Given:

$T_1= 300K, P_1 = 0.1 MPa = 1 bar, V_1/V_2 (Comp Ratio) = 18$

Solution:

At 1, applying Ideal Gas Equation,

$P_1V_1 = RT_1$

$10^5 × V_1 = 287 × 300$

$V_1 = 0.861m^3/kg$

Therefore,

$V_2 = V_1/18 = 0.0478m^3/kg$

Now, for process 1-2,

$P_1V_1^y = P_2V_2^y$

$10^5 × 0.861^14 = P_2 × 0.0478^14$

$P_2 = 57.198 bar$

At 2, applying ideal gas equation,

$P_2V_2 = RT_2$

$57.198 × 10^5 × 0.0478 = 287 × T_2$

$T_2 = 953.3K$

For process 2-3 ,

$\frac{P_3}{P_2} = \frac{T_3}{T_2} = 1.5$

Therefore,

$T_3 = 1429.95K$
$P_3 = 85.79bar$
$V_3 = V_2 = 0.0478m^3/kg$

For process 3-4,

$\frac{V_4}{V_3} = \frac{T_4}{T_3} = 1.2$

Therefore,

$T_4 = 1715.94K$

$V_4 = 0.0574m^3/kg$

$P_4 = P_3 = 85.797 bar$

Now,

$V_5 = V_1 = 0.861m^3/kg$

Now, for process 4-5,

$P_5V_5^y = P_4V_4^y$

$P_5 × 0.861^{1.4} = 85.797 × 10^5 × 0.0574^{1.4}$

$P_5 = 1.93617bar$

At 5, applying Ideal Gas Equation,

$P_5V_5 = RT_5$

$1.93617 × 10^5 × 0.861 = 287 × T_5$

$T_5 = 580K$

Now,

$Q_v = C_v(T_3 - T_2) = 0.718(1429.95 - 953.3) = 342.2347kJ/kg$

$Q_p = C_p(T_4 - T_3) = 1.005(1715.94 - 1429.95) = 287.42kJ/kg$

$Q_{out} = C_v(T_5 - T_1) = 0.718(580 - 300) = 201.04kJ/kg$

$W = Q_v + Q_p - Q_{out} = 428.61kJ/kg$

Now, Efficiency is given as,

$η = \frac{W}{Q} = \frac{428.61}{629.6547} = 0.6807 = 68.07%$

The Mean Effective Pressure is given as,

$MEP = \frac{W}{V_1 - V_2} = \frac{428.61 × 10^3}{0.861 - 0.0478} = 5.2707 × 10^5 = 5.2707 bar$

Swept volume,$V_x = 0.0053m^3$

Clearance volume, $V_e = V_3 = V_2 = 0.00035m^3$

Maximum pressure , $P_3 = P_4 = 65bar$

Initial temperature, $T_1 = 80 + 273 = 353K$

Initial pressure , $p_1 = 0.9bar$

$Π_{dual} = ?$

The efficiency of a dual combustion cycle is given by

$Π_{dual} = 1 - \frac{1}{(r)^y-1}[\frac{β.p^y - 1}{(β - 1) + βγ(p - 1)}]......(i)$

Compression ratio,

$r = \frac{V_1}{V_2} = \frac{V_x - V_c}{V_c} = \frac{0.0053 + 0.00035}{0.00035} = 16.14$

$V_2 = V_3 = V_c = clearance \ \ \ volume$

Cut-off ratio,

$p = \frac{V_4}{V_3} = \frac{\frac{5}{100}V_x + V_c}{V_3} = \frac{0.05V_x + V_c}{V_c} ......V_2 = V_3 = V_c$

$= \frac{0.05 × 0.0053 + 0.00035}{0.00035} = 1.757 \ \ say \ \ 1.76$

Also during the compression operation 1-2

$p_1v_1^y = p_1v_2^y$

$\frac{p_2}{p_1} = (\frac{V_1}{V_2})^y = (16.14)^3.4$

$p_2 = p_1 × 49.14 = 0.9 × 49.14 = 44.22 bar$

Pressure or explosion ratio, $β = \frac{p_3}{p_2} = \frac{65}{4422} = 1.47$

Putting the value of r, and p and β in eqn. (i), we get

$Π_{dual} = 1 - \frac{1}{(16.14)^{14 - 1}}[\frac{1.47 × (1.76)^{14} - 1}{(1.47 - 1) + 1.47 × 1.4(1.76 - 1)}]$

$Π_{dual} = 7 - 0.328 [\frac{3.243 - 1}{0.47 + 1.564}] = 0.6383 \ \ or \ \ 63.83$