Page 543 - Standard Handbook Petroleum Natural Gas Engineering VOLUME2

Flow of Fluids 497

pod = 1.915cP

KO, = Aptd
A = 10.715(Rs + = 10.715(947.3 +

A = 0.2983
B = 5.44(Ra + 150)-0.55s = 5.44(947.3 + 150)-O.””

B = 0.5105
pOs = 0.2983 x 1.915°.5105 0.416 cp
=
10. Determine the average water viscosity. No water is in the example.
11. Calculate the liquid mixture viscosity:

1

12. Find the liquid mixture surface tension:

1 WOR

13. Find Bo at p, and Tq.

Bo = 0.972 + 0.00147F’~175

a5 a5
+
+
0’752
F = R8( 2) 1.25% [T(OF)] = 947.3 - 1.25(90) = 1052.9
(0.763)
Bo = 1.495 bbl/stb
14. Find the turbine flow area %:

15. Find the liquid viscosity number:

as
N, = 0.15’13x (0.5) ((28&.6) = 2.05 x 10”

16. Find vd (assume Bw = 1.0):