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).
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