ESP: 2018 #42

Problem 42. Designing an ESP installation from given data, the TDH (ft) is closest to: 
Current production: 500 BFPD at 500 psi drawdown.
Desired production: 4X (negligible free gas, annulus friction). 
Pressures (psi) at 2M BFPD: Pump-intake/wellhead = 1M/200. 
Depth (ft) of pump at perfs = 6M; Friction loss = 41.7 ft/1M ft tbg. 
WC: 80%, 1.0625 SG; OC: 20%, 0.91 SG & 24 API. 
(A) 4,460 (B) 4,660 (C) 4,860 (D) 5,060 

See Guidebook 7 PRD 3-4, or the ESP Sizing Guide reference listed there.
Note: this problem was updated on Kindle after 10/12/2018 as the original given well data was nonsensical (even though it led to the same answer selection). Please use the well info shown above:

1. PI = dq/dp  = 500/500 = 1 bfpd/psi.
2. SGw = 0.8(1.0625) = 0.850; SGo = 0.2(.91) = 0.182; so SGf = 0.85 + 0.182 = 1.032.
4. NDL = Pdepth - PIP(2.31)/SGliq = (6,000 – (1,000*2.31)/1.032 = 3,762 psi ft
5. Hf = 6,000 ft and 41.7 ft/1,000 ft = 250 ft.
6. Hwh = 200 * 2.31 / 1.032 = 448 ft.
7. TDH = 3762 + 250 + 448 = 4460 ft (A).

Because the answer is the lowest TDH option given, any lower value one calculates will give the correct answer.