Superheated steam enters a steady-state, adiabatic throttling valve at T = 200 °
ID: 1718968 • Letter: S
Question
Superheated steam enters a steady-state, adiabatic throttling valve at T = 200 °Cand P = 1 bar. The pressure of the fluid leaving the valve is P = 0.5 bar. It has been suggestedthat perhaps it would make more sense to replace the valve with a turbine – the desired pressuredrop will still occur, but some work would be obtained from the turbine.
a. Determine the physical state of the fluid leaving the valve.
b. Determine the rate at which entropy is generated in the valve per kilogram ofsteam.
c. Determine the maximum work that could be produced by a turbine per kilogramof entering steam if the stream entering the turbine was identical to the streamentering the valve, and the pressure leaving the turbine was P = 0.5 bar.
d. Compare the physical state of the fluid leaving the “idealized” turbine describedin part c to the physical state of the fluid leaving the valve as described in part a.Are they identical? Can you rationalize why they are, or are not, identical?e. Compare your answers to this problem to your answers to Problem 2. How is theoutcome for steam different from the outcome of a similar process for liquidwater?
Explanation / Answer
At 1 bar and 200 C
h1 = 2875.475 kJ/kg , s1 = 7.8355 kJ/kg K
Exit enthalpy h2 = h1 = 2875.475 kJ/kg K
a) at 0.5 bar exit fluid is in super heated state at 199 C
b) exit entropy = 8.15 kJ/kg K
Rate of entropy generation = (s2 - s1)
= -7.8355 + 8.15
= 0.3145 kJ/kgK
c) if we use a turbin then s2 = s1 = 7.8355 kJ/kg K
therefore h2 = 2740 kJ/kg K
Work = h1 - h2 = (-2740 + 2875.475 ) = 135.475 kJ/kg
d) No they are not identical
exit fluid in both cases is superheated vapour but in case of valve temperature is 199 C and in case of turbine exit temperature is 128 C
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