during the cycle is limited to $750^0C$. The pressure ratio is 4. Calculate specific output, overall efficiency of cycle. Take γ=1.4 and neglect all losses.

Mumbai University > Mechanical Engineering > Sem 7 > Power Plant Engineering

Marks : 10M

Year: Dec 2015

Isentropic Efficiency = 88%, Inlet air temperature = $20^0C$, Max Temperature = $750^0C$, Pressure ratio = 4, Gama = 1.4 ,

To Find: Specific output ,& overall efficiency.

T1 = 293 K ,& T3 = 1023 K

$\frac{T_2}{T_1} = \frac{P_2}{P_1 }^\frac{γ-1}{γ} = (4)^\frac{1.4-1}{1.4}$ = 1.4859 x 293 K

T’2 = 435.3667 K

$\frac{T_4}{T_3} = \frac{P_4}{P_3 }^\frac{γ-1}{γ} = (4)^\frac{1.4-1}{1.4}$ = 1.4859 x 1023

T’4 = 688.471 K

Isentropic Efficiency for process 1-2 n_isen= $\frac{T'_2-T_1}{T_2-T_1}$

0.88 =$\frac{435.36-293}{T_2-293}$

T2 = 454.77 K

Isentropic Efficiency for process 3-4 n_isen= $\frac{T_3-T_4}{T_3-T'_4}$

0.88 = $\frac{1023-T_4}{1023-688.47}$

T4 = 728.61 K

Work done net = $(m_(g ) ) ̇C_pg (T_3-T_4 )- (m_a ) ̇C_pa (T_2-T_1 )$

= 1 x 1.11 (1023 – 728.61 ) – 1 x 1.005 ( 454.77– 293)

$W_net$ = 164.194 kJ/ kg of air

Over all thermal efficiency n_thermal= $(T_3-T_4) - \frac{T_2-T_1}{T_3-T_2}$

= $((1023-728.61)-\frac{454.77-293}{1023-454.77}$

nthermal= 23.33 %