Question

6.12 WP Steam enters a turbine operating at steady state at 1 MPa, 200°C and exits at 40°C with a quality of 83%. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the power developed by the turbine, in kJ per kg of steam flowing, (b) the change in specific entropy from inlet to exit, in kJ/K per kg of steam flowing.

Answer 1

The answer is:

P/m = 612.97 kJ/kg

Δs = 0.2566 kJ/kgK

Step-by-step explanation:

In this case we have the following values and calculate:

STEP A:

water --> p₁ = 1 MPa and T₁ = 200 °C

saturated liquid and vapor --> T₂ = 40 °C

h₂ = h₁(40)+x₂(hv(40)-h₁(40))

h₂ = 167.57+0.83*(2574.30-167.57) = 2165.13 kJ/kg

s₂ = s₁(40)+x₂(sv(40)-s₁(40))

s₂ = 0.5725+0.83*(8.2570-0.5725) = 6.9506 kJ/kgK

STEP B:

Now we have all of the values required for making calculations. But we now the power is required, too

`(P)/(m)` = h₁-h₂

= 2827.9-2165.13

= 612.97 kJ/kg

Δs = 0.2566 kJ/kgK