An engine working on dual cycle uses a compression of 14. The intake pressure and temperature are 1 bar and 330k. The explosion ratio is 1.4. The heat supplied during constant pressure process is twice that at the constant volume process. -

Mumbai university > MECH > SEM 3 > THERMO

Marks: 10M

Year: May 2015

Given:

.r= 14; T1 =330K and P1 = 1bar;;er = 1.4

Qp=2Qv

$\frac{T_2}{T_1} = (\frac{v_1}{v_2})^{k - 1} , \ \ \ \ T_2 = T_1r^{k - 1}$

$T_2 = 300 × 14^{1.4 - 1} = 948$

$\frac{T_2}{T_1} = (\frac{v_1}{v_2})^{k} , \ \ \ \ P_2 = P_1r^k$

$P_2 = 1 × 14^{1.4 - 1} = 948K$

1.4 = P3 /40;

P3 = 56 bar

P3/T3 = P2/T2

56/T3 = 40/948

T3 = 1327K

Qp = 2 Qv

Cp(T4-T3) = 2 Cv (T3-T2)

1000 (T4-1327) = 2 × 720 × (1327 – 948)

T4 = 1872K

V1 = RT1/P1 = 0.924 = V5

V2 = RT2/P2 = 0.066 = V3

V4 = RT4/P4 = 0.094

$\frac{T_2}{T_1} = (\frac{v_1}{v_2})^{k - 1}$

$T5/T4 = (V4/V5)^{(k-1)}$

$T5 = 1872 × (0.094/0.924)^{0.4} = 750K$

I) cut off ratio

rc = V4/V3 = T4/T3 = 1872/1327 = 1.41

II) Work done

W = Qin – Qout = Cp(T4 - T3) + Cv (T3 -T2) – Cv (T5 - T4) = 1000(1872 - 1327) + 720(1327 - 948) – 720

(750-330) = 515 kJ/kg

III) Efficiency

Thermal efficiency = 1 – {(T5 – T1)/(T3 – T2) + k(T4 – T3)}

Thermal efficiency = 1 – {(750 – 330)/(1327 – 948) + 1.4(1872 – 1327)} = 0.63