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ruiwp13
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Homework Statement
Liquid water at 212ºF and 1atm has a specific internal energy of 180.02Btu/lbm. The water is transformed into steam at 400ºF and the pressure of 100psi where it's specific volume is 4.937ft^3/lb and it's specific enthalpy is 1228.4Btu/lb. Determine the variation of internal energy and the variation of enthalpy of the process.
Homework Equations
ΔH=ΔU+(ΔPV)
h=u+Pv
The Attempt at a Solution
So I know the specific internal energy in the liquid state (180.02Btu/lbm), and I can calculate the specific internal energy in steam (h=u+Pv ⇔ 1228.4=u+6.804*4.937) which gives me 1194.80. But now I'm stuck... I have the variation of the specific volume (vsteam-vliquid) , the variation of pressure (6.804atm-1atm) and the variation of specific internal energy. So, I can get the variation of the specific enthalpy(h). But how do I get the mass to get the variation of enthalpy(ΔH)? (if what I said is correct of course).
Thank you for your time and patience