Question 3.2: Find the gain in thermal efficiency due to the expansive act...

Initial pressure of steam,                                    p_{1} = 17.5 bar
Dryness fraction,                                                  x_{1} = 1
Pressure of steam after expansion,                    p_{3} (= p_{b})  = 0.7 bar
Refer Fig. 30.

Work obtained from the engine taking steam for \frac{1}{5} th of the stroke = area ‘512345’.
Work obtained from the engine taking steam for the whole stroke = area ‘51′345’.
Let us consider that 1 kg of steam is supplied to the engine during one full stroke. Hence for the engine taking steam for \frac{1}{5} th of stroke, only \frac{1}{5} th of the steam will be admitted.

Case 1. Engine taking steam for \frac{1}{5} th of stroke :
Considering the p-V diagram 51234, we have :

Mean effective pressure,                                p_{m}  = \frac{p1}{r}(1+  \log_{e}  r) –  p_{b}

Here expansion ratio,                                            r = \frac{V_{2}}{V_{1}}  =  \frac{V_{2}}{\frac{V_{2}}{5} }   =  5

∴                                                  p_{m}  =  \frac{17.5}{5} (1+ \log_{e}   5)  –   0.7  =  8.43  bar

At 17.5 bar.  Specific volume,         v = 0.113 m³/kg.
Work done by \frac{1}{5}   th of steam               = p_{m}  ×  v

= \frac{ 8.43 ×  10^{5}  ×  0.113}{10^{3}}   = 95.26 kJ/kg.

Work done per kg of steam                     = 95.26 × 5 = 476.3 kJ

Heat supplied per kg                                =    h_{1}   – h_{f_3}                                   [At  17.5  bar .   h_{1}  =  h_{g_1}  = 2794.1    kJ/kg
At   0.7 bar .   h_{f_3}  = 376.8  kJ/kg   ]

= 2794.1 – 376.8
= 2417.3 kJ/kg

Thermal efficiency =  \frac{Work    done }{Heat    supplied}

= \frac{476.3 }{2417.3 } = 0.197 or 19.7%.

Case 2. Engine taking steam for the whole stroke :
Considering the p-V diagram 51′ 34.
Work done/kg of steam                                    = area 51′ 345

= (p_{1} –  p_{b} ) v_{2} =   \frac{(17.5  –  0.7 ) ×  10^{5}  ×  0.113}{10^{3}} = 189.84 kJ/kg

Heat supplied per kg                                           =   h_{1}   – h_{f_3} = 2794.1   –   376.8

= 2417.3  kJ/kg

Thermal efficiency                                 =  \frac{189.84 }{2417.3} = 0.0785 = 7.85%

Gain in thermal efficiency = 19.7 – 7.85 = 11.85%.
This gain is an increase in thermal efficiency by more than 150 per cent due to the expansive working of steam.