Thermodynamics: h vs u and Quality?

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SUMMARY

The discussion focuses on a thermodynamics problem involving two insulated tanks containing superheated steam and a saturated water mixture. The analysis reveals that, upon reaching thermodynamic equilibrium, the final state contains no liquid due to the specific volume being in the superheated regime. The correct formula for internal energy is clarified as Y = Yf + X Yfg, which differs from the initial misunderstanding regarding the use of quality in calculations. The discussion also emphasizes the application of the first law of thermodynamics, leading to the conclusion that internal energy is the appropriate measure in this scenario, rather than enthalpy.

PREREQUISITES
  • Understanding of thermodynamic properties: internal energy, enthalpy, and specific volume
  • Familiarity with the first law of thermodynamics and its application in control volumes
  • Knowledge of saturated and superheated states in thermodynamics
  • Ability to perform calculations involving quality and fluid properties
NEXT STEPS
  • Study the derivation and application of the first law of thermodynamics in control volumes
  • Learn about the differences between internal energy and enthalpy in thermodynamic systems
  • Explore the properties of saturated and superheated steam using steam tables
  • Investigate the implications of quality in phase change calculations in thermodynamics
USEFUL FOR

Mechanical engineering students, thermodynamics practitioners, and anyone involved in analyzing thermal systems and fluid properties will benefit from this discussion.

swraman
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Hi,

I am trying to understand the solution to a quiz I had, and am having trouble.

The question was:
Two well-insulated rigid tanks are connected by a valve. Tank A contains 5 kg of superheated steam at 800oC and 800kPa. Tank B contains 1 kg of saturated water mixture at 150oC and 30% quality. The valve is opened and the two tanks eventually come to thermodynamic equilibrium. Perform a thermodynamics analysis based on conservation of mass and energy to determine if there is any liquid in the final state.

The solution says:
KE=PE=W=Q=0
Conservation of Energy: Ei-Ef=ΔEsys=0 Therefore Ei=Ef. This becomes Ui=Uf or MA*UA+MB*UB=Mf*Uf

Tank A: VA=MA*vA=5kg x 0.618 m3/kg =3.09 m3
UA = 5kg x 3662.5 kJ/kg =18312.5 kJ

Tank B: VB=MB*vB=1kg x(0.3* 0.3924 m3/kg+0.7* 0.00109 m3/kg)=0.1184 m3
UB = 0.3x 2559.1 kJ/kg +0.7 *631.66 kJ/kg =1209.89 kJ

Total volume at equilibrium = 3.208 m3
Total mass= 6 kg

Specific volume= 3.208m3/6kg=0.5346 m3/kg

Total internal energy= 19522.39 kJ specific energy=3253.7 kJ/kg
(v, u) states falls into the superheated regime; there will be no liquid in the final state.

First off, why (when they solve for internal energy and specific volume) do thy use the formula (Quality)*Uvapor + (1-quality)*Uliquid ? This makes sense to me in my head, but it says no our text that the generic equation for u,h,or v is:
Y = Yliq + (Quality)*Ysat.vapor

so why don't they use it in this scenario?

Second, why do they use internal energy and not enthalpy?

Thanks

--a lost Mechanical engineering student
 
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swraman said:
First off, why (when they solve for internal energy and specific volume) do thy use the formula (Quality)*Uvapor + (1-quality)*Uliquid ? This makes sense to me in my head, but it says no our text that the generic equation for u,h,or v is:
Y = Yliq + (Quality)*Ysat.vapor

so why don't they use it in this scenario?

Second, why do they use internal energy and not enthalpy?

Thanks

--a lost Mechanical engineering student
The equation you're trying to write: Y = Yliq + (Quality)*Ysat.vapor

... is off just a bit. The correct equation is: Y = Yf + X Yfg
Where Yf is the fluid property at 100% saturated liquid
and
Yfg is the difference between the fluid property at 100% saturated gas and 100% saturated liquid.

Regarding why not enthalpy, if you rewrite the first law it helps to clear this up. Normally, we write:
dU = dQ + dW

But for any control volume with flow in and out, we can also rewrite it:
dU = dQ + dW + dH
where
dH = (Uin + pVin) - (Uout + pVout)
Note that gas flowing into or out of a control volume does work pV so we say H = U + pV
(See ref: http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_4.htm )

So if we apply this to the system with two separated tanks that has a control surface around both tanks, we find no heat or work crossing the control surface (dQ=0 and dW=0), nor do we find any flow across the control surface, so dH = 0. All we're left with is dU = 0 (or U1 + U2 = U3).
 
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