1. Compare the measured permeability in Permeability of Liquid method and permeability of sand from table in the your references. 2. Discus the effect of viscosity on permeability in Gas Permeametermethod ?
1. Compare the measured permeability in Permeability of Liquid method and permeability of sand from table in the your references. 2. Discus the effect of viscosity on permeability in Gas Permeametermethod ?
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
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1. Compare the measured permeability in Permeability of Liquid method and permeability of sand from table in the your references.
2. Discus the effect of viscosity on permeability in Gas Permeametermethod ?
Transcribed Image Text:p= 0.9979
H = 0.9544
This Liquid permeability
A= T/4 d? (cm²)
A= T/4(9.9)2
A=76.97687
AP=p gh (atom)
AP=0.9979(9.8)(25)
AP=244.4855
q=Q/t
q=600/187=3.20855 cc/sec
quL
Α ΔΡ
k
%3D
K=3.20855*0.9544*25/76.97687*244.4855
K=0.00406786
Transcribed Image Text:To calculate the permeability value (K)
D= 3.76cm
L= 4.76cm
= 0.0176 Cpu
A =" d? + A = 11.09cm²
4
- P1+P2 P.
2
18.06+14.36
P
2
= 16.21 PsiP
2 0.0176 x4.76 x 14.36
q xux L x Patm
AxP x AP
-K1
11.09x 16.21x (18.13–14.36)
= 0.000177508mdK1
P1+P2
P2
2
-
This permeability of sand
28.03+14.36
P2
2
= 21.195 PsiP2
18
qxμxLx Patm
2 0.0176 x4.76 x 14.36
60
K2
11.09 x 21.195x (28.03–14.36)
Ax PT x ΔΡ
= 0.000112321 mdK2
P1+P2
P3
46.16+14.36
P3
2
= 30.26 PsiP3
35
X 0.0176 x4.76 x 14.36
60
qxμxLxPatm
AxP x AP
K3
11.09 x 30.9x (46.16–14.36)
= 0.0000643982 mdK3
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