Question

Air enters a steady-flow turbine. The conditions of the air entering and leaving the

turbine are as follows: inlet, 300 kPa and 52°C; exit, 100 kPa and 12°C. The mass

flow rate is 10 kg/s. Heat transfer from the turbine to the surroundings and the

kinetic and potential energy effects are negligible. Calculate the power developed

by the turbine. Determine whether the process in the turbine is reversible. If not,

determine the isentropic efficiency of the turbine.

Answer 1

Given information-

P₁ = 300 kPa P₂ = 100 kPa

T₁ = 52 deg C = 325 K T₂ = 12 deg C = 285 K

mass flow rate m = 10 kg/s

Using steady flow energy equation and neglecting kinetic and potential energy and heat loss , we get :

h₁ = h₂ + W

Work done, W = h₁ - h₂ = C_P * (T₁ - T₂)

= 1.005 * (325 - 285) = 40.2 kJ/kg

Power developed , P = m * W

= 10 * 40.2 = 402 kW

For checking reversibility we will check the entropy change for the system since this is an adiabatic system so for a reversible process the entropy change must be zero -

S₂ - S₁ = C_P ln (T₂/T₁) - R ln (P₂/P₁)

= 1.005 ln (285/325) - 0.287 ln (100/300)

= 0.183 kJ/kg

Since entropy change is not equal to zero the process is irreversible.

Considering ideal process (1-2s) -

Applying reversible adiabatic equation to find out the ideal temp 2S -

T_2S/T₁ = (P₂/P₁)^^((Y-1)/Y)

T_2S/325 = (100/300)^^{((1.4-1)/1.4)}

T_2S = 237.44 K

Isentropic efficiency-

n = W_actual / W_ideal

= (C_P(T₁-T₂)) / (C_P(T₁-T_2S))

= (T₁-T₂) / (T₁-T_2S)

= 0.4568 = 45.68%