TRG-TRC003-EN 43
Answers14 a 10°F [5.6 °C] (temperature rise from % to &)
b14°F [7.8°C] (temperature drop from ) to *)
c68.7 Btu/lb [159.8 kJ/kg] (enthalpy difference between $ and &)
d22% refrigerant vapor
% of Refrigerant Vapor at AEnthalpy at AEnthalpy at H–
Enthalpy at BEnthalpy at H–
=
% of Refrigerant Vapor at A41.6 Btu/lb 22.7 Btu/lb–
108.5 Btu/lb 22.7 Btu/lb–
22%==
% of Refrigerant Vapor at A96.8 kJ/kg 52.8 kJ/kg–
252.4 kJ/kg 52.8 kJ/kg–
22%==