1- (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.
2- 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.
Thermodynamic properties of refrigerants and water in a rigid tank
1. 1- (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.
2- 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
