# Viscosity for steam in the Reynolds number

In calculating the Reynolds number for a flow of steam in a pipe, this is the general formula I am trying to use:

$$Re=\frac{\rho d v}{\mu}$$

with density $\rho$, pipe diameter $d$, flow steam $v$ and viscosity $\mu$. I have trouble finding the viscosity $\mu$ for the steam. In tables and articles I find, only liquid viscosity is considered.

Is viscosity not defined for steam? And in that case, how does the expression for the Reynolds number look?

I make my calculations with Engineering Equation Solver (EES). Trying to obtain the viscosity with mu=Viscosity(steam;T=T_1;p=p_1) gives an error saying that there is no viscosity data available for these conditions (the steam state).

• How hard did you search?. The first link has a pretty detailed answer. – tpg2114 Apr 5 '15 at 18:09
• And the equation for Reynolds number doesn't change with multicomponent flow. You just need to compute your density and viscosity based on the species present. – tpg2114 Apr 5 '15 at 18:10

Steam is just gaseous H2O. Here is a Chemkin format curvefit:

H2O                               V1C1  Curve-fit from ChemKin
V-0.1450E-04 0.3855E-06 0.6725E-11-0.2230E-13 0.6445E-17-0.8578E-21 0.4490E-25
C 0.1921E+03 0.5710E+01 0.8708E-02-0.3621E-05 0.7376E-09-0.8109E-13 0.3761E-17


and a NASA curvefit:

H2O                               V3C3  SENGERS & WATSON (1986)  SVEHLA (1994)
V  373.2   1073.2   0.50019557E+00-0.69712796E+03 0.88163892E+05 0.30836508E+01
V 1073.2   5000.0   0.58988538E+00-0.53769814E+03 0.54263513E+05 0.23386375E+01
V 5000.0  15000.0   0.64330087E+00-0.95668913E+02-0.37742283E+06 0.18125190E+01
C  373.2   1073.2   0.10966389E+01-0.55513429E+03 0.10623408E+06-0.24664550E+00
C 1073.2   5000.0   0.39367933E+00-0.22524226E+04 0.61217458E+06 0.58011317E+01
C 5000.0  15000.0  -0.41858737E+00-0.14096649E+05 0.19179190E+08 0.14345613E+02


I have no idea where those came from originally, it's just what we happen to have in our input deck.

• Those curve fits actually come from tabulated calculations of Gibbs free energy minimization where the Gibbs energy is written in terms of the partition function. Some poor soul back in the day would have had to carry out these equilibrium calculations (by computer of course) in order to build the tabulated curve fit relations. Organizing all of the experimental data for the rotational, vibrational, and electronic excitation for all of those species would have been brutal. It is amazing what we take for granted these days! – TRF Jan 9 '17 at 2:16

Under 1 atm, in the design thermal system book - see the properties of gases at 1 atm, table A-16 for water vapor H2O.

• Hello, and welcome to Stack Exchange. A few more details (e.g. a link to info about the book you reference) would make your question a lot more accessible. – Daniel Griscom Dec 11 '15 at 20:41