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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 147 3.1.2007 8:54pm Compositor Name: SJoearun

TRANSPORTATION SYSTEMS 11/147
z
of the z’s with z ¼ (z 1 þ z 2 )=2, where z 1 and z 2 are f M ¼ 0:01223
obtained at p 1 and p 2 , respectively. On the other hand, r ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2
should p 1 and p 2 lie in the range where z is not linear with 3:23(520) 1 (600 200 )(12:09) 5
2
pressure (double-hatched lines), the proper average would q h ¼ 14:7 0:01223 (0:7)(540)(0:9188)(200)
result from determining the area under the z-curve and
dividing it by the difference in pressure: ¼ 1,148,450 scfh
Ð
p 2
zdp Second trial:
z z ¼ p 1 , (11:102)
( p 1 p 2 ) q h ¼ 1,148,450 cfh
where the numerator can be evaluated numerically. Also, z ¯ 0:48(1,148,450)(0:7)
can be evaluated at an average pressure given by N Re ¼ ¼ 3,224,234
(0:0099)(12:09)
3
2 p p 3
p p ¼ 1 2 : (11:103) 1 21:25
2
3 p p 2 2 p ﬃﬃﬃﬃﬃﬃ ¼ 1:14 2 log 0:00005 þ
1
(3,224,234) 0:9
f M
Regarding the assumption of horizontal pipeline, in actual
practice, transmission lines seldom, if ever, are horizontal, f M ¼ 0:01145
so that factors are needed in Eq. (11.101) to compensate
for changes in elevation. With the trend to higher operat- r ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
5
2
2
ing pressures in transmission lines, the need for these q h ¼ 3:23(520) 1 ð 600 200 Þ(12:09)
factors is greater than is generally realized. This issue of 14:7 0:01145 (0:7)(540)(0:9188)(200)
correction for change in elevation is addressed in the next
section. ¼ 1,186,759 scfh
If the pipeline is long enough, the changes in the kinetic- Third trial:
energy term can be neglected. The assumption is justified
for work with commercial transmission lines. q h ¼ 1,186,759 scfh
0:48(1,186,759)(0:7)
Example Problem 11.5 For the following data given for a N Re ¼ ¼ 3,331,786
3
horizontal pipeline, predict gas flow rate in ft =hr through (0:0099)(12:09)
the pipeline. Solve the problem using Eq. (11.101) with the
trial-and-error method for friction factor and the p 1 ﬃﬃﬃﬃﬃﬃ ¼ 1:14 2 log 0:00005 þ 21:25
Weymouth equation without the Reynolds number– f M (3,331,786) 0:9
dependent friction factor:
f M ¼ 0:01143
d ¼ 12.09 in. s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
L ¼ 200 mi 3:23(520) r ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð 600 200 Þ(12:09) 5
2
2
1
e ¼ 0.0006 in. q h ¼
T ¼ 80 8F 14:7 0:01143 (0:7)(540)(0:9188)(200)
g g ¼ 0:70 ¼ 1,187,962 scfh,

T b ¼ 520 R
p b ¼ 14:7 psia which is close to the assumed 1,186,759 scfh.
p 1 ¼ 600 psia
p 2 ¼ 200 psia B. Using the Weymouth equation:
s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
Solution The average pressure is 18:062(520) (600 200 )(12:09) 16=3
2
2
q h ¼
14:7 (0:7)(540)(0:9188)(200)
p p ¼ (200 þ 600)=2 ¼ 400 psia:
¼ 1,076,035 scfh
With p ¯ ¼ 400 psia, T ¼ 540 8R and g g ¼ 0:70, Brill-Beggs-
Z.xls gives Problems similar to this one can be quickly solved with the
spreadsheet program PipeCapacity.xls.
z z ¼ 0:9188:
With p ¯ ¼ 400 psia, T ¼ 540 8R and g g ¼ 0:70, Carr- 11.4.1.2.2 Weymouth Equation for Non-horizontal
Kobayashi-BurrowsViscosity.xls gives Flow Gas transmission pipelines are often nonhorizontal.
Account should be taken of substantial pipeline elevation
changes. Considering gas flow from point 1 to point 2 in a
m ¼ 0:0099 cp:
nonhorizontal pipe, the first law of thermal dynamics gives
Relative roughness:
ð 2 ð 2
g f M u 2
vdP þ Dz þ dL ¼ 0: (11:104)
e D ¼ 0:0006=12:09 ¼ 0:00005 2g c D
g c
1 1
A. Trial-and-error calculation: Based on the pressure gradient due to the weight of gas column,
First trial: dP r g
¼ , (11:105)
q h ¼ 500,000 scfh dz 144
and real gas law, r g ¼ p(MW) a ¼ 29g g p , Weymouth (1912)
0:48(500,000)(0:7) zRT zRT
N Re ¼ ¼ 1,403,733 developed the following equation:
(0:0099)(12:09) s ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2
s 2
q h ¼ 3:23T b ( p e p )d 5 , (11:106)
1
2
1 21:25 f M g g T

TzzL
p ﬃﬃﬃﬃﬃﬃ ¼ 1:14 2 log 0:00005 þ p b
(1,403,733) 0:9
f M
where

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