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.

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%

answer:

jomamais ay

explanation:

answer: excel provides a method to add or to remove borders around and between cells. instead of selecting cells first and then formatting, you can choose the formatting tool that you want to use, and then apply the format to one or more cells.

explanation: plz brainliest,im 1 more brainliest from ranking up< hop icould