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 Author Topic: What is the largest adiabatic expansion of steam without a phase change?  (Read 29010 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
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 « Embed this message on: December 06, 2009, 04:29:45 am » posted from:Plattsburgh,New York,United States

I'm having trouble working this knowing that if I allow too much adiabatic expansion, the water vapor will condense to a liquid. I believe that the Clapeyron Equation is used to keep the steam in its phase.  (dP / dT) = L / (T * dV)

Steam at 180 psi is piped into an 80cm cylinder with a piston that has a total travel length of 1.8 m. At what distance along the cylinder should I stop the addition of steam and allow the adiabatic expansion to take place if I want to have the Carnot Engine operating at it's highest efficiency? By highest efficiency I mean the largest adiabatic expansion possible without converting the steam into liquid. This is of course balanced against the initial volume of steam added.

Could someone point me in the direction of solving this problem?
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: December 06, 2009, 11:51:37 am » posted from:Taipei,T\'ai-pei,Taiwan

Please check out http://en.wikipedia.org/wiki/Triple_point
And look for phase diagram. You will find out the pressure-temperature region for gaseous state, i.e. below the blue line.

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 « Embed this message Reply #2 on: December 06, 2009, 11:05:15 pm » posted from:Plattsburgh,New York,United States

My understanding of this problem is that I must find the volume of initial steam to add so that the adiabat is as large as possible without condensing the steam.  Point B to Point C on the following image is that adiabatic transition that I am talking about.

Using a steam table from http://www.energysolutionscenter.org/BoilerBurner/Eff_Improve/Primer/SteamTables.pdf I got the following values.  The ɣ is from the adiabatic index on Wikipedia.

Vgas=3.5*10^-3 m^3/mol
Vliq=2.0*10^-5 m^3/mol
Volume of full cylinder = 0.90735 m^3
P = 992736Pa
ɣ=1.13
T=453K
P = 992736Pa
ΔHvap=36260J/mol

The following is my attempt to solve for the initial amount of steam added:

Equation 4 is the pressure and temperature of steam at which a phase change to liquid water will begin.  Equation 5 is a rearranged form of equation 3 giving the final pressure after expansion.  Equation 6 Is Equation 2 with pressure from equation 5 and the number of moles based on the amount of volume added.  Equation 7 is equation 4 with equation 5 and equation 6 plugged into it.  Volume is the only variable and I have attempted to solve for it, but Maple 11 will not solve for me.  Here is a picture of what happens when I attempt to solve in Maple:

Is this the right approach?
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: December 07, 2009, 08:23:15 pm » posted from:Taipei,T\'ai-pei,Taiwan

The equation 4 you found did not match the P-T curve in the previous post.

I would suggest you find P-T curve, and draw new curves for carnot cycle (usually it was dwan on P-V diagram).

The two constant temperature curves will become two vertical lines in P-T curve.
With ideal gas law: $PV=nRT$,
the adiabatic curve $PV^\gamma=C$ (constant can be reduced to $P (\frac{nPT}{P})^\gamma=C$
or $P=C' T^{\gamma/(\gamma-1)}$
Draw starting point with your initial condition: $P_i,T_i$, then follow the above new PT curve before it touch the PT condensation curve.

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