Property lines on the 2d projections

AI Thread Summary
The discussion revolves around the existence of water as a vapor or liquid at -40°C and the interpretation of constant volume lines on a phase (P-T) diagram. It is clarified that water can exist as both a liquid and vapor at this temperature, depending on pressure, although it generally does not exist as a liquid due to the expansion upon freezing. The conversation also touches on the use of software like CoolPack for visualizing phase diagrams, which can help in understanding these properties. Additionally, there is a mathematical exploration of enthalpy and its relationship to pressure and specific volume, leading to confusion about the validity of certain equations in polytropic processes. The participants emphasize the importance of accurately interpreting phase diagrams and thermodynamic properties.
nanunath
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Hi there...I need ur help on these questions:
1) Can H2O exist as a vapor at -40oc,As a liquid?
Why?
2) What would be the general nature of Constant volume lines on Phase (P-T) diagram?

Both questions are from Engg. Thermodynamics-Moran,Shaprio(Things engg. think abt section)

Also, does anyone know a program (like in Matlab) or a pro-e,etc model for the 3D Phase diagram for any substance(preferably not in the log scale)...I want it to visualize and cut it through various planes...and see the various const. property lines on the 2d projections
Thanx..

something like this...with ability to cut section at desired plane
[PLAIN]http://www.geology.iastate.edu/gccourse/hydro/aspects/images/diagram.gif
 
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http://www.et.web.mek.dtu.dk/Coolpack/UK/Index.html
During my thermo. class I used CoolPack exclusively - It is a Danish developed application that will draw the diagrams for different fluids. T.s, H.s, Log-P.H etc.
And it IS in English :)
 
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This problem tests basic ability to read a P vs. T phase diagram. Since the state of the substance depends on both temperature and pressure, think about fixing your temperature and then seeing what happens as you alter pressure.
 


M.D.G said:
This problem tests basic ability to read a P vs. T phase diagram. Since the state of the substance depends on both temperature and pressure, think about fixing your temperature and then seeing what happens as you alter pressure.

Ya...thanks so much
So my ans would be...
It requires use of a more detailed phase diagram...
though I guess approximately by the P-T diagram of book,that it won't exist as a liquid at -40 deg celsius , however high the pressure be(as water expands on freezing);
but can exist as a vapor (Solid-vapor phase change at very low pressures)
 
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Claws said:
http://www.et.web.mek.dtu.dk/Coolpack/UK/Index.html
During my thermo. class I used CoolPack exclusively - It is a Danish developed application that will draw the diagrams for different fluids. T.s, H.s, Log-P.H etc.
And it IS in English :)

Thanks so much..I'll check it out...
 
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nanunath said:
Ya...thanks so much
So my ans would be...
It requires use of a more detailed phase diagram...
though I guess approximately by the P-T diagram of book,that it won't exist as a liquid at -40 deg celsius , however high the pressure be(as water expands on freezing);
but can exist as a vapor (Solid-vapor phase change at very low pressures)

That was a bad guess ..I discovered after I noted the critical temp and pressure of water...
water could exist as a liquid...and a vapor at -40oC ...I suspect...(without referring to and actual co-ordinate axis labeled Phase diagram..to scale!
 


hi..I got another doubt..please help me with this friends,
Enthalpy is defined as:
h= u + p*v
u-internal energy
p-pressure (abs)
v-specific volume

Books say :
"Enthalpy is a property as all the terms in the above expression are properties..."

That is:
dh = d [u + (p*v)] = du + d(p*v) ...(I) => Enthalpy is a property

But now, if I write the 2nd term on RHS "d(p*v)" as:

d(p*v)= p*dv + v*dp
Integrating this and interpreting graphically it would mean:
p*dv=>Area under p curve
v*dp=>Area under v curve
Addition of two (actually the 2nd term would be -ve in a normal polytropic process like pvn=Constant which actually leads to the substraction of two rectangular areas above the axes)
The net result I think then would be the integration of d(p*v) would be a constant..graphically atleast...

But how do I arrive at this mathematically...i.e p*dv+v*dp=0
[Is anything wrong with the following few lines:
for a polytropic process
v*dp + (n*p*dv) = 0 ..(a)
and not
v*dp + p*dv = 0 .. (b)
where as graphically it "seems" to be in favor a (b)
..
So I don't get where I made the mistake...Is writing d(p*v) = v*dp + p*dv valid here..or what else is wrong...

Thanks...
Any help is appreciated..:smile:
 
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nanunath said:
hi..I got another doubt..please help me with this friends,

[Is anything wrong with the following few lines:
for a polytropic process
v*dp + (n*p*dv) = 0 ..(a)
and not
v*dp + p*dv = 0 .. (b)
where as graphically it "seems" to be in favor a (b)
..:

ohh..
Got the mistake...2day
Really I don't believe I get confused to such extents..the above quoted lines r wrong
 

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