(a) Consider a steam turbine with a inlet conditions 8 MPa and 460 o C. What is
ID: 483305 • Letter: #
Question
(a) Consider a steam turbine with a inlet conditions 8 MPa and 460oC. What is the entropy (kJ/kg-K) of the inlet stream?
(b)What is the enthalpy (kJ/kg) of the inlet steam?
(c) What is the state of the reversible outlet stream at 0.6 MPa?
Vapor and Liquid
All Liquid
All Vapor
(d) What is the outlet enthalpy (kJ/kg) of a reversible turbine with an outlet pressure of 0.6 MPa?
(e) What is the change in enthalpy (kJ/kg) across the turbine if it is reversible?
(f) What is the change in enthalpy (kJ/kg) across the turbine if it is 0.72 efficient?
(g) What is the actual enthalpy (kJ/kg) of the turbine exit stream?
(h) What is the state of the actual outlet stream?
All Vapor
Vapor and Liquid
All Liquid
Explanation / Answer
8MPa = 8000Kpa, the saturation temperature corresponding to 8Mpa is 295 deg.c. But the given temperature is 460 deg.c. Hence the conditions of steam at the inlet to the turbine is super heated. The entropy at this temperature corresponding to super heated steam pressrure of 8000 Kpa is 6.62 Kj/Kg.K. Enthalpy is 3305.5 Kj/kg
The expansion across the turbine is isentropic, at 600 Kpa the entropy should match with the entropy of the inlet stream. let xf is vapor fraction, 1-xf is liquid fraction,
x*6.7575+(1-x)*1,9308 = 6.62 ( 6.7575 and 1.9308 are entropy of saturated liquid and saturated vapor respectively
x*6.7575+1.9308-1.9308x= 6.62
4.8267x= 6.62-1.9308
x >1. Suggesting the stream is completely vapor.
the enthalpy of steam at the exit is 2755 Kj/Kg
Change in enthalpy = 3305.5-2755=550.5 Kj/Kg
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