In an air-standard dual cycle, the pressure and temperature are 0.1 MPa and 27°C. The compression ratio is 18. The pressure ratio for the constant volume part of heating process is 1.5 and the volume ratio for the constant pressure part of heating is 1.2. - Determine: - - (i) thermal efficiency. - (ii) Mean effective pressure in MPa. -

Mumbai university > MECH > SEM 3 > THERMO

Marks: 12M

Year: Dec 2013

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

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

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