Answer:
(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[tex]_s[/tex])
s₁ = 6.727 kJ/kgK
h₂[tex]_s[/tex] = 289.229+ 2335.32* (6.8235 - 6.727 )/6.8235 = 322.26 kJ/kg
Therefore, h₂ = h₁ - (h₁ - h₂[tex]_s[/tex])*0.9
h₂ = 3399 - (3399 - 322.26 )*0.9 = 629.934 kJ/kg
The power output = [tex]\dot m[/tex](h₁ - h₂) = 3 *(3399 - 629.934) = 8307.2 kJ/s =
8.3 MW.
