The dependence on diameter is non-dimensional, because that part of the chart speaks in terms of relative roughness, which is roughness divided by diameter. So as long as roughness has the same units as diameter, you are good to go.
Laminar flow problems are a lot easier to solve, because the relationship between flow rate and frictional losses can be solved in closed-form from the first principles of viscosity. The relationship between Reynolds' number and friction factor is therefore much simpler. In fact, if you see the Moody diagram, there is a straight line on the double log scale for laminar flow, which implies a simple power function. In the case of circular pipes & tubes, f = 64/Re for laminar flow with Re
It looks like you are using 80 degree temperature for a 60 degree temperature table as well. This video still shows you how to use the table but the reference book seems to only give the 60 degree table.
A pump lifts water from a large tank at a rate of 30L/s . If the input power is 10kN and the is operating at an efficiency of 40% . find the maximum height to which it can raise water if the delivery pipe is vertical, with diameter 100mm and friction factor 0.015. ( please help)
Hello Sean, under each of my videos, I have links to books that I recommend for the FE exam. For FE civil, I usually recommend the FE review manual and FE civil practice problems by Linderburg. I hope this helps and thank you for watching. Please share my channel with your friends who might find it helpful and good luck with your studying!
Hello Frankie, so if it is more than 4000, which what books specify, (on the reference manual, they specify greater than 10,000), the flow is turbulent and so you need to get the friction factor from the moody chart. If it is less than 2000, then the flow is laminar and the friction factor is complicated but not retrieved from the moody chart. I know this might be confusing and that is why I will make a video on this to clarify any confusions. Thank you for watching and good luck with your studying!
Framkiee, I hope you are having nice weekend, just power though, you are almost there and don't forget to rest a little bit tomorrow, your brain will thank you for it later. Good luck on your exam Monday!
@@frankieeog1625 There is no theoretical basis for most of the Moody chart, except the region of laminar flow, where friction factor is a simple equation of 64/Re. The entire domain of turbulent flow is all curve-fitting to experimental results. That's why you get such wild equations that aren't user-friendly for most of the chart, and a simple reciprocal function for laminar flow.
Can u explain the gravity in the very end, I did not get it, I think we need to multiple the 9,81with 10,76 to convert to (square feet ) , so why was it 32,2?
"Head" is a topic that is defined to equal the form of energy per unit mass in the Bernoulli equation, divided by g. Even forms of energy that have nothing to do with gravity, get divided by g, by convention. The only form of energy that doesn't divide by g to get head, is gravitational potential energy, i.e. the only form of energy that has anything to do with g. In this problem, solving it without a "head" form of the Bernoulli equation, we set up the following: P1/rho + v1^2/2 + g*z1 = P2/rho + v2^2/2 + g*z2 Let z2 = 0, both sides are exposed to the atmosphere, so P1 and P2 are both zero. v1 is negligibly small. This reduces the equation to the following, in an ideal world: g*z1 = v2^2/2 Then we add the frictional losses term to the right hand side: g*z1 = v2^2/2 + 1/2*L/d*f*v2^2 And solve for v2: v2 = sqrt(2*g*z1/(1 + f*L/d)) This shows where dependence on gravity really is, if you were solving it with consistent SI units.
Hello Syed, thank you for watching, I actually talk about that in this video, ruclips.net/video/XdINNQLVwCk/видео.html. I hope this helps and good luck with your studying!
There is a minor head loss as the water converges into the pipe. That needs to be added, but it could be that it is much smaller than the friction loss. I'd want my students to at least show that.
It is exposed to the atmosphere. Since both sides are exposed to the atmosphere, and this equation only cares about difference in pressure, you can simply work in gauge pressure to solve it, and both pressures are therefore zero.
As Carl points out, they are both zero psig and 14.7 psia. You don't even need to know this - only that they are both open to the atmosphere so they cancel out.
Wait, I think the diameter is supposed to be in inches for the moody chart?
The dependence on diameter is non-dimensional, because that part of the chart speaks in terms of relative roughness, which is roughness divided by diameter. So as long as roughness has the same units as diameter, you are good to go.
Your videos are so helpful!
Excellent run down...Toujours!
1:40 So what are we supposed to do if it's laminar flow? You told us what to do if it's turbulent flow.
Laminar flow problems are a lot easier to solve, because the relationship between flow rate and frictional losses can be solved in closed-form from the first principles of viscosity. The relationship between Reynolds' number and friction factor is therefore much simpler. In fact, if you see the Moody diagram, there is a straight line on the double log scale for laminar flow, which implies a simple power function.
In the case of circular pipes & tubes, f = 64/Re for laminar flow with Re
It looks like you are using 80 degree temperature for a 60 degree temperature table as well. This video still shows you how to use the table but the reference book seems to only give the 60 degree table.
She's my hope for passing my FE. I'll imagine her voice while taking it.
When you calculated roughness coefficient, why did you convert your diameter to feet? Doesn't the chart say diameter should be in inches?
My question as well
@@rowana2177 the surface roughness coefficient, e, is in ft.
If epsilon not given should we take average of the range of values for a particular material from the moody chart
A pump lifts water from a large tank at a rate of 30L/s . If the input power is 10kN and the is operating at an efficiency of 40% . find the maximum height to which it can raise water if the delivery pipe is vertical, with diameter 100mm and friction factor 0.015. ( please help)
what happened to the minor head losses part: entrance from tank to the tube part?
could you technically just find relative roughness to find the friction factor f in this problem?
Genie… did you get the PE?
Are you planning to make any PE videos or know any good resource to study for it?
What book would you recommend with practice problems for someone preparing for FE civil
Hello Sean, under each of my videos, I have links to books that I recommend for the FE exam. For FE civil, I usually recommend the FE review manual and FE civil practice problems by Linderburg. I hope this helps and thank you for watching. Please share my channel with your friends who might find it helpful and good luck with your studying!
why greater than 4000 to use the moody chart? i couldnt find that value in the manual. i really appreciate all of your videos :)
Hello Frankie, so if it is more than 4000, which what books specify, (on the reference manual, they specify greater than 10,000), the flow is turbulent and so you need to get the friction factor from the moody chart. If it is less than 2000, then the flow is laminar and the friction factor is complicated but not retrieved from the moody chart. I know this might be confusing and that is why I will make a video on this to clarify any confusions. Thank you for watching and good luck with your studying!
Framkiee, I hope you are having nice weekend, just power though, you are almost there and don't forget to rest a little bit tomorrow, your brain will thank you for it later. Good luck on your exam Monday!
enGENIEer thank you. I appreciate your comment.will do 💪 thanks again best vibes to you too
@@frankieeog1625 There is no theoretical basis for most of the Moody chart, except the region of laminar flow, where friction factor is a simple equation of 64/Re. The entire domain of turbulent flow is all curve-fitting to experimental results. That's why you get such wild equations that aren't user-friendly for most of the chart, and a simple reciprocal function for laminar flow.
Can u explain the gravity in the very end, I did not get it, I think we need to multiple the 9,81with 10,76 to convert to (square feet ) , so why was it 32,2?
"Head" is a topic that is defined to equal the form of energy per unit mass in the Bernoulli equation, divided by g. Even forms of energy that have nothing to do with gravity, get divided by g, by convention. The only form of energy that doesn't divide by g to get head, is gravitational potential energy, i.e. the only form of energy that has anything to do with g.
In this problem, solving it without a "head" form of the Bernoulli equation, we set up the following:
P1/rho + v1^2/2 + g*z1 = P2/rho + v2^2/2 + g*z2
Let z2 = 0, both sides are exposed to the atmosphere, so P1 and P2 are both zero. v1 is negligibly small. This reduces the equation to the following, in an ideal world:
g*z1 = v2^2/2
Then we add the frictional losses term to the right hand side:
g*z1 = v2^2/2 + 1/2*L/d*f*v2^2
And solve for v2:
v2 = sqrt(2*g*z1/(1 + f*L/d))
This shows where dependence on gravity really is, if you were solving it with consistent SI units.
What if the reynold's number falls under laminar flow? How do you determine the Friction Factor for Laminar Flow?
Hello Syed, thank you for watching, I actually talk about that in this video, ruclips.net/video/XdINNQLVwCk/видео.html. I hope this helps and good luck with your studying!
enGENIEer Thanks.
There is a minor head loss as the water converges into the pipe. That needs to be added, but it could be that it is much smaller than the friction loss. I'd want my students to at least show that.
Thank you
why is the pressure at point 2 zero? thanks!
It is exposed to the atmosphere. Since both sides are exposed to the atmosphere, and this equation only cares about difference in pressure, you can simply work in gauge pressure to solve it, and both pressures are therefore zero.
As Carl points out, they are both zero psig and 14.7 psia. You don't even need to know this - only that they are both open to the atmosphere so they cancel out.
Hey, thanks for the video, but check your calculation - The answer should be 56.25 m.
It seems like there are many mistakes, gravity at the end should be multiplied with 10,76 , and the diameter should be in inch not feet
@@saraism4557 32.2 is the convertion for ft/sec2 and the diameter is in feet because epsilon is in feet … is this correct ? @enGENIEer
isn't it Area of the pipe = pi x d^2? Can you please explain why Area = pi/4 x d^2?
Binh Truong area of pipe is pi x r ^2
When using radius
If you are using the diameter it is pi/4 x d^2
Jonathan Perez explained it well! Hope it makes sense and make sure you add that to your cheat sheet, so you remember it for the exam!
Your major is only Mechincal and there is no videos for Electrical and computer
Civil, actually
Thank you