Fluid Mechanics: The Momentum Equation

[@anuragrajbhar5622]

Thx you sir given me free knowledge ❤

[@josephtraverso5355]

Well presented, thank you Caspar!

[@-HenguljyotiGayan]

Thanks a lot sir.

[@pippakolking1998]

How do we determine the direction of pressure force?

[@CasparHewettFluids]

We are considering the forces on the control volume, so the pressure force is acting inwards on that control volume (there is an equal and opposite pressure force outward, but that isn't relevant to the calculation) thus it is acting to the right on the left hand side and to the left on the right hand side.

[@liambond3569]

Would the vertical pressure force still be oriented downward if the curved pipe was facing down not up?

[@CasparHewettFluids]

No, the pressure force always acts inwards towards the control volume so it would be acting upwards if the pipe were facing downward.

[@Thookudhurai]

Let the the fluid has uniform flow throughout the pipe. Is that mean there is no momentum change in the pipe?

[@CasparHewettFluids]

Yes, that is correct. Momentum is conserved.

[@kaursingh637]

EXCELLENT-- WHICH UNIVERSITY ? -WHICH COUNTRY ?

[@CasparHewettFluids]

Newcastle University, UK

[@kaursingh637]

@@CasparHewettFluids thank u sir

[@lightspd714]

Why isn’t there a Q1 and Q2? In other words why can we assume that A1u1=A2u2=Q ?

[@abdulkabiraduragba5240]

law of continuity, Q1 = Q2 = Q

[@lightspd714]

@@abdulkabiraduragba5240 Thanks!

[@amanullahhayat8487]

P2A2 and pQu2 have got opposite direction then why same sign (-)?? i mean why pQu2 have (+) mark??

[@ghaithaq501]

kzbin.info/door/gnO_q7xjxSYotvyUMgwzEAvideos

[@ronald_plus6]

same question

[@CasparHewettFluids]

P2A2 is force due to pressure acting on the control volume while pQu2 is momentum out of the control volume

[@danielhakimzambri892]

Hello, what is ubar in the bousinesque equation

[@danielhakimzambri892]

Is it like the average velocity or what?

[@CasparHewettFluids]

@@danielhakimzambri892 yes, it is the average velocity 😉

[@danielhakimzambri892]

@@CasparHewettFluids thanks

[@mohsen9828]

if we consider a 90-degree bend pipe, with initial conditions as follows: inlet velocity 15 m/s, no change in cross-section area (let's say radius of 0.2 m), and we only know outlet pressure is equal to atmospheric pressure. Is Bernoulli's equation satisfied? does it mean outlet and inlet pressure are the same? in that case, there's no flow?!

[@CasparHewettFluids]

Bernoulli equation is satisfied within the pipe so pressure will be constant if the pipe is horizontal, resulting in a jet at the outlet. The pressure in the pipe can't be atmospheric pressure if the water has a velocity! Conditions are different once water reaches the outlet. Thus, the problem you pose can't be solved. There is insufficient information. You would need to know the pressure in the pipe.

[@ghbbllk4095]

Why is the momentum force in the opposite direction as the pressure force? Seems like they would both be acting in the same direction, no?🤔

[@CasparHewettFluids]

There are two elements of the momentum. The momentum into the control volume, which acts in the same direction as the pressure and the momentum out of the control volume which is in the opposite direction to the pressure

[@CasparHewettFluids]

The pressure forces are those acting on the control volume, which is why they are both inward

[@punsthedisgrace]

is the momentum of a fluid equal to mass flow rate * velocity ???

[@forresternick]

Mass * velocity = momentum, so mass/time * velocity = momentum/time, which just means that a change in momentum over time = a force :)

[@forresternick]

You can verify this with the units: (kg/s)*(m/s) = (kg*m/s^2)

[@ghaithaq501]

kzbin.info/door/gnO_q7xjxSYotvyUMgwzEAvideos

[@blessingamosmatunya8908]

On the summation of y direction forces, pQu is it not a negetive sir.

[@CasparHewettFluids]

If you mean in the example of the vertical bend you are correct, the outward momentum (not force) is in positive y direction, but you need to remember that for the momentum equation in the y-direction it is momentum out (i.e. is on the right hand side of the equation).

[@mpt2878]

​@@CasparHewettFluids omg true

[@comment8767]

grrrreat!