Sunday, June 24, 2018

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.

7 comments:

  1. David,
    For the CB given (18065), I have a bigger difference than the answer. My CB-CBE=-626. Is there some sort of conversion I need to do with the Measured Counterbalance?

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    Replies
    1. Detailed numbers step by step.
      #1: Wr = 1.833
      #2: Fo = 5,756 lb
      #15: W = 16,130 lb
      #16: Wrf = 14,365 lb
      #27: CBE = 18,277 lb
      Which step are you different?

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    2. Thanks for the reply. I had most of the numbers right, I just had Wrf and W off by just a bit and it ended up adding up to just barely the wrong answer...I'll have to be more careful with my calculations! Thanks

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  2. I cannot figure out why D is 2.25. Shouldn't it be 1.5? If you use 76 API and plunger diameter of 1.5 to get Wr = 1.833 then wouldn't D be 1.5? or am I missing something?

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    Replies
    1. I apologize; D is the plunger diameter (1.5"). I just squared it (2.25) & stuck it in the formula (RP 11L does it that way; I'm still stuck in the past).

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    2. I saw it in the practice exam that way too and I didn't realize it was already squared! whoops. Got it now, thanks!

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    3. I saw it in the practice exam that way too and I didn't realize it was already squared! whoops. Got it now, thanks!

      Delete