(a) A single-cylinder, two-stroke, high-speed diesel engine working on 'dual cycle' has a compression ratio of 15 : 1. The engine takes in atmospheric air at 1 bar and 27 °C. The maximum temperature in the cylinder is 1312 K. The cutoff ratio is 1·09

Question

(a) A single-cylinder, two-stroke, high-speed diesel engine working on 'dual cycle' has a compression ratio of 15 : 1. The engine takes in atmospheric air at 1 bar and 27 °C. The maximum temperature in the cylinder is 1312 K. The cutoff ratio is 1·093. The engine has a bore of 200 mm and stroke length of 250 mm. The engine speed is 3000 r.p.m. Calculate the following:

(i) The cycle efficiency
(ii) The net work output per unit mass of air
(iii) The power output
(iv) The mean effective pressure
(v) The ratio of heat added in the constant-pressure process to that in the constant-volume process

Draw the P-V and T-s diagrams of the cycle.

For air, γ = 1·4, R = 0·287 kJ/kg-K. (20 marks)

(b)

An ideal regenerative steam power cycle operates so that steam enters the turbine at 30 bar and 350 °C, and exhausts at 0·1 bar. A single open feedwater heater is employed which operates at 5 bar. Compute the thermal efficiency of the cycle. The steam properties at 30 bar and 350 °C are h = 3115·3 kJ/kg and s = 6·7427 kJ/kg-K.

Use Steam tables given at the end to get other properties. 20 marks

Use Steam tables given at the end to get other properties. (20 marks)

(c) Why is vapour absorption refrigeration system considered to be a better option in comparison to vapour compression refrigeration system for large capacity refrigeration and air-conditioning plants? (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i) Air-standard dual-cycle analysis.
Given: r = 15, ρ = 1.093, T₁ = 27 °C = 300 K, T₄ = 1312 K, γ = 1.4, R = 0.287 kJ/kg-K.
Isentropic compression 1–2: T₂ = T₁ r^(γ−1) = 300 × 15^0.4 = 886.25 K.
Constant-pressure process 3–4: ρ = V₄/V₃ = T₄/T₃, so T₃ = 1312/1.093 = 1200.37 K.
Isentropic expansion 4–5: T₅ = T₄(ρ/r)^(γ−1) = 1312 × (1.093/15)^0.4 = 460.20 K.
cv = R/(γ−1) = 0.287/0.4 = 0.7175 kJ/kg-K, cp = γcv = 1.0045 kJ/kg-K.
qin = cv(T₃−T₂) + cp(T₄−T₃)
qin = 0.7175(314.11) + 1.0045(111.63) = 225.38 + 112.14 = 337.51 kJ/kg.
qout = cv(T₅−T₁) = 0.7175(160.20) = 114.94 kJ/kg.
η = 1 − qout/qin = 1 − 114.94/337.51.
η = 0.6594 = 65.94 %.

(a)(ii) Net work per unit mass:
wnet = qin − qout = 337.51 − 114.94.
wnet = 222.57 kJ/kg.

(a)(iii) Swept volume:
Vs = (π/4)D²L = (π/4)(0.2)²(0.25) = 0.007854 m³.
V₁ = Vs r/(r−1) = 0.007854 × 15/14 = 0.008415 m³.
Mass per cycle: m = P₁V₁/(RT₁) = 100 × 0.008415/(0.287 × 300) = 0.009773 kg.
For a two-stroke engine, cycles/s = N/60 = 3000/60 = 50.
Power = m wnet × cycles/s = 0.009773 × 222.57 × 50.
Power = 108.76 kW.

(a)(iv) Mean effective pressure:
MEP = wnet per cycle / Vs = (0.009773 × 222.57)/0.007854 = 276.97 kPa.
MEP = 276.97 kPa = 2.77 bar.

(a)(v) Ratio of constant-pressure heat to constant-volume heat:
Qp/Qv = cp(T₄−T₃)/[cv(T₃−T₂)]
= 1.0045 × 111.63/(0.7175 × 314.11).
Qp/Qv = 0.4976.

Diagrams: P-V: 1→2 isentropic compression, 2→3 constant-volume heat addition, 3→4 constant-pressure heat addition, 4→5 isentropic expansion, 5→1 constant-volume heat rejection.
T-s: 1→2 vertical isentropic compression, 2→3 constant-volume heat addition (steeper up-right), 3→4 constant-pressure heat addition (less steep up-right), 4→5 vertical isentropic expansion, 5→1 constant-volume heat rejection (down-left).

(b) Ideal regenerative Rankine cycle with one open feedwater heater at 5 bar.
At 5 bar (0.5 MPa), from steam tables: h_f = 640.23 kJ/kg, h_fg = 2108.5 kJ/kg, s_f = 1.8607 kJ/kg-K, s_fg = 4.9606 kJ/kg-K, v_f = 0.001093 m³/kg.
At 0.1 bar (10 kPa): h_f = 191.81 kJ/kg, h_fg = 2392.8 kJ/kg, s_f = 0.6492 kJ/kg-K, s_fg = 7.5010 kJ/kg-K, v_f = 0.001010 m³/kg.
State 1: h₁ = 3115.3 kJ/kg, s₁ = 6.7427 kJ/kg-K.
At 5 bar: x₂ = (6.7427 − 1.8607)/4.9606 = 0.9842.
h₂ = 640.23 + 0.9842 × 2108.5 = 2715.3 kJ/kg.
At 0.1 bar: x₃ = (6.7427 − 0.6492)/7.5010 = 0.8124.
h₃ = 191.81 + 0.8124 × 2392.8 = 2135.6 kJ/kg.
Condensate pump to heater: h₅ = h₄ + v₄(P₅ − P₀.₁) = 191.81 + 0.001010(500 − 10) = 192.30 kJ/kg.
Open FWH energy balance: yh₂ + (1−y)h₅ = h₆ = h_f at 5 bar = 640.23 kJ/kg.
y = (640.23 − 192.30)/(2715.3 − 192.30) = 0.1775.
Feed pump to boiler: h₇ = h₆ + v₆(3000 − 500) = 640.23 + 0.001093 × 2500 = 642.96 kJ/kg.
qin = h₁ − h₇ = 3115.3 − 642.96 = 2472.34 kJ/kg.
Turbine work: wT = (h₁ − h₂) + (1−y)(h₂ − h₃) = 399.98 + 0.8225 × 579.70 = 876.76 kJ/kg.
Pump work: wP = (1−y)(h₅ − h₄) + (h₇ − h₆) = 0.407 + 2.733 = 3.14 kJ/kg.
η = (wT − wP)/qin = (876.76 − 3.14)/2472.34.
η = 0.3534 = 35.34 %.

(c) Vapour absorption refrigeration is preferred for large-capacity plants because:
It uses low-grade heat, such as waste heat, solar heat, turbine exhaust, or process heat, instead of costly high-grade mechanical work.
It has very few moving parts; the compressor is replaced by a pump, giving less noise, vibration, wear, and maintenance.
It is reliable and long-lasting, which is important in large continuous air-conditioning and refrigeration plants.
The refrigerants used, such as water–LiBr or ammonia–water, are relatively cheap and environment-friendly; water–LiBr has no ozone-depletion or global-warming problem.
It reduces electrical peak demand and operating cost where waste heat is available.
It scales well for large capacities, where the size and power consumption of a vapour-compression compressor become very high.
Limitation: its COP is usually lower than vapour compression, and it needs a heat source and cooling water, so it is best where cheap heat is available.

(a) A single-cylinder, two-stroke, high-speed diesel engine working on 'dual cycle' has a compression ratio of 15 : 1. The engine…

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