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

(a) Take states: 1 compressor suction, 2 compressor discharge, 3 condenser outlet, 4 after throttle. The cycle is ideal: compression 1-2 is isentropic, throttling 3-4 is isenthalpic, condensation and evaporation are isobaric.

Given: T_cond = 26.7°C, T_evap = −6.7°C. From table at 26.7°C: h_f3 = 125.6 kJ/kg, h_g2 = 1297.9 kJ/kg, s_g2 = 4.334 kJ/kg K. Since vapour is dry at end of compression, state 2 is dry saturated: h₂ = 1297.9 kJ/kg, s₂ = 4.334 kJ/kg K.

For ideal isentropic compression, s₁ = s₂ = 4.334 kJ/kg K. At −6.7°C, ammonia is wet. Dryness fraction: x₁ = (s₁ − s_f1)/(s_g1 − s_f1) = (4.334 − (−0.113))/(4.752 − (−0.113)) = 4.447/4.865 = 0.9141. Then h₁ = h_f1 + x₁ h_fg1 = −29.3 + 0.9141 × 1293.8 = 1153.34 kJ/kg.

No undercooling: state 3 is saturated liquid at 26.7°C, so h₃ = h_f3 = 125.6 kJ/kg. Throttling is isenthalpic: h₄ = h₃ = 125.6 kJ/kg.

Refrigeration effect: q_L = h₁ − h₄ = 1153.34 − 125.6 = 1027.74 kJ/kg.

Compressor work: w_c = h₂ − h₁ = 1297.9 − 1153.34 = 144.56 kJ/kg.

COP = q_L / w_c = 1027.74 / 144.56 = 7.109.

COP = 7.11. Valid for an ideal saturated vapour compression cycle with isentropic compression, isenthalpic expansion, and negligible pressure drops in the heat exchangers.

(b) Let m = mass flow rate into HP stage (kg/s), B = bled mass flow rate (kg/s). State 1: HP inlet at 1 MPa, 200°C: h₁ = 2827.9 kJ/kg, s₁ = 6.6939 kJ/kg K. HP expansion is ideal, so s₂ = s₁ = 6.6939 kJ/kg K at 0.3 MPa.

From steam tables at 0.3 MPa: h_f2 = 561.47 kJ/kg, h_fg2 = 2163.8 kJ/kg, s_f2 = 1.6718 kJ/kg K, s_g2 = 6.9919 kJ/kg K. Dryness fraction at state 2: x₂ = (s₂ − s_f2)/(s_g2 − s_f2) = (6.6939 − 1.6718)/(6.9919 − 1.6718) = 5.0221/5.3201 = 0.9440. h₂ = h_f2 + x₂ h_fg2 = 561.47 + 0.9440 × 2163.8 = 2604.07 kJ/kg.

Bled steam leaves heat exchanger as saturated liquid at 0.3 MPa: h₅ = h_f2 = 561.47 kJ/kg. Heat exchanger energy balance: B(h₂ − h₅) = 500 kW B = 500/(2604.07 − 561.47) = 500/2042.60 = 0.2448 kg/s.

LP stage receives (m − B) and expands isentropically to 40 kPa, so s₃ = s₂ = 6.6939 kJ/kg K. From steam tables at 40 kPa: h_f3 = 317.58 kJ/kg, h_fg3 = 2318.8 kJ/kg, s_f3 = 1.0261 kJ/kg K, s_g3 = 7.6700 kJ/kg K. x₃ = (6.6939 − 1.0261)/(7.6700 − 1.0261) = 5.6678/6.6439 = 0.8531. h₃ = h_f3 + x₃ h_fg3 = 317.58 + 0.8531 × 2318.8 = 2295.71 kJ/kg.

Turbine power balance: m(h₁ − h₂) + (m − B)(h₂ − h₃) = 1000 kW. m(2827.9 − 2604.07) + (m − 0.2448)(2604.07 − 2295.71) = 1000. m(223.83) + (m − 0.2448)(308.36) = 1000. m(223.83 + 308.36) − 0.2448 × 308.36 = 1000. m(532.19) = 1000 + 75.48 = 1075.48. m = 1075.48/532.19 = 2.021 kg/s.

Required HP-stage mass flow rate = 2.02 kg/s. Bled flow = 0.2448 kg/s.

(c) Both processes are steady-flow adiabatic throttling, so enthalpy remains constant, pressure falls, entropy increases, and available energy is destroyed. The difference arises from the initial phase of the fluid.

(i) Throttling of steam at inlet to turbine for governing: This is vapour-phase throttling. The steam is usually superheated or dry saturated. On a T-s diagram, the initial state lies in the superheated region (or on the saturated vapour line). At constant h, the pressure is reduced to the lower admission pressure. The state moves down and to the right because entropy increases. Temperature falls slightly. If the steam is superheated, it remains superheated; if it is wet, its dryness fraction increases. The main phenomenon is a loss of available work: the enthalpy drop across the turbine is reduced, so the turbine output is controlled. This is an irreversible process, but no phase change occurs if the steam is superheated.

(ii) Throttling of condensate in closed feed heater trap exit: This is liquid-phase throttling. The condensate is saturated liquid at the higher heater pressure. On a T-s diagram, the initial state lies on the saturated-liquid line. At constant h, it is throttled to a lower pressure. Since h at the higher pressure is greater than h_f at the lower pressure, the fluid cannot remain all liquid. The state moves down and to the right into the two-phase dome. Temperature drops to the lower saturation temperature. The observed phenomenon is flashing: part of the condensate evaporates. The quality after throttling is x = (h_before − h_f_low)/h_fg_low. This flash mixture is sent to the next lower-pressure heater or condenser, recovering some heat. Unlike turbine-inlet throttling, a phase change occurs.