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 $\frac{1}{5}$ th of the stroke.
Assume negligible clearance and hyperbolic expansion.

Question

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 [latex] \frac{1}{5} [/latex]th of the stroke.
Assume negligible clearance and hyperbolic expansion.

Accepted Answer

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

Work obtained from the engine taking steam for [latex] \frac{1}{5} [/latex]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 [latex] \frac{1}{5} [/latex]th of stroke, only [latex] \frac{1}{5} [/latex]th of the steam will be admitted.

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

Mean effective pressure, [latex] p_{m} = \frac{p1}{r}(1+ \log_{e} r) - p_{b} [/latex]

Here expansion ratio, r = [latex]\frac{V_{2}}{V_{1}} = \frac{V_{2}}{\frac{V_{2}}{5} } [/latex] = 5

∴ [latex] p_{m} = \frac{17.5}{5} (1+ \log_{e} 5) - 0.7 = 8.43 bar [/latex]

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

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

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

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

= 2794.1 – 376.8
= 2417.3 kJ/kg

Thermal efficiency = [latex] \frac{Work done }{Heat supplied} [/latex]

= [latex] \frac{476.3 }{2417.3 } [/latex] = 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

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

Heat supplied per kg = [latex] h_{1} - h_{f_3} = 2794.1 – 376.8 [/latex]

= 2417.3 kJ/kg

Thermal efficiency = [latex] \frac{189.84 }{2417.3} = 0.0785 = 7.85% [/latex]

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.

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

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