Pumping: 2021 #29

A casing string is to be cemented at 10,000 ft in a well containing 10-lbm/gal mud, which will be displaced from the annulus by 500 ft of 8 lbm/gal mud flush, 1,500 ft of 12.5 lbm/gal filler cement, and 1,000 ft of 16.5 lbm/gal cement. The pump pressure (Mpsig) required to completely displace the cement from the casing with 9-lbm/gal brine is closest to: A) 1; B) 1.1; C) 1.2; D) 1.3.

Solution:
10000-500-1500-1000 = 7000 ft 9-lbm/gal.
0.052*((500*8)+(1500*12.5)+(1000*16.5)+(10*7000))=5681psig.
10,000*9=4680 psig.
5946 – 4680=1001 psig, “A”.
(TS12 P186; GB 3 HYD 2; RG P41, P67).

Cement Displacment: 2016 #2

Given: 10,000' casing in 10.7 ppg mud.
Pump into casing, up annulus:
100' 8.5 ppg mud = 44 psi hydrostatic.
1,550' 12.9 ppg cement = 1,040 psi hydrostatic.
1,500' 15.4 ppg cement = 1,201 psi hydrostatic.
9.1 ppg brine until cement is displaced.
Pump pressure required to displace?

Annulus (remaining):
6,850' 10.7 ppg mud = 3,811 psi hydrostatic.
Total hydrostatic pressure in annulus: 6,096 psi.

Hydrostatic pressure in casing: 9.1 ppg brine = 4,732 psi

Pump pressure required to displace?
6,096 psi - 4,732 psi = 1,364 psi (A).

This is a simple displacement problem. It's common in industry and textbooks both (good examples are in TS2, TS12, Mian, etc...and if you have a chance to take Wines' class, Bing has the best examples of this problem type I've seen). I referenced this problem using TS12 but as mentioned there are a dozen just like it. All you need is the basic hydrostatic equation (0.052(D)MW) and to understand how cement is generally pumped:

Weight Up: 2018 #64

Problem 64: Drill 10.1M to 10.2M ft; pressure 5M psig. At 12M ft ROP increases. Surface casing 5M ft; LOT calculated fracture MW 14.5 ppg. Standpipe pressure is 650 psi. Using a 200-psi trip margin the barite (sacks) needed to weight up the 800 bbl mud system from 9.8 ppg to the KMW? 

Res pressure: Pr = 9.8(0.052)12,000 ft + 650 psi = 6,765 psi.
KWM: = (6765+200)/(0.052*12,000) = 11.16 lbm/gal.
(35 – W1)/(35 – W2) = (11.16 – 9.8)/(35 – 11.16) = 1.057 bbl bar/bbl mud.
Volume = 1.057(800) = 845.6 bbl
Sacks: 845.6 bbl (1/sk/100 lbm)(35 lbm/gal)(42 gal/bbl) = 671 sacks (C).

Note this problem has the wrong answer in the answer key, as does the original SPE source for the problem; I should not have been so trusting.... 

UPDATE: The 2019 Update SPE Petroleum Engineering Reference Guide (provided during the exam) shows the following:

 

I've shown traditional way of solving above, but using the provided equation gets the same thing: 

1470*((11.16-9.8)/(35-11.16)) = 83.9*8 = 671 sacks.

 Note I would read TS12 on P105 to get familiar with this sort of problem. One of the issues with the reference guide is it gives only a few specific equation options for calculating and doesn't show how the equation is a function of Wf/Wi...and the exam most certainly could expect you to understand how this all works for any situation: (1. Weight up add V. 2. Weight up same V. 3. Dilute & dump). I'll include a few types of these problems using the Reference Guide on the 2021 problem set. 

Source: GB 1 DRL 1, 4 MUD 1; A Guide to Prof. Reg. for PEs, SPE, 1991.