(a) A four-cylinder diesel engine with swept volume of 0·98 litre is tested on a performance bed. The engine running at a speed of 2500 r.p.m. against a brake with arm of 0·3 m produces brake load of 190 N with fuel consumption of 6·8 litres/hr. The

Question

(a) A four-cylinder diesel engine with swept volume of 0·98 litre is tested on a performance bed. The engine running at a speed of 2500 r.p.m. against a brake with arm of 0·3 m produces brake load of 190 N with fuel consumption of 6·8 litres/hr. The calorific value of fuel is 45000 kJ/kg and specific gravity of fuel is 0·82. A Morse test is carried out on the engine by cutting off the fuel supply of individual cylinder in the order 1, 2, 3, 4 with corresponding brake loads 131 N, 135 N, 133 N and 137 N, respectively. Calculate the b.p., b.m.e.p., brake thermal efficiency, b.s.f.c., i.p., mechanical efficiency and i.m.e.p. of the engine at test speed. (20 marks)

(b) A power plant operates on a regenerative steam cycle with one closed feedwater heater. Steam enters the first turbine stage at 125 bar, 500 °C and expands to 10 bar, where some of the steam is extracted and diverted to the closed feedwater heater. Condensate exiting the feedwater heater as saturated liquid at 10 bar passes through a trap into the condenser. The feedwater exits the heater at 120 bar with a temperature of 170 °C. The condenser pressure is 0·06 bar. Assuming isentropic turbine and pump work, determine the thermal efficiency of the cycle. At 125 bar, 500 °C for steam, h = 3343·6 kJ/kg and s = 6·4651 kJ/kg-K. [Use steam tables provided at the end of this Paper] (20 marks)

(c) Explain NH₃-water vapor absorption refrigeration system with a neat diagram. What are the desired properties of refrigerant-absorber combination? (10 marks)

UPSC Answer Check · Accepted Answer

(a)
Torque, T = brake load × arm = 190 × 0.3 = 57 N m.

Brake power, B.P. = 2πNT/60 = 2π × 2500 × 57/60 = 14922.6 W = 14.9226 kW.

Swept volume, V_s = 0.98 litre = 0.98 × 10⁻³ m³. For a four-stroke engine:
b.m.e.p. = B.P. × 2 × 60/(V_s × N) = 14.9226 × 120/(0.98 × 10⁻³ × 2500) = 730.90 kPa = 7.309 bar.

Fuel mass flow, m_f = 6.8 × 0.82 = 5.576 kg/h = 0.0015489 kg/s.
Heat supplied, Q_in = m_f × CV = 0.0015489 × 45000 = 69.70 kW.

Brake thermal efficiency, η_bt = B.P./Q_in = 14.9226/69.70 = 0.2141 = 21.41%.

b.s.f.c. = m_f/B.P. = 5.576/14.9226 = 0.3737 kg/kWh = 373.7 g/kWh.

Morse test: if one cylinder is cut, its indicated power = B.P. − B.P. with that cylinder cut. Since speed and arm are unchanged, B.P. ∝ brake load.

Total I.P. = B.P. × [(190−131)+(190−135)+(190−133)+(190−137)]/190
= 14.9226 × (59+55+57+53)/190 = 14.9226 × 224/190 = 17.5929 kW.

Mechanical efficiency, η_m = B.P./I.P. = 190/224 = 0.8482 = 84.82%.

i.m.e.p. = b.m.e.p. × I.P./B.P. = 7.309 × 224/190 = 8.617 bar = 861.7 kPa.

(b)
State 1: 125 bar, 500 °C, h₁ = 3343.6 kJ/kg, s₁ = 6.4651 kJ/kg-K.

At 10 bar, from steam tables:
h_f10 = 762.81 kJ/kg, h_fg10 = 2015.3 kJ/kg, s_f10 = 2.1387, s_fg10 = 4.4478 kJ/kg-K.

Isentropic expansion 1→2:
x₂ = (s₁ − s_f10)/s_fg10 = (6.4651 − 2.1387)/4.4478 = 0.9727.
h₂ = h_f10 + x₂ h_fg10 = 762.81 + 0.9727 × 2015.3 = 2723.10 kJ/kg.

At 0.06 bar:
h_f0 = 151.53 kJ/kg, h_fg0 = 2415.9 kJ/kg, s_f0 = 0.5209, s_fg0 = 7.8095 kJ/kg-K.

Isentropic expansion 2→3 at s = 6.4651:
x₃ = (6.4651 − 0.5209)/7.8095 = 0.7611.
h₃ = 151.53 + 0.7611 × 2415.9 = 1990.39 kJ/kg.

Pump work, 0.06 bar → 120 bar:
w_p = v_f0(P_120 − P_0.06) = 0.0010064 × (12000 − 6) kPa = 12.07 kJ/kg.
h_fw,in = h_f0 + w_p = 151.53 + 12.07 = 163.60 kJ/kg.

Feedwater exits heater at 120 bar, 170 °C. Using compressed-liquid tables, h_fw,out ≈ 731.7 kJ/kg.

Closed feedwater heater heat balance:
y(h₂ − h_f10) = h_fw,out − h_fw,in
y = (731.7 − 163.60)/(2723.10 − 762.81) = 0.2898.

Turbine work:
w_t = (h₁ − h₂) + (1 − y)(h₂ − h₃)
= (3343.6 − 2723.10) + 0.7102(2723.10 − 1990.39)
= 620.50 + 520.37 = 1140.87 kJ/kg.

Net work, w_net = w_t − w_p = 1140.87 − 12.07 = 1128.80 kJ/kg.
Heat input, q_in = h₁ − h_fw,out = 3343.6 − 731.7 = 2611.9 kJ/kg.

Thermal efficiency, η_th = w_net/q_in = 1128.80/2611.9 = 0.4322 = 43.2%.

(c)
In an NH₃-water vapour absorption refrigeration system, ammonia is the refrigerant and water is the absorbent.

Schematic flow:
Generator: heat supplied separates NH₃ vapour from strong aqua.
Rectifier: removes water vapour carried with NH₃ vapour.
Condenser: NH₃ vapour condenses, rejecting heat.
Expansion valve: pressure of liquid NH₃ is reduced.
Evaporator: NH₃ absorbs heat from the refrigerated space and vaporises.
Absorber: NH₃ vapour from evaporator is absorbed in weak aqua coming from generator, forming strong aqua; heat of absorption is rejected.
Pump: strong aqua is pumped to generator pressure.
Solution heat exchanger: hot weak aqua preheats strong aqua, improving COP.
Pressure-reducing valve: weak aqua returns from generator pressure to absorber pressure.

Flow path:
Generator NH₃ vapour → Rectifier → Condenser → Expansion valve → Evaporator → Absorber.
Strong aqua: Absorber → Pump → Solution heat exchanger → Generator.
Weak aqua: Generator → Solution heat exchanger → Pressure-reducing valve → Absorber.

Desired properties of refrigerant-absorber combination:
Refrigerant should have high latent heat, low boiling point, high critical temperature, and low specific volume.
Absorbent should have high affinity for refrigerant, low volatility, low viscosity, low specific heat, and low vapour pressure at absorber temperature.
The pair must be chemically stable, non-corrosive, non-toxic, non-flammable and inexpensive.
Refrigerant should be more volatile than absorbent so it can be separated easily in the generator.
The mixture should not form solids or polymers in the operating range and should give high COP with low circulation ratio.

(a) A four-cylinder diesel engine with swept volume of 0·98 litre is tested on a performance bed. The engine running at a speed…

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