Page 176 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 172 4.1.2007 2:43pm Compositor Name: SJoearun

12/172 ARTIFICIAL LIFT METHODS
Let us restrict our discussion to conventional units. index and production rate. A reasonable estimate may be a
Then Eq. (12.39) becomes value that is twice the production drawdown.
2
W r D r SN M Volumetric efficiency can decrease significantly due
dl o ¼ (ft): (12:40) to the presence of free gas below the plunger. As the
A r E 70,471:2 fluid is elevated and gas breaks out of solution, there is
Equation (12.40) can be rewritten to yield dl o in inches. W r is a significant difference between the volumetric displace-
ment of the bottom-hole pump and the volume of the
W r ¼ g s A r D r
fluid delivered to the surface. This effect is denoted by
3
2
6
and g S ¼ 490 lb=ft with E ¼ 30 10 lb=m . Eq. (12.40) the shrinkage factor greater than 1.0, indicating that
becomes the bottom-hole pump must displace more fluid by some
2
2
dl o ¼ 1:93 10 11 D SN M(in:), (12:41) additional percentage than the volume delivered to the
r
surface (Brown, 1980). The effect of gas on volumetric
which is the familiar Coberly expression for overtravel efficiency depends on solution–gas ratio and bottom-hole
(Coberly, 1938). pressure. Down-hole devices, called ‘‘gas anchors,’’ are
Plunger stroke is approximated using the above expres- usually installed on pumps to separate the gas from
sions as the liquid.
In summary, volumetric efficiency is mainly affected by
S p ¼ S dl r dl t þ dl o
the slippage of oil and free gas volume below plunger.
or Both effects are difficult to quantify. Pump efficiency can
12D
S p ¼ S vary over a wide range but are commonly 70–80%.
E

2
1 1 SN M W r
W f þ (in:): (12:42) 12.5.3 Power Requirements
A r A t 70,471:2 A r
The prime mover should be properly sized to provide
If pumping is carried out at the maximum permissible adequate power to lift the production fluid, to overcome
speed limited by Eq. (12.34), the plunger stroke becomes friction loss in the pump, in the rod string and polished
12D rod, and in the pumping unit. The power required for
S p ¼ S lifting fluid is called ‘‘hydraulic power.’’ It is usually ex-
E
pressed in terms of net lift:
c
1 1 1 þ LW r
h
6
W f þ c (in:): (12:43) P h ¼ 7:36 10 qg l L N , (12:45)
A r A t 1 A r
h
c where
For the air-balanced unit, the term 1þ h c is replaced by its
1 h
reciprocal. P h ¼ hydraulic power, hp
q ¼ liquid production rate, bbl/day
12.5.2 Volumetric Efficiency g l ¼ liquid specific gravity, water ¼ 1
Volumetric efficiency of the plunger mainly depends on L N ¼ net lift, ft,
the rate of slippage of oil past the pump plunger and the
solution–gas ratio under pump condition. and
Metal-to-metal plungers are commonly available with p tf
plunger-to-barrel clearance on the diameter of 0.001, L N ¼ H þ , (12:46)
0:433g l
0.002, 0.003, 0.004, and 0.005 in. Such fits are re-
ferred to as 1, 2, 3, 4, and 5, meaning the plunger where
outside diameter is 0.001 in. smaller than the barrel inside H ¼ depth to the average fluid level in the annulus, ft
diameter. In selecting a plunger, one must consider the p tf ¼ flowing tubing head pressure, psig.
viscosity of the oil to be pumped. A loose fit may be
acceptable for a well with high viscosity oil (low 8API The power required to overcome friction losses can be
gravity). But such a loose fit in a well with low viscosity empirically estimated as
oil may be very inefficient. Guidelines are as follows:
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P f ¼ 6:31 10 W r SN: (12:47)
a. Low-viscosity oils (1–20 cps) can be pumped with a
plunger to barrel fit of 0.001 in. Thus, the required prime mover power can be expressed as
b. High-viscosity oils (7,400 cps) will probably carry sand P pm ¼ F s (P h þ P f ), (12:48)
in suspension so a plunger-to-barrel fit or approxi-
mately 0.005 in. can be used. where F s is a safety factor of 1.25–1.50.
An empirical formula has been developed that can be Example Problem 12.2 A well is pumped off (fluid
used to calculate the slippage rate, q s (bbl/day), through level is the pump depth) with a rod pump described in
the annulus between the plunger and the barrel: Example Problem 12.1. A 3-in. tubing string (3.5-in. OD,
2:9 2.995 ID) in the well is not anchored. Calculate (a)
k p d b d p d b þ d p Dp
q s ¼ , (12:44) expected liquid production rate (use pump volumetric
m d 0:1 L p
b efficiency 0.8), and (b) required prime mover power (use
where safety factor 1.35).
k p ¼ a constant
d p ¼ plunger outside diameter (in.) Solution This problem can be quickly solved using the
d b ¼ barrel inside diameter (in.) program SuckerRodPumpingFlowrate&Power.xls.The
Dp ¼ differential pressure drop across plunger (psi) solution is shown in Table 12.3.
L p ¼ length of plunger (in.)
m ¼ viscosity of oil (cp).
6
6
The value of k p is 2:77 10 to 6:36 10 depending on 12.6 Procedure for Pumping Unit Selection
6
field conditions. An average value is 4:17 10 . The value The following procedure can be used for selecting a pump-
of Dp may be estimated on the basis of well productivity ing unit: