Why is the force on the fluid not equal to the change in momentum per unit time?

[andyrk]

There is a pipe which is bent in the shape of a right angle as shown. The cross section area of the pipe is A, the velocity of the water inside is v, and the density of water is ρ. The momentum flowing per second towards the right corner is thus: Av²ρ towards right and momentum flowing per second downwards from the corner is Av²ρ downwards. Then why is the force acting on the liquid the vector difference of the flows of momentum per second? Isn't force equal to change in momentum per unit time? But we never calculated change in momentum, we calculated change in flow of momentum per unit time. This leaves me confused.

 

[Simon Bridge]

Force is the rate of change of momentum.
When you calculate the change in the flow of momentum, what do the units come out as?

 

[andyrk]

I know it is dimensionally correct but it doesn't make logical sense since it isn't equal to rate of change of momentum, it is equal to change in flow of momentum.

 

[Simon Bridge]

How would you work out the rate of change of momentum?

 

[andyrk]

Change in momentum per unit time

 

[Simon Bridge]

Yes - so do it: work it out ... do the maths.

 

[andyrk]

But we don't know what the momentum is. We know the flow of momentum ( that is already momentum per unit time). And we don't know that how long would it take for the change to take place.

 

[Simon Bridge]

You are telling me that the change in momentum flow rate is not the same as the rate of change in momentum. Prove it!
You are totally free to set up the problem however you like - but you do need to express the momentum in terms of a density, crossectional area, and an appropriately chosen length.

 

[andyrk]

Because, rate of change of means the the difference in momentum's with a slight increase in time (assuming momentum is a function of time, i.e. p(t)). But change in momentum flow rate simply means: momentum flow rate A - momentum flow rate B which does not give the difference of momentum's with respect to slight increase in time.

 

[Simon Bridge]

The time won't increase because the time to go around the corner is the time to go around the corner and cannot be different.

Also: the corner drawn is very sharp. This is key.
The time to go around the corner is very small.
Thus: time for a section of fluid length x to go around the corner is t=x/v.
So work out the change in momentum for a mass $m=\rho Ax$ in going around the corner - divide it by the time it took to go around the corner.

 

[Delta2]

The time won't increase because the time to go around the corner is the time to go around the corner and cannot be different.

Also: the corner drawn is very sharp. This is key.
The time to go around the corner is very small.
Thus: time for a section of fluid length x to go around the corner is t=x/v.
So work out the change in momentum for a mass $m=\rho Ax$ in going around the corner - divide it by the time it took to go around the corner.

Ok the math seem to work because velocity is constant in magnitude and also density is constant in space and time. What happens if velocity doesn't have constant magnitude and density also varies?

 

[Chestermiller]

OK. I'm going to give it my best show at explaining this. I hope it works for you.

The figure above shows the situation in your original figure. I have indicated a control volume in the figure (between the red dashed lines) into which fluid is flowing in, and out of which fluid is flowing out. The rate of flow in is equal to the rate of flow out.

I have also shown a Before picture in which a small parcel of fluid is about to flow into the control volume at time t. I have also shown an After picture at time t + Δt in which the entering parcel has just fully flowed into the control volume, and another small parcel of equal size has just flowed out. So the net effect during the time interval Δt is that a mass of fluid equal to ρvAΔt has flowed in, and a mass of ρvAΔt has flowed out. The horizontal momentum of the mass flowing into the control volume during the time interval Δt is ρv²AΔt, and the horizontal momentum of the mass flowing out of the control volume during the time interval is zero. The vertical momentum of the mass flowing into the control volume during the time interval Δt is zero, and the vertical momentum of the mass flowing out of the control volume during the time interval is -ρv²AΔt. So the rate of change of momentum of the fluid passing through the control volume over the time interval is:

Rate of change of horizontal momentum of fluid = $\frac{0-ρv^2AΔt}{Δt}=-ρv^2A$

Rate of change of vertical momentum of fluid = $\frac{-ρv^2AΔt-0}{Δt}=-ρv^2A$

These rates of change of momentum require forces to bring them about. These forces are applied to the fluid by the pipe wall. The pipe wall in the L pushes backwards against the fluid in the negative x direction with a force equal to the rate of change of horizontal momentum. The pipe wall in the L pushes downward on the fluid in the negative y direction with a force equal to the rate of change of vertical momentum.

I hope this helps. If not, I've given it my best shot.

Chet

 

[Simon Bridge]

Ok the math seem to work because velocity is constant in magnitude and also density is constant in space and time. What happens if velocity doesn't have constant magnitude and density also varies?

... one of the assumptions in the theory is that the fluid is incompressible - so its density won't change. If it is compressible then you have extra thermodynamics to take into account like if the fluid were a gas.
The derivation assumes laminar flow, a sharp corner, and that there are no forces besides those from the walls of the tube.

For the velocity to change, other things must change too - like the crossectional area of the tube. These are related via Bernoulli's equation. You can do the math yourself to see what effect this has. You can also try the math for the case that the corner is not sharp.

 

[andyrk]

So Chestermiller, what would be the overall change in momentum per unit time?

 

[Chestermiller]

So Chestermiller, what would be the overall change in momentum per unit time?

$-\rho v^2A\vec{i}_x-\rho v^2A\vec{i}_y$

Chet

 

[boneh3ad]

@andyrk, are you familiar with the material derivative (also called the convective derivative or a total derivative)?

 

[Tom_K]

$-\rho v^2A\vec{i}_x-\rho v^2A\vec{i}_y$

Chet

Can you just add the x and y forces in quadrature, as in this example?

Example - Resulting force on a bend due to mass flow and flow velocity
The resulting force on a 45o bend with

• diameter 114 mm = 0.114 m
• water with density 1000 kg/m³
• flow velocity 20 m/s

can be calculated by as
Resulting force in x-direction:
Rx = (1000 kg/m³) π ((0.114 m) / 2)² (20 m/s)² (1 - cos(45))
= 1196 (N)
Resulting force in y-direction:
Ry = (1000 kg/m³) π ((0.114 m) / 2)² (20 m/s)² sin(45)
= 2887 (N)
Resulting force on the bend
R = (1196 (N)² + 2887 (N)²)^1/2
= 3125 (N)
Note - if β is 90othe resulting forces in x- and y-directions are the same.

 

[Delta2]

Shouldnt we add $pA$ to the result for the x and y component?(p the pressure which we assume constant). I mean the rest of the fluid exerts pA force at the control volume along the dashed lines doesn't it?

 

[Chestermiller]

Can you just add the x and y forces in quadrature, as in this example?

Example - Resulting force on a bend due to mass flow and flow velocity
The resulting force on a 45o bend with

• diameter 114 mm = 0.114 m
• water with density 1000 kg/m³
• flow velocity 20 m/s

can be calculated by as
Resulting force in x-direction:
Rx = (1000 kg/m³) π ((0.114 m) / 2)² (20 m/s)² (1 - cos(45))
= 1196 (N)
Resulting force in y-direction:
Ry = (1000 kg/m³) π ((0.114 m) / 2)² (20 m/s)² sin(45)
= 2887 (N)
Resulting force on the bend
R = (1196 (N)² + 2887 (N)²)^1/2
= 3125 (N)
Note - if β is 90othe resulting forces in x- and y-directions are the same.

If you did the arithmetic right (which I haven't checked), this looks correct. (I don't know what you mean by adding the forces in quadrature.

 

[Chestermiller]

Shouldnt we add $pA$ to the result for the x and y component?(p the pressure which we assume constant). I mean the rest of the fluid exerts pA force at the control volume along the dashed lines doesn't it?

The momentum change calculation presented here using the control volume approach corresponds to the mass times acceleration side of the equation in Newton's 2nd law. This mass times acceleration is the result of the contact forces acting on the fluid in the control volume. The contact forces include the inlet and outlet pressure forces and the force exerted by the pipe wall on the fluid in the control volume. That's the other side of the equation.

Chet

 

[Delta2]

The momentum change calculation presented here using the control volume approach corresponds to the mass times acceleration side of the equation in Newton's 2nd law. This mass times acceleration is the result of the contact forces acting on the fluid in the control volume. The contact forces include the inlet and outlet pressure forces and the force exerted by the pipe wall on the fluid in the control volume. That's the other side of the equation.

Chet

Yes sorry i meant the x and y component of the force from the pipe walls. So for the x component of the force exerted by the pipe $F_x$ will be such that $pA+F_x=-\rho v^2A$ correct?

 

[Chestermiller]

Yes sorry i meant the x and y component of the force from the pipe walls. So for the x component of the force exerted by the pipe $F_x$ will be such that $pA+F_x=-\rho v^2A$ correct?

Yes, if p is the pressure at the inlet to the control volume.

Chet

 

[andyrk]

@andyrk, are you familiar with the material derivative (also called the convective derivative or a total derivative)?

No. I think that's way beyond my knowledge.

 

[Tom_K]

If you did the arithmetic right (which I haven't checked), this looks correct. (I don't know what you mean by adding the forces in quadrature.

Really? I do get that a lot, but it always surprises me when people say they don't know what I mean by http://www.researchgate.net/post/How_do_we_know_when_to_add_quantities_in_quadrature .

I suppose it is second nature to me from my background in optical surveying with a theodolite.

 

[Chestermiller]

Really? I do get that a lot, but it always surprises me when people say they don't know what I mean by http://www.researchgate.net/post/How_do_we_know_when_to_add_quantities_in_quadrature .

I suppose it is second nature to me from my background in optical surveying with a theodolite.

Well, I must say that, even though I have worked in many different areas in my >50 year career (see my Mentor bio), I have never heard the term "adding in quadrature."

 

[Vibhor]

Hi Chet ,

Please clarify my doubts .

$-\rho v^2A\vec{i}_x-\rho v^2A\vec{i}_y$

Yes sorry i meant the x and y component of the force from the pipe walls. So for the x component of the force exerted by the pipe $F_x$ will be such that $pA+F_x=-\rho v^2A$ correct?

Yes, if p is the pressure at the inlet to the control volume.

Chet

So , is the horizontal force exerted by pipe/bend on the fluid $-\rho v^2A$ or $-\rho v^2A - pA$ ??

So , is the vertical force exerted by pipe/bend on the fluid $-\rho v^2A$ or $-\rho v^2A - pA$ ??

 

[Chestermiller]

Hi Chet ,

Please clarify my doubts .

So , is the horizontal force exerted by pipe/bend on the fluid $-\rho v^2A$ or $-\rho v^2A - pA$ ??

So , is the vertical force exerted by pipe/bend on the fluid $-\rho v^2A$ or $-\rho v^2A - pA$ ??

Which do you think, and why?

 

[Vibhor]

Initial Horizontal momentum/sec of fluid = $\rho v^2A$

Final Horizontal momentum/sec of fluid = 0

Horizontal Force exerted by pipe = rate of change of momentum in x direction = $- \rho v^2A$

Similarly Vertical force exerted by pipe = rate of change of momentum in y direction = $- \rho v^2A$

Net force exerted by pipe on fluid = $\sqrt{2} \rho v^2A$ in downward direction at angle 45° with vertical .

This is what I know and , this is what is written in the textbook .

Your posts #12 and #15 make sense to me . But then introduction of pA term (post#21 and post#22 ) has confused me . Interestingly post#21 and post#22 also seem logical . This made me wonder what is the correct expression for the force .

I am not looking for easy answers .Almost everything has been answered by you in this thread. I have posted in this thread only because a lot of work was already done by you in this thread , so the need to start everything from scratch could be avoided

Thanks

 

[Chestermiller]

Initial Horizontal momentum/sec of fluid = $\rho v^2A$

Final Horizontal momentum/sec of fluid = 0

Horizontal Force exerted by pipe = rate of change of momentum in x direction = $- \rho v^2A$

Similarly Vertical force exerted by pipe = rate of change of momentum in y direction = $- \rho v^2A$

Net force exerted by pipe on fluid = $\sqrt{2} \rho v^2A$ in downward direction at angle 45° with vertical .

If you're doing a force balance on a body, do you set just one of the forces acting on the body equal to ma, or do you set the resultant of all the forces acting on the body equal to ma? The body in this analysis is the fluid contained within the control volume at a given instant.

This is what I know and , this is what is written in the textbook .

Your posts #12 and #15 make sense to me . But then introduction of pA term (post#21 and post#22 ) has confused me . Interestingly post#21 and post#22 also seem logical . This made me wonder what is the correct expression for the force .

Suppose that the fluid were not flowing. Would the fluid within the control volume be exerting a horizontal force on the pipe bend if the fluid pressure is p?

 

[Vibhor]

Chet , I understand your point . Just help me understand why in the books it is written that horizontal force exerted by fluid on pipe is $\rho v^2A$ . Are there any assumptions involved or am I misinterpreting something ?

 

[Chestermiller]

Chet , I understand your point . Just help me understand why in the books it is written that horizontal force exerted by fluid on pipe is $\rho v^2A$ . Are there any assumptions involved or am I misinterpreting something ?

Maybe they are assuming that the fluid pressure in the pipe is just atmospheric pressure (zero gage), or maybe they just inadvertently left out the pressure force.