Page 194 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 190 3.1.2007 9:07pm Compositor Name: SJoearun

13/190 ARTIFICIAL LIFT METHODS
g g ¼ gas-specific gravity, air ¼ 1 Because this is greater than 6, more than one-stage com-
p 1 ¼ suction pressure of the gas, psia pression is required. Using two stages of compression gives
p 2 ¼ pressure of the gas at discharge point, psia. 1=2
1,165
r ¼ ¼ 3:41:
When the deviation from ideal gas behavior is appre- 100
ciable, Eq. (13.24) is empirically modified. One such modi-
fication is The gas is compressed from 100 to 341 psia in the first
" # stage, and from 341 to 1,165 psia in the second stage.
Z 1 (k 1)=k
k 53:241T 1 p 2 Based on gas-specific gravity, the following gas property
w ¼ 1 (13:25)
k 1 g g p 1 data can be obtained:

or, in terms of power, T c ¼ 358 R
" # p c ¼ 671 psia
k 3:027p b Z 1 (k 1)=k
p 2
Hp MM ¼ T 1 1 , (13:26) T r ¼ 1:42
k 1 T b p 1
p r,1 ¼ 0:149 at 100 psia
where
p r,2 ¼ 0:595 at 341 psia
Hp MM ¼ required theoretical compression power, hp/ z 1 ¼ 0:97 at 70 F and 100 psia

MMcfd

z 1 ¼ compressibility factor at suction conditions. z 2 ¼ 0:95 at 70 F and 341 psia:
The theoretical adiabatic horsepower obtained by the First stage:
proceedingequationscan beconverted to brakehorsepower 1:25 14:7 h i
(Hp b ) required at the end of prime mover of the compressor Hp MM ¼ 3:027 530 (3:41) 0:97(0:25=1:25) 1
using an overall efficiency factor, E o . The brake horsepower 0:25 520
is the horsepower input into the compressor. The efficiency ¼ 61 hp=MMcfd
factor E o consists of two components: compression
efficiency (compressor-valve losses) and the mechanical Second stage:
efficiency of the compressor. The overall efficiency of a 1:25 14:7 h i
compressor depends on a number of factors, including Hp MM ¼ 0:25 3:027 520 530 (3:41) 0:95(0:25=1:25) 1
design details of the compressor, suction pressure, speed of
the compressor, compression ratio, loading, and general ¼ 59 hp=MMcfd
mechanical condition of the unit. In most modern compres-
sors, the compression efficiency ranges from 83 to 93%. The Total theoretical compression work ¼ 61 þ 59 ¼ 120 hp=
mechanical efficiency of most modern compressors ranges MMcfd.
from 88 to 95%. Thus, most modern compressors have Required brake horsepower is
an overall efficiency ranging from 75 to 85%, based on (32)(120)
the ideal isentropic compression process as a standard. Hp b ¼ ¼ 4,800 hp:
(0:8)
The actual efficiency curves can be obtained from the
manufacturer. Applying these factors to the theoretical Number of moles of gas is
horsepower gives 1,000,000
3
n G ¼ (32) ¼ 2:640 10 (32)
q MM Hp MM 378:6
Hp b ¼ , (13:27) 6
E o ¼ 84 10 lb-mole=day:
where q MM is the gas flow rate in MMscfd. Gas temperature after the first stage of compression is
The discharge temperature for real gases can be calcu- T 2 ¼ (530)(3:41) 0:97(0:25=1:25) ¼ 670 R ¼ 210 F:

lated by 210 þ 70

The average cooler temperature is ¼ 140 F.
z 1 (k 1)=k 2
p 2
T 2 ¼ T 1 : (13:28) btu
p 1 C p at 140 F and 341 psia ¼ 9:5 :

lb mol F
Calculation of the heat removed by intercoolers and after- 3
Intercooler load ¼ 2:640 10 (32)(9:5)(210 70)
coolers can be accomplished using constant pressure-
specific heat data: 6
¼ 55:67 10 btu=day:
DH ¼ n G C p DT, (13:29) Final gas temperature:
where T d ¼ (530)(3:41) 0:95(0:25=1:25) ¼ 669 R ¼ 209 F

n G ¼ number of lb-mole of gas
C p ¼ specific heat under constant pressure evaluated at It can be shown that the results obtained using the analyt-
cooler operating pressure and the average tem- ical expressions compare very well to those obtained from
perature, btu/lb-mol-8F. the Mollier diagram.
ThecomputerprogramReciprocatingCompressorPower.xls
Example Problem 13.3 For data given in Example can be used for computing power requirement of each
Problem 13.2, assuming the overall efficiency is 0.80, stage of compression. The solution given by the program
calculate the theoretical and brake horsepower required to for the first stage of compression in this example problem
compress the 32 MMcfd of a 0.65-specific gravity natural is shown in Table 13.2.
gas from 100 psia and 70 8F to 1,165 psia. If intercoolers
cool the gas to 70 8F, what is the heat load on the
intercoolers and what is the final gas temperature?
13.4.3.2 Centrifugal Compressors
Solution The overall compression ratio is Although the adiabatic compression process can be assumed
in centrifugal compression, polytropic compression process
1,165 is commonly considered as the basis for comparing centrifu-
r ov ¼ ¼ 11:65:
100 gal compressor performance. The process is expressed as