = -10.1 deg C,
internal energy u = 171 kJ/kg, enthalpy h = 185 kJ/kg
Total internal energy U = m*u = 3.5*171 = 598.5 kJ
Total enthalpy H = m*h = 3.5*185 = 647.5 kJ
2)
For H2O properties, see http://www.irc.wisc.edu/properties/
Density = m/V = 0.5 / 0.25 = 2 kg/m^3
At 1 MPa and density = 2 kg/m^3, we get quality x = superheated vapor, Temperature T = 812
deg C, internal energy u = 3680 kJ/kg, enthalpy h = 4180 kJ/kg
Total internal energy U = m*u = 0.5*3680 = 1840 kJ
Total enthalpy H = m*h = 0.5*4180 = 2090 kJ