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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 169 4.1.2007 2:43pm Compositor Name: SJoearun
SUCKER ROD PUMPING 12/169
the rods. The minimum load is at or near the top of the or
stroke. Neglecting the weight of the plunger and friction 1 2
term, the minimum PRL is T ¼ SW f þ 2SN W r (in:-lb): (12-30)
4 70,471:2
W r
PRL min ¼ S f (62:4) þ W r W r F 2 , Because the pumping unit itself is usually not perfectly
g s
balanced (C s 6¼ 0), the peak torque is also affected by
which, for 50 8API oil, reduces to structure unbalance. Torque factors are used for correc-
PRL min ¼ 0:9W r F 2 W r ¼ (0:9 F 2 )W r , (12:22) tion:
1
where for the conventional units T ¼ 2 ½ PRL max (TF 1 ) þ PRL min (TF 2 ) , (12:31)
2
c
SN (1 ) 0:93
F 2 ¼ h (12:23) where
70,471:2
and for air-balanced units TF 1 ¼ maximum upstroke torque factor
TF 2 ¼ maximum downstroke torque factor
c
2
SN (1 þ )
F 2 ¼ h : (12:24) 0.93 ¼ system efficiency.
70,471:2
For symmetrical conventional and air-balanced units,
TF ¼ TF 1 ¼ TF 2 .
12.4.3 Counterweights There is a limiting relationship between stroke length
To reduce the power requirements for the prime mover, a and cycles per minute. As given earlier, the maximum
counterbalance load is used on the walking beam (small value of the downward acceleration (which occurs at the
units) or the rotary crank. The ideal counterbalance load C top of the stroke) is equal to
is the average PRL. Therefore, SN g 1 c
2
a max = min ¼ h , (12:32)
1
C ¼ (PRL max þ PRL min ): 70,471:2
2
(the + refers to conventional units or air-balanced units,
Using Eqs. (12.19) and (12.22) in the above, we get see Eqs. [12.9] and [12.12]). If this maximum acceleration
1 1 divided by g exceeds unity, the downward acceleration of
C ¼ W f þ 0:9W r þ (F 1 F 2 )W r (12:25) the hanger is greater than the free-fall acceleration of the
2 2
rods at the top of the stroke. This leads to severe pounding
or for conventional units when the polished rod shoulder falls onto the hanger
1 SN 2 c (leading to failure of the rod at the shoulder). Thus, a
C ¼ W f þ W r 0:9 þ (12:26) limit of the above downward acceleration term divided
2 70,471:2 h
by g is limited to approximately 0.5 (or where L is deter-
and for air-balanced units mined by experience in a particular field). Thus,
1 SN 2 c SN 1 c
2
C ¼ W f þ W r 0:9 : (12:27) h #L (12:33)
2 70,471:2 h 70,471:2
The counterbalance load should be provided by structure or
unbalance and counterweights placed at walking beam s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(small units) or the rotary crank. The counterweights can N limit ¼ 70,471:2L : (12:34)
c
be selected from manufacturer’s catalog based on the cal- S(1 )
h
culated C value. The relationship between the counterbal-
ance load C and the total weight of the counterweights is For L ¼ 0.5,
187:7
r d 1 N limit ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : (12:35)
C ¼ C s þ W c , S(1 )
c
c d 2 h
where The minus sign is for conventional units and the plus sign
for air-balanced units.
C s ¼ structure unbalance, lb
W c ¼ total weight of counterweights, lb
r ¼ distance between the mass center of counter- 12.4.5 Tapered Rod Strings
weights and the crank shaft center, in. For deep well applications, it is necessary to use a tapered
suckerrodstringstoreducethePRLatthesurface.Thelarger
diameter rod is placed at the top of the rod string, then the
12.4.4 Peak Torque and Speed Limit next largest, and then the least largest. Usually these are in
The peak torque exerted is usually calculated on the most sequences up to four different rod sizes. The tapered rod
severe possible assumption, which is that the peak load strings are designated by 1/8-in. (in diameter) increments.
(polished rod less counterbalance) occurs when the ef- Taperedrodstringscanbeidentifiedbytheirnumberssuchas
fective crank length is also a maximum (when the crank
8
arm is horizontal). Thus, peak torque T is (Fig. 12.5) a. No. 88 is a nontapered ⁄ 8 - or 1-in. diameter rod string
7
b. No. 76 is a tapered string with ⁄ 8 -in. diameter rod at
d 2
6
T ¼ cC (0:9 F 2 )W r : (12:28) the top, then a ⁄ 8 -in. diameter rod at the bottom.
½
d 1
c. No. 75 is a three-way tapered string consisting of
Substituting Eq. (12.25) into Eq. (12.28) gives 7 ⁄ 8 -in. diameter rod at top
6
1 ⁄ 8 -in. diameter rod at middle
T ¼ SC (0:9 F 2 )W r (12:29) 5 ⁄ 8 -in. diameter rod at bottom
½
2
d. No. 107 is a four-way tapered string consisting of
or 10 ⁄ 8 -in. (or 1 ⁄ 4 -in.) diameter rod at top
1
1
10
1 1 1 9 ⁄ 8 -in. (or 1 ⁄ 8 -in.) diameter rod below ⁄ 8 -in. diameter rod
T ¼ S W f þ (F 1 þ F 2 )W r 8 ⁄ 8 -in. (or 1-in.) diameter rod below ⁄ 8 -in. diameter rod
9
2 2 2
7 8
⁄ 8 -in. diameter rod below ⁄ 8 -in. diameter rod