# Incompressible fluid acquiring momentum

by Philip Wood
Tags: acquiring, fluid, incompressible, momentum
P: 747
Suppose we have the set up shown in the thumbnail, in which an incompressible fluid moves from left to right, trapped by converging streamlines.

The rate of acquisition, F, of momentum by the fluid is surely
$$F = \rho A_2 v_2^2 - \rho A_1 v_1^2 = \rho A_1 v_1^2 \left\{\frac{A_1}{A_2} - 1\right\}$$
in which I've eliminated v2 using the continuity equation, $A_1v_1 = A_2v_2$.

My question is whether there's any simple physical argument which connects this with the Bernoulli relationship for the same sample of fluid, namely:
$$p_1 -p_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$
that is, again using the continuity equation,
$$p_1 -p_2 = \frac{\rho}{2} v_1^2 \left\{\frac{A_1^2}{A_2^2} - 1\right\}$$
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P: 2,939
 Quote by Philip Wood Suppose we have the set up shown in the thumbnail, in which an incompressible fluid moves from left to right, trapped by converging streamlines. The rate of acquisition, F, of momentum by the fluid is surely $$F = \rho A_2 v_2^2 - \rho A_1 v_1^2 = \rho A_1 v_1^2 \left\{\frac{A_1}{A_2} - 1\right\}$$ in which I've eliminated v2 using the continuity equation, $A_1v_1 = A_2v_2$. My question is whether there's any simple physical argument which connects this with the Bernoulli relationship for the same sample of fluid, namely: $$p_1 -p_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$ that is, again using the continuity equation, $$p_1 -p_2 = \frac{\rho}{2} v_1^2 \left\{\frac{A_1^2}{A_2^2} - 1\right\}$$
Yes. I think I understand what you are asking. F is the rate of change of momentum, so it must represent the net force acting on the fluid within the control volume. But, at first glance, it appears that the only forces acting on the fluid in the control volume are the pressure forces at the inlet and exit areas. This doesn't seem to be compatible with the Bernoulli equation. However, there is an additional force present. The channel formed by the converging streamlines is not uniform, so there are pressure forces at the wall of the channel with components in the main flow direction (i.e., the pressure acts normal to the wall of the channel, and the normal to the wall has a component in the flow direction). The force exerted by the wall of the stream channel on the fluid in the control volume works out to be:$$\int_{x_1}^{x_2}{p\frac{dA}{dx}dx}$$
So the force F is related to the pressure variation by:
$$F=p_1A_1-p_2A_2+\int_{x_1}^{x_2}{p\frac{dA}{dx}dx}$$
I hope this helps.

Chet
P: 747
Many thanks, Chet, for replying to this. I agree with your approach and had indeed realised that we must also consider the pressure across the 'walls' of the tube. Indeed I looked at the simple case of a wedge-shaped 'tube of fluid' (see thumbnail), fondly believing that the resultant force on the wedge would be
$$p_1 A_1 - p_2 A_2 + \frac{p_1 + p_2}{2} (A_2 - A_1)$$.
This assumes a linear change of pressure with distance. It simplifies to
$$(p_1 - p_2)\frac{A_1 + A_2}{2}$$.
If we equate this to the rate of change of momentum of the fluid, we get
$$(p_1 - p_2)\frac{A_1 + A_2}{2} = \rho \left\{A_2 v_2^2 - A_1 v_1^2\right\}.$$
I don't think this is consistent with the Bernouilli equation, even if we use $A_1 v_1 = A_2 v_2$.

So what's gone wrong? Perhaps my assumption of a linear pressure gradient. I thought I might evade this difficulty by considering just a thin slice $\Delta x$ of fluid, but it was still wrong. Also I'm neglecting 'sideways' momentum acquired by the fluid.
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## Incompressible fluid acquiring momentum

 Quote by Philip Wood Many thanks, Chet, for replying to this. I agree with your approach and had indeed realised that we must also consider the pressure across the 'walls' of the tube. Indeed I looked at the simple case of a wedge-shaped 'tube of fluid' (see thumbnail), fondly believing that the resultant force on the wedge would be $$p_1 A_1 - p_2 A_2 + \frac{p_1 + p_2}{2} (A_2 - A_1)$$. This assumes a linear change of pressure with distance. It simplifies to $$(p_1 - p_2)\frac{A_1 + A_2}{2}$$. If we equate this to the rate of change of momentum of the fluid, we get $$(p_1 - p_2)\frac{A_1 + A_2}{2} = \rho \left\{A_2 v_2^2 - A_1 v_1^2\right\}.$$ I don't think this is consistent with the Bernouilli equation, even if we use $A_1 v_1 = A_2 v_2$. So what's gone wrong? Perhaps my assumption of a linear pressure gradient. I thought I might evade this difficulty by considering just a thin slice $\Delta x$ of fluid, but it was still wrong. Also I'm neglecting 'sideways' momentum acquired by the fluid.
Hi Philip,

I think I can deliver what you are looking for. You are trying to show that the equation you wrote down for the net force F is consistent and compatible with the Bernoulli equation.

Let me start out with the equation I wrote for F, expressed in terms of the pressure and area variation:
$$F=p_1A_1-p_2A_2+\int_{x_1}^{x_2}{p\frac{dA}{dx}dx}$$
If I carry out the integration by parts, I obtain:
$$F=\int_{x_1}^{x_2}{A\left(-\frac{dp}{dx}\right)dx}$$
Now, from the Bernoulli equation, we have:
$$\left(-\frac{dp}{dx}\right)=ρv\frac{dv}{dx}$$
Substituting the continuity equation into this relationship yields:
$$\left(-\frac{dp}{dx}\right)=\frac{ρv_1^2A_1^2}{A}\frac{d(\frac{1}{A})}{dx}$$
If we now substitute this back into the equation for F, we obtain:
$$F=\int_{x_1}^{x_2}ρv_1^2A_1^2\frac{d(\frac{1}{A})}{dx}dx$$
Evaluating the integral in this relationship gives:
$$F=ρv_1^2A_1^2\left(\frac{1}{A_2}-\frac{1}{A_1}\right)$$
This matches your equation for F. Therefore, your equation for F is fully compatible with the Bernoulli equation.

Chet
P: 747
Chet. Enormous thanks for this. I take the liberty of modifying the last stages of your derivation to produce the following - which is exactly what I wanted!

 Quote by Chestermiller $$F=p_1A_1-p_2A_2+\int_{x_1}^{x_2}{p\frac{dA}{dx}dx}$$ If I carry out the integration by parts, I obtain: $$F=\int_{x_1}^{x_2}{A\left(-\frac{dp}{dx}\right)dx}$$ Now, from the Bernoulli equation, we have: $$\left(-\frac{dp}{dx}\right)=ρv\frac{dv}{dx}$$
Hence
$$F = \int_{x_1}^{x_2} A \rho v \frac{dv}{dx} dx.$$
But, from the continuity equation, $Av$ is constant throughout, so
$$F = \rho A v \int_{x_1}^{x_2} \frac{dv}{dx} dx.$$
so
$$F = \rho A v (v_2 - v_1).$$
But $Av = A_1v_1 = A_2v_2$, so
$$F = \rho (A_2 v_2^2 - A_1 v_1^2).$$
So net force on fluid = rate of change of momentum of fluid.

This analysis has cleared up a difficulty I've had with the textbook derivation of Bernoulli. It's usually stated as obvious that the force on a slice of fluid is $-A dp$ in which dp is the difference in pressure on either side of the slice. I thought it ought to be $-d(Ap)$, because A is changing as the streamlines converge or diverge. But this neglects to consider the forces acting on the sides of the slice, whose component in the x-direction is $pdA$. So the resultant force on the slice is $-d(Ap) + pdA$. And this is simply $-A dp$.

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