Thank you. When you combine the conservation of energy and the second law formulae, and then multiply both numerator and denominator by 2·g·rho·v², where did the rho in the numerator go? I can account for every other variable except that one and the gamma that was in the denominator. The only thing I can think of is that rho interacted with gamma (since gamma disappeared as well). It would be nice to know where those two variable characters went.
You can see from the mg=gamma.V equation that gamma=mg/V so mg=gamma.V, so g/gamma is just= V/m which is 1/rho, that’s why they combine to produce rho in the denominator?
I love this video series. However, here I think the introduction of head loss is unclear. Since you invoked the energy equation it would've been better to use the heat and the internal energy terms as they appear in the conservation equation and then lump them together as the head loss term. Also, this approach was very configuration-specific. What if there's a turbine or a pump. Would the results still hold? Thanks.
When calculating the head loss through fittings, is the same "K" value used no matter what the viscosity? Everything I've read, states the "K" values are for based on water. Wouldn't there be bigger losses in a fitting for a more viscous liquid?
Here is a video discussing loss coefficients (K values). ruclips.net/video/6DFe8eUrbcI/видео.html In short, K values can vary based on a lot of factors, including the Reynolds number (which contains viscosity).
Hi .. The video was really useful. But can u tell me why velocity at inlet and outlet are the same when calculating head loss. Is it because once the flow is fully developed, velocity profile remains constant?
From conservation of mass, mass flowrate in = mass flowrate out. If the fluid density and area at the inlet and outlet are the same, the speed has to be the same at those two locations as well.
2(g)rhoV^2 / 2(g)rhoV^2 is equal to 1. You are always allowed to multiply by 1 because you get back the expression you started with. Similarly, you are always allowed to add zero to an expression.
![[MAE 242] Pipe flow with major and minor head losses](http://i.ytimg.com/vi/WH1fn6dMYiw/mqdefault.jpg)
Thank god, this is exactly what I needed for my final year!!!
This fluid mechanics playlist is the best there ever is
Thanks!
Dear Mr. Paul,
This video was excellently presented, thanks to you. Timely help.
Wonderful.
Thank you. When you combine the conservation of energy and the second law formulae, and then multiply both numerator and denominator by 2·g·rho·v², where did the rho in the numerator go? I can account for every other variable except that one and the gamma that was in the denominator. The only thing I can think of is that rho interacted with gamma (since gamma disappeared as well). It would be nice to know where those two variable characters went.
You can see from the mg=gamma.V equation that gamma=mg/V so mg=gamma.V, so g/gamma is just= V/m which is 1/rho, that’s why they combine to produce rho in the denominator?
I love this video series. However, here I think the introduction of head loss is unclear. Since you invoked the energy equation it would've been better to use the heat and the internal energy terms as they appear in the conservation equation and then lump them together as the head loss term. Also, this approach was very configuration-specific. What if there's a turbine or a pump. Would the results still hold? Thanks.
Turbines and pumps are covered in other videos. They have an efficiency that covers losses in those devices.
nicely done
When calculating the head loss through fittings, is the same "K" value used no matter what the viscosity? Everything I've read, states the "K" values are for based on water. Wouldn't there be bigger losses in a fitting for a more viscous liquid?
Here is a video discussing loss coefficients (K values).
ruclips.net/video/6DFe8eUrbcI/видео.html
In short, K values can vary based on a lot of factors, including the Reynolds number (which contains viscosity).
This was a really good explanation. Thank you!
You're welcome!
excellent explanation. Thank you sir!
Hi .. The video was really useful. But can u tell me why velocity at inlet and outlet are the same when calculating head loss. Is it because once the flow is fully developed, velocity profile remains constant?
From conservation of mass, mass flowrate in = mass flowrate out. If the fluid density and area at the inlet and outlet are the same, the speed has to be the same at those two locations as well.
So can the little cylinder be a sort of a valve?
What is 2(g)rhoV^2? Where is that coming from? Great videos by the way!
2(g)rhoV^2 / 2(g)rhoV^2 is equal to 1. You are always allowed to multiply by 1 because you get back the expression you started with. Similarly, you are always allowed to add zero to an expression.
Ok. but why 2 (g)rhoV^2?
Why not?
Actually, it allows us to come up with a convenient formula -- the Darcy-Weisbach equation.
Sir can you explain how to calculate pipe size if natural gas (Density - 0.6) flowing vertical through pipe. Is there any formula ?
At 6:39 why is hl negative (-)? All the books and videos I´ve seen always present it as positive, why is that?
Perhaps the textbooks have the inlet and outlet terms on opposite sides of the equation?
great video
Thanks!
finally. thanks
what happened to sin
very good video
Thanks!
head losses short btao Hindi me
why you talking so soft??
WHAT?!?!!?