A double-acting single-cylinder steam engine runs at 250 r.p.m. and develops 30 kW. The pressure limits of operation are 10 bar and 1 bar. Cut-off is 40% of the stroke. The L/D ratio is 1.25 and the diagram factor is 0.75. Assume dry saturated steam at inlet, hyperbolic expansion and negligible effect of piston rod. Find :
(i) Mean effective pressure, ( ii) Cylinder dimensions, and
(iii) Indicated thermal efficiency. (AMIE Summer, 2001)

Question

Accepted Answer

Speed of the steam engine, N = 250 r.p.m.
Power developed, P = 30 kW
Pressure limits of operation : 10 bar ([latex] p_{1} [/latex]), 1 bar ([latex] p_{b} [/latex])
Cut-off ratio, r =[latex] \frac{1}{0.4} [/latex] = 2.5
L/D ratio = 1.25
Diagram factor, D.F. = 0.75
Condition of steam at inlet to the engine = Dry saturated.

(i) Mean effective pressure, [latex] p_{m} [/latex] :

[latex] p_{m} = D.F. [\frac{p_{1}}{r} (1 + \log _{e} r ) - p_{ b}][/latex]

= 0.75 [latex] [\frac{10}{2.5} (1 + \log _{e} 2.5) - 1 ] \simeq [/latex] 5.0 bar.

(ii) Cylinder dimensions, L and D :

Indicated power, I.P. = [latex] \frac{10 p_{m} LAN }{3} [/latex] kW

or 30 = [latex] \frac{10 × 5.0 × 125D × \frac{π}{4} D² × 250}{3} = 4090.6 D³ [/latex]

or D³ = [latex] \frac{30}{4090.6 } or Cylinder dia., D = (\frac{30}{4090.6 } )^{1 / 3 }[/latex]

= 0.194 m or 194 mm.

Length of stroke, L = 1.25 D = 1.25 × 194 = 242.5 mm.

(iii) Indicated thermal efficiency, [latex] η_{th(I)} [/latex]:

Mean flow rate of steam, [latex] \dot{m}_{s} = \frac{ \frac{π}{4} D² × L × \frac{1}{r} × 2 × N}{v_{g}} [/latex]

=[latex] \frac{ \frac{π}{4} × (0.194)² × 0.2425 × \frac{1}{2.5} × 2 × \frac{250}{60}}{0.194} [/latex] = 0.1231 kg/s

[where [latex] v_{g} [/latex] = specific volume of dry saturated steam at 10 bar = 0.194 m³/kg (from steam tables)]

∴ Indicated thermal efficiency, [latex] η_{th(I).} = \frac{I.P.}{\dot{m}_{s} (h_{1} - h_{f} ) } [/latex]

= [latex] \frac{30}{0.1231 (2776.2 - 417.5) } [/latex] = 0.1033 or 10.33%.

[where, [latex]h_{1} [/latex] = Enthalpy of dry saturated steam at 10 bar = 2776.2 kJ/kg, and
[latex] h_{f} [/latex] = Enthalpy of water at 1 bar = 417.5 kJ/kg (from steam tables)]

Question 3.15: A double-acting single-cylinder steam engine runs at 250 r.p...

Related Answered Questions

Find the gain in thermal efficiency due to the expansive action of steam considering the work done per kg of steam in two engines receiving dry steam at 17.5 bar and exhausting at 0.7 bar, one of them taking steam for whole stroke and other taking steam for 1/5th of the stroke. Assume negligible ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

The cylinder of a non-condensing steam engine is supplied with steam at 11.5 bar. The clearance volume is 1 /10 th of the stroke volume and the cut-off takes place at 1 /4 th of the stroke. If the pressure at the end of compression is 5.4 bar, compute the value of mean effective pressure of the ...

Verified Answer:

A steam engine having swept volume 0.034 m³ and clearance volume 20% of the piston displacement, consumes 2725 kg of steam per hour at 250 r.p.m. The engine is double acting and the pressure of steam near cut-off is 8.8 bar, when piston has traversed 30% of working stroke. If the compression ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

A steam engine is governed by varying the initial steam pressure, used 6 kg/kWh when developing 60 kW and 4.65 kg/kWh when developing 150 kW. If steam supplied is 96% dry at 15 bar and a back pressure of 0.36 bar, determine the steam consumption per hour and indicated thermal efficiency when ...

Verified Answer: