1. Calculating Residual Properties of Enthalpy and Entropy Brittany Schmidt Stephanie Veselka Ryan Terry Brian Fagala

2. What are Residual Properties When values are calculated using the general means given throughout most of thermodynamics, such as the ideal gas equation, an exact answer is not generated. This is because gases are not actually ideal.

3. What are Residual Properties A residual property is the difference between the ideal, and the actual. For example: X R = X-X ig Where X is either Volume, Entropy, Enthalpy, internal energy, or Gibbs.

4. What are Residual Properties The simplest calculation for a residual property is for Volume, we will use it as an example here: V V-V ig PV ig = RT V ig = RT/P PV = ZRT (Recall, Z is the compressabilty factor) V= ZRT/P Thus V R (Z-1)RT/P (through basic algebra)

5. What is Z? A Significant amount of data is contained in Z, the compressibility factor. It is used throughout calculations for Residual properties, so it is important to touch on its calculation here. Recall that Z is the general variable added to the Ideal Gas equation to make it true for non-ideal circumstances. The equation to calculate Z is: Z=Z 0 – wZ 1 (Where w is the accentric factor) Z 0 = 1+ B 0 (P R /T R ) Z 1 = B 1 (P R /T R ) Recall that P R = P/P c And T R =T/T c Once you have P R and T R you can also find the Z 0 and Z 1 on tables E1-4

6. Residual Enthalpy The equation to find Residual Enthalpy is below. The only Scary term in this entire equation is the Partial Derivative of Z with Respect to T, however this term can be found fairly easily (it’s a constant)

7. dZ/dT? The Following graph is from Dr. Holtzapple’s Lecture 6.1 notes.

8. Residual Entropy Once again, the only scary term here is dZ/dT, and it is found in exactly the same way as in the previous problem.

9. Sample Problem Problem 6.47: From steam-table, estimate values for residual properties V R, H R, and S R for steam at 498.15K and 1600kPa, and compare with values calculated from the generalized correlation equations.

10. Sample Problem Taken from Table F.2 at 498.15K and 1600kPa H = 2856.3 J/g S = 6.5503 J/gK V = 132.85 cm 3 /g Table F.2 Ideal Gas values: 498.15K and 1kPa H ig = 2928.7 J/g S ig = 10.0681 J/gK V R = V – (RT)/(molecular weight * P) (plugging in the values) V R = -10.96 cm 3 /g

11. Sample Problem H R = H – H ig H R = -72.4 J/g S ig = (-R/molecular weight) * ln(P/P 0 ) S R = S – (S ig + S ig ) S R = -0.11 J/gK This simplifies to:

12. Sample Problem Find T c P c and in Table B1 T r = T/T c = 0.76982 P r = P/P c = 0.072546 B 0 = 0.083 - B 0 = -0.558 B 1 = 0.139 – (0.172/T r 4.2 ) = -0.377 Z = 1 + (B 0 + B 1 ) * (P r /T r ) = 0.935

13. Sample Problem V R = [R*T*(Z-1)]/(P*molwt) = -9.33 cm 3 /g H R = = -53.4 J/g S R = = -0.077 J/gK