Temperature Force: 2016 #52

3.5 inch, 9.2 lb/ft tubing: 

X-section area = 2.59 sq in (Baker Packer Handbook, or Redbook w/ OD & ID calc.). On the CBT exam, these dimensions would be given in a text box, or even provided in a tricky way as part of a table where you have to be familiar with tubing.

1) ST = 50 dF; BHT = 150 dF: Average = 100 dF.

2) ST = 75 dF; BHT = 75 dF: Average = 75 dF. 

dT = 75 - 100 = - 25 dF.

Force applied by dT: 207(Ats)dT = 207(2.59)-25 = - 13,403 lb tension up. (A)

Temperature Force: 2005 #21 (similar)

The governing equation for the temperature force on tubing that is latched into a packer is 207ATs(Tf-Ta). It's found on Guidebook page 6 DTC 9.

ATs is merely the area of the x-section; for example 9.60 - 7.01 = 2.59 in^2. Get these numbers from the given problem, or from a packer handbook.

The temperature difference is confusing. Start with the average temperature for the well at start (Ta); easy enough to calculate (surface + bottom hole)/2. In this case, assume 145 deg F.

However, the Tf is confusing. This is the average temperature of the string when cooled to the maximum condition while injecting. It is dependent on the fluid temperature plus the injection rate & duration. Just roll with whatever a problem gives you; if given "final pumping conditions", use that. In this case let's assume 100 deg F.

Working the numbers it's 207(2.59)(100 - 145) = -24.1 Mlbf.
Note the negative sign reflects an "up" force (highlighted in the Guidebook variable box).

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.