R-134a flows in a pipe at 30oC with a specific volume of 0.04434 m3/kg. The internal energy and enthalpy of R-134a are 250.8 kJ/kg and 273.0 kJ/kg, respectively. Determine the pressure of the refrigerant in MPa. in 2 decimal places
V=0.04434 m3/kgU=250.8 kJ/kgh=273.0 kJ/kgh=U+pV273.0=250.8+P×0.04434P=500.67 kPa=0.50 MPaV=0.04434 \;m^3/kg \\ U = 250.8 \;kJ/kg \\ h= 273.0 \;kJ/kg \\ h=U+pV \\ 273.0 = 250.8 + P \times 0.04434 \\ P = 500.67 \;kPa = 0.50 \;MPaV=0.04434m3/kgU=250.8kJ/kgh=273.0kJ/kgh=U+pV273.0=250.8+P×0.04434P=500.67kPa=0.50MPa
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