Page 191 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 187 3.1.2007 9:07pm Compositor Name: SJoearun

GAS LIFT 13/187

D v where
p c,v ¼ p c,s 1 þ , (13:9)
40,000 m ¼ gas viscosity at in situ temperature and pressure, cp.
which gives Equation (13.12) indicates that the upstream pressure is
independent of downstream pressure under sonic flow
p c,v
p c,s ¼ (13:10) conditions. If it is desirable to make a choke work under
D v
1 þ : sonic flow conditions, the upstream pressure should meet
40,000
the following condition:
Neglecting the pressure losses between injection choke and
p dn
the casing head, the pressure downstream of the choke p up ¼ 1:82p dn (13:15)
( p dn ) can be assumed to be the casing surface injection 0:55
pressure, that is, Once the pressure upstream of the choke/orifice is deter-
mined by Eq. (13.15), the required choke/orifice diameter
p dn ¼ p c,s :
can be calculated with Eq. (13.12) using a trial-and-error
approach.
13.4.2.3.2 Subsonic Flow Under subsonic flow con-
13.4.2.3 Pressure Upstream of the Choke ditions, gas passage through a choke can be expressed as
The pressure upstream of the injection choke depends on
flow condition at the choke, that is, sonic or subsonic flow. q gM ¼ 1,248C c Ap up
"
#
Whether a sonic flow exists depends on a downstream-to- v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u

2
kþ1
upstream pressure ratio. If this pressure ratio is less than a u k p dn k p dn k : (13:16)
t
critical pressure ratio, sonic (critical) flow exists. If this (k 1)g g T up p up p up
pressure ratio is greater than or equal to the critical pres-
sure ratio, subsonic (subcritical) flow exists. The critical If it is desirable to make a choke work under subsonic flow
pressure ratio through chokes is expressed as conditions, the upstream pressure should be determined
k from Eq. (13.16) with a trial-and-error method.
2 k 1
R c ¼ , (13:11)
k þ 1 13.4.2.4 Pressure of the Gas Distribution Line
where k ¼ C p =C v is the gas-specific heat ratio. The value The pressure at the inlet of gas distribution line can be
of the k is about 1.28 for natural gas. Thus, the critical calculated using the Weymouth equation for horizontal
pressure ratio is about 0.55. flow (Weymouth, 1912):

Pressure equations for choke flow are derived based on v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
2
an isentropic process. This is because there is no time for 0:433T b t p p 2 up D 16=3
L
heat to transfer (adiabatic) and the friction loss is negli- q gM ¼ , (13:17)
TzzL
p b g g T g
gible (assuming reversible) at choke.
where
13.4.2.3.1 Sonic Flow Under sonic flow conditions, T b ¼ base temperature, 8R
the gas passage rate reaches and remains its maximum p b ¼ base pressure, psi
value. The gas passage rate is expressed in the following p L ¼ pressure at the inlet of gas distribution line, psia
equation for ideal gases: L g ¼ length of distribution line, mile
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u !
u k 2 kþ1 Equation (13.17) can be rearranged to solve for
k 1
t
q gM ¼ 879C c Ap up , (13:12) pressure:
g g T up k þ 1 s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
TzzL
q gM p b g g T g
2
where p L ¼ p þ (13:18)
up
0:433T b D 16=3
q gM ¼ gas flow rate, Mscf/day
p up ¼ pressure upstream the choke, psia
A ¼ cross-sectional area of choke, in: 2 Example Problem 13.2 Anoilfieldhas16oilwellstobegas
T up ¼ upstream temperature, 8R lifted. The gas lift gas at the central compressor station is
g g ¼ gas specific gravity related to air firstpumpedtotwoinjectionmanifoldswith4-in.ID,1-mile
C c ¼ choke flow coefficient. lines and then is distributed to the wellheads with 4-in. ID,
0.2-mile lines. Given the following data, calculate the
The choke flow coefficient C c can be determined using required output pressure of compression station:
charts in Figs. 5.2 and 5.3 (Chapter 5) for nozzle- and
orifice-type chokes, respectively. The following correlation Gas-specific gravity (g g ): 0.65
has been found to give reasonable accuracy for Reynolds Valve depth (D v ): 5,000 ft
4
6
numbers between 10 and 10 for nozzle-type chokes (Guo Maximum tubing pressure at valve
and Ghalambor, 2005): depth ( p t ): 500 psia
Required lift gas injection rate per well: 2 MMscf/day
d 0:3167 Pressure safety factor (S f ): 1.1
C ¼ þ 0:6 þ 0:025[ log (N Re ) 4], (13:13)
D d Base temperature (T b ): 60 8F
Base pressure ( p b ): 14.7 psia
D
where Solution Using Dp v ¼ 100 psi, the injection pressure at
d ¼ choke diameter, inch valve depth is then 600 psia. Equation (13.10) gives
D ¼ pipe diameter, in.
N Re ¼ Reynolds number p c,s ¼ p c,v ¼ 600 ¼ 533 psia:
D v 5,000
and the Reynolds number is given by 1 þ 40,000 1 þ 40,000
20q gM g g
N Re ¼ , (13:14) Neglecting the pressure losses between the injection choke
md and the casing head, pressure downstream of the choke