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Find:- (i) Air standard efficiency (ii) Mean effective pressure

In an air standard cycle pressure at the beginning of compression is 1 bar, while temperature is 310 K. Compression ratio is 10:1, Heat added is 2800 KJ/Kg of charge. The maximum pressure limit is 70 bar. If heat is added partially at constant volume partially at constant pressure,

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Given: $T_1= 310K, P_1 = 1 bar, V_1/V_2 (Comp Ratio) = 10, Q = 2800 kJ/kg, P_3 = P_4 = 70 bar$.

Solution:

At 1, applying Ideal Gas Equation,

$P_1V_1 = RT_1$

$10^5 × V_1 = 287 × 310$

$V_1 = 0.8897m^3/kg$

Therefore

$V_2 = V_1/10 = 0.08897^{1.4}$

Now, for process 1-2,

$P_1V_1^y = P_2V_2^y$

$10^5 × 0.8897^{1.4} = P_2 × 0.08897^{1.4}$

$P_2 = 25.1188 bar$

At 2, applying ideal gas eqaution,

$P_2V_2 = RT_2$

$25.118 × 10^5 × 0.08897 = 287 × T_2$

$T_2 = 778.6828K$

For Process 2-3,

$\frac{P_3}{P_2} = \frac{T_3}{T_2}$

$\frac{70}{25.1188} = \frac{T_3}{778.6828}$ $T_3 = 2170K$

$V_3 = V_2 = 0.08897m^3/kg$

Now, $Q_v = C_v(T_3 - T_2) = 0.718(2170 - 778.6828) = 998.965kJ/kg$

$Q_p = Q - Q_v = 2800 - 998.965 = 1801.034kJ/kg$

Therefore

$Q_p = C_p(T_4 - T_3) = 1.005(T_4 - 2170)$

$T_4 = 3962.07K$

For process 3-4,

$\frac{V_3}{V_4} = \frac{T_3}{T_4}$

$\frac{0.08897}{V_4} = \frac{2170}{3962.07}$

$V_4 = 0.162445m^3/kg$

Now,

$V_5 = V_1 = 0.8897m^3/kg$

Now, for process 4-5,

$P_5V_5^y = P_4V_4^y$

$P_5 × 0.8897^{1.4} = 70 × 10^5 × 0.162445^{1.4}$

$P_5 = 6.4736 bar$

At 5, applying Ideal Gas Equation,

$P_5V_5 = RT_5$

$6.4736 × 10^5 × 0.8897 = 287 × T_5$

$T_5 = 2006.816K$

Therefore

$Q_{out} = C_v(T_5 - T_1) = 0.718(2006.816 - 310) 1218.313kJ/kg$

$W = Q - Q_{out} = 1581.68kJ/kg$

Now, efficiency is given as,

$η = \frac{W}{Q} = \frac{1581.68.68}{2800} = 0.5648 = 56.48 %$

The Mean Effective Pressure is given as,

$MEP = \frac{W}{V_1 - V_2} = \frac{1581.68 × 10^3}{0.8897 - 0.08897} = 19.7529 × 10^5 = 19.7529 bar$

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