Vogel IPR: 2005 #64 (similar)

Vogel IPR problems generally involve: 1) calculating Vogel IPR using the qo/qmax equation (see 7 PRD 1 on how to calculate qmax from a given well test, in this case say, 600 BOPD), 2) calculate qo for a range of FBHP that crossover the TBG curve, and 3) note that the well's natural flow rate is where the IPR & TBG curves cross. That's it.

From the calculated and plotted IPR example below, it's easy to see the roughly 338 psi crossover. For this example, the TBG curve is simply given and the IPR calculated (using as few points as possible, just the crossover area).

For a good explanation of this problem type, see Well Performance by Golan (P29), or Production Optimization by Beggs (P142). Both these resources are excellent (I own both even though they have quite a bit of overlap).

Note that neither of these sources are SPE. Personally, I find the lack of example problems for nodal analysis (or total system analysis) to be a major gap in the SPE Handbook and the SPE Textbook Series. For this reason, I've never spent much time on this problem type. However, I get continual questions about it so I'm showing a detailed similar solution and how to use the Guidebook's applicable section for it. Also, for anyone interested in more explanation the SPE 6th Edition (1991) has an excellent example problem they walk you through as well.

However: Vogel IPR as a "concept" is definitely fair game and is found in HS IV P1-40 (albeit with a lack of example problems or number examples). So understand IPR (including Vogel, Fetkovich, Jones, and Wiggins, who wrote HS IV C1). I'll try and fit an IPR-style problem into the Guidebook Companion 2018 41-80 to help with reviewing this problem type.

IPR Vogel Equation
bopd psia
259 1600
329 1400
392 1200

Given TBG curve 
bopd psia
200 1450
300 1390
400 1350

 

Economics Present Worth: 2005 #27 (similar)

Assume a $700M investment, a lease abandonment cost of $50M, and a 15% discount rate. Then assume the following years of revenue (minus expenses & taxes):  $650M, $330M, $150M. For Present Worth, use the standard NPV formula:

-700M + 565M + 250M + $99M - $33M = $180M

This is painfully easy except for the abandonment cost part, which must be both subtracted and discounted (it's easy to space out on either of these; the typical problem uses salvage value, which is added not subtracted). Watch for the answer options to offer solutions for either mistake.

I can't implore enough to warily check these "easy" economics problems. The number of smart guys who miss these problems is myriad. I triple check these because they are so easy yet so easy to slip up on. Because overconfidence. 

Vogel & Flow Efficiency: 2005 #26 (similar)

Vogel problems often involve flow efficiency (FE). Why? They both need Pwf & Pr.

Example: Say Pwf = 1M and Pr = 2M psi at q = 490 BOPD. What is qmax?
Go to the Vogel table (7 PRD 1) with Pwf/Pr = 0.5; note qo/qmax = 0.7.
So qmax = qo/0.7 = 490/0.7 = 700 BOPD.

But what then if FE is 0.7 and we stimulate to an FE = 1? See 12 WLT 2:

Another way to describe FE: the percentage of well fluid producing at a given drawdown compared to what it would produce with zero skin (FE = 1).

So at FE = 1.7M/0.7 = 1,000 BOPD.

Hydrostatic: 2005 #24 (similar)

Sand at 9,000 ft. Pr = 4,000 psi. Two fluids. Which A-B-C option below is 500 psi underbalanced?

(A) 3,000 ft 0.1 ppf N2 cushion, 6,000 ft 10 ppg, 80-psi surface pressure.
(B) 2,000 ft 0.1 ppf N2 cushion, 7,000 10 ppg, 160-psi surface pressure.
(C) 2,500 ft 0.1 ppf N2 cushion, 6,500 10 ppf, 400-psi surface pressure.

Use the hydrostatic pressure equation (2 DRL 1). Calculating pressures p = 0.052(D)MW) + gradient(D) + surface pressure - reservoir pressure for:

(A) 300 + 3,120 + 80 - 4,000 = -500 psi
(B) 200 + 3,640 + 160 - 4,000 = 0 psi
(C) 250 + 3,380 + 400 - 4,000 = 30 psi

Expect this sort of problem with any fluid, depths, or pressures. Just remember to line them up in an orderly fashion; speed is of the essence.

Net Piston Force: 2005 #23 (similar)

Tubing movement problems often ask for a specific force (Guidebook 6 DTC 9).
Here the "Net Piston Effect" is shown.

Say Csg & Tbg pressure changes are:
     dPcp = 500 psi (often given as pre-post job csg pressure).
     dPtp = 3,000 psi (often calculated by hydrostatic).
Packer, Tbg, & Csg areas: Apb, Ati, Ato = 7.1, 7.0, 9.2 sq in (given or from dia).

The packer (7.1) is larger than tbg id (7); so the GB predicts a negative (up) force:
Fp = dPCp(Apb-Ato) - dPTp(Apb - Ati)
Fp = 500(7.1-9.2)-3,000(7.1 - 7.0) = -1,385 lbf (up).
Fp = 500(-2.1) - 3,000(0.1)
Fp = -1,050-300 = -1,385
Fp = 500(7.1-7) - 3,000(7.1-9.2) = -1,385 lbf (up). This checks.

To calculate tbg length change? Say the tubing is 10,000'; tbg area is 9.2 - 7 = 1.8 sq in:
LtF/EAt = (10,000 ft*1,385 lbf)/(30M*1.8) = 0.15'*12 = -1.8 inches (up)

The entire tubing move section is a single page, with all the variables listed to the right.

Balloon Force: 2005 #22 (similar)

6 DTC 9 is one of my favorite Guidebook pages (TBG Move). It shows all the forces you will need on a single page. It was a labor of love.

So if you are asked for the force from say the ballooning effect, it's just a glance: Keep in mind the dPt and dPc are pressure changes, so find the initial and final pressures in both the casing and the tubing.

You typically chase down the BHTP using hydrostatic. For example at 10,000 ft, 8 ppg:
Initial Tubing: 0.052(8)10,000 = 4,160 psi. If surface: 0 psi, dPTa1 = 2,100 psi
Final Tubing: Surface & BHP say 6,000 & 7,900 psi you average: dPTa2  = 6,950 psi
From initial & final tubing psi, find the change by subtracting: 6,950 - 2,100 = 4,850 psi

Casing pressure change is often just given as an increase, say 1,000 psi.
Calculate or look up the tubing ID & OD area. That's easy; here we will use 7.0 & 9.6 in.
Thus: -0.6 [(4850*7) - (1000*9.6)] = -14,600 psi (up).

The units are negative (note the sign in front) which means tension.
Since pressure increased, we should indeed see tension if the packer is fixed.
These problems can be very confusing. Go slow; be sure the numbers make sense. Once you've done a few, it's fairly easy.

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. 

PBU: 2005 #16 & #68 (similar)

The standard pressure buildup (PBU) question wants skin, pressure drop, or radius of investigation (ROI). 12 WLT 12:

Variables:
B FVF bbl/STB
ct compressibility 1/psi
h thickness reservoir ft
k permeability md
MTR middle time region
p pressure psia
pwf pressure well flow psia
q flow rate well last STB/hr
ri radius investigation ft
rw radius wellbore ft
tp time produced pre-SI hr
Δt time new well SI after tp hr
ϕ porosity x.xx
μ viscosity cp

Example: q = 100 BO/D, h = 50 ft, Bo = 1.4 RB/STB, μ = 0.8 cp, & a graph of psi vs (tp-Δt)/Δt with an MTR of m = 300 psi (use one log cycle):

k = [162.6(100 BO/D)1.4 rb/STB(0.8 cp)]/[300*50 ft] =  2.25 md

From here, the ROI from SI pressure transient (say after 2 days) calculates from a simple equation on 12 WLT 6. A few more variables like ϕ (say 12%) and ct (say 4E-6 /psi) are needed:

ri = [2.25 md(48 hr)]/[948(0.12)(0.8 cp)4.3E-6 psi]^0.5 = 540 ft

Skin is calculated using tp (say 72 hr) and PBU data (tp+Δt)/Δt. Use Δt = 1 hr (easier division); this means (tp+Δt)/Δt = (72 hr+1 hr)/1 hr = 73 hrs. Now: back-extrapolate MTR to 73 hrs for pressure; 2,500 psi is a typical value. Subtract from given Pwf for drawdown (say 1,000 psi) and then divide by slope m over one log cycle.This is the skin equation's first term. 6.5 would be typical.

Using our our prior data, plus the well diameter (assume 4 in diameter) we can calculate skin:

s = 1.151[(P1hr-Pfbhp)/m]-log[k/[(por)(vis)ct(rw^2)]+3.23
   = 1.151[6.5-log[2.25[(0.12*0.8*4.3E-6*0.125^2]+3.23
   = 1.151[6.5-6.7+3.23] = 3.5

Archie Cementation Factor m: 2005 #11 (similar)

Given a formation with known porosity (say 9%), Ro (say 25 ohmm), and Rw from a core test (say 0.25 ohmm): calculate the formation cementation factor (or exponent m; this may be called either).

Two equations for F are on 15 LOG 1, with 2 variables each. Given 3 of the variables (we are missing only m) this is easy to solve. The Guidebook actually gives the equation for m here so it's just plug-and-chug.

Problems like this may show many sample points; if so, it doesn't matter which you choose. But you can check your work using a second set if you have the time.

F = Ro/Rw = 25/.25 = 100
F = 1/por^m = 1/0.09^m
(m)log(por) = log(1/F)
m = log(1/F)/log(por)
m = log(1/100)/log(.09)
m = -2/-1.05 = 1.9

Be aware on other problems you may be asked for (or given) a tortuosity factor "a" (F=a/por^m) rather than a = 1 (the norm). It's all on the 15 LOG 1 page if this unlikely event arises.

Mud Cleaning: 2005 #7 (similar)

One must often select the most cost effective mud cleaning equipment given a desired particle size. Those problems are easy using 4 MUD 6 of the Guidebook (below is the relevant part):

Solid Control Equipment Order:
1. Shale Shaker: down to 75 μm particle size
Mesh ex: 70x30: 70 openings/in one direction, 30 in perpendicular
Degasser (vacuum pump) or Gas Separator (no pump)
2. Mud tank:
Mud Agitator: prevents “steeling”
Mud Gun: flushes tanks
3. Desander: down to 45 μm, Desilter down to 15 μm
…fine screens can replace desander/desilter for power savings
Mud Cleaner: parts 1-3
4. Hydrocyclone: changes flow path; solids into cones
5. Centrifuge: rotating drum; rotation speed variable
6. Additives: Chemical, Bentonite, Water

75 + μm? Shaker.
75 - 45 μm? Shaker + Desander.
45 - 15 μm? Shaker + Desander + Desilter.

Pump HP: 2005 #3 (similar)

Calculating drilling pump horsepower? Total pressure drop is needed (pipe, bit, annulus, surface).
We will assume annulus & surface pressure drop is 0.
All 5 equations needed are on a single Guidebook page, 3 HYD 1.

1) For pipe pressure drop (say MW=10ppg, PV=40cp, V=14 fps, 500 gpm, 10M ft):
 dP/dL= (ρ^0.75 V^1.75 μ^0.25)/1,800 d^1.25
dPd/10,000 = (10^0.75 14^1.75 40^0.25)/1,800*3.8^1.25
dPd = 1,500 psi.

2) Not given pipe ID? Merely back-calculate ID from your 14 fps & 500 gpm:
d=[q/2.448V]^1/2 = 500/[2.448*14)] = 3.8 in.

3) For bit pressure loss (given nozzle area = 0.39 si):
dPb = 8.311E-5 MW q²/[C_D² A_t²] dPb = 8.311E-5(10)500²/[0.95² 0.39²d
Pb = 1,500 psi.

4) So total pressure drop over the pumping system is 1,500 + 1,500 = 3,000 psi.

5) Enter this into the standard HP equation:
[dP q]/1,714 = 3,000*500/1714 = 875 HP.

This is a tricky problem; step 2 is especially mean. Yet the equations do logically progress on a single Guidebook page, making it easier.

Collapse; Tension & Pressure: 2005 #2 (similar)

Given: 7 in. P-110 casing (D/t = 7/0.59) with axial tension of 50M & internal pressure of 11M psi. Collapse pressure (psi)?

Combined tension and pressure? Complex. Use 6 DTC 4:

1) (σz + pi)/σyield = (50M + 11M)/110M = 0.554
2) Chart: --> 0.544 ---> -0.60 = (pi - pcrr)/pcr (note negative sign for collapse)
3) pcrr = pi - (-0.60)pcr) = pi + 0.6(pcr) = 11M + 0.6(pcr)

We're here in less than 2 minutes but still need pcr. The Redbook shows 7 in. P-110 casing's collapse rating is 16,990 psi. Of course you can calculate it (from D & t; see the formula on 6 DTC 2) but it's faster to use the Redbook. This allows us to calculate collapse pressure in this situation:

pcrr = pi - (-0.60)pcr) = pi + 0.6(16,990) = 11M + 10.1M = 21.1M.

3 minutes. Not bad! We've lost at least half our engineers by now on an exam. Last but not least: do a quick mental check; does internal pressure strengthen or weaken collapse? Clearly strengthen, and that's what the equation shows. Just be careful; it's easy to make a sign mistake.

Petroleum PE Problems 2018: 1-40

The 2018 Guidebook Companion is available on Amazon. I only publish these on Kindle to keep the cost <$10.

The sample problems look great on a smartphone or computer using the free Kindle app. The format displays two problems per page; all you need is scratch paper.

These problems reference the 2018 Guidebook exclusively (which has new sections and additional material). It also leans heavy on the SPE Handbook (although every problem can be solved using the 2018 Guidebook alone).

These practice problems were reviewed by three different 2017 PE exam takers, each giving it the thumbs-up. Harder than the 2016 and 2017 versions in my opinion (with some new twists) so I think they offer a challenge to nearly everyone. However, I personally find the 2016 version the most applicable to reality, and many agree with me. YMMV. I've been surprised at the diversity of opinion out there: one person's yawn is another person's bane.

My intent is to have problems 41-80 out by September; wish me luck.

ESP

Some interesting points regarding ESPs: the older SPE Handbook (Bradley 1987) has two seemingly contradictory quotes on the same page (7-1):

a) The ESP has the broadest producing range of any artificial lift method.

b) The major disadvantage of the ESP is that it has a narrow producing rate range compared with other artificial lift forms.

What is correct?  Here's what the newer SPE Handbook says regarding ESPs:

1. 200 to 20M B/D typical (30M max).

2. High-Volume Lift Capacity excellent

3. Low-Volume Lift Capacity generally poor: low efficiency & high operation costs <400 BFPD. 

4. Limited by needed horsepower.

5. Can be restricted by casing size.

Hard to know what Bradley was trying to say. Just be aware of both SPE sources, and let this be a lesson on how language can make an otherwise simple question more difficult. I like to underline these "money quotes" in pencil in my Handbook when I run across them.

SPE Petroleum Engineering Certification and PE License Exam Reference Guide (Ghalambor, 2014)

Below is my Amazon review. I'm posting this here since I get so many questions about this book. As you can see, I like the book, but not for the PE Exam. Why? It's designed for a different exam, lacks number examples for the equations, and is not easy to search to find the correct formula. Feel free to ask questions or make your own assessment in the comment section below.

To be clear about what this book is: it’s a list of equations, graphs, and tables. They are broken up into the following six subjects:

Reservoir Engineering
Drilling Engineering
Formation Evaluation
Production Engineering
Facilities
Petroleum Economics

Strong points:

1. Comprehensive. I would add a few here and there, but one must draw the line somewhere.
2. Clear print. Big graphs.
3. Each variable listed after the equation.

Weak points:

A. No numbers shown with equations! This makes many hard to use, even if you understand them.
B. No explanations! You better understand these equations before you use them.
C. Let’s be clear: this book is just a list of equations/graphs/tables, nothing more.
D. No easy way to find what you need besides the six chapters. You have to know this book well to make it useful on an exam.
E. Ring bound. This is both a plus and minus, just be aware of it.

In summary: if you want a comprehensive book of equations without numbers or examples, this book is for you. I’ve found it makes for a useful office reference. Just don’t expect much more except some graphs and tables (which are not comprehensive, but pretty complete for general use). It is the primary and only allowed reference for the Certification Exam, but I wouldn't bring it to the Professional Engineering Exam unless I knew it very well and supplemented it with notes.

SPE Petroleum Engineering Handbook (Bradley 1987)

There are only a few useful SPE books for the PE Exam. Bradley's SPE Petroleum Engineering Handbook is one. On the used market it goes for $100 to $200. My Amazon review is here.

I bring this up because I'm often returning to Bradley. I'm amazed at how concise and well-organized this text is. I'm currently updating my reservoir section with a few "money quotes" from Bradley. Remember, this is still an SPE reference, and thus it's fair game on the PE Exam.

In fact, I know several people who used it exclusively for their PE Exam and did well. One of the reasons it still shines for the exam is how tight it is: no wasted words. Clear explanations. Simple format for quick reference. And it has a lot of practical, work-related stuff the newer Handbook series leaves out for some reason.

In summary: because the new Handbook Series is out, people are selling their old Bradley Handbooks thus making them at least "somewhat" affordable. Back in the day, it was a collector's item and very hard to even find. So while quite dated it's worth another look as a primary reference.