Page 148 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 143 3.1.2007 8:54pm Compositor Name: SJoearun

TRANSPORTATION SYSTEMS 11/143
where DHp m is mechanical power losses, which is usually Because z 2 did not change, q 1 remains the same value of
taken as 20 horsepower for bearing and 30 horsepower for 7,977 cfm.
seals. Calculate gas horsepower:
The proceeding equations have been coded in the
spreadsheet CentrifugalCompressorPower.xls for quick (7,977)(250) 0:97 þ 0:77 2:4 0:2662 1
calculations. Hp g ¼ (229)(0:727) 2(0:97) 0:2662
¼ 10,592 hp
Example Problem 11.3 Size a centrifugal compressor for
the following given data: Calculate gas apparent molecular weight:
Gas-specific gravity: 0.68
Gas-specific heat ratio: 1.24 MW a ¼ (0:68)(29) ¼ 19:72
Gas flow rate: 144 MMscfd at 14.7 psia and Calculated gas constant:
60 8F
Inlet pressure: 250 psia 1,544

3
Inlet temperature: 100 8F R ¼ 19:72 ¼ 78:3 psia-ft =lb m - R
Discharge pressure: 600 psia
Polytropic efficiency: E p ¼ 0:61 þ 0:03 log (q 1 ) Calculate polytropic head:

0:97 þ 0:77 2:4 0:2662 1
Solution Calculate compression ratio based on the inlet H g ¼ (78:3)(560)
and discharge pressures: 2 0:2662
600 ¼ 37,610 lb f -ft=lb m
r ¼ ¼ 2:4
250 Calculate gas horsepower requirement:
Calculate gas flow rate in scfm: Hp b ¼ 10,592 þ 50 ¼ 10,642 hp:
144,000,000
q ¼ ¼ 100,000 scfm
(24)(60)
11.4 Pipelines
Based on the required gas flow rate under standard condi- Transporting petroleum fluids with pipelines is a continu-
tion (q), estimate the gas capacity at inlet condition (q 1 )by ous and reliable operation. Pipelines have demonstrated
ideal gas law: an ability to adapt to a wide variety of environments
(14:7) (560) including remote areas and hostile environments. With
q 1 ¼ (100,000) ¼ 6,332 cfm very minor exceptions, largely due to local peculiarities,
(250) (520)
most refineries are served by one or more pipelines,
Find a value for the polytropic efficiency based on q 1 : because of their superior flexibility to the alternatives.
E p ¼ 0:61 þ 0:03 log (6,332) ¼ 0:724 Pipelinescan bedivided intodifferentcategories,including
the following:
Calculate polytropic ratio (n–1)/n:
n 1 1:24 1 1 . Flowlines transporting oil and/or gas from satellite wells
R p ¼ ¼ ¼ 0:2673 to manifolds
n 1:24 0:724
. Flowlines transporting oil and/or gas from manifolds to
Calculate discharge temperature: production facility
. Infield flowlines transporting oil and/or gas from

T 2 ¼ (560) (2:4) 0:2673 ¼ 707:7 R ¼ 247:7 F between production facilities
Estimate gas compressibility factor values at inlet and . Export pipelines transporting oil and/or gas from
discharge conditions (spreadsheet program Hall- production facilities to refineries/users
Yaborough-z.xls can be used): The pipelines are sized to handle the expected pressure and

z 1 ¼ 0:97 at 250 psia and 100 F fluid flow on the basis of flow assurance analysis. This
section covers the following topics:

z 2 ¼ 0:77 at 600 psia and 247:7 F 1. Flow in oil and gas pipelines
2. Design of pipelines
Calculate gas capacity at the inlet condition (q 1 )by
3. Operation of pipelines.
real gas law:
(0:97)(14:7) (560)
q 1 ¼ (100,000) ¼ 7,977 cfm 11.4.1 Flow in Pipelines
(0:77)(250) (520)
Designing a long-distance pipeline for transportation of
Use the new value of q 1 to calculate E p : crude oil and natural gas requires knowledge of flow
E p ¼ 0:61 þ 0:03 log (7,977) ¼ 0:727 formulas for calculating capacity and pressure require-
ments. Based on the first law of thermal dynamics,
Calculate the new polytropic ratio (n–1)/n: the total pressure gradient is made up of three distinct
components:
n 1 1:24 1 1
R p ¼ ¼ ¼ 0:2662 dP g f M ru 2 rudu
n 1:24 0:727 ¼ r sin u þ þ , (11:78)
dL g c 2g c D g c dL
Calculate the new discharge temperature:
where

T 2 ¼ (560) (2:4) 0:2662 ¼ 707 R ¼ 247 F g
g c r sin u ¼ pressure gradient due to elevation or
Estimate the new gas compressibility factor value: potential energy change
f M ru 2

z 2 ¼ 0:77 at 600 psia and 247 F 2g c D ¼ pressure gradient due to frictional losses