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Force the phase. ThermodynamicData for saturated liquid *and* vapor.

Posted 12 years ago

This returns the density of saturated liquid 958.367 kg/m3:

ThermodynamicData["Water", "Density", {"Pressure" -> Quantity[1, "Atmosphere"], 
"Temperature" -> Quantity[ThermodynamicData["Water","LiquidVaporPhaseBoundary", 
{"Pressure" -> Quantity[1, "Atmosphere"]}][[1]], "Kelvin"]}]

Other than doing something ugly like this

ThermodynamicData["Water", "Density", {"Pressure" ->Quantity[1, "Atmosphere"], 
"Temperature" -> Quantity[ThermodynamicData["Water","LiquidVaporPhaseBoundary", 
{"Pressure" -> Quantity[1.001, "Atmosphere"]}][[1]], "Kelvin"]}]

how can I get the density of saturated vapor 958.3675 kg/m3?

Thanks.

POSTED BY: Glenn Carlson
5 Replies
Posted 1 year ago

A blast from the past.

FWIW, Paul's suggested form of ThermodynamicData does not work in Mathematica v11.3, which is what I currently have. (Upgrades became too expensive.) Mathematica returns "ThermodynamicData::para: Parameter list includes parameters other than Pressure and Temperature."

Glad to hear Wolfram has since provided this functionality to ThermodynamicData.

Regards to all.

Glenn

POSTED BY: Glenn Carlson
Posted 1 year ago

It's now 2024. Any changes in Mathematica that address this important question?

POSTED BY: Paul Nielan
Posted 1 year ago

Some (re)reading of the documentation suggests that

ThermodynamicData["Water", "InternalEnergy", {"Pressure" -> 
   Quantity[0.6*10^6, "Pascals"], "Quality" -> 0}]

works. I specify 0 for saturated liquid and 1 for saturated vapor.

Does that make sense? My answers check with Reynolds and Perkins (2nd edition SI) on page 108.

POSTED BY: Paul Nielan
Posted 1 year ago

enter image description here

POSTED BY: Paul Nielan
POSTED BY: Jihad AlSadah
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