PBU Wellbore Storage: 2005 #15 (similar)

Straight out of TS1, wellbore storage during a PBU is a must-know. The Guidebook summary (with units) is on 12 WLT 13. A typical problem: wellbore storage time (say for the below well data, choose the closest estimate: A=0.1 hr, B=0.5 hr, C=1 hr, D=2 hr):

Well data:
Wellbore skin and volume (say skin=2 & 30 bbl).
Reservoir thickness & perm (say 10 ft & 60 md).
Fluid viscosity & comprehensibility (say 1 cp and 0.0002 /psi).

12 WLT 13:
Cs = CwbVwb = 0.0002(30 bbl) = 0.006
t (wbs approx) = [170,000(Cs)(e^(0.14*s))]/[kh/u]
t (wbs approx) = [170,000(0.006)(e^(0.14*2))]/[(60*10)/1] = ~2.2 hr

Note that Cs can be calculated using different inputs; the Guidebook shows each of the three equations on one page for immediate use. Note also some of these equations are approximate; problem wording should reflect this and thus hint as to the proper equation to use.

Crossplots: 2005 #72 (similar)

Crossplots are on Guidebook 15 LOG 5.

X-Plots Advantages:
   Avoids calculating Sw for long logs.
   Can find Sw w/out m, Rw, or a porosity tool.
   Finds m, Rw, Δtma, ρma, rock types.
   Rw, and m can be varied w/out long calcs.

X-Plots Limitations:
   Large porosity range required.
   Cementation exponent m required.
   Matrix properties, & Rw must be constant.

Recognize these plots on-sight (the Guidebook has clear examples).
Hingle: saturation lines run from x-axis northeast
     matrix density is where lines converge to x-axis
Pickett: saturation lines run southeast to northwest
      𝐑𝐭 = Rw on 100% saturation line is when por = 1

An example of a Picket plot set a of data points (GB 15 LOG 5):



1) Find the 100% saturation line (through the most SW points); pick 2 points on line.
In this example, the line will pass through (porosity, Rt) points of (0.04, 5) and (0.1, 1).
2) Calculate the [log] of each point:
     Porosity = 0.04 [-1.4], Rt = 5 [0.7]
     Porosity = 0.1 [-1]; Rt = 1  [0]
3) Interpolate [log] values to Porosity = 1 (all water, no rock, so Rw = Rt).
     Por = 1 [0], Rt = 0.02 [-1.7]. Therefore, Rt = Rw = 0.02 ohm-m.

Given log-log paper? Plot & extend the 100% saturation line to porosity = 1 for Rw. It's easier.

Natural Gas: 2005 #61 (similar)

The Guidebook doesn't include complicated graphs; it uses tables for speed. There are several graphs that were too complex to convert into table format that you should bring to the exam.

One (naturally) is a z-chart. I use the one in TS8 (or TS1). Another is the equilibrium water vapor content for natural gas (I use Fig 5.7 in HS3-199). You should own and bring (at a minimum) TS1 and HS3, so just use those charts. Practice using them until you are fast.

A typical natural gas problem? Water vapor (lbs) to be removed from a saturated gas (MMscf).

Just go to the chart at any temperature and pressure to get saturated lbs water/ MM gas (at say 100 deg F & 1,000 psig you get about 60 lbm water). If given the minimum water allowed for sales (say 10 lbs/MM) you must thus remove 50 lbm for each 1 MM sold (for my example). Simple.

Always glance over your answer choices before finding "exact" chart numbers. Work fast and estimate. Remember, it's all about speed. This is why tables are much faster; just scan down the table and make estimates between numbers.

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!

Archie: 2005 #66 (similar)

Given the following, what is oil saturation?

Rw = 0.02, Rt = 30 ohm-meter
a = 0.6, m & n = 2, Porosity = 8%.

Classic Archie. Guidebook 15 LOG 1:
Sw = [(a/Por^m)(Rw/Rt)]^0.5
Sw= [(0.6/0.08^2)(0.2/30)]^0.5 =  0.25
Oil Saturation: 1 - 0.25 = 0.75 = 75%

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.