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

Speed of the steam engine,                                    N = 250 r.p.m.
Power developed,                                                    P = 30 kW
Pressure limits of operation : 10 bar ( p_{1} ), 1 bar ( p_{b} )
Cut-off ratio,                                                             r = \frac{1}{0.4}   = 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, p_{m} :

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

(ii) Cylinder dimensions, L and D :

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

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

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

= 0.194 m or 194 mm.

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

(iii) Indicated thermal efficiency, η_{th(I)} :

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

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

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

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

=  \frac{30}{0.1231  (2776.2  –   417.5) }   =  0.1033 or 10.33%. 

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