Question

Propane vapor initially at 7.0 bar and 50C (State 1) is contained within a piston-cylinder device. The refrigerant is cooled at constant volume until its temperature reaches -10C (State 2) and is then compressed isothermally to a pressure of 6.0 bar (State 3). (a) Locate the state points on appropriately labeled p-v diagram. (b) Determine the specific volume (in m3 /kg), internal energy (in kJ/kg) and enthalpy (in kJ/kg) at each state point.

Answer 1

a) We will be plotting Pressure vs Volume states in P-V diagram.

That is plotted in attached excel file.

b) We will be calculating specific volume (in m3 /kg), internal energy (in kJ/kg) and enthalpy (in kJ/kg) at each state point. T will be converted to Kelvin.

T	V (state)	P
50	1	7
-10	2	7
-10	3	6

Specific Volume

v = RT/ P

State 1

v = RT/P = ((8.314 * 10 ^-5) * 323.15) / 7

                = 0.003838 m3 /kg

State 2

v = RT/P = ((8.314 * 10 ^-5) * 263.15) / 7

                = 0.003125 m3 /kg

State 3

v = RT/P = ((8.314 * 10 ^-5) * 263.15) / 6

                = 0.003646 m3 /kg

Internal energy

State 1

E = 3/2 RT

   = 3/2 * 8.314 * 323.15

   = 4030 * 10^-3 kJ/kg

State 2

E = 3/2 RT

   = 3/2 * 8.314 * 263.15

   = 3281.74 * 10^-3 kJ/kg

State 3

E = 3/2 RT

   = 3/2 * 8.314 * 263.15

   = 3281.74 * 10^-3 kJ/kg

Enthalpy

H = E + PV as V is constant take it as 1.

State 1

H = E + PV

    = 4030 * 10^-3 + 7 * 1

    = 11.03 kJ/kg

State 2

H = E + PV

    = 3281.74 * 10^-3 + 7 * 1

    = 10.28 kJ/kg

State 3

H = E + PV

    = 3281.74 * 10^-3 + 7 * 1

    = 10.28 kJ/kg