Question

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow rate is 5.5 kg/s, and the power developed is 1120 kW. Stray heat trans- fer and kinetic and potential energy effects are negligible. Determine:
a. The temperature of the air at the turbine exit, in K.
b. The isentropic turbine efficiency.

Answer 1

Answer:

a. T₂ = 837.2 K

b. ηt = 91.4%

Explanation:

Assuming k = 1.4

KNOWN: Air expands adiabatically through a turbine operating at steady state. Operating data  are known.

FIND: Determine the exit temperature and the isentropic turbine efficiency.

ENGINEERING MODEL (See the pics):

(1) The control volume is at  state.

(2) For the control volume,  Qcv = 0 and kinetic and  potential energy effects can be neglected.

(3) The air is modeled  as an ideal gas with constant specific heats: k = 1.4.

ANALYSIS:

(a) Mass and energy rate balances reduce to give:

0 =  - Wcv + m*(h₁ – h₂).

With   h₁ – h₂ = cp*(T₁ – T₂)

then

cp = k*R/(k-1) = 1.4*(8.314/28.97)/(1.4 – 1) = 1.004 kJ/kg∙K    and

T₂ = T₁ - (Wcv/(m*cp))  

⇒  T₂ = 1040 K – (1120 kW)/[(5.5 kg/s)(1.004 kJ/kg∙K)(1 kJ/s/ 1 kW)

⇒  T₂ = 837.2 K

b) The isentropic efficiency is

ηt = (h₁ – h₂) / (h₁ – h₂s) = cp*(T₁ – T₂) / (cp*(T₁ – T₂s)).

To get T₂s we  note that for an isentropic process of an ideal gas with constant specific heats

(T₂s / T₁) = (p₂ / p₁)∧((k-1)/k)  ⇒  T₂s = T₁*(p₂ / p₁)∧((k-1)/k)

⇒  T₂s = (1040 K)*(120 / 278)∧((1.4-1)/1.4) =  818.1 K

Thus, the isentropic efficiency is

ηt = (1040 – 837.2)/(1040 – 818.1) = 0.914

ηt = 91.4%