Answer:
The temperature of the steam at turbine exit is
= 1190.26 K
Step-by-step explanation:
Given data
°c = 773 K
Mass flow rate = 3.2
![(kg)/(s)](https://img.qammunity.org/2021/formulas/engineering/college/4v7v8g06cf8zwqtjgaxilsklmhetrs237f.png)
Power output = 2.5 MW = 2500 KW
From steady flow energy equation
![h_(1) + KE_(1) + PE_(1) + Q = h_(2) + KE_(2) + PE_(2) + W\\](https://img.qammunity.org/2021/formulas/engineering/college/3xvvk5g90abqj0gbuddkk6i9x20it5bxuv.png)
Since Kinetic & potential energy changes are negligible.
![h_(1) + Q = h_(2) + W](https://img.qammunity.org/2021/formulas/engineering/college/a57lm0r42fw0fssgn1r2t2ux1qtz8rs9ah.png)
Since turbine is adiabatic so Q = 0
![h_(1) = h_(2) + W](https://img.qammunity.org/2021/formulas/engineering/college/subxex8dxilnspoafdbma0bnf94w4x1hq0.png)
![h_(1) - h_(2) = W](https://img.qammunity.org/2021/formulas/engineering/college/3tcaoxsqndql1xefsvvq7b55owxdw4xhhj.png)
![m C_(p) (T_(2) - T_(1) ) = W](https://img.qammunity.org/2021/formulas/engineering/college/3o6aq8v3gzeq9p2oj2ejhn0u9flkzd3tjh.png)
Put all the values in above equation
3.2 × 1.8723 × (
- 773 ) = 2500
- 773 = 417.26 K
= 1190.26 K
This is the temperature of the steam at turbine exit.