# Using thermodynamic tables to solve for unknowns?

Need help with thermodynamics.

I don't understand how to use thermodynamic tables to solve for unknowns!

I don't get the concept of interpolations.

Anyone know any good sites that explains this, or if you can do so yourself.

Thanks!

Relevance

Suppose you have a steam flow exiting a turbine and you know the temperature and enthalpy but not the pressure.

Go to your steam tables, find the temperature, look in the H column until you find that H, then read the pressure.

However, if that exact H is not there, find where it would be between the two values above and below it. Then figure out how far from one to the other the exact value would be. Go to the P column and move that far [proportionately] between the P values to find the interpolated P value for the exact H value.

• Login to reply the answers
• This was cute enough of a problem for me to work this out in detail. Since ordinarily f(x,y) = 0 doesn't have just a single root {x,y}, this root must be an extrema, so we partial differentiate as follows: ∂/∂x(√(x² + 1) + √((24 - x - y)² + 4) + √(y² + 16) - 25) = x/(√(x² + 1)) - z/(√(z² + 1)) = 0 where z = 24-x-y. Solving this gets x = z/2 and from that we get y = 24 - 3x. We do this again: ∂/∂y(√(x² + 1) + √((24 - x - y)² + 4) + √(y² + 16) - 25) = y/(√(y² + 16)) - z/(√(z² + 1)) = 0 where z = 24-x-y as before. Solving this gets y = 2z, and from that we get x = 24 - (3/2)y. From these 2 linear equations, we come up with: {x,y} = {24/7, 96/7}, confirming Rita's result. Of course, this was based on guessing that it's an extrema, I wouldn't recommend this as a general method of finding roots. But this is Dragan K's poser, and so here we are.

• Login to reply the answers
• Download the software Engineering Equation Solver. It does all this for you.

• Login to reply the answers