Fluid Mechanics: Topic 8.3 - Pressure drop and head loss in pipe flow

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CPPMechEngTutorials

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@ellyvatedaf 3 жыл бұрын

This fluid mechanics playlist is the best there ever is

@CPPMechEngTutorials Жыл бұрын

Thanks!

@gumball2077 Ай бұрын

Thank god, this is exactly what I needed for my final year!!!

@chandramoule 7 жыл бұрын

Dear Mr. Paul, This video was excellently presented, thanks to you. Timely help.

@CPPMechEngTutorials 7 жыл бұрын

Wonderful.

@lefthandeddoghouse3586 3 жыл бұрын

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.

@leoyao7598 3 жыл бұрын

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?

@nikan4now 4 жыл бұрын

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.

@CPPMechEngTutorials Жыл бұрын

Turbines and pumps are covered in other videos. They have an efficiency that covers losses in those devices.

@yosephnega5774 Жыл бұрын

nicely done

@russelayochok6813 2 күн бұрын

What happened to sin(fee)?

@karimazafar9101 4 жыл бұрын

So can the little cylinder be a sort of a valve?

@allenswartz1853 6 жыл бұрын

This was a really good explanation. Thank you!

@CPPMechEngTutorials 6 жыл бұрын

You're welcome!

@gig777 7 жыл бұрын

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?

@CPPMechEngTutorials 6 жыл бұрын

Here is a video discussing loss coefficients (K values). kzbin.info/www/bejne/bHWplmubiteVmas In short, K values can vary based on a lot of factors, including the Reynolds number (which contains viscosity).

@enesprtc115 Жыл бұрын

excellent explanation. Thank you sir!

@snowpro5ryusmc 7 жыл бұрын

What is 2(g)rhoV^2? Where is that coming from? Great videos by the way!

@CPPMechEngTutorials 7 жыл бұрын

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.

@snowpro5ryusmc 7 жыл бұрын

Ok. but why 2 (g)rhoV^2?

@CPPMechEngTutorials 7 жыл бұрын

Why not? Actually, it allows us to come up with a convenient formula -- the Darcy-Weisbach equation.

@nikhileshkrishna724 4 жыл бұрын

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?

@CPPMechEngTutorials 4 жыл бұрын

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.