water H2o density and bulk modulus at 194 C 180 Bar abs 1

[MarcTroy]

how can I calculate bulk modulus and density of water at 194 C 180 Bar ?
From my papers I get a density of about 880 Kg/m3 but I would need an accurate value

 

[katmar]

See the NIST data at

http://webbook.nist.gov/chemistry/

Click on Formula or Name and enter H2O or water and select the data you want.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[PaoloPemi]

the bulk modulus is the inverse of compressibility, you can calculate isothermal compressibility as K=-(1/V)*(dV/dP), I use Prode Properties, there is a free version, see

http://www.prode.com

which includes also the IAPWS95 formulation and I get 883.176 Kg/M3 for density and 7.53922E-10 (1/Pa) for isothermal compressibility

 

[MarcTroy]

PaoloPemi, is the bulk modulus = 1/(7.53922E-10) Pa ? I have values about 1.4×10E9 Pa which is not much different

 

[PaoloPemi]

it is the value calculated according IAPWS95 , don't know the NIST value, you can check as suggested by Katmar

 

[25362]

The NIST value at 194^oC and 180 bar absolute for density is 883.18 kg/m³

 

[MarcTroy]

thanks Paolopemi & 25362, I need to calculate density and bulk modulus at different temperatures and pressures so your suggestions have been appreciated.

 

[25362]

Use the table by NIST as suggested by Katmar. One equation you can apply for the modulus E is:

E = [Δ]P[×][ρ]_average[÷][Δ][ρ]​

Another, although less exact, is:

E = c²[×][ρ]​

where

c = speed of sound
[ρ] = density
[Δ][ρ]= density difference
[Δ]P = pressure difference

I suggest to never use more than three significant figures, for example, at the conditions mentioned above, 1.36[×]10⁹.

 

[PaoloPemi]

25362, you are suggesting to use numerical differentiation to get an approximation for isothermal compressibility but you can calculate compressibility directly (see above) and get much better accuracy (se your note about three significant digits).

 

[delittle]

my copy of IAPWS 1995 steam tables give 883.176 kg/m3 and 13263.97 (Bar) in agreement with values calculated by paolopemi

 

[25362]

PaoloPemi, by selecting two pressures 2 bar apart enclosing the pressure in question one can use the equation above. Alternatively, just by reading the densities at two pressures just 1 bar apart, at any selected temperature, the calculation reduces to:

E = ?_average ÷ [Δ]?​

I don't have IAPWS 1995 steam tables, but I can tell the exactness of the data by NIST is beyond reproach.

 

[PaoloPemi]

25382, the suggested numerical differentiation gives an average value for the first derivative (in this case dRho/dP) in the selected range p1,p2 , usually one selects the minimum step possible with the available precision (in your case the 5 digits for density from NIST site) with the hope that the second derivative is constant in that range which is not the case here (altough I think it changes little) if you wish to get accurate values you need analytic derivatives.
BTW I do not discuss the quality of NIST data which I appreciate only suggest different ways to get reliable data.

 

[25362]

Is it warranted when considering the probable errors in thermometry and pressure measurements ?

 

[katmar]

Paolo, how do you get dV/dP if not by numerical methods?

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[PaoloPemi]

Katmar, as far as I know Prode Properties (as similar IAPWS 1995 based codes) solves numerically the dimensionless form of Helmholtz free energy then it includes analytical derivatives to get all derived properties including isothermal compressibility, the same tool exports derivatives of fugacity, enthalpy, entropy, volume vs. p,t,composition, REFPROP also exports derivatives as other tools, quite usual stuff....

 

[katmar]

Paolo, that's pretty much what I expected and I don't see any real difference between that and what 25362 has proposed. If you analytically differentiate an equation that was derived by numerically fitting the experimental data then I would expect the results to be very much the same as if you simply draw a straight line through the two nearest points straddling the desired point that will give reasonable precision.

Anyway, we are splitting hairs and MarcTroy has got the information he needed - and confirmed from several sources.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[PaoloPemi]

splitting hairs may appear (at some reader) a bit too optimistic considering the importance of the matter but I respect all opinions and I agree that one can process data in different ways.