Question 3.3: The cylinder of a non-condensing steam engine is supplied wi...

Admission pressure,                                  p_{1} = 11.5 bar
Clearance volume,                                      V_{c}  =  \frac{1}{10}th of stroke volume V_s
Cut-off                                                           = \frac{1}{4} th of the stroke
Pressure at the end of compression,        p_{5} = 5.4 bar
Back pressure,                                              p_{b}  = p_{3}   = 1.1 bar.
Mean effective pressure, p_{m} :
The theoretical mean effective pressure, when clearance and compression are considered, is given by (eqn. 9) :

p_{m(th.)}  =  p_{1}  [\frac{1}{r}  +   \left\lgroup c +  \frac{1}{r}  \right\rgroup  \log _{e} \left\lgroup \frac{c+1}{c + \frac{1}{r} } \right\rgroup ]  –  p_{b}   [ (1 –  α)  +  (c + α)  \log _{e}\left\lgroup \frac{α  + c }{c}  \right\rgroup ]                                    …(i)

r = \frac{1}{¼}   =  4          or          \frac{1}{r} = \frac{1}{4}                                     and                c  =  \frac{V_{s} / 10 }{V_{s}}  = 0.1.

To find α (ratio of volume between points of compression and admission to the swept volume   V_{s} ) applying hyperbolic law between the point of beginning and end of compression:

i.e.,                                    1.1  ×   ( V_{c} + α  V_{s})   =  5.4   ×  V_{c}  

∴                                        α =  \frac{5.4  V_{c}  –    1.1  V_{c} }{1.1 V_{s}}

= \frac{5.4  V_{c}  –    1.1  V_{c}}{11  V_{c}}   = 0.39

[ ∵    V_{s}  =  V_{c}          and          p_{4}  =  p_{b}   =  1.1   bar   and
∵            α  =  \frac{V_{4}     –  V_{c}  }{V_{s}}
∴             V_{4} =  V_{c}  +  α  V_{s} 
Also      \frac{V_{c}}{V_{s}}  =  \frac{1}{10}
∴              V_{s} =   10 V_{c}   ]

Now inserting various values in eqn. (i), we get

– 1.1 [ (1 – 0.39 )  +  (0.1  +  0.39 )  \log _{e}\left\lgroup \frac{0.39  +  0.1  }{0.1  }  \right\rgroup ]

= 11.5 × 0.65 – 1.1 × 1.388 = 5.95 bar

i.e.,                                      Mean effective pressure of steam = 5.95 bar.