engineering, thermodynamics

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#1
question:
In a CHP power plant superheated steam in a back-pressure steam turbine is expanded adiabatically from 100 bar and 500°C to saturated vapour at 10 bar. Neglecting the kinetic and potential energy
terms, if the turbine power is 𝑊dot = 6 MW, what is the mass flow rate of the steam?
(Ans. 10.08 kg/s)

so the equation i have is -Wdot = Mdot* change in enthalpy, since kinetic energy potential energy is neglected and process is adiabetic so no Qdot
-Wdot = 6 * 10^6
and i got the first enthalpy as 3373 * 10^3, and 2nd enthalpy as 763*10^3. both units in J/kg
so i end up with (-6*10^6)/(10^3(763-3373)) = 2.3298
i dont know where im going wrong
i write 'dot' to show the variable account for mass flow rate
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1 year ago
#2
Alright, so everything is fine until the 2nd enthalpy value. First enthalpy value is correct! To get the 2nd enthalpy value use the saturated water and steam tables and find the row where P = 10 bar (for the 2nd state). Now since it is saturated vapour it lies on the VLE curve of saturated steam and hence has a dryness fraction (x value) of 1. Now, for the steam tables the formula for specific enthalpy is: h = hf + x (hg - hf). Now since x is 1, specific enthalpy for state 2 is just hg value from the table. Put that in instead of 763 and you should get 10.08. Hope that helps!
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