Liquid line sizing & pumping downhill
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- Опубликовано: 4 июн 2024
- I don't think I was ever shown how to size a pipeline from nothing but a problem statement. If you haven't either - here it is. We also look at why it would be necessary to use a pump, even though our pipeline runs downhill.
The calculations I do are free to access here:
docs.google.com/spreadsheets/...
If you spot anything I've messed up, let me know!
00:00 Designing from scratch
01:16 What are we pumping?
04:04 Nominal pipe sizes
05:23 Velocity
06:24 Reynolds number, friction factor & pressure drop
10:49 System curves
16:35 A note on compressible flow
Process with Pat is the place to come for perspective and to ask stupid questions. I want you to leave more knowledgeable, confident, motivated, and most importantly, curious. I also want to invigorate a field that seems tired and uninspiring, at least if you get your perspective from internet forums. These are not lectures. This is a place for you to leave thinking “Oh! That’s why...”
This channel is not only for chemical engineers - anyone who works with processes should be able to find something of value here.
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Great stuff. For long pipelines, this is great. But in systems within a building, the pressure drop in the length of the pipe is trivial compared to all the other 'minor losses' (elbows, valves, fittings...). Perhaps a video discussing these factors?
A really good video. Thanks for sharing and keep doing more videos
To date I am yet to find a 125NB / 5.0" pipe. Thank you for an awesome reminder of why the rules of thumb guide you to proper design boundaries
I've met DN125 pipe during my career, but I'm always surprised. It's when they used it in the past and we have to connect to it during a modification.
The reason it's not common is that its price is almost the same as DN150, so the go with DN150.
(at least here in C-E Europe)
Excellent video, very clear explanation and very helpful! Thank you from a fellow South African :)
Glad you think so! All the best…
Beautifully explained.
Please do explain steam with a simple example.
You are a great teacher.
Thank you very much for your efforts and teaching.
That’s very kind of you to say. I talk about steam and do some simple demos in my Thermodynamics video, but I can appreciate that that one is a little long to be able to catch.
Great video Pat, very informative.
Can I ask if you have done a video on two pumps in parallel and the effect of pressure and flow rate.....
Thanks in advance
Great video 👍🏻
You wanted a bit longer ;)
The equation that you used on the video to calculate the friction factor is very different from the one in Churchill's article. Can you explain the transformation of the original equation into the one you used?
Hi Pat, I just discovered your channel so, you might have covered this topic, but here it is:
So you mentioned compressible flow, and I have a problem involving that.
I have high pressure steam source at 30 bar and a 30 m long DN250 pipeline.
The other end of the line is 20 bar. If I want to determine the max flow rate through the line, I generally use similar equations as in this video, but for compressible flow and take into account the lower density and the 1,5 times big velocity. OK.
But what happens, when I blow to the atmosphere? The pressure difference is greater so the mass flow has to be at least as great as in the fist place, right?
But due to the 30 times more velocity, at the end of the pipe, the gas goes with something like 4000 m/s. Not possible.
If it was a valve, then chocking would occur, and the flow rate could be calculated from Kv. The max velocity is the speed of sound. But it is a long pipe not a sudden contraction.
So what happens here? There is a pressure drop of ~10 bar at the exit point? Or the last meter determines the flow with the max velocity - meaning the flow is smaller than in the 20 bar case? Or 4000 m/s is correct, since it is just a couple of mm-s at the end? How would you solve this?
I would expect optimization curve may be 2 parellel 4 inch pipe will be cheaper
Hi Pat! Looking again your video two fundamental questions came to my mind and I would like to see your reply.
1) A pump creates pressure or flow?
2) When the speed of a pump increases why the flowrate of the pump increases?
The pressure vs. flow question is one that comes up often, and I am going to look into it properly. I just need to come up with a good way to address it.
On increasing the speed of a pump - well for a positive displacement pump it should be obvious - you are displacing more fluid per unit time. But I guess you mean centrifugal. Centrifugal pumps impart energy onto a fluid - so increasing the speed increases the rate at which you impart energy. That energy is then translated into an increased velocity/flow.
Pumps create flow. Packing that flow into a confined space, the pipe, creates pressure. Other than that, the distinction is relatively immaterial.
My advice is Do not worry about such things. They always occur together, so there is no advantage to trying to separate them and there are more important things to worry about.
I am having a doubt pumping from 15m to ground level let's say 0 m ,so elevation difference you have taken 0-15=-15m
If I am pumping from 5m to 10 m elevation means =10-5= 5m right?
Yes. Positive number means you are pumping uphill.
Hi Pat! As always I love your videos and I have a small comment.
The expression of ΔPsystem that you have in excel includes three terms (friction+elevation+pressure at the lowest point).
If I apply the Bernoulli equation and let's say point 1 is the start of downhill and point 2 the end of downhill (where the pressure is 1 bar).
By applying bernoulli there P1+ρgh1+(1/2)ρ(u1)^2=P2+ρgh2+(1/2)ρ(u2)^2+htotal,
where u1=u2, you can eliminate them and then you have:
P1+ρgh1=P2+ρgh2+htotal
where htotal=2fρu^2L/D
h2=0
Why do you consider as ΔPsystem=P2-ρgh1+2fρu^2L/D? Instead I would say that P1=P2-ρgh1+2fρu^2L/D.
What is your opinion about it?
Hey Panos! I would say that ΔP_system is the amount of pressure that needs to be overcome by the pump (which is why if it is negative it is “free” flow that doesn’t need to come from a pump).
I would say that the full expression is:
ΔP_system = (P2-P1) + ρg(h2-h1) + 0.5ρ(v2² - v1²) + ΔP_friction.
I agree the velocity terms are negligible. So if P2 is greater than P1 and h2 is greater than h1, this requires more pressure from my pump. Again, in my case h2 is less than h1 which is some “free” (no pump needed) flow. But P2 is higher than P1 (because P1 is zero/atmospheric), but P1 could be higher if the tank was pressurised, which would also require less from my pump. That is why I disagree with your definition - because you are saying P1 is a function of the other terms, but P1 could have its own value if the tank/supply pressure was some other value that actually helps the flow along, requiring a smaller pump.
One note: my h2 is NOT zero. I define h1 as zero, and h2 as -15 m.
Does that makes sense?
@@ProcesswithPat hi Pat, I had missed your reply.
Actually with your description almost everything is crystal clear.
The only unclear thing is why P1=0 bar?
@@ProcesswithPat I'm looking forward to your reply! ☺️
Oh crap! I totally forgot about this one - sorry for not replying!
I think I just understood the confusion in our conversation. You are considering P1 to be the discharge pressure of the pump, in which case I understand. In my head, P1 is the suction of the pump. So increasing P1 helps the pumps (lower system resistance, in a way).
But if you consider P1 to be the discharge pressure of the pump then I think your original equation (in your first comment) is correct.
Actually, I was considering the P1 as suction pressure. But, I don't get why you consider it as zero!?!
Hii..pat can u find the chocolate pipe size
Flow - 2000 kg/hr
Temperature - 40 °c
Velocity - as u consider
Please find ..
I charge by the hour ;)
Please share the excel
If I use haaland equation my friction factor is a lot different about .026 and if I use Colebrook I get .026027. Am I doing something different or incorrect
Am guessing that it is because both those equations calculate the Darcy friction factor which is 4 times greater than the Fanning friction factor which is what I use. I mention this at 08:30 in the video. I’m not sure which conditions you’re doing your calculation for, but check out whether dividing by 4 gets you to what I get… Let me know if this was indeed the issue or not.
I have doubt about friction equation used in excel sheet, where you used (1/-4log(XX))^2 where I think it should be (1/-2log(XX))^2
or -0.25/log(XX)^2
en.wikipedia.org/wiki/Darcy_friction_factor_formulae