angie249
12.08.2020 •
Engineering
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
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Ответ:
(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.
Ответ:
your answer is A or b
Step-by-step explanation: