Thank you! This helped me a lot. I needed to know this for a case when there was a pressure difference
@professorbehrang
8 ай бұрын
You are welcome.
@rezanoormohammadi6914
5 ай бұрын
It was very interesting, Iranian professor👌
@itsfikree
4 ай бұрын
Wonderful explanation sir 🤩, relieved some pressure 😁
@sharp_guy2310
4 ай бұрын
Good explanation sir. I expect a video from you for same parallel flow with valves of different Cv value installed in 2 pipes. Also valves installed in series too if possible.
@aravindkumar-vn7ox
Жыл бұрын
Thank you Prof. Thats really a nice video. Just curious to know the gadgets you have used for the video. They are really awesome. If its sharable, what kind of hardware and software you are using for this?
@professorbehrang
Жыл бұрын
Thanks for watching. I used Microsoft oneNote with a Microsoft surface for this video.
@professorbehrang
Жыл бұрын
For my newest videos I have been using Microsoft Surface Studio
@claudiosanchez4479
3 күн бұрын
Nice video professor! Thank you! I just noticed that you used the Darcy-Weisbach for the calculation of head loss of pipes. I would just like to ask, would this still be applicable if I would use the Hazen-Williams equation for head loss?
@professorbehrang
2 күн бұрын
@@claudiosanchez4479 the head loss in both branches is the same. Whatever valid equation you have you can use. I think the equation you mentioned is an empirical equation used for water systems.
@claudiosanchez4479
2 күн бұрын
@@professorbehrang I see. Thank you very much professor! :)
@JamesRogerLackore
11 ай бұрын
I have the same question as one of the other commenters. To calculate J, you need to know both FA and FB. You glossed over determining this in the video. Unless I am mistaken determining these values require you to know the velocity in each branch. But we don't know the velocity in each branch because we don't yet know the flow in each branch. Seems like a chicken and egg problem. Can you explain what I am missing?
@professorbehrang
11 ай бұрын
No, You are not missing anything. We need to know friction factors to be able to solve this problem. However, there are 2 ways to solve the problem if you don't know the friction factors. Case 1: sometimes you can guess the values of friction factors correctly. For example, if the pipes are really rough and the flow is fully turbulent f values on the moody chart become independent of Reynolds number and can be picked from the Moody chart. Case 2: You will need to add 2 equations for friction factor (Colebrook equation) to your set of equations in a code or something and have it iterate until it converges.
@onsite2406
12 күн бұрын
@@professorbehrang I'm having this same issue, where Qtot = 20 gpm, branches A and B are the same diameter (1") and material, but different lengths. At such a low flow rate, can Fa and Fb be assumed to be equivalent?
@professorbehrang
7 күн бұрын
@@onsite2406 f_a and f_b are not necessarily the same. I responded in my last comment. There are 2 cases.
@tipsandlearnings9421
Жыл бұрын
Well done 👍 Sir
@professorbehrang
Жыл бұрын
Thanks
@ming91947
4 ай бұрын
hello prof! nice video. can you help me understand again in a case where the 2 pipe in path A and B are made of different material and size will have the same head loss again?
@professorbehrang
4 ай бұрын
As long as the process is steady state and then both rejoin again yes!
@MathHandsOnwithPython
Жыл бұрын
Thanks for the nice video Prof. Please prepare a video for the velocity of fluid flow in a curved tube
@professorbehrang
Жыл бұрын
Even if the tubes are curved, it won’t change anything. The flowrate distribution will be exactly the same as this one.
@robertamarinei170
Жыл бұрын
Thanks for the nice explanation! One thing I do not understand is why the pressures (P2) in the two branches are equal when you write the bernoulli equation for the two branches?
@professorbehrang
Жыл бұрын
P2 is the same because there’s only one point that has the pressure of P2.
@TP-iv3os
9 ай бұрын
Professor Behrang, I have a 24" line that I need to measure flow. The pipe is liquid full, the pressure varies but is measured and always known, temperature is pretty constant but is also measured and known. The current method being used to measure the flow is an ultrasonic meter which is not working properly due to the fluid causing internal scaling. The scaling has a negative effect on the meter. Taking the line out of service to install an inline meter such as a mag meter or an orifice plate is not an option. I am contemplating hot tapping the pipe so I can install a small 2" lined pipe with a Coriolis meter. My question is can I simply tap and install a 2” line in parallel to the 24” line, and measure the flow in the 2” and be able to determine the total flow in the 24” line?DeDear Professor Behrang, I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it. Can I measure the flow in the 2" line and determine the total flow in the 24" line? Thank you.ar Professor Behrang, I have a liquid-filled 24" pipe that requires flow measurement. The pressure fluctuates but is always known, while the temperature remains relatively constant and is also measured and known. Currently, we are using an ultrasonic meter, but it is not functioning correctly due to internal scaling caused by the fluid, which has a negative impact on the meter. Unfortunately, we cannot shut down the line to install an inline meter such as a mag meter or an orifice plate. Therefore, I am considering hot tapping the pipe to install a small 2" lined pipe with a Coriolis meter. My question is whether I can tap the 24" line and install a 2" line in parallel to it, then measure the flow in the 2" line and determine the total flow in the 24" line? Thank you.
@TP-iv3os
9 ай бұрын
If I measure the flow in the smaller line (Qa) then I should be able to determine the total flow by using the following equation Qa + Qa/J. WOuld this be the correct understanding?
@professorbehrang
9 ай бұрын
That’s true. But when you add the 2” pipe, the flow in the 24” pipe would not be the same as when the 2” pipe didn’t exist.
@DespoinaAnastasopoulou
Жыл бұрын
Thank you Prof. Could please explain what would happen if the two branches wouldn't be connected and had different lengths?
@professorbehrang
Жыл бұрын
In this problem, they do have different lengths but do you mean they are not connected on both ends? Or they are only connected on one end? Can you elaborate?
@robertamarinei170
Жыл бұрын
@@professorbehrang I have in fact the same curiosity. Say pipes A and B have arbitratry lengths, LA and LB, and they simply have open ends?
@professorbehrang
Жыл бұрын
@@robertamarinei170 Open end would be a different problem. More info on the pressure on the open side is needed.
@wang-ni8tq
9 ай бұрын
would this apply if you have 3 or 4 more pipes in parallel?
@professorbehrang
8 ай бұрын
The principle is no matter how many pipes you have in parallel after the whole reaches becomes steady state, the head loss in every pipe becomes the same.
@blessingmuningwi7353
Жыл бұрын
if i add say a turbine in one of these branches is hLA still equal to hLB?
@professorbehrang
Жыл бұрын
If you add a turbine in one of the branches, let’s say branch A, then: h_L_A + h_turbine = h_L_B.
@wang-ni8tq
9 ай бұрын
would this be the same if you have 3 or 4 parallel pipes?
@professorbehrang
8 ай бұрын
If it is steady state the head loss in every branch becomes the same. But those pipes need to be connected to the same pressure nodes on both ends.
@shahin.kimiya
Жыл бұрын
very well
@tipsandlearnings9421
Жыл бұрын
Also make video in case of more than 2 parallel tubes
@professorbehrang
Жыл бұрын
Good suggestion.
@davemorgan8349
Жыл бұрын
@@professorbehrang I thought that was left as an exercise for the reader ;)
@professorbehrang
Жыл бұрын
@@davemorgan8349 All the head losses will be the same in every branch.
@ninara2317
Жыл бұрын
Thanks doctor! How can I contact with you please?
@professorbehrang
Жыл бұрын
How can I help you?
@karthickarya2419
Жыл бұрын
how to know friction in each pipe without velocity
@professorbehrang
Жыл бұрын
Good question. In some cases you can use a guess value: For example, if we know that the flow will be fully rough and fully turbulent, the friction factor on moody chart levels off and becomes independent of velocity. That value can be picked from the moody chart.
@TP-iv3os
9 ай бұрын
I thought that Q total = Qa + Qa/J. Therefore, knowing Qa, I could calculate Q total.
@professorbehrang
9 ай бұрын
That's true but after the process becomes steady state . Adding another branch will change the total steady state solution if the upstream pumping power stays the same. A new branch after becomes steady state will change the total head loss. In other words, without an added branch you have one flowrate in the pipe. Once you add the branch you won't have the same flowrate in the same pipe. You might think it won't matter because you are looking for the total flowrate. But if the pumping power upstream stays the exact same, adding another branch will also change the total steady state flow rate of the system.
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