Conservation of Momentum Fluids Question

[member 428835]

Homework Statement

I am studying for an upcoming exam and stumbled upon a website with a bunch of practice problems. I would typically state the question but this one is so long and requires a picture so here is the hyperlink to it: http://web.mit.edu/2.25/www/5_10/5_10.html

Homework Equations

Bernoulli, mass/momentum conservation.

The Attempt at a Solution

I'll write out the equations I have (I'm not interested in actually solving them, just want to know I can set it up). Applying conservation of mass yields $$2 \pi r^2 V_1 = \pi R^2 V_2$$. Bernoulli yields $$\frac{P_1}{\rho}+\frac{V_1^2}{2} = \frac{P_2}{\rho}+\frac{V_2^2}{2}$$ Conservation of momentum is $$2 \pi r^2 V_1^2 \hat{r} + \pi R^2 V_2^2 \hat{x} = -2 \pi r^2 P_1\hat{r} - \pi R^2 P_2\hat{x} - F_{wall} \hat{x}$$

where $F_{wall}$ is the force the wall exerts on the control volume. Let me know if I should explain anything I wrote. Thanks so much for looking this over!

 

[haruspex]

I do not understand your momentum equation. $\hat r$ is not a single direction. What is the total momentum of the inflow in the x direction?
Why are you adding the inflow and outflow momenta?
Is the minus sign in front of the F supposed to be an equals sign? Or maybe = - F ...?

 

[member 428835]

You're right; since $\hat{r}$ changes wrt $\theta$ it cannot simply come out of the integral (brain fart on my part). How about this: a force balance on the described control volume assuming inviscid, stead, constant density yields $$\int_{A_1} \rho (-V_1 \hat{r}) (-V_1 \hat{r} \cdot \hat{r}) \,dA_1 + \int_{A_2} \rho (V_2 \hat{x}) (V_2 \hat{x} \cdot \hat{x}) \,dA_2 = -\int_{A_1} P_1 \hat{r} \,dA_1 -\int_{A_2} P_2 \hat{x} \,dA_2-F_{wall} \hat{x}$$
where the lhs is the convective term and the rhs is the pressure force and the force of the wall acting on the control volume.

 

[haruspex]

where the lhs is the convective term and the rhs is the pressure force and the force of the wall acting on the control volume

How are you defining A₁ and A₂? Don't the two pressures act over the same area?

 

[member 428835]

I've defined $A_1$ as the hemispherical surface to the left side of the wall, as sketched in the figure. $A_2$ is the circular output that allows flow to go from the left side to the right side.

 

[haruspex]

I've defined $A_1$ as the hemispherical surface to the left side of the wall, as sketched in the figure. $A_2$ is the circular output that allows flow to go from the left side to the right side.

In that case I agree with your equation.
Edit: I think there is a problem with your P₂ integral. This needs to be over a circle radius r (the large radius), not just over the hole (A₂). And I assume this will come out to be opposite in sign to the P₁ integral.
By the way, there is an easy way to do the P₁ integral.

 

[member 428835]

In that case I agree with your equation.
Edit: I think there is a problem with your P₂ integral. This needs to be over a circle radius r (the large radius), not just over the hole (A₂). And I assume this will come out to be opposite in sign to the P₁ integral.
By the way, there is an easy way to do the P₁ integral.

Now if I made the control volume touch the wall inside the left side and had it going down and then following the hole of radius $R$ (so the CV looks kind of like a mushroom) would I still integrate over the area of the large radius $r$ or could I then integrate over the area of small $R$?

How are you suggesting doing the $P_1$ integral?

 

[haruspex]

Now if I made the control volume touch the wall inside the left side and had it going down and then following the hole of radius $R$ (so the CV looks kind of like a mushroom) would I still integrate over the area of the large radius $r$ or could I then integrate over the area of small $R$?

How are you suggesting doing the $P_1$ integral?

The question asks for the net force on the wall over a circle radius r, no? So you need to consider the pressure over that area on both sides.

For the integral of P₁, imagine the situation with the hole blocked, so the pressure on the left is P₁ everywhere.

 

[member 428835]

For the integral of P₁, imagine the situation with the hole blocked, so the pressure on the left is P₁ everywhere.

I'm not sure what you're saying here? Could you elaborate?

 

[haruspex]

I'm not sure what you're saying here? Could you elaborate?

If the hole were blocked, the whole of the left side would be at pressure P₁. Consider the hemisphere of fluid radius r. The net horizontal force from the pressure P₁ acting on the curved surface must balance the net horizontal force from the same pressure acting on the flat surface. What would that equal?

 

[member 428835]

If the hole were blocked, the whole of the left side would be at pressure P₁. Consider the hemisphere of fluid radius r. The net horizontal force from the pressure P₁ acting on the curved surface must balance the net horizontal force from the same pressure acting on the flat surface. What would that equal?

This would be $\pi r^2 P_1$.

 

[haruspex]

This would be $\pi r^2 P_1$.

Exactly.

 

[member 428835]

So you're saying we could replace the spherical integral with the $\pi r^2 P_1$ (for constant $P_1$)? This seems like you're almost using Stoke's theorem (same surface boundary but different surface). Is it at all related?

 

[haruspex]

So you're saying we could replace the spherical integral with the $\pi r^2 P_1$ (for constant $P_1$)? This seems like you're almost using Stoke's theorem (same surface boundary but different surface). Is it at all related?

No, it has nothing to do with Stokes' Theorem. That sort of theorem relates integrals over manifolds of different dimensions. It the same as the principle by which the relationship between surface tension, pressure and bubble size is found. Integral of pressure over bubble cross section equals net force on each hemisphere.

 

[member 428835]

No, it has nothing to do with Stokes' Theorem. That sort of theorem relates integrals over manifolds of different dimensions. It the same as the principle by which the relationship between surface tension, pressure and bubble size is found. Integral of pressure over bubble cross section equals net force on each hemisphere.

So are you referring to the young-laplace equation?

 

[Chestermiller]

Josh,

You are going to have to use spherical coordinates to solve this problem. Let Q be the volumetric throughput rate. The velocity in the left compartment is in the radial spherical coordinate direction, and is equal to $\frac{Q}{2\pi r^2}$. The pressure at r is $$p=p_1-\frac{\rho}{2}\left(\frac{Q}{2\pi r^2}\right)^2$$ What is the mass flow rate per unit area into the control volume? If $\theta$ is the angle between the x-axis and the spherical surface at r, what is the differential area of control surface between $\theta$ and $\theta + d\theta$?

 

[member 428835]

Josh,

You are going to have to use spherical coordinates to solve this problem. Let Q be the volumetric throughput rate. The velocity in the left compartment is in the radial spherical coordinate direction, and is equal to $\frac{Q}{2\pi r^2}$. The pressure at r is $$p=p_1-\frac{\rho}{2}\left(\frac{Q}{2\pi r^2}\right)^2$$ What is the mass flow rate per unit area into the control volume? If $\theta$ is the angle between the x-axis and the spherical surface at r, what is the differential area of control surface between $\theta$ and $\theta + d\theta$?

The surface area would be $r \, d\phi \,r \cos \phi \, d\theta : \phi \in [-\pi,\pi], \theta \in [-\pi,\pi]$ (wrote like this so you can see my logic). I've defined $\phi$ as the angle the radius makes coming out of the plane with the x-y plane taken from the origin (so slightly different than convention).

How did you arrive at this pressure term $p(r)$? It is obvious your velocity is volumetric flow rate per hemisphere area.

And based on the expression for velocity, isn't mass flow rate per unit area $V \rho$.

 

[Chestermiller]

The surface area would be $r \, d\phi \,r \cos \phi \, d\theta : \phi \in [-\pi,\pi], \theta \in [-\pi,\pi]$ (wrote like this so you can see my logic). I've defined $\phi$ as the angle the radius makes coming out of the plane with the x-y plane taken from the origin (so slightly different than convention).

The system is axisymmetric, so the longitudinal direction doesn't need to be considered. The differential area I was envisioning was $2\pi (r\sin{\theta})rd\theta$. My $\theta$ is the acute angle that any line through the origin makes with the negative x-axis. It is the standard $\theta$ coordinate that is used in spherical coordinates.

How did you arrive at this pressure term $p(r)$? It is obvious your velocity is volumetric flow rate per hemisphere area.

I applied Bernoulli equation to the points at radius = infinity (p = p1) and radius = r (pressure=p). Regarding the (radial) velocity, yes it is the volumetric flow rate per unit area.

And based on the expression for velocity, isn't mass flow rate per unit area $V \rho$.

Yes. The mass flow rate per unit area is $\frac{\rho Q}{2\pi r^2}$. So, the mass flow rate through the differential area between $\theta$ and $\theta + d\theta$ is:
$$2\pi (r\sin{\theta})rd\theta\frac{\rho Q}{2\pi r^2}=\rho Q sin{\theta}d\theta$$The x-compoent of fluid velocity is $\frac{Q}{2\pi r^2}\cos{\theta}$. So what is the amount of x-momentum entering the control volume through the portion of the control surface between $\theta$ and $\theta + d\theta$?

 

[member 428835]

The system is axisymmetric, so the longitudinal direction doesn't need to be considered. The differential area I was envisioning was $2\pi (r\sin{\theta})rd\theta$. My $\theta$ is the acute angle that any line through the origin makes with the negative x-axis. It is the standard $\theta$ coordinate that is used in spherical coordinates.

I applied Bernoulli equation to the points at radius = infinity (p = p1) and radius = r (pressure=p). Regarding the (radial) velocity, yes it is the volumetric flow rate per unit area.

Yes. The mass flow rate per unit area is $\frac{\rho Q}{2\pi r^2}$. So, the mass flow rate through the differential area between $\theta$ and $\theta + d\theta$ is:
$$2\pi (r\sin{\theta})rd\theta\frac{\rho Q}{2\pi r^2}=\rho Q sin{\theta}d\theta$$The x-compoent of fluid velocity is $\frac{Q}{2\pi r^2}\cos{\theta}$. So what is the amount of x-momentum entering the control volume through the portion of the control surface between $\theta$ and $\theta + d\theta$?

I am getting that total momentum entering through that side of the sphere is $-\pi r^2 V^2 \rho = -\pi r^2 (\frac{\rho Q}{2\pi r^2})^2 \rho$. I know this isn't what you asked for (yet?) but I tried going ahead. What I did was $$\iint_{A_1} \rho \vec{V} (\vec{V} \cdot \hat{n}) \, dS = \int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi (-V \hat{r})(-V\hat{r} \cdot \hat{r}) \, d\theta \, d\phi \implies \\ M_x=\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi V^2 (\hat{r} \cdot \hat{x}) \, d\theta \, d\phi\\ =\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 V^2 \cos \theta \sin^2 \phi \, d\theta \, d\phi$$
How does this look?

 

[Chestermiller]

I am getting that total momentum entering through that side of the sphere is $-\pi r^2 V^2 \rho = -\pi r^2 (\frac{\rho Q}{2\pi r^2})^2 \rho$. I know this isn't what you asked for (yet?) but I tried going ahead. What I did was $$\iint_{A_1} \rho \vec{V} (\vec{V} \cdot \hat{n}) \, dS = \int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi (-V \hat{r})(-V\hat{r} \cdot \hat{r}) \, d\theta \, d\phi \implies \\ M_x=\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi V^2 (\hat{r} \cdot \hat{x}) \, d\theta \, d\phi\\ =\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 V^2 \cos \theta \sin^2 \phi \, d\theta \, d\phi$$
How does this look?

It doesn't make sense to me. As I said, $\phi$, the longitudinal angle should integrate out because the flow is symmetric about the x axis. I think you are having trouble applying spherical coordinates. $\theta$ should only run from zero to pi/2. $\phi$ should run from 0 to 2pi. Please draw a diagram showing your understanding of the angle $\theta$. What do you get for the rate of x momentum entering if you use my relationship, and how does it compare with your result?

 

[member 428835]

I drew it as best I could. I included pressure on each side so you have more reference.

 

[member 428835]

The x-compoent of fluid velocity is $\frac{Q}{2\pi r^2}\cos{\theta}$.

So this is what I don't understand. It seems to me the magnitude of velocity $V = \frac{Q}{2\pi r^2} \implies \vec{V} = -\frac{Q}{2\pi r^2} \hat{r}$ (since it enters radially, and $\hat{r}$ is an outward unit normal in the radial direction). Then the x-component is $\vec{V} \cdot \hat{i}= -\frac{Q}{2\pi r^2} \hat{r} \cdot \hat{i} = -\frac{Q}{2\pi r^2} \cos \theta \sin \phi$.

 

[Chestermiller]

x component of velocity is $\frac{Q}{2\pi r^2}\cos{\theta}$. This is because theta is the angle any spherical coordinate radial line (any line drawn from the origin) makes with the x-axis. The surfaces of constant $\theta$ are cones.

 

[member 428835]

But the force balance is integrated over $\rho \vec{V} (\vec{V} \cdot \hat{n})$. I am unable to reason through your approach using this technique. Do you have a vector expression for $\vec{V}$ in terms of cartesian unit vectors?

 

[Chestermiller]

But the force balance is integrated over $\rho \vec{V} (\vec{V} \cdot \hat{n})$. I am unable to reason through your approach using this technique. Do you have a vector expression for $\vec{V}$ in terms of cartesian unit vectors?

Hi Josh,

I see now what our difficulty on communicating on this problem is. You are interpreting the x-direction on the figure literally as the usual x Cartesian coordinate direction typically employed in relating spherical coordinates to Cartesian coordinates. I am viewing it conceptually as if the labels on the axes had been switched. Suppose we were to switch the axis labels such that the direction of the jet were taken as the z direction (rather than x), and we were trying to determine the force in this z direction. In this case, the wall in the figure would be the x-y plane (perpendicular to the z-axis). The fluid flow would then be axi-symmetric with respect to the z axis. If, as is usually done in spherical coordinates, the angle $\theta$ were measured relative to the positive z direction, then $\theta$ would run from $\pi/2$ to $\pi$ in the left-hand part of the figure. Does this make any sense to you now? Would you be able to more easily perform the momentum balance using this framework?

Chet

 

[member 428835]

It shouldn't matter the result though as long as we do the physics/math right. I did it your way though (which I agree is mathematically simpler) and it looks something like this:
$$\iint_{A_1} \rho \vec{V} (\vec{V} \cdot \hat{n}) \, dS = \int_0^{\pi/2} \int_0^{2\pi} \rho r^2 \sin \phi (-V \hat{r})(-V\hat{r} \cdot \hat{r}) \, d\theta \, d\phi \implies \\ M_z=\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi V^2 (\hat{r} \cdot \hat{z}) \, d\theta \, d\phi\\ =\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 V^2 \cos \phi \sin \phi \, d\theta \, d\phi$$ which equals the same result I posted earlier, namely $-\pi r^2 V^2 \rho$.

 

[Chestermiller]

It shouldn't matter the result though as long as we do the physics/math right. I did it your way though (which I agree is mathematically simpler) and it looks something like this:
$$\iint_{A_1} \rho \vec{V} (\vec{V} \cdot \hat{n}) \, dS = \int_0^{\pi/2} \int_0^{2\pi} \rho r^2 \sin \phi (-V \hat{r})(-V\hat{r} \cdot \hat{r}) \, d\theta \, d\phi \implies \\ M_z=\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 \sin \phi V^2 (\hat{r} \cdot \hat{z}) \, d\theta \, d\phi\\ =\int_0^\pi \int_{\pi/2}^{3\pi/2} \rho r^2 V^2 \cos \phi \sin \phi \, d\theta \, d\phi$$ which equals the same result I posted earlier, namely $-\pi r^2 V^2 \rho$.

Great. So we both agree that the rate of x-momentum flow into the control volume is $\rho \frac{Q^2}{4\pi r^2}$.

Next, what do you get for the total pressure force on the hemispherical control surface (This force is oriented in the x direction)?

 

[member 428835]

Great. So we both agree that the rate of x-momentum flow into the control volume is $\rho \frac{Q^2}{4\pi r^2}$.

Except I'm getting a negative, though I think it should be positive since momentum is entering our CV.

Next, what do you get for the total pressure force on the hemispherical control surface (This force is oriented in the x direction)?

Pressure is $$P_{total} = -\int_0^{2 \pi}\int_{\pi/2}^\pi P \hat{r}\cdot\hat{z} r^2 \sin\phi \, d \phi \, d \theta\\= \left(P_1-\frac{\rho}{2}\left(\frac{Q}{2 \pi r^2}\right)^2\right)\pi r^2$$ where $P$ was found via Bernoulli, as you suggested earlier.

 

[Chestermiller]

OK. Now how about the rate of momentum leaving the control volume and the backward pressure force.

 

[member 428835]

OK. Now how about the rate of momentum leaving the control volume and the backward pressure force.

Momentum leaving would be $$\iint_{A_2} \rho \vec{V} (\vec{V} \cdot \hat{n})\, dS = \rho \pi R^2 V^2 = \frac{\rho Q^2}{\pi R^2}$$. The backward pressure force would be $$-\iint_{A_2}P_2 \, \vec{dS} = -P_2 \pi r^2\hat{z}$$ where $r$ is the radius of the hemisphere.

Do you agree? Sorry for the late response; I'd been busy the last few days.

 

[Chestermiller]

Momentum leaving would be $$\iint_{A_2} \rho \vec{V} (\vec{V} \cdot \hat{n})\, dS = \rho \pi R^2 V^2 = \frac{\rho Q^2}{\pi R^2}$$. The backward pressure force would be $$-\iint_{A_2}P_2 \, \vec{dS} = -P_2 \pi r^2\hat{z}$$ where $r$ is the radius of the hemisphere.

Do you agree? Sorry for the late response; I'd been busy the last few days.

Yes, I do agree. So now, what is the momentum balance?

 

[member 428835]

Yes, I do agree. So now, what is the momentum balance?

The momentum balance is $$-\pi r^2 V^2 \rho+\frac{\rho Q^2}{\pi R^2} = -P_2 \pi r^2+ \left(P_1-\frac{\rho}{2}\left(\frac{Q}{2 \pi r^2}\right)^2\right)\pi r^2 - F_{wall}$$. How does that look?

 

[Chestermiller]

The momentum balance is $$-\pi r^2 V^2 \rho+\frac{\rho Q^2}{\pi R^2} = -P_2 \pi r^2+ \left(P_1-\frac{\rho}{2}\left(\frac{Q}{2 \pi r^2}\right)^2\right)\pi r^2 - F_{wall}$$. How does that look?

This looks good. But in the "momentum-in" term, I would have replaced V with $\frac{Q}{2\pi r^2}$.

Now, to complete the solution to this problem, you need one additional relationship to eliminate $Q^2$ and replace it with something involving $(p_1-p_2)$. Any ideas?

Chet

 

[member 428835]

This looks good. But in the "momentum-in" term, I would have replaced V with $\frac{Q}{2\pi r^2}$.

Now, to complete the solution to this problem, you need one additional relationship to eliminate $Q^2$ and replace it with something involving $(p_1-p_2)$. Any ideas?

Chet

Let's apply Bernoulli at $x\to-\infty$ and $x=0$ along the x-axis. Then we have $$P_1 + \lim_{r\to\infty}\frac{\rho Q}{4 \pi r^2} = P_2 + \frac{\rho Q}{4 \pi R^2} \implies\\ Q = \frac{(P_1-P_2)4\pi R^2}{\rho}$$

think we're done now, right?

 

[Chestermiller]

Let's apply Bernoulli at $x\to-\infty$ and $x=0$ along the x-axis. Then we have $$P_1 + \lim_{r\to\infty}\frac{\rho Q}{4 \pi r^2} = P_2 + \frac{\rho Q}{4 \pi R^2} \implies\\ Q = \frac{(P_1-P_2)4\pi R^2}{\rho}$$

think we're done now, right?

I get $$P_1=P_2+\frac{\rho}{2}\left(\frac{Q}{\pi R^2}\right)^2$$