Dynamometer Card: 2005 #56 & #57 (similar)

The Dynomometer Card is covered on Guidebook 7 PRD 11.  It shows the Surface Card (top total load) and Pump Card (bottom fluid load). Note the graph is Load (lb) versus Position (in).

Peak PRL & Min PRL are the highest and lowest point on the y-axis in inches, multiplied by the lbf/in conversion.

So if PPRL is about 4.4" and MPRL is 1" with a conversion of 5,000 lbf/in you've got about 22,000 lbf and 5,000 lbf for PPRL & MPRL.

Of concern is the max stress on the top rod of each section. So PPRL is important. For a quick reminder on axial stress, see GB 2 DRL 6; it's just tension / area.

So given a rod size of 86, you have 1" top rod diameter (see the rod name/size explanation on 7 PRD 10; a 86 rod = 1" dia top rod with an area of 3.1415(1)/4 = 0.785 in^2). So stress is about 22,000/0.785 = 28,000 lbf/in^2.

WLT Pseudo-steady-state: 2005 #63 (similar)

Well test problems are often pseudo-steady-state. The Guidebook covers these on 12 WLT 2-5.

Several tricks to be aware of:
1) If given reservoir volume, assume it's circular to find the radius.
           Example: A reservoir is 100 acres, re = [(43,560 ft/ac)100 ac(1/pi)]0.5 = ~1,200 ft
2) If given a range of B & viscosity, average the start and end values over flow time.
          Typical values are 1.4 and 0.5.
3) If reservoir pressure falls below the BP, remember it's no longer single phase flow. This qualifier can be used to calculate the drawdown.
4) Chose the right formula: there are three types: with skin, skin pressure drop, or adjusted k. The GB has all three on the same page.

In this example, q is requested. I'll use typical values: skin (+1) and drawdown (1,000 psi), k (50 md), h (45 ft), and rw (0.5).



Q is easily found:
1,000 = 141.2q(1.4)0.5(1/50)(1/45) [ln(0.472*(1,000/0.5)) + 1]
q = 2,900 STB/D

Metering: 2005 #60 (similar)

Gas metering? Lots of word problem potential.
Guidebook notes on metering are on 8 FAC 3-4:

1) Flange taps are 1" on either side of the plate.
2) Pipe taps location to plate: 2.5 and 8 pipe diameters upstream & downstream
3) 27.67 inches of H20 is 1 psi
4) Gas orifice plates are installed with beveled edge upstream downstream.

ESP Cost: 2005 #58 (similar)

ESP economic analysis: 7" vs 5" casing & pumps to 3,400 ft.
7" vs 5": casing $14 vs $12/ft & ESP $41 vs $28/stg (motors $25/hp).
TDH of 1,860' & 3,000 bbls of H20 required.

1) Given ft/stage (or on provided chart): 5" is 16'/stage & 7" is 44'/stage
2) 5" & 7" Stages: 1860'/16'/stage = 117 stages & 1860'/44'/stage = 43 stages (round up)
3) 5" & 7" Costs:
a) Casing: 3400'($12/ft) & 3400'($14/ft) = $41M & $48M.
b) Pump: 117 stages($28/stage) & 43($41/stage) = $3M & 2M (use curves as needed).
c) Motor: 117($0.58 BHP/stage)$25/HP & 43($1.4/stage) BHP/stage)$25/HP = $2M & $2M (use curves as needed).

Total 5" or 7" Costs: $41+$3+$2 or $48+$2+$2 = $46M or $52M.

Fishing Risk: 2005 #76 (similar)

This problem is just like the Guidebook, you merely enter the numbers as given into the equation; $400M sidetrack, $100M cost of fish,  $50 daily cost of rig and fishing. Then apply the even-money risk factor of 50%. Note these problems can get a lot more complex. Don't let that worry you, just follow the logic and keep the basic equation intact.

[($400M+$100M)/($40M+$10M)]0.5 = $500M/$50M[0.5] =  5 days.

Frac: 2005 #59 (similar)

Fracture treatments are simple hydraulics, but there are a million ways to mess it up. The Guidebook has a single frac page, 7 PRD 2, in an attempt to keep it simple.

Getting a frac gradient from instant or initial shut-in pressure (ISIP) is a common calculation:



In the GB example above, ISIP is 1,800 psi, depth is 10,000 ft, and MW = 10 ppg. Hydrostatic pressure becomes 5,200 psi, and bottom hole frac pressure = 1,800 + 5,200 = 7,000 psi.

Frac gradient: 7,000 psi / 10,000 ft = 0.7 psi/ft

Frac problems are word problem heaven for test writers. Make sure you understand what's happening with both SI and dynamic pressures, and read the "hints" section in the Guidebook.

MBE: 2005 #51 (similar)

An example of a saturated, below BP (gas cap) MBE problem typically requires Rp and m.

Also usual is to be given a reservoir oil and gas volume to calculate m:

Rp = Gp/Np = 468 MMCF / 260 M bbl = 1.8
m = Vgas/Voil = 3,000 ac-ft / 10,000 ac-ft = 0.3

Next assume the following reservoir properties.
Bgi & Bg = 1.2 & 1.4 bbl/STB
Boi & Bo = 1.45 & 1.4 bbl/STB
Rsi & Rs = 900 & 800 scf/STB

The single page required in the Guidebook is 13 RES 3.
It's critical to list out your known values WITH PROPER UNITS.
On the exam, look at the variable list in the sidebar and convert to those units.
---------------------------------------------------

----------------------------------------------------- 
Bt = 1.4 bbl/STB + (900-800 scf/STB)0.0014 bbl/scf = 1.4 + 0.14 = 1.54 bbl/STB
Bti = 1.45 bbl/STB (note Bti = Boi)

N = 260,000 STB [1.54 bbl/STB+((1.8-0.9 MCF/STB)1.4 bbl/MCF)] = 730M bbl
              ...divided by...
1.54 bbl/STB - 1.45 bbl/STB + 0.3(1.45 bbl/STB)[(1.4 - 1.2)/1.2)] = 0.16 bbl/STB
N =  4.5 MMSTB

Consider skipping detailed problems like this. Dozens of ways to go wrong.
The difficulty? To figure this out in 30 seconds before investing too much time!

Dry Gas Reservoir: 2005 #49 (similar)

The gas MBE is very simple; see 13 RES 2:
 ----------------------------------------------------------------------------

-----------------------------------------------------------------------------
You are usually asked to solve for water influx, since this is the only term you don't know from production volumes. For example, given:

G = 10,000 MMCF, Bgi 1 M RB/MMCF
Gp 1,000 MMCF, Bg 1.3 M RB/MMCF
2,200 MSTB water, Bw = 1 RB/STB

G(Bg-Bgi) = GpBg + WpBw - We

10,000 MMCF(1.3 - 1 M RB/MMCF) =
1,000 MMCF(1.3 M RB/MMCF) + 2,200 M STB(1 M RB/M STB) - We

3,000 M RB = 1,300 M RB + 2,200 M RB - We
3,000 M RB = 3,500 M RB - We

We = 500 M RB

With gas it's all about the units.

Wet Gas: 2005 #48 (similar)

Equations for wet gas are found on 13 RES 9.

Q1) Given API or SG of oil, what is the molecular weight?

A1) Given API, MW = 5954/(API - 8.8).

Q2) Given the GOR and SG of gas and oil, along with the Mol wgt (say 50,000 scf/STB, 0.7, 0.83, and 197), what is the SG of the fluid mix back in the reservoir?

Q3) What is the "Gaseous Equivalent" of a stock-tank liquid?
A3) 133,300(SG_oil / MW_oil)

The numbers for Q2 above are easy to play with. If your GOR was 15,000, SG 0.67 & 0.76, and a standard mol wgt, you calculate a SG of the reservoir wet gas in the mid 0.8's. This sort of problem is easy if you check your units. And don't panic when seeing terms like condensate, wet gas, etc. 

MBE: 2005 #45 (similar)

The Guidebook uses Slider's MBE format (13 RES 3) for speed and simplicity.

The SPE Textbook Series #8 reservoir book (Towler) uses different MBE terminology. No matter what you use personally, be familiar with Expansion (Et) and Withdrawal (Fo) terms (included in the GB on 13 RES 3 below):


These terms are often used for MBE line plots. There are lots of possible plot types.
For example, F/Et vs We/Et creates a straight line. Note Et is in both denominators; therefore Et measurement error would still produce a line.

Not so regarding We (influx) or F (withdrawal). When these are in error, the line would curve. If We is low, the line turns up; if We is high, it turns down (follow the math). It's opposite for withdrawal (low, the line turns down, high, it turns up).

Most questions, however, revolve around the reservoir (e.g. influx) since we can't measure it directly.

MBE: 2005 #44 (similar)

Know your material balance equation. In the Guidebook, it's 13 RES 3.



At pressures above the bubble point, pore volume drive is a dominate denominator term. Compressibility is important at pressures > BP, for both liquid and formation.

Water drive, however, is unaffected by the BP, and Rp (produced GOR cumulative) changes very slowly over the life of the field.

MBE: 2005 #43 (similar)

Know your material balance equation. In the Guidebook, this is 13 RES 3.

Note We is negative, so with influx N calculates to be less.
Forget to include influx? Calculated MBE N will be higher than it actually is in real life.

PVT STB vs bbl: 2005 #41 (similar)

Stock Tank Barrels STB represent fluid at standard conditions with vapor released. The oil "shrinks" when gas leaves. Hence the term "shrinkage factor". Shrinkage factors are often used when the fluid is not fully at STB. Like at a separator. Units are typically STB/bbl.

So if given "oil shrinkage at the separator", 0.9 STB/sep bbl could be a typical number. Considering "total" oil shrinkage (reservoir to surface) 0.7 STB/res bbl is more likely. Note this latter number is just the formation volume factor inverted; a typical value would be 1.4 res bbl/STB.

GOR (gas/oil) can be measured at any location, and thus use any units. For example, separator GOR could be sep SCF / sep bbl, while sales GOR could use SCF/STB.

Given a separator GOR of 700 sep SCF/STB (a typical value where I'm from) and also a separator shrinkage factor of 0.9 STB/sep bbl (typical), dimensional analysis gives 630 sep SCF/sep bbl. Again, these are common numbers/units I've seen.

Use your units to get whatever answer you need; just know what your units mean.

Separator PVT: 2005 #40 (similar)

Separator PVT problems are found on 9 PVT 8. Understand:
1) Flash & Differential liberation in regard to reservoir bubble point (BP).
2) Formation Volume Factor (FVF) & Shrinkage Factor (SF).
3) Gas/Oil Ratio (GOR).

A common question: from known OOIP what will you get at your separator?
If the reservoir is >BP, you need Flash data only (if <BP, you will need both flash & differential).

For the typical example, say your Relative Volume was 0.99 and flash Bo was 1.5. These are typical numbers (rounded for simplicity, of course). Bo would calculate to 0.99(1.5) =  1.49 res bbl/STB.

If separator shrinkage is given at say 0.9 STB/sep bbl, it's easy to find res bbl/sep bbl:
1.49 res bbl/STB (0.9 STB/sep bbl) = 1.33 res bbl/sep bbl

Last, if given OOIP (say 2, 2.25, or 2.5 MM res bbl) you can get sep BBL (say 1.5, 1.7, or 1.9 MM).

Separator problems intimidate but they are almost always all bark and no bite.

Waterflood Volume: 2005 #38 (similar)

Waterflood has a whole chapter (14 WFL) in the Guidebook. There are a lot of "special" equations.

Yet many (even most) "waterflood" questions are just simple reservoir volumetrics, so you never need to leave the 13 RES chapter. But know the basic volume equations well, such as N = 7,758 Vb phi (1-Swi)/Boi. Note also that 1 - Swi can include 1 - Swi - Sgr - Sor.

A standard problem uses initial water saturation. More complex problems include residual gas and residual oil. For an example:

Vb = 1,400 acre-ft
Porosity = 20%
Swi = 25%, Sgr = 10%, & Sor = 30%
Bo = 1.3 RB/STB, Bw = 1 RB/STB

These are all typical values. Then calculate the max oil that can be removed from the reservoir, by waterflood or otherwise:

[7,758(1,400)0.2(1 - 0.25 - 0.1 - 0.3)]/1.3 = 580 MSTB

Gas Reservoir Volumetrics: 2005 #36 (similar)

Gas reservoir volumetrics are on 13 RES 1 and 2. The basic equation is:

G(Bg - Bgi) = WpBw + GpBg - We

Looks simple enough. But the devil is in the units. I like to stick with MCF.
For example, assume a gas reservoir with IGIP and all production known.
Reservoir water influx is easily calculated (I'll use round numbers):

G = 10E6 MCF; Gp = 6E6 MCF
Bg = 5 rcf/MCF; Bgi = 4.5 rcf/MCF
Wp = 2 MSTB = 11 MCF; Bw = 1 rcf/CF

G(Bg - Bgi) = GpBg + WpBw - We
10E6 MCF (5 - 4.5 cf/MCF) = 6E6 MCF (5 cf/MCF) + 11 MCF (1 Mcf/MCF) - We
5E6 cf = 30E6 cf + 0.1E6 cf - We
We = 25 MMcf = 4.5 MM Bbl water.

With gas it's all about the units.