Steam is admitted to an engine for 35% of the stroke with a pressure of 7.2 bar, the law of expansion followed is pV^1.1 = constant. Compression starts at 55% of return stroke and follows the law pV^1.2 = constant, the clearance volume is 20% of the displacement volume and the back pressure is 660

Question

Steam is admitted to an engine for 35% of the stroke with a pressure of 7.2 bar, the law of expansion followed is [latex] pV^{1.1}[/latex] = constant. Compression starts at 55% of return stroke and follows the law [latex] pV^{1.2} [/latex] = constant, the clearance volume is 20% of the displacement volume and the back pressure is 660 mm of mercury vacuum when barometer reads 760 mm of mercury. Estimate the mean effective pressure and indicated power of a double-acting engine with cylinder diameter 320 mm, stroke 480 mm and speed 200 r.p.m.

Accepted Answer

Cylinder diameter, D = 320 mm = 0.32 m
Stroke length, L = 480 mm = 0.48 m
Admission steam pressure, [latex] p_{1} [/latex] = 7.2 bar
Cut-off = 0.35 [latex] V_{s} [/latex]
Law of expansion, [latex] pV^{1.1}[/latex] = constant
Law of compression, [latex] pV^{1.2}[/latex] = constant
Clearance volume, [latex] V_{c} = 0.2 V_{s} [/latex]
Back pressure, [latex] p_{b} [/latex] = 660 mm of Hg
Barometer reading = 760 mm of Hg
Engine speed, N = 200 r.p.m.
Refer Fig. 36.

Back pressure , [latex] p_{b} = \frac{760 - 660}{760} × 1.013 = 0.133[/latex] bar

Considering expansion curve, [latex] pV^{1.1}[/latex] = constant

[latex] p_{1}V_{1}^{1.1} = p_{2}V_{2}^{1.1}[/latex]

[latex] p_{2} = p_{1} ( \frac{V_{1}}{V_{2}} )^{1.1} = 7.2 [\frac{(0.35 V_{s} + 0.2 V_{s})}{( V_{s} + 0.2 V_{s})}]^{1.1} [/latex]

= [latex] 7.2 (\frac{0.55}{1.2}) ^{1.1} [/latex] = 3.05 bar

Considering the compression curve, [latex] pV^{1.2}[/latex] = constant

[latex] p_{4}V_{4}^{1.2} = p_{5}V_{5}^{1.2}[/latex]

[latex] p_{5} = p_{4} × (\frac{V_{4}}{V_{2}}) ^{1.2} = 0.133 × [\frac{0.45 V_{s} + 0.2 V_{s}}{0.2 V_{s}}] ^{1.2} [/latex]

= [latex]0.133 × ( \frac{0.65}{0.2} ) ^{1.2} [/latex] = 0.55 bar.

Also, mean effective pressure, [latex] p_{m} = \frac{Area of the indicator diagram }{Stroke volume} [/latex]

= [latex] \frac{area ‘61ab6’ + area ‘12ca1’ - area ‘43cd4’ - area ‘54db5’ }{V_{s}} [/latex]

= [latex] \frac{[(7.2 × 0.35 V_{s} ) \left\{\frac{7.2 × 0.55 V_{s} - 3.05 × 1.2 V_{s}}{1.1 - 1} \right\} ] }{V_{s}} - \frac{(0.133 × 0.55 V_{s}) + (\frac{ 0.55 × 0.2 V_{s} - 0.133 × 0.65 V_{s} }{1.2 - 1} )}{V_{s}} [/latex]

= [latex] \frac{[2.52 + \frac{3.96 - 3.66}{0.1}] - [0.073 + \frac{0.11 - 0.086}{0.2}]}{1} [/latex]

= [latex] \frac{(2.52 + 3.0) - (0.073 + 0.12)}{1} [/latex] = 5.33 bar

∴ Indicated power, I.P. = [latex] \frac{10 p_{m} LAN }{3} [/latex] [∵ Engine is double-acting]

= [latex] \frac{10 × 5.33 × 0.48 × π / 4 × 0.32 ² × 200 }{3} [/latex]

= 137.17 kW.

Question 3.9: Steam is admitted to an engine for 35% of the stroke with a ...

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