(a) An ammonia vapour compression refrigeration system works between temperature limits of −6·7°C and 26·7°C. The vapour is dry…

(a) COP of the ammonia vapour-compression machine

State 2 (end of compression) is dry saturated at 26.7°C: h₂ = 1297.9 kJ/kg and s₂ = 4.334 kJ/kg K. With isentropic compression, s₁ = 4.334 at −6.7°C, which is below s_g = 4.752, so the suction state is wet.

x₁ = (4.334 + 0.113)/(4.752 + 0.113) = 0.914.

h₁ = −29.3 + 0.914 × 1293.8 = 1153.3 kJ/kg.

There is no undercooling, so h₃ = h_f(26.7°C) = 125.6 kJ/kg, and throttling gives h₄ = 125.6 kJ/kg.

Refrigerating effect = h₁ − h₄ = 1153.3 − 125.6 = 1027.7 kJ/kg. Compressor work = h₂ − h₁ = 1297.9 − 1153.3 = 144.6 kJ/kg.

COP = 1027.7/144.6 = 7.11.

(b) Cogeneration plant: steam flow into the HP stage

State 1: 1 MPa, 200°C, h₁ = 2827.9 kJ/kg and s₁ = 6.6939 kJ/kg K. Ideal processes are isentropic.

HP exit at 0.3 MPa: x₂ = (6.6939 − 1.6718)/(6.9919 − 1.6718) = 0.944 and h₂ = 561.5 + 0.944 × 2163.8 = 2604.1 kJ/kg.

LP exit at 40 kPa: x₃ = (6.6939 − 1.0261)/(7.670 − 1.0261) = 0.853 and h₃ = 317.6 + 0.853 × 2318.8 = 2295.7 kJ/kg.

Heat exchanger, with saturated liquid at 0.3 MPa leaving (h = 561.5 kJ/kg): B(2604.1 − 561.5) = 500, so B = 0.245 kg/s is bled.

Turbine power: ṁ(h₁ − h₂) + (ṁ − B)(h₂ − h₃) = 1000 kW, so

ṁ(223.8) + (ṁ − 0.245)(308.4) = 1000, giving 532.2ṁ = 1075.5.

ṁ = 2.02 kg/s into the HP stage.

(c) Throttling at turbine inlet and at a heater trap exit

Both are steady, adiabatic throttling: enthalpy stays constant, pressure falls and entropy rises, so available energy is lost.

(i) Governing throttle at turbine inlet: the steam is superheated or dry. On the T–s diagram the state moves along a constant-enthalpy line to lower pressure and higher entropy, with only a small fall in temperature. The steam stays single-phase (superheated), and the loss is a reduction of the available enthalpy drop across the turbine, which is how load is controlled.

(ii) Heater drain trap: the condensate is saturated liquid at the higher heater pressure. At constant enthalpy, its enthalpy exceeds that of saturated liquid at the lower pressure, so the state moves into the two-phase region: some liquid flashes to vapour and the temperature drops to the lower saturation temperature. The dryness fraction after the trap is x = (h_before − h_f,low)/h_fg,low.

So the first is a single-phase process with a small temperature change; the second is a phase-change process (flashing) with a large temperature drop. The T–s sketches below contrast the two.