2017 #46

Which specific operation would typically incur the largest ton-miles using a 1-1/8” drilling line (assume a typical US onshore rig, typical BHA and drillpipe, and the same well)? 

(A) Drilling from 4,000 to 5,000 ft using a 8.5 inch PDC bit. 
(B) A round trip at 5,000 ft. 
(C) Core from 5,000 ft to 5,080 ft. 
(D) Running 7 inch 26 lb/ft casing to 5,080 ft. 

See 1 RIG 5. After calculating RTTM multiply the following factors based upon the actions: 1) Trip/csg: x1; 2) Coring x2; 3) Drilling x3. This makes sense; drilling has 3x the drill line action due to tripping/wiper/drilling, coring 2x due to no wiper, and tripping is one-way.

But in this problem the multipliers are moot. A) 1,000 ft drilling, 2) 5,000 ft round trip or 10,000 ft, C) 80 ft core, running casing 5,080 ft. The round trip dominates all multipliers, or (B).

Reservoir Volume: 2017 #44

A single well is drilled in a small isolated volumetric oil reservoir with a total compressibility factor 0.00002 1/psi and an oil FVF of 1.3 bbl/STB.

The well flows at 100 BOPD until pseudosteady-state flow is reached and maintained until BHP falls 125 psi over 4 days. The reservoir volume is closest to (Mrb): (A) 420; (B) 320; (C) 220; (D) 120.

This is plug-and-chug using 12 WLT 3: -0.234qB/(ct dpwf/dt) = 0.234(100)1.3 / (2E-5(125/4*24), which at 5.615 cf/bbl converts to 208 Mrb, or (C).

Just be careful with the units, and know the provided CBT reference locations for unit lookup.

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).

Gas Meters: 2017 #40

A gas meter orifice:
(A) Typically has less than 10% of an effect...
(B) Should be installed with the beveled edge upstream. 
(C) Should be installed with the sharp edge downstream. 
(D) Typically has less than 20% of an effect...

The Guidebook (8 FAC 3 & 4) states gas orifice plates are installed with beveled edge downstream. So we know B & C are false.

The same page also says an orifice installed backwards reads about 15% low. So we know A is false.

Finally, since we know backwards is the greatest error it can have due to orientation (15%), D must be true.

LOT: 2017 #37

A LOT at 10,000 ft with 10 ppg mud records the following volumes of fluid and pressures after pumping: 3 bbl, 600 psi; 4 bbl, 800 psi; 4.5 bbl, 900 psi; 5 bbl, 950 psi. If 100 psi was lost overcoming the mud gel strength, the formation fracture pressure is closest to:

This is the Guidebook example. It's also a fair question because it tests one's understanding of a LOT. One should have the needed page in a few seconds using the TOC under "LOT".

The hydrostatic pressure calculates to 5,200 psi, 900 psi is added, and 100 psi is subtracted making 6,000 psi. Simple and fast if you understand the process and have the equation. Just be prepared to graph it out on scratch paper.