Example: Oil and gas flow simultaneously in a vertical well. The well does not produce water. At a certain location in the tubing, the pressure is 1700 psig. Calculate the pressure loss gradient at this location with Hagedorn-Brown method. Q=10000 stb/d; po at surface = 54.7 lbm/ft³; Bo = 1.2 bbl/stb; μ in situ = 0.97 CP; Qg= 10 mmscf/d; pg in situ = 5.88 lb/ft3; Sg= 0.65; Bg= 0.009 ft³/scf; μg in situ = 0.016 CP; oil-water interfacial tension = 8.41 dyn/cm; GOR = 280 scf/stb; tubing ID = 6 in ; tubing roughness = 0.00005 ft ; Re=1488 PsVmD Bo = Hs PSTO +0.01357× RsYgs POR

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
icon
Related questions
Question
Solving this question step by step in corrct way
Example: Oil and gas flow simultaneously in a vertical well. The well does not
produce water. At a certain location in the tubing, the pressure is 1700 psig.
Calculate the pressure loss gradient at this location with Hagedorn-Brown
method. Q=10000 stb/d; po at surface = 54.7 lbm/ft³; Bo = 1.2 bbl/stb; μ in situ
= 0.97 CP; Qg= 10 mmscf/d; pg in situ = 5.88 lb/ft3; Sg= 0.65; Bg= 0.009 ft³/scf;
μg in situ = 0.016 CP; oil-water interfacial tension = 8.41 dyn/cm; GOR = 280
scf/stb; tubing ID = 6 in ; tubing roughness = 0.00005 ft ;
Re=1488 PsVmD
Bo
=
Hs
PSTO
+0.01357× RsYgs
POR
Transcribed Image Text:Example: Oil and gas flow simultaneously in a vertical well. The well does not produce water. At a certain location in the tubing, the pressure is 1700 psig. Calculate the pressure loss gradient at this location with Hagedorn-Brown method. Q=10000 stb/d; po at surface = 54.7 lbm/ft³; Bo = 1.2 bbl/stb; μ in situ = 0.97 CP; Qg= 10 mmscf/d; pg in situ = 5.88 lb/ft3; Sg= 0.65; Bg= 0.009 ft³/scf; μg in situ = 0.016 CP; oil-water interfacial tension = 8.41 dyn/cm; GOR = 280 scf/stb; tubing ID = 6 in ; tubing roughness = 0.00005 ft ; Re=1488 PsVmD Bo = Hs PSTO +0.01357× RsYgs POR
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Process Dynamics and Control, 4e
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:
9781119285915
Author:
Seborg
Publisher:
WILEY
Industrial Plastics: Theory and Applications
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The