Conservation of Momentum Question

[member 428835]

Homework Statement

An incompressible fluid of density $\rho$ flows steadily through a 2D infinite row of fixed shapes. The vertical distance between shapes is $a$. Define station 1 as the space where velocity enters and station 2 where it exits. Also, velocity and pressure are constant along stations 1 and 2, and are given by $v_1$, $v_2$, $p_1$, $p_2$. The angle the flow makes with the horizontal direction at station 1 is $\beta_1$, and it is $\beta_2$ at station 2. Compute the reactions $R_x$ and $R_y$ necessary to keep one vane in place.

Homework Equations

Conservation of momentum

The Attempt at a Solution

Horizontal forces at station 1 are $F_{x1} = ap_1+\rho a^2v_1^2 \cos^2 \beta_1$ (since 2D I assume pressure is force per unit line and density is mass/square unit). I only wrote $a^2$ to make sure the dimensions agree. I have no intuition as to why it should be there though, although I do understand it for the pressure. Horizontal forces at station 2 are $F_{x2} = -ap_2+\rho a^2v_2^2 \cos^2 \beta_2$. Then the reaction would be $R_x = F_{x1} - F_{x2}$. How does this look so far?

 

[haruspex]

I find the description much too vague. Is there a diagram with this? The horizontal is mentioned, so is one of the two dimensions vertical or are they both horizontal?

 

[member 428835]

Here's the picture. Please let me know if I can be clearer.

 

[haruspex]

Yes, that certainly clarifies it.
Your work so far looks right to me, except that I see no reason for the extra factor of a. The dimensions were fine without it, on the basis that, as you say, force is per unit depth. I think your mistake is in also saying density is per unit area. It might help to introduce some arbitrary length in the direction perpendicular to the page.
I believe some simplification can be done. Since the flow is incompressible, there is a relationship between v1, v2, β1 and β2.

 

[member 428835]

Yes, that certainly clarifies it.
Your work so far looks right to me, except that I see no reason for the extra factor of a. The dimensions were fine without it, on the basis that, as you say, force is per unit depth. I think your mistake is in also saying density is per unit area. It might help to introduce some arbitrary length in the direction perpendicular to the page.
I believe some simplification can be done. Since the flow is incompressible, there is a relationship between v1, v2, β1 and β2.

Okay, so let's let $W$ be the width coming out of the page. Then a force balance implies $R_x+p_1aW + \rho V_1^2 a W \cos \beta_1 = p_2aW + \rho V_2^2 a W \cos \beta_2$. Does this look correct? Here $R_x$ is the force exerted on the shapes.

Is the relationship you're referring to Bernoulli's?

 

[haruspex]

Here R_x is the force exerted on the shapes.

The way you have written the equation, it would be the force the shapes exert on the flow. Other than that, it looks right.

Is the relationship you're referring to Bernoulli's?

No, it is the equation which expresses non-compressibility, i.e. volumetric flow out = volumetric flow in.

 

[member 428835]

The way you have written the equation, it would be the force the shapes exert on the flow. Other than that, it looks right.

Hmmm can you explain how you knew this? I'm wanting to know so I can do this on my own next time. Also, looking at the equation I wrote, shouldn't the pressure force at station two be negative when placed on the right side of the equation, since it is acting to the left, in the negative-x direction?

No, it is the equation which expresses non-compressibility, i.e. volumetric flow out = volumetric flow in.

Do we need this since density, angles, and both velocities are given? Isn't conservation of volume $\rho a W v_1 \cos \beta_1=\rho a W v_2 \cos \beta_2$?

 

[haruspex]

can you explain how you knew this?

Just look at the two pressure terms. p₁ applies a force left to right, p₂ applies a force right to left. The net force to the right is therefore the p₁ term minus the p₂ term.

Do we need this since density, angles, and both velocities are given? Isn't conservation of volume $\rho a W v_1 \cos \beta_1=\rho a W v_2 \cos \beta_2$?

Yes, that's the equation. As I wrote, it provides a simplification to the net horizontal force.

 

[member 428835]

Just look at the two pressure terms. p₁ applies a force left to right, p₂ applies a force right to left. The net force to the right is therefore the p₁ term minus the p₂ term.

Shoot, so I guess what I should have written is the total force exerted on the shapes equals the momentum force and the pressure force. Then we would have $R_x = p_1 a W - p_2 a W + a W v_1^2 \cos \beta_1-a W v_2^2 \cos \beta_2$? This seems like I've written something wrong though since the momentum force at station 2 is going in the opposite direction as the pressure force at station 2. What are your thoughts?

Yes, that's the equation. As I wrote, it provides a simplification to the net horizontal force.

Gotcha!

 

[haruspex]

the momentum force at station 2 is going in the opposite direction as the pressure force at station 2.

It is? The station 1 terms there are both positive, and the station 2 terms both negative. It all looks consistent to me.

 

[member 428835]

It is? The station 1 terms there are both positive, and the station 2 terms both negative. It all looks consistent to me.

Hmmmm, at station 2 I thought momentum was leaving to the right yet pressure was acting to the left. Is this true? If so, wouldn't they have opposite signs since they are acting in opposite directions?

Thanks for your help by the way!

 

[haruspex]

First, I only just noticed that in going from post #1 to post #5 you dropped the powers of 2 on the cos terms. Is that a typo? I believe they should be squared. (If not squared, the simplification I mentioned is not that dramatic.)

Hmmmm, at station 2 I thought momentum was leaving to the right yet pressure was acting to the left. Is this true? If so, wouldn't they have opposite signs since they are acting in opposite directions?

Thanks for your help by the way!

With the above corrected, there is no change to horizontal momentum, so it becomes academic - the terms cancel.
Overlooking that, the drop in horizontal momentum, will be momentum₁-momentum₂. That would be a result of a leftward force from the vanes. The corresponding rightward force the flow exerts on the vanes would therefore equal momentum₁-momentum₂.

 

[member 428835]

First, I only just noticed that in going from post #1 to post #5 you dropped the powers of 2 on the cos terms. Is that a typo? I believe they should be squared. (If not squared, the simplification I mentioned is not that dramatic.)

Yes, that is a typo. Thanks for pointing this out.

With the above corrected, there is no change to horizontal momentum, so it becomes academic - the terms cancel.
Overlooking that, the drop in horizontal momentum, will be momentum₁-momentum₂. That would be a result of a leftward force from the vanes. The corresponding rightward force the flow exerts on the vanes would therefore equal momentum₁-momentum₂.

Okay, so that I understand you we have $\Sigma {F} = m {a}$ (since mass is not time dependent here) which implies the x-direction balance becomes $-R_x + \rho a W v_1^2 \cos^2\beta_1-\rho a W v_2^2 \cos^2\beta_2 + p_1aW-p_2aW=0$. Conservation of mass implies flow in = flow out, which implies $\rho a W v_1^2 \cos^2\beta_1=\rho a W v_2^2 \cos^2\beta_2$. This simplifies the equation to $-R_x + p_1aW-p_2aW=0 \implies R_x=p_1aW-p_2aW$. This is what you were saying, right?

 

[haruspex]

Yes, that is a typo. Thanks for pointing this out.

Okay, so that I understand you we have $\Sigma {F} = m {a}$ (since mass is not time dependent here) which implies the x-direction balance becomes $-R_x + \rho a W v_1^2 \cos^2\beta_1-\rho a W v_2^2 \cos^2\beta_2 + p_1aW-p_2aW=0$. Conservation of mass implies flow in = flow out, which implies $\rho a W v_1^2 \cos^2\beta_1=\rho a W v_2^2 \cos^2\beta_2$. This simplifies the equation to $-R_x + p_1aW-p_2aW=0 \implies R_x=p_1aW-p_2aW$. This is what you were saying, right?

Yes.

 

[member 428835]

Yes.

Cool! So how would we deal with the vertical force? I don't think pressure is considered, and conservation of mass implies vertical volumetric flow is zero too, right?

 

[haruspex]

Cool! So how would we deal with the vertical force? I don't think pressure is considered, and conservation of mass implies vertical volumetric flow is zero too, right?

There is a change to the vertical momentum of the fluid.

 

[member 428835]

There is a change to the vertical momentum of the fluid.

So I am now wondering if the correct conservation of volume constraint should instead be written as $v_1 \cos \beta_1 + v_1 \sin \beta_1=v_2 \cos \beta_2 + v_2 \sin \beta_2$? If so then $F_x$ is not as simple, but then we have $-F_y +a W \rho v_1^2 \sin^2 \beta_1-a W \rho v_2^2 \sin^2 \beta_2=0$, where I did not include pressure since I think it acts only horizontally. What do you think?

 

[haruspex]

So I am now wondering if the correct conservation of volume constraint should instead be written as $v_1 \cos \beta_1 + v_1 \sin \beta_1=v_2 \cos \beta_2 + v_2 \sin \beta_2$?

By what reasoning?

 

[member 428835]

By what reasoning?

Shoot, now I'm confusing myself. Perhaps you could explain to me why this is not correct? Sorry for dragging this out, I just really want to understand this.

 

[haruspex]

Shoot, now I'm confusing myself. Perhaps you could explain to me why this is not correct? Sorry for dragging this out, I just really want to understand this.

Consider the flow in. If the speed is v₁, how would you calculate the volumetric flow? What do you need to multiply by?

 

[member 428835]

Consider the flow in. If the speed is v₁, how would you calculate the volumetric flow? What do you need to multiply by?

Would have to be the two space dimensions orthogonal to velocity (in this case $a$ and $W$). Then volumetric flow is $a W v_1$.

 

[haruspex]

Would have to be the two space dimensions orthogonal to velocity (in this case $a$ and $W$). Then volumetric flow is $a W v_1$.

a is not the width orthogonal to v₁. Consider the angle.

 

[member 428835]

a is not the width orthogonal to v₁. Consider the angle.

Shoot, yea I made a typo. Should be (in the x-direction) $a W v_1 \cos \beta_1$ at station 1 and the y-direction should be $W v_1\sin\beta_1$ but what length to introduce? I assume this is what you were getting at? Would we just define a length from station 1 to station 2, call that length $H$. Then $H W v_1\sin\beta_1$.

 

[haruspex]

Should be (in the x-direction) $aW v_1 \cos \beta_1$ at station 1 and the y-direction should be $aW v_1\sin\beta_1$

Right, but the y direction is not relevant here. We just want the volumetric flow rate through each station, which is the horizontal component.
Not sure what you are asking about a length. The expression already has the right dimensions for volumetric flow.

 

[member 428835]

Not sure what you are asking about a length. The expression already has the right dimensions for volumetric flow.

Yea, I was asking about how to perform a mass balance in the $y$ direction. Would it just be with velocity and not pressure since pressure acts horizontally? In other words, $-F_y + aW\rho v^2_1\sin^2 \beta_1-aW\rho v^2_2\sin^2 \beta_2=0$?

 

[haruspex]

Yea, I was asking about how to perform a mass balance in the $y$ direction. Would it just be with velocity and not pressure since pressure acts horizontally? In other words, $-F_y + aW\rho v^2_1\sin^2 \beta_1-aW\rho v^2_2\sin^2 \beta_2=0$?

Yes, depending on how you are defining F_y. Is that the force the flow exerts on a blade (R_y) or the force a blade exerts on the flow?

 

[member 428835]

I thought it was the force the blade exerts on the flow (from how I set it up). You agree?

 

[haruspex]

I thought it was the force the blade exerts on the flow (from how I set it up). You agree?

Yes.

 

[member 428835]

Could you explain why these velocities don't cancel out like they did in the $x$ direction? We know $A W v_1 \cos \beta_1=A W v_2 \cos \beta_2$. Why is this not true for the $y$ component?

 

[haruspex]

Could you explain why these velocities don't cancel out like they did in the $x$ direction? We know $A W v_1 \cos \beta_1=A W v_2 \cos \beta_2$. Why is this not true for the $y$ component?

In the x direction, we know that the volumetric rate of flow into the blade system from the left must equal the flow out on the right. In the y direction there is no requirement that volumetric rate of flow up the left side of the blades must equal that on the right side.

 

[member 428835]

In the x direction, we know that the volumetric rate of flow into the blade system from the left must equal the flow out on the right. In the y direction there is no requirement that volumetric rate of flow up the left side of the blades must equal that on the right side.

Ok, I think I have it! I wrote up a proof, starting with the momentum equation, assuming incompressible, steady, non-viscous flow and neglecting body forces. I'll attach it since typing it took some time. Let me know what you think, as I have one question with what I wrote.

 

[haruspex]

Ok, I think I have it! I wrote up a proof, starting with the momentum equation, assuming incompressible, steady, non-viscous flow and neglecting body forces. I'll attach it since typing it took some time. Let me know what you think, as I have one question with what I wrote.

Yes, that all looks correct to me.

 

[member 428835]

Yes, that all looks correct to me.

Awesome, thanks so much for your help! I really appreciate it!

 

[Chestermiller]

I get the opposite signs than you have obtained for the components of the external force necessary to hold the vane in place.

 

[haruspex]

I get the opposite signs than you have obtained for the components of the external force necessary to hold the vane in place.

Good point. I read the diagram as indicating Rx and Ry are the forces the flow exerts on the vanes, and did not notice that wording.