Problem 31. 10 ppg, 20 cp fluid pumped through 697 ft, 3.5-inch OD tubing rising 12 ft...100 ft...down 1.5 ft...203 ft down 10.5 ft into a 10x10x10 ft open tank filling 1 ft/min. Pump pressure (psig)? (A) 800; (B) 810; (C) 820; (D) 830.
This is a basic Mechanical Energy Balance (8 FAC 6) problem. It's simple if you don't forget the kinetic energy; of course, the wrong answer is conveniently waiting for you as a choice if you do forget it (in this case, "B"):
1. HS vertical: 12 – 1.5 – 10.5 = 0 vertical ft
2. Friction: 697 + 100 + 203 = 1,000 ft
3. 34.14 ft/s (2x GB 3 HYD 1) = 0.809 psi/ft(1000 ft) = - 809 psi
4. KE: 8.074E-4(10)(34.14^2-0^2) = - 9.4 psi
5. Total: 818 psi.
How did you get velocity to be 34.4 ft/s?
ReplyDeleteFrom 1 ft/min? A
Please explain more on step 3. Are you using Friction pressure loss in Turbuinet flow?
See 3 HYD 1, second equation; it uses the same numbers just 2X. Make sense?
DeleteAlmost. The ID used there is 2.992. Is that just common for 3.5" tubing? There are a lot of IDs based on the weight of the pipe...
DeleteV, that was my evil way of making you try the high & low 3.5" pipe weights. Options for 3.5" range from 8.5 to 15.5 ppf. Which gives a range of 32.6 to 45.1 fps then 8.6 to 16.4 psi. Added to 809, this ranges from 818 to 825 psi. Options are 800/810/820/830. Every pipe wgt but one gives you 820 (C) so that's the answer.
DeleteI doubt the real exam is this evil. But my goal is to get you puzzled during the problem and getting used to "working through it" fast and not being overconfident.
To solve this one, I would have grabbed the high & low, run the calc, seen 90% of the options give 820, and rolled with it. Do enough of these types of problems you will know the game like the back of your hand on the real exam and eat the competition alive. At least, that's my goal :-).
m I'm stuck on rate. How did you know it's 100 ft3/min.. is that 10x10x1 ft/min?
ReplyDelete1. 10x10x10 ft tank filling 1 ft/min.
Delete2. The tank area is 100 ft2.
3. The fluid height rising at 1 ft/min.
4. Thus the volume is 100 ft^2*1 ft/min = 100 ft^3/min.
Does this check out?
I find this sort of "misdirection" the hardest part of timed exams. The only solution to it I've found is to do LOTS of well-written practice problems in order to 1) get used to figuring odd info fast, and 2) eventually see every type of misdirection pre-exam. There really isn't that many ways to get clever and eventually you will see most of them if you do enough problems...
DeleteOn this problem, why is the left side of the mechanical energy balance (p2 - p1) assumed to be 0?
ReplyDeleteThe tank is 0 gage pressure, the pump pressure is calculated solution. So it's p2 - p1 = 808 (pump) - 0 (tank). Check?
Deletemake that 818 not 808.
DeleteHow do you know the tank is 0 gauge. Unless specified on the test, should we assume that the tank is open to the atmosphere (making it 0 gauge)?
DeleteI'm sure they would say open tank; I'll change it.
DeleteHi m, I wonder why 34.14 ft/s becomes 0.809 psi/ft?
ReplyDeleteDavid, I wondered the same question. I lost the train of thought from 34.14 ft/s to .809 psi/ft
DeleteHe got it from the pressure loss equation
DeleteCould you explain the KE portion a little more?
ReplyDeleteI was able to get the Friction portion (0.81 psi/ft x 1000ft), got the Velocity piece to the KE portion 34.14 ft/s
I don't know much about the 8.074constant... not sure if I'm having a units issue?
8.074e^-4 x (MW) x (V^2) --- 8.074e^-4 x (10) x (34.14^2) --- I'm not even close to 9 lbs (1723???)
Do you have the Guidebook? Or TS2? These have the best explanation IMO. Just remember: testers LOVE to test KE, in school or any exam, because it's so odd people miss it. Just look carefully at the units in TS2 or the GB...let me know if you don't have these and we can go into more detail here but make absolutely sure you understand this problem. Remember the KE is a very small percentage so often the A-B-C answers will be close but just enough to catch you if you forget the KE like so many people do.
DeleteBy the way, thank you for putting all of this together. I plan to work all of these practice exams. This is a great tool to refresh and learn new material I'm not exposed to on a regular basis. Thank you
DeleteI have the GuideBook 2018 #3, Facilities section (8FAC6) -- Mechanical Energy Balance
ReplyDelete8.074e^-4 x (10 lb/gal) x (34.14 ft/s)^2 ---> constant x MW x Velocity^2
When you are referring to TS2 are you saying this is in the Applied Drilling Book (Red)?
Yep, TS stands for SPE "Textbook Series". If you have the GB doesn't the GB calc check out? I'm currently away from my home so I can't look anything up...
DeleteHow would you solve this using the latest Reference Guide.
ReplyDeleteI show this on #15 on the 2021 practice test.
DeleteI can get the 34ft/sec. I don't understand the friction part. You have 697 + 100 + 203 = 1,000 ft. How is that a friction factor? Do you use the Moody Chart?
ReplyDelete