Rod Pump: 2016 #46

A 160-173-54 PU. 16 SPM. 200 BFPD. Rods have no significant stretch. 1.5 inch diameter pump. Anchored. Fluid SG 0.9 at 4,500 ft. Lift system volumetric efficiency?

The PD equation is on 7 PRD 9 (note we can use S for Sp since there is assumed no rod stretch):
Pump Displacement (ideal) = 0.1166(54 in)(16 spm)(1.5^2 in^2) = 227 BFPD
Pump Displacement (actual) = 200 BFPD 200/227 = 88% (B) 

The provided Reference Guide can only hurt you on this problem; any time spent looking in chapter 4 (rod pumps, pg 124-131) is only wasted time. IMO, this is a big weakness for this reference; what good is a PE reference without the rod pump displacement equation? UPDATE: the SPE Reference Guide does have 9702.03 ci/bbl under Constants. and a % Clearance Volume for a Single-acting Cylinder equation (page 151 under Facilities) that can be manipulated so as to be close to the PD equation. Note 9702.03/127 is 0.1169 not 0.1166, but 1% error is not going to matter.

Regardless: know all terms and how the pump designation works (HS V4 pg 489) plus the 0.1166 volumetric conversion constant (HS V4 pg 471). Note you cannot use a direct volume conversion (cubic inches/bbl & min/day) as the number will be too large. 0.1166 is an easy number to memorize though, and the rest is just pump volume.

Rod Pump VE: 2017 #42

Problem 42. The volumetric efficiency of the lift system in Problem 41 above (19 bbl oil+ 188 bbl water) is closest to...

On 7 PRD 9, we see: PD = 0.1166(64)10(1.75^2) = 228 bpd (prior problem).
Volumetric Efficiency = (Produced Volume)/(Pump Displaced Volume) = 19 + 188 = 207 so:
VE = 207/228 = 91%.

Beam Counterbalance: 2017 #41

Problem #41: 25 bbl of 40 API oil and 188 bbl of water (SG = 1.1) are produced from a pumping unit operating with...The counterweight required is nearest to: (A) 9,900 lbs; (B) 10,100 lbs; (C) 10,300 lbs; (D) 10,500 lbs.

I've shown several examples of this type of problem on the blog. Why? They are quick, common rod pump problems because they don't require lengthy charts or tables. So expect them and know how to do them fast. Everything you need to solve this problem is in the Guidebook on two pages 7 PRD 8-9. Solution (note the corrections to the fluids above do not effect this problem's solution):

7/5 rods; 1.75” pump@5000: Wr = 1.732 lb/ft & L = 5000 (7 PRD 10).
W = Wr*L = 1.732(5,000 ) = 8,660 lbs.
Find G: (WC)SGw+(OC)SGo = (0.9)1.1+(0.1)0.825 = 1.07 (7 PRD 1).
Wrf = W(1-0.128G) = 8660(1-(0.128*1.07)) = 7,474 (7 PRD 9).
Fo = 0.34*1.07*1.75^2*4000 = 4,457.
CBE = 1.06(Wrf+(0.5*Fo)) = 1.06(7474+(0.5*4457)) = 10,285 lbs (C).

Rod Pump Counterbalance: 2016 #45

A 160-173-54 pumping unit runs at 16 SPM producing 179 BOPD and 21 BWPD out of 2 inch anchored tubing. The 7/6 rods ...counterweight required? 

Counterbalance problems are meat & potatoes on exams. Why? No fancy charts are needed like most rod pump problems. This can be solved using just two pages in the Guidebook and on the EBT any equations/data not found in the provided reference are easy to give for the specific problem. Now that's not saying there isn't plenty of room for error. I make mistakes on these problems all the time. For this one:

1) 7/6 1.5” rods; Wr = 1.833 lb/ft (7 PRD 10) 
2) W = 1.833(5,000) = 9,165 lbs 
3) Wfr = W(1-0.128G) = 9,165(1-(0.128*0.9)) = 8,109 (7 PRD 9) 
4) Fo=0.34*0.9*1.5^2*4500 = 3098 
5) CBE = 1.06(8,109 + (0.5*3,098)) = 10,238 (C).

Dynamometer Card: 2005 #56 & #57 (similar)

The Dynomometer Card is covered on Guidebook 7 PRD 11.  It shows the Surface Card (top total load) and Pump Card (bottom fluid load). Note the graph is Load (lb) versus Position (in).

Peak PRL & Min PRL are the highest and lowest point on the y-axis in inches, multiplied by the lbf/in conversion.

So if PPRL is about 4.4" and MPRL is 1" with a conversion of 5,000 lbf/in you've got about 22,000 lbf and 5,000 lbf for PPRL & MPRL.

Of concern is the max stress on the top rod of each section. So PPRL is important. For a quick reminder on axial stress, see GB 2 DRL 6; it's just tension / area.

So given a rod size of 86, you have 1" top rod diameter (see the rod name/size explanation on 7 PRD 10; a 86 rod = 1" dia top rod with an area of 3.1415(1)/4 = 0.785 in^2). So stress is about 22,000/0.785 = 28,000 lbf/in^2.

Beam Counterbalance: 2005 #20 (similar)

Counterbalance is calculated fairly easily. No fancy charts; it's all in the Guidebook.
Rod Pump equations are found on a single page 7 PRD 9.
The counterbalance equation (CBE) is step #27. It references steps #1, #5, #15-16.  Again, easy:

#1: Wr is found on 7 PRD 10 in C3, and is the Wgt Rods / ft in air (lbf/ft)
#5: Fo = 0.340(G)D^2(H) which is Fluid Weight On Pump (lbf)
#15: W = Wr∙L which is Weight Rods total in air (lbf)
#16: Wrf = W(1−0.128 G) which is the Weight Rods total in fluid (lbf)
#27: CBE = 1.06(Wrf + 0.5 Fo)

The GB makes this calculation simple plug-and-chug. Just march through the steps.
Here, I'll assume standard 76 rods, 1.5 plunger, & SG = 0.85 (get this from oil API as needed).

#1: Wr = 1.833 (7 PRD 10 column 3 for 76 rods)
#2: Fo = 0.34(0.85)1.5^2(8,800') =  5,850 lb
#15: W = 1.833 lb/ft(8,800') = 16,500 lb
#16: Wrf = 16,500 lb [1-(0.128*0.85)] = 14,700 lb
#27: CBE  1.06 (14,350 + 0.5*5720) = 18,700 lb.

In real life, you may measure CB to find the difference between calculated CBE and actual CB.
For example, if measured CB was 18,500 lb, you are 200 lb short.

Be careful on these problems to keep it simple. Don't panic at all the possible options from extra data like dyno cards, unit type, etc. Just assume it's going to be simple, focus, and march through the steps.

Every item on this type of problem can be found in the GB: Wr, Fo, W, CBE. It's in tables on 7 PRD 9, in order, with a typical number.

Rod Pump Volumetric Efficiency: 2005 #17 (similar)

Volumetric Efficiency for a rod pump is an easy problem. So do it fast.

7 PRD 9 is the Guidebook page for all Rod Pump equations. It's got everything; no flipping pages! For a quick example from the page:

1) Pump volume displaced equation: PD = 0.1166(Sp)D^2(N).
2) Total fluid pumped is usually given: say 370 BFPD.
3) Given pump data: Sp = 90"*, D = 2", N = 10 SPM.
        *Sp, or stroke, is often presented as the pump third number (see 7 PRD 9).

Calculation: 0.1166(90)2^2(10) = 420 BPD
Volumetric Efficiency = BFPD/PD = 380/420 = 90%

Rod Pump: 2005 #18 (similar)

Q: What pumping unit change reduces torque the least? 1) Reducing stroke length or speed, 2) Changing rotation direction, or 3) Installing larger rods?

A: Reducing stroke length or speed lowers torque. Larger rods? This doesn't "reduce" torque at all. So that would be my choice.

On a sidenote: changing direction is a bit more tricky to consider. Why? Class III (& Mark II) API specs recommend the crank arm rotate in the counterclockwise direction only. Class I can operate in either direction but this is not recommended for many reasons. Greater torque spikes clockwise is one of them.