Question

A mass of 0.3 kg of saturated refrigerant-134a is contained in a piston-cylinder device at 240 kPa.
Initially, 70 percent of the mass is in the liquid phase. Now heat is transferred to the refrigerant at
constant pressure until the cylinder contains vapor only.
(a) show the process on a P-v and T-v diagrams with respect to saturation lines. Determine;
(b) the volume occupied by the refrigerant initially,
(c) the work done, and
(d) the total heat transfer.

Answer 1

Ans-

Mass of the refrigerant, m = 0.3 kg

Quality of the refrigerant at state 1, x₁ = 1 - 0.7 = 0.3

(a) The  process on P-v and T-v diagrams with respect to saturation lines are shown below:

(b) At 240 kPa,

v_f = 0.0007618 m³/kg and v_g = 0.0834 m³/kg

Therefore, volume occupied by the refrigerant initially is

V₁ = m(v_f + x₁(v_g - v_f ))

= 0.3(0.0007618 + 0.3(0.0834 - 0.0007618))

= 0.00767 m³ (ans)

(c) Volume occupied by the refrigerant finally ,

V₂ = mv_g

  = 0.3 * 0.0834

= 0.02502 m³

Therefore,

Work done, W = P(V₂ - V₁)

= 240 * (0.02502 - 0.00767)

= 4.16 KJ (ans)

(d)

Now, at 240 kPa

h_f = 42.95 KJ/kg and h_g = 244.09 KJ/kg

Specific enthalpy of refrigerant initially,

h₁ = h_f +  x₁( h_g - h_f)

= 42.95 + 0.3(244.09 - 42.95)

= 103.3 KJ/kg

Specific enthalpy of refrigerant finally,

h₂ = h_g = 244.09 KJ/kg

Therefore,

Total heat transfer, Q = m(h₂ - h₁ )

= 0.3(244.09 - 103.3)

= 42.24 KJ (ans)