A diesel engine contains 0.1 m³ of air at 0.98 bar and 30°C at the beginning of compression. The compression ratio is 15 and the volume at cut-off is 0.0125 m³. Determine for the corresponding air standard cycle :
(i) The cut-off ratio ; (ii) The per cent clearance ;
(iii) The work done ; (iv) The air standard efficiency.
Take $c_{p}$ = 1.005 kJ/kg K and γ = 1.4. (GATE)

Question

A diesel engine contains 0.1 m³ of air at 0.98 bar and 30°C at the beginning of compression. The compression ratio is 15 and the volume at cut-off is 0.0125 m³. Determine for the corresponding air standard cycle :
(i) The cut-off ratio ; (ii) The per cent clearance ;
(iii) The work done ; (iv) The air standard efficiency.
Take [latex] c_{p}[/latex] = 1.005 kJ/kg K and γ = 1.4. (GATE)

Accepted Answer

Fig. 43. shows the diesel cycle on p-v diagram.
Given : [latex]V_{1} = 0.1 m³ ; T_{1} = 30 + 273 = 303 K ; [/latex]

[latex]p_{1} [/latex] = 0.98 bar .

Compression ratio, r = [latex] \frac{V_{1}}{V_{2}} [/latex] = 15

∴ Clearance volume, [latex] V_{2} = \frac{V_{1}}{15} = \frac{0.1}{15} [/latex] = 0.006667 m³

(i) Cut-off ratio, ρ = [latex] \frac{V_{3}}{V_{2}} [/latex]

= [latex] \frac{0.0125}{0.006667} [/latex] = 1.875.

(ii) The per cent clearance = [latex] \frac{1}{15} [/latex] = 0.06667

or 6.667%.

(iii) The work done, W :
[latex]T_{2} = T_{1} × (r)^{γ–1} = 303 × (15)^{1.4 – 1} [/latex] = 895.1 K
[latex]T_{3} = ρ × T_{2}[/latex] = 1.875 × 895.1 = 1678.3 K

[latex]T_{4} = T_{3} × (\frac{ ρ}{r})^{γ–1} = 1678.3 × (\frac{1.875 }{15})^{1.4 – 1} [/latex] = 730.5 K

Heat supplied, [latex]Q_{2–3} = c_{p}(T_{3} – T_{2})[/latex] = 1.005(1678.3 – 895.1) = 787.1 kJ/kg
Heat rejected, [latex]Q_{4–1} = c_{v} (T_{4} – T_{1})[/latex]

= 0.718 (730.5 – 303) = 306.9 kJ/kg (∵ [latex]c_{v} = \frac{c_{p} }{γ} = \frac{1.005}{1.4}[/latex] = 0.718)

∴ Work done, W = [latex]Q_{2–3} – Q_{4–1}[/latex] = 787.1 – 306.9 = 480.2 kJ/kg.

(iv) The air standard efficiency, [latex]η_{air-standard}[/latex] :

∴ [latex]η_{air-standard} = \frac{W}{Q_{2–3}} = \frac{480.2}{787.1}[/latex] = 0.61 or 61%.

[Check : [latex]η_{diesel} = 1 - \frac{ρ^{γ} - 1}{γ (r)^{γ - 1 } ( ρ - 1) } = 1 - \frac{(1.875)^{1.4 } - 1}{1.4 × (15)^{1.4 - 1} × (1.875 - 1)}[/latex] = 0.61 or 61% ]

Question 8.41: A diesel engine contains 0.1 m³ of air at 0.98 bar and 30°C ...

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