Refrigerant R410a is contained in a cylinder with a leakproo

Refrigerant R410a is contained in a cylinder with a leakproof frictionless piston. A linear spring acts on the piston, such that when the pressure is 100 kPa the volume is 0.01 m^3, and when the pressure is 1.0 MPa the volume is 0.035m^3. The cylinder initially contains 0.02 m^3 of saturated R410aat -30 degree C, of which 1.% by volume is saturated liquid. The R410a is then heated to a final pressure of 600 kPa. Draw this process on a P-v diagram, showing lines of constant temperature at the initial and final state. Also show on this diagram the critical point, the superheated vapour region, and the compressed liquid region.. Calculate the final volume in the cylinder. Calculate the work done in kJ. Calculate the heat transfer in kJ and the final temperature.

Solution

b)

From the spring data, we get that P = (1000 - 100) / (0.035 - 0.01) *(V - 0.01) + 100

P = 36000*(V - 0.01) + 100

When P2 = 600 kPa, we get 600 = 36000*(V2 - 0.01) + 100

V2 = 0.0239 m^3

c)

Work done = Integral P dV from V1 to V2

= Integral [36000*(V - 0.01) + 100] dV from V1 = 0.02 m^3 to V2 = 0.0239 m^3

= [18000*V^2 - 260*V] from V1 = 0.02 m^3 to V2 = 0.0239 m^3

= 18000*(0.0239^2 - 0.02^2) - 260*(0.0239 - 0.02)

= 2.068 kJ

d)

For R-410a at -30 deg C, we have v_f = 0.000781 m^3/kg, v_g = 0.0948 m^3/kg

We have volume of liquid = 0.01*0.02 = 0.0002 m^3

Volume of vapor = 0.02 - 0.0002 = 0.0198 m^3

Mass of liquid = 0.0002 / 0.000781 = 0.256 kg

Mass of vapor = 0.0198 / 0.0948 = 0.209 kg

Total mass = 0.256 + 0.209 = 0.465 kg

Quality x1 = 0.256 / 0.465 = 0.55

At -30 deg C and x1 = 0.55, we get u1 = 141 kJ/kg, v1 = 0.053 m^3/kg

m = V1/v1 = 0.02 / 0.053 = 0.377 kg

v2 = V2/m = 0.0239 / 0.377 = 0.0634 m^3/kg

At P2 = 600 kPa and v2 = 0.0634 m^3/kg, we get u2 = 317 kJ/kg

Q - W = m(u2 - u1)

Q - 2.068 = 0.377*(317 - 141)

Q = 68.42 kJ

At P2 = 600 kPa and v2 = 0.0634 m^3/kg, we get T2 = 74.6 deg C