Steam enters an adiabatic turbine at 8 MPa and 500 C with a mass flow rate of 3 kg/s andleaves at 30 kPa. The isentropic efficiency of the turbine is 0.90. Neglecting the kinetic energy change of the steam, determine (a) the tem-perature at the turbine exit and (b) the power output of theturbine

(a) The exit temperature = 69.1°C

(b)  The power output = 8.3 MW

Explanation:

(a) The enthalpy of steam at 8 MPa and 500°C is given (from online sources) as h₁ = 3399 kJ/kg

The exit temperature = The saturation temperature at 30 kPa = T₄ = 69.096 + 273.15 = 342.246 K

The exit temperature = 69.1°C

(b) The mass flow rate = 3 kg/s

The isentropic efficiency = 0.9 = (h₁ - h₂)/(h₁ - h₂)

s₁ = 6.727 kJ/kgK

h₂ = 289.229+ 2335.32* (6.8235 - 6.727 )/6.8235 = 322.26 kJ/kg

Therefore, h₂ = h₁ - (h₁ - h₂)*0.9

h₂ = 3399  - (3399  - 322.26 )*0.9 = 629.934 kJ/kg

The power output = (h₁ - h₂) = 3 *(3399 - 629.934) = 8307.2 kJ/s =

8.3 MW.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

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