Page 140 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 135 3.1.2007 8:54pm Compositor Name: SJoearun

TRANSPORTATION SYSTEMS 11/135
where The work per one rotation of crank is

P ¼ pressure, lb=ft 2 W 2 ¼ P pD 2 1 L þ P pD 2 1 pD 2 2 L (1) ft lbs :
L ¼ stroke length, ft 4 4 4 rotation
D ¼ piston diameter, ft. (11:13)
Thus, for a triplex pump, the theoretical power is Thus, for a duplex pump, the theoretical power is

pD 2 ft lb Power ¼ 2
Power ¼ 3P LN , (11:1)
4 min 2 2 2
pD pD pD ft lbs
P 1 L þ P 1 2 L N :
where N is pumping speed in strokes per minute. The 4 4 4 min
theoretical horsepower is
(11:14)
3P pD 4 2 The theoretical horsepower is
HP th ¼ LN (hp) (11:2) n h i o
550(60) pD 2 pD 2 pD 2
2 P 4 1 L þ P 4 1 4 2 L N
or HP th ¼ (hp)
550(60)
3P pD 2 or
4
HP th ¼ LN (hp): (11:3) n h i o
33,000 p 2 pD 2 pD 2
2 P 4 D L þ P 4 1 4 2 L N
1
The input horsepower needed from the prime mover is HP th ¼ : (11:15)
33,000
3P pD 4 2 The input horsepower needed from the prime mover is
HP i ¼ LN (hp), (11:4) n h i o
33,000e m p 2 pD 2 pD 2
2 P 4 D L þ P 4 1 4 2 L N
1
where e m is the mechanical efficiency of the mechanical HP i ¼ (hp): (11:16)
system transferring power from the prime mover to 33,000e m
the fluid in the pump. Usually e m is taken to be about The theoretical volume output from the double-acting
0.85. duplex pump per revolution is
The theoretical volume output from a triplex pump per pD 2 pD 2 pD 2 N
3
revolution is Q th ¼ 2 1 L þ 1 2 L ft =sec : (11:17)
4 4 4 60
pD 2 LN
3
Q th ¼ 3 ft =sec : (11:5) The theoretical output in gals/min is thus
4 60
pD 2 pD 2 pD 2
The theoretical output in bbl/day is thus q th ¼ 2 1 L þ 1 2 L
4 4 4
q th ¼ 604LND 2 bbl : (11:6) N
day (gal= min): (11:18)
0:1337
If we use inches (i.e., d [in.] and l [in.]), for D and L, then
If we use inches (i.e., d [in.] and l [in.]), for D and L,
bbl then
q th ¼ 0:35lNd 2 : (11:7)
day pd 2 pd 2 pd 2 N
q th ¼ 2 1 l þ 1 2 l (gal= min): (11:19)
The real output of the pump is dependent on how effici- 4 4 4 231
ently the pump can fill the chambers of the pistons. Using The real output of the pump is
the volumetric efficiency e v in Eq. 11.7 gives
pd 2 1 pd 2 1 pd 2 2 N
q r ¼ 0:35e v lNd 2 bbl (11:8) q r ¼ 2 4 l þ 4 4 l 231 e v (gal= min)
day
or
or
2
2
2
2
q r ¼ 0:01e v d lN (gal=min), (11:9) q r ¼ 0:0068 2d d lNe v (gal= min), (11:20)
1
that is,
where e v is usually taken to be 0.88–0.98.
As the above volumetric equation can be written in d q r ¼ 0:233 2d d lNe v (bbl=day): (11:21)
2
2
and l, then the horsepower equation can be written in d, l, 1 2
and p (psi). Thus, As in the volumetric output, the horsepower equation can
also be reduced to a form with p, d 1 , d 2 , and l
3p pd 4 2 12 l N 2 2
HP i ¼ (11:10) HP i ¼ p 2d d lN : (11:22)
1
2
33,000e m
252,101e m
reduces to Returning to Eq. (11.16) for the duplex double-action
2
pd lN pump, let us derive a simplified pump equation. Rewriting
HP i ¼ : (11:11) Eq. (11.16), we have
168,067e m n h pD 2 pD 2 i o
2
2 P p 4 D L þ P 4 1 4 2 L N
1
HP i ¼ : (11:23)
11.2.2 Duplex Pumps 33,000e m
The work per stroke cycle is expressed as The flow rate is
2 2 2
pD 2 pD 2 pD 2 pD pD pD 3
W 1 ¼ P 1 L þ P 1 2 L(ft lbs): (11:12) Q th ¼ 2 1 L þ 1 2 L N (ft = min ), (11:24)
4 4 4 4 4 4