(a) The following data refers to a single-stage vapour compression refrigeration system :

Refrigerant = R134a
Condenser temperature = 35 °C
Evaporator temperature = – 10 °C
Compressor motor speed = 2800 r.p.m.
Clearance ratio = 0·03
Swept volume = 2

Question

(a) The following data refers to a single-stage vapour compression refrigeration system :

Refrigerant = R134a
Condenser temperature = 35 °C
Evaporator temperature = – 10 °C
Compressor motor speed = 2800 r.p.m.
Clearance ratio = 0·03
Swept volume = 269·4 cm³
Expansion index = 1·12
Compression isentropic efficiency = 75%
Condensate subcooling in condenser = 5 °C

Draw P-h diagram and determine the following :

(i) The capacity of the plant in TR
(ii) The power required in kW
(iii) The COP
(iv) The heat rejection to condenser
(v) The second law efficiency

The properties of R134a are given in the table :

| T (°C) | P (bar) | Specific volume of saturated vapour v_g (m³/kg) | Enthalpy (kJ/kg) h_f | h_g | Entropy (kJ/kg-K) s_f | s_g |
|--------|---------|-----------------------------------------------|----------------------|-----|----------------------|-----|
| – 10 | 2·014 | 0·0994 | 186·7 | 392·4 | 0·9512 | 1·733 |
| 35 | 8·870 | — | 249·1 | 417·6 | 1·1680 | 1·715 |

Assume specific heat of liquid and vapour at 8·87 bar as 1·458 kJ/kg-K and 1·1 kJ/kg-K, respectively. The refrigerant at entry to compressor is in dry saturated state. (20 marks)

(b) (i) What are the important properties of lubricants used in IC engines? Discuss their significance. (10 marks)

(ii) What are the advantages and disadvantages of the Indirect Injection (IDI) swirl chamber over the open-type Direct Injection (DI) combustion chamber in CI engines? (10 marks)

(c)

Calculate the critical pressure and throat area per unit mass flow rate of a convergent-divergent nozzle, expanding steam from 10 bar and dry saturated, down to atmospheric pressure of 1 bar. Assume that the inlet velocity is negligible and that the expansion is isentropic.

Use Steam tables given at the end to get the steam data.

The value of isentropic expansion index may be taken as 1·135. 10 marks

Use Steam tables given at the end to get the steam data.

The value of isentropic expansion index may be taken as 1·135. (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i) – (v)
P–h diagram: P vertical, h horizontal. Mark 1 at 2.014 bar, dry saturated vapour; 2s at 8.870 bar after isentropic compression; 2 at 8.870 bar after actual compression; 3 after condensation and 5 °C subcooling; 4 after isenthalpic throttling at 2.014 bar. Draw 4–1 evaporation, 1–2 actual compression, 1–2s dashed isentropic, 2–3 condensation/subcooling, 3–4 throttling.

State 1: h₁ = 392.4 kJ/kg, s₁ = 1.733 kJ/kg·K, v₁ = 0.0994 m³/kg.
Clearance volumetric efficiency, η_v = 1 + C − C(p₂/p₁)^(1/n).
p₂/p₁ = 8.870/2.014 = 4.4042.
η_v = 1 + 0.03 − 0.03(4.4042)^(1/1.12) = 1.03 − 0.03(3.7573) = 0.9173.

Swept volume V_s = 269.4 × 10⁻⁶ m³. N = 2800/60 = 46.667 r.p.s.
ṁ = η_v V_s N / v₁ = 0.9173 × 269.4 × 10⁻⁶ × 46.667 / 0.0994 = 0.1160 kg/s.

For isentropic compression, s₂s = s₁ = 1.733. At 8.870 bar, s_g = 1.715, cp_v = 1.1 kJ/kg·K.
T₂s = 308.15 exp[(1.733 − 1.715)/1.1] = 313.23 K.
h₂s = 417.6 + 1.1(313.23 − 308.15) = 423.19 kJ/kg.
h₂ = h₁ + (h₂s − h₁)/η_isen = 392.4 + (423.19 − 392.4)/0.75 = 433.45 kJ/kg.
Compressor work w_c = h₂ − h₁ = 41.05 kJ/kg.

Subcooling: h₃ = h_f35 − cp_l ΔT = 249.1 − 1.458 × 5 = 241.81 kJ/kg.
Throttling: h₄ = h₃ = 241.81 kJ/kg.
Refrigerating effect q_e = h₁ − h₄ = 392.4 − 241.81 = 150.59 kJ/kg.

(i) Capacity
Q_e = ṁ q_e = 0.1160 × 150.59 = 17.47 kW.
1 TR = 3.51685 kW.
Capacity = 17.47/3.51685 = 4.97 TR.

(ii) Power required
W = ṁ w_c = 0.1160 × 41.05 = 4.76 kW.
Power = 4.76 kW.

(iii) COP
COP = q_e/w_c = 150.59/41.05 = 3.67.
COP = 3.67.

(iv) Heat rejection
Q_c = ṁ(h₂ − h₃) = 0.1160(433.45 − 241.81) = 22.23 kW.
Also Q_c = Q_e + W = 17.47 + 4.76 = 22.23 kW.
Heat rejection = 22.23 kW.

(v) Second-law efficiency
COP_Carnot = T_L/(T_H − T_L) = 263.15/(308.15 − 263.15) = 5.848.
η_II = COP/COP_Carnot = 3.67/5.848 = 0.627.
Second-law efficiency = 62.7%.

(b)(i)
Important lubricant properties in IC engines:
Viscosity: maintains oil film; too high causes pumping loss, too low causes wear.
Viscosity index: low change of viscosity with temperature; ensures cold starting and hot protection.
Pour point: must be low for cold-weather flow.
Flash and fire point: high for safety and low evaporation loss.
Oxidation and thermal stability: resist sludge, varnish and thickening at high temperature.
Detergency and dispersancy: keep soot and deposits suspended, keeping engine clean.
Anti-wear and extreme-pressure properties: protect cams, bearings and piston rings in boundary lubrication.
Corrosion and rust inhibition: neutralise acidic combustion products.
Foam resistance: prevents air entrainment and oil-film breakdown.
Seal compatibility: prevents swelling, shrinkage and leakage.
Alkalinity/TBN: neutralises acids and extends drain interval.
Significance: together these ensure lubrication, cooling, sealing, cleaning, corrosion protection and longer engine life.

(b)(ii)
Advantages of IDI swirl chamber over open DI: lower injection pressure is needed; better fuel-air mixing at low speed and part load; smoother pressure rise, lower combustion noise; lower NOx due to lower peak temperature; less sensitive to fuel quality; good cold starting with glow plug; wider speed range for small engines.
Disadvantages: higher specific fuel consumption, often 10–20% penalty, because of heat loss and throttling between chambers; lower thermal efficiency and power output; more complex cylinder head and higher manufacturing cost; greater heat loss and thermal loading; limited to smaller engines; poorer full-load economy than open DI.

(c)
For steam, use isentropic expansion index n = 1.135.
Critical pressure ratio:
p*/p₁ = (2/(n+1))^(n/(n−1)) = (2/2.135)^(1.135/0.135) = 0.5774.
p₁ = 10 bar, so p* = 10 × 0.5774 = 5.774 bar.
Critical pressure = 5.77 bar.

From saturated steam tables at 10 bar dry saturated: v₁ = v_g = 0.1944 m³/kg.
Critical velocity:
c* = √[2n/(n+1) p₁v₁]
= √[(2×1.135/2.135) × 10×10⁵ × 0.1944]
= √(1.0632 × 1.944×10⁵) = 454.6 m/s.

Specific volume at throat:
v* = v₁(p₁/p*)^(1/n) = 0.1944(1/0.5774)^(1/1.135) = 0.3155 m³/kg.
Throat area per unit mass flow rate:
A*/ṁ = v*/c* = 0.3155/454.6 = 6.94×10⁻⁴ m² per (kg/s).
**A*/ṁ = 6.94 cm² per kg/s.**

(a) The following data refers to a single-stage vapour compression refrigeration system : Refrigerant = R134a Condenser…

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