Question 3.8: The following readings were taken during a trial of a single...

(i) Indicated power (I.P.) :

I.P.  =  \frac{10  p_{m_1} LA_{1} N}{6}   +  \frac{10  p_{m_2} LA_{2} N}{6}                                                   …(i)

∴                              p_{m_1}  =  \frac{A_{1} S}{L}  =  \frac{1570  ×   0.12 }{53}  =  3.55   bar   

and                          p_{m_2}  =  \frac{A_{2} S}{L}  =  \frac{1440  ×   0.12 }{53}  =  3.26   bar   

A_{1} = π/4 D² = π/4 × 0.25² = 0.04908 m²
A_{2} = π/4 (D² –  d²) = π/4 (0.25²  –   0.05²) = 0.04712 m²
Inserting these values in eqn. (i), we get

I.P.  =  \frac{10 ×  3.55  ×  0.35  ×  0.04908 × 240 }{6}   +  \frac{10 ×  3.26  ×  0.35  ×  0.04712 × 240}{6}

= 24.4 + 21.5 = 45.9 kW.

(ii) Brake power (B.P.) :

B.P.  =  \frac{(W  –  S)  π (D  +  d ) N}{60  × 10³}   =  \frac{(1900  –  190)  ×  \left\lgroup  5.25  +   \frac{62.5}{1000}  \right\rgroup × 240 }{60  ×  1000}   = 36.34 kW. 

(iii) Mechanical efficiency,  η_{mech} :                             [∵     pD  =  5.25  m;    π d  =     \frac{62.5}{1000}   m                                                           …Given] 

 η_{mech}   =  \frac{B.P. }{I.P.}   =  \frac{36.34}{45.9}   =   0.7917    =     79.17%.

(iv) Specific steam consumption :
Steam consumption per hour,   m_{s} = Steam consumed/revolution × r.p.m. × 60
= [(steam consumption of cover end/stroke + steam consumption of crank end/stroke) × r.p.m. × 60]

= [ \frac{π / 4  D²  × L }{r}  +  \frac{π / 4  (D² –  d²)   L }{r}  ]  × \frac{r.p.m  ×   60  }{specific    volume   of    steam}

where r = cut-off ratio.

∴                                m_{s}  =  \frac{π}{4} . \frac{L}{r} [D²  +  (D²   –  d²)]  × \frac{240  ×   60}{x v_{g}}

= \frac{π}{4} × \frac{0.35}{3.33}  [0.25²  +  (0.25²  –  0.05²)]  ×  \frac{240  ×   60}{0.92 ×  0.185 }

[∵           r = \frac{1}{0.3}  = 3.33      and      v_{g}  = 0.185   m³ / kg    at    10.5    bar    from    steam   tables]

= 855.56 kg/h

Specific steam consumption on I.P. basis

\frac{855.56}{45.9}    =  18.64 kg/kWh.

Specific fuel consumption on B.P. basis

\frac{855.56}{36.34}    =  23.54 kg/kWh.

(v) Brake thermal efficiency:
Brake thermal efficiency,

Here                      h _{1}  =  h_{f_1}   +   x _{1} h_{fg_1}                                    [At 10.5 bar, from steam tables :  h_{f_1}  =  772  kJ / kg,    h_{fg}  =  2005.9  kJ / kg] 

= 772 + 0.92 × 2005.9 = 2617.4 kJ/kg

h_{f_2}  =  0,     since engine is non-condensing.

∴      η_{th(B)}   =  \frac{36.34}{\frac{855.56}{3600}  ×  2617.4} =  0.0584 or 5.84%.