Fluid Dynamics - Rate of Change of Momentum question

[Tom Hardy]

Homework Statement

A vertical jet of water of 3cm in diameter impinges upon a horizontal flat plate with a velocity of 4m/s, and all of the water spreads out horizontally. Find the force exerted on the plate.

Homework Equations

The Attempt at a Solution

I am not 100% sure how to tackle this, from what I can understand understand it could have something to do with $$F=dp/dt$$ (p being momentum) but I am not sure how to work out the change in momentum. If we consider an element that is about to hit the plate with dimensions Area = $$\pi*(1.5*10^{-2})^2$$ volume = $$4*(\pi*(1.5*10^{-2})^2)$$ mass is the same value as volume so momentum would be $$16*(\pi*(1.5*10^{-2})^2)$$ That's where I get up to, I mean the water goes sideways so how do I figure out it's momentum after, how does continuity play out here? :S Should I just take it after to be 0?

Many thanks

 

[Chestermiller]

What you've calculated so far is the mass flow rate of water against the plate. But this is not the rate of momentum flow against the plate. To get that, you need to multiply by the velocity again. Check the units to be sure that this indeed comes out to force.

You are correct that, since the flow after hitting the plate is zero, that vertical momentum is zero.

Chet

 

[Tom Hardy]

What you've calculated so far is the mass flow rate of water against the plate. But this is not the rate of momentum flow against the plate. To get that, you need to multiply by the velocity again. Check the units to be sure that this indeed comes out to force.

You are correct that, since the flow after hitting the plate is zero, that vertical momentum is zero.

Chet

Hi Chet,

Thank you for replying. If I take a look at the units then:

We start with area $\pi(1.5*10^{-2})^2$ which has units $m^2$, then we times by a velocity 4 to get volumetric flow rate with units $\frac{m^3}{s}$, then we multiply the volumetric flow rate by the density of water to get a mass flow rate of units $\frac{kg^3}{s}$ so to get a force shouldn't I just multiply that by the velocity again to get $\frac{kg^3m}{s}$?

So I end up with $( \pi(1.5*10^{-2})^2*4*999.7*4 ) \approx 16000\pi(1.5*10^{-2})^2$, the answer that I get from that (11.3) is one of the multiple choice answers but I am still not 100% sure if it is correct.

Many thanks for your help!

 

[Chestermiller]

Hi Chet,

Thank you for replying. If I take a look at the units then:

We start with area $\pi(1.5*10^{-2})^2$ which has units $m^2$, then we times by a velocity 4 to get volumetric flow rate with units $\frac{m^3}{s}$, then we multiply the volumetric flow rate by the density of water to get a mass flow rate of units $\frac{kg^3}{s}$ so to get a force shouldn't I just multiply that by the velocity again to get $\frac{kg^3m}{s}$?

Yes. But that power of 3 on the kg needs to be corrected. It should just be kg. And, when you multiply by the velocity again, you get kg-m/s^2 = Newtons.

So I end up with $( \pi(1.5*10^{-2})^2*4*999.7*4 ) \approx 16000\pi(1.5*10^{-2})^2$, the answer that I get from that (11.3) is one of the multiple choice answers but I am still not 100% sure if it is correct.

Many thanks for your help!

I haven't checked the arithmetic, but the methodology is correct.

Chet

 

[Tom Hardy]

Yes. But that power of 3 on the kg needs to be corrected. It should just be kg. And, when you multiply by the velocity again, you get kg-m/s^2 = Newtons.

I haven't checked the arithmetic, but the methodology is correct.

Chet

Woops,that was a typo, thanks a lot for your help.

 

[Chestermiller]

Woops,that was a typo, I meant to say meter cubed. Thanks a lot for help.

Actually, you should only have m to the first power in your final answer. The units have to be Newtons = kg-m/s^2.

Chet

 

[Tom Hardy]

Actually, you should only have m to the first power in your final answer. The units have to be Newtons = kg-m/s^2.

Chet

Yeah, I edited it straight after I posted it :P