Thermo properties for Water Phase

[Santosh Shanbhogue]

Hi,

If I execute this snippet:

import cantera as ct

w = ct.Water()
w.TX =273.16,0.0

I get this:

  water:

       temperature          273.16  K

          pressure         610.157  Pa

           density         999.772  kg/m^3

  mean mol. weight          18.016  amu

    vapor fraction               0

                          1 kg            1 kmol

                       -----------      ------------

          enthalpy    -1.59708e+07       -2.877e+08     J

   internal energy    -1.59708e+07       -2.877e+08     J

           entropy         3519.98        6.342e+04     J/K

    Gibbs function    -1.69323e+07       -3.051e+08     J

 heat capacity c_p    nan              nan              J/K

 heat capacity c_v    nan              nan              J/K

this pressure and specific volume is different from steam tables and even in Bill Reynolds' book (which is referenced in liquidvapor.cti), where the value is 611.3 kPa. Is this an interpolation error or something else?

Also, what are the reference state for enthalpy or internal energy for this phase? It does not seem to be 298.15K either

Thanks!

Santosh

[Ray Speth]

Santosh,

If I had to guess, I think this is related to the tolerances used on the iterative method used to find the saturated states. Unfortunately, these aren’t user-accessible, so there’s not a quick way to verify this hypothesis. I would note that properties calculated away from the saturation curve agree more exactly with the Reynolds book, which suggests that the coefficients and the equation of state are implemented correctly.

The reference state for the enthalpy is the usual one used within Cantera, i.e. the enthalpy of the pure elements in their stable form at 298.15 K is zero. Thus, if you set the mixture as

w.TX = 298.15, 0

you will see that the molar enthalpy is the heat of formation of liquid water.

Regards,
Ray

[Santosh Shanbhogue]

Thanks Ray! 

The reference state is indeed 298.15K (the JANAF standard); but I was looking for the value, relative to which the absolute enthalpy and entropy matches values found in steam tables. Maybe I should have called it the datum state

Turns out for water it is a value close to the triple point (or if I had to wager, what Bill Reynolds found to be the triple point when he wrote the book that provides the data for Cantera) -- 273.16K and 611.3 Pa

So, if I flip to the back of a thermo textbook and randomly lookup a value from the saturated water pressure table, I see that:

@ P = 3.0 kPa and Tsat = 24.08C, hf = 100.98 KJ/kg and hg = 2544.8 kJ/kg. To get cantera to spit out these values, I have to do this:

In [1]: import cantera as ct

In [2]: gas = ct.Water()

In [3]: gas_tp = ct.Water()

In [5]: gas.TP = (273+24.08), 3000

In [6]: gas_tp.TP = (273.16), 611.3

In [7]: (gas.enthalpy_mass - gas_tp.enthalpy_mass)/1000

Out[7]: 100.39136582351476

In [9]: gas.TX = None, 1.0

In [10]: (gas.enthalpy_mass - gas_tp.enthalpy_mass)/1000

Out[10]: 2545.2323348881846

Same procedure for entropy as well

For all the default pure substance that ship with cantera (except R134a and Heptane) the datum state is a value close to the triple point. For Heptane, it is  300K, whereas its triple point is in the vicinity of 182.5K

R134a is a whole another story and I am still wrapping my head around what it is doing. Will post my thoughts on it when I figure things out

Regards,

Santosh

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[Santosh Shanbhogue]

For RFC134a, turns out the datum state is -40C. So for posterity, here are the datum states for common PureFluids included with Cantera:

Water - 273.16 K, 611.3 kPa

R134a - 233.15K, 51.25 kPa

CO2 - 216.54 K, 517.3 kPa

For others, the values are in Bill Reynold's book for the interested user

Santosh

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[Ray Speth]

Santosh,

I'm not sure I fully understand what you're trying to determine. The values you state are indeed the reference states used in the Reynolds book (except for R134a, which is more recent than the book). But these values are independent of what Cantera does. Indeed, Cantera specifically changes the reference state to the one based on the pure elements at 298.15 K and 1 atm, since this makes the most sense when you are considering multiple different species. And of course, if you subtract the value of the enthalpy computed by Cantera at the reference state of the original equation of state, you will be able to obtain the original values for the enthalpy. This would be true no matter what reference state Cantera used.

Regards,

Ray

[Santosh Shanbhogue]

Ray,

I was attempting to reproduce a textbook homework-style problem in Cantera (compressor/ turbine stuff). I wanted to check at a given point if the enthalpy value that I get from Cantera matches what is listed in a steam table.

Because Cantera uses a different reference state than textbook steam tables, I could not readily do an absolute value comparison - e..g. why, at exit of a compressor, does the enthalpy value as output by EES differ from Cantera's (the densities and specific volumes would match). I thought that text book reference states are always the triple point, but that logic broke down because R134a -- a popular substance -- follows a different standard. This is where I was getting tripped. 

Hence my question. Its resolved now. 

As always, thanks for your tips!

Regards,

Santosh

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