Three-process gas cycle: net work and heat

Take the gas as ideal. Then U₂ = U₁ means T₂ = T₁, so process 1-2 (PV = constant) is isothermal. Process 3-1 is at constant pressure, P₃ = P₁ = 800 kPa.

(a) Sketch

First find the volumes. Process 3-1: W₃₁ = P(V₁ − V₃) = −1200 kJ, so V₁ − V₃ = −1200/800 = −1.5 m³. With V₃ = V₂ = 2 m³:

V₁ = 2 − 1.5 = 0.5 m³ P₂ = P₁V₁/V₂ = 800 × 0.5/2 = 200 kPa

States: 1 (0.5 m³, 800 kPa), 2 (2 m³, 200 kPa), 3 (2 m³, 800 kPa). Since T ∝ PV, T₃ = 4T₁.

 P-V diagram
 P (kPa)
 800 |  1<-----------3
     |   \           ^
     |    \  PV = C  |
     |     `-.       |
 200 |        `------2
     +------------------> V (m³)
       0.5           2

 T-V diagram
 T
 4T1 |               3
     |            /  ^
     |         /     |
     |      /        |
  T1 |   1---------->2
     +------------------> V (m³)
       0.5           2

 1->2 isothermal expansion; 2->3 heating at constant V;
 3->1 compression at constant P (cooling; T ∝ V, so a straight line on T-V)

The path 1→2→3→1 runs anticlockwise on the P-V diagram.

(b) Net work

W₁₂ = P₁V₁ ln(V₂/V₁) = 800 × 0.5 × ln 4 = 400 × 1.3863 = 554.5 kJ W₂₃ = 0 (constant volume) W₃₁ = −1200 kJ

Net work = 554.5 + 0 − 1200 = −645.5 kJ (work is done on the gas).

(c) Net heat

For a cycle, ΣQ = ΣW, so net heat = −645.5 kJ (net heat is rejected).

(d) Heat for process 1-2

ΔU₁₂ = 0, so Q₁₂ = W₁₂ = 554.5 kJ (added).

(e) Heat for process 3-1

Q₂₃ = ΔU₂₃ + W₂₃ = 300 + 0 = 300 kJ. For the cycle ΔU = 0: ΔU₃₁ = −(ΔU₁₂ + ΔU₂₃) = −300 kJ. Q₃₁ = ΔU₃₁ + W₃₁ = −300 − 1200 = −1500 kJ (rejected).

Check: 554.5 + 300 − 1500 = −645.5 kJ ✓.

(f) Type of cycle

The net work is negative (work input) and the net heat is negative, with the cycle running anticlockwise on the P-V diagram. So it is a refrigeration (heat pump) cycle, not a power cycle.