Bernoulli vs. Reynold Transport Theorem Problem for Raincap

Hi,

I am trying to map out the forces that act on a raincap installed at the end of a pipe, but I have ran into a problem where I am getting 2 different equations depending on the method that I use. I'm definitely missing something here and I was hoping someone is about to point it out to me.

Consider the case where there is a hinged raincap at the end of a pipe with a known mass air flow going through the pipe. Assuming no viscous effects, I was hoping to find the force of the airflow acting on the raincap as it exits.

For my first scenario, I considered the cap fully closed and wanted to find the stagnation pressure of the airflow on the inside of the endcap (I was interested to see the worst-case scenario to see if the airflow would be capable of opening the raincap). To do this, I used Bernoulli's equation and assumed that all of the dynamic pressure of the airflow is converted to static pressure. I then took that pressure multiplied by the area of the cap to find force.

I then wanted to consider the case where the raincap is open at some angle theta. To do this, I used a simplified Reynolds Transport Theorem (RTT), took an angled control surface such that the flow entering from the pipe and the angled flow leaving after being deflected by the raincap are both perpendicular to the control surface. Assuming no losses (I'm looking for a relatively rough value here), I took the flow entering the control volume as being equal to the flow exiting. For the flow entering, there is only a y-component (v), for the flow existing, I found the x and y components with respect to theta. Completing the equation, I get a very similar result to what I had when using Bernoulli's principle, however there is a factor of 1/2 missing. This is more obvious if you assume theta is very small and sin(theta) approaches 0.

This doesn't make sense to me; why would the pressure be greater when assuming that all of the dynamic pressure becomes static versus when the flow still continues to move. I have definitely done something wrong here.Attached is my work.

Magic865 said:

TL;DR Summary: I am trying to find the forces acting on a raincap mounted on a pipe with a known mass airflow flowing through the pipe, but I am getting 2 different equations depending on whether I use Bernoulli's equation or Reynold Transport Theorem. Scenario 1 is with the cap closed, scenario 2 is with the cap open at angle theta. Attached is my work.

Hi,

I am trying to map out the forces that act on a raincap installed at the end of a pipe, but I have ran into a problem where I am getting 2 different equations depending on the method that I use. I'm definitely missing something here and I was hoping someone is about to point it out to me.

Consider the case where there is a hinged raincap at the end of a pipe with a known mass air flow going through the pipe. Assuming no viscous effects, I was hoping to find the force of the airflow acting on the raincap as it exits.

For my first scenario, I considered the cap fully closed and wanted to find the stagnation pressure of the airflow on the inside of the endcap (I was interested to see the worst-case scenario to see if the airflow would be capable of opening the raincap). To do this, I used Bernoulli's equation and assumed that all of the dynamic pressure of the airflow is converted to static pressure. I then took that pressure multiplied by the area of the cap to find force.

I then wanted to consider the case where the raincap is open at some angle theta. To do this, I used a simplified Reynolds Transport Theorem (RTT), took an angled control surface such that the flow entering from the pipe and the angled flow leaving after being deflected by the raincap are both perpendicular to the control surface. Assuming no losses (I'm looking for a relatively rough value here), I took the flow entering the control volume as being equal to the flow exiting. For the flow entering, there is only a y-component (v), for the flow existing, I found the x and y components with respect to theta. Completing the equation, I get a very similar result to what I had when using Bernoulli's principle, however there is a factor of 1/2 missing. This is more obvious if you assume theta is very small and sin(theta) approaches 0.

This doesn't make sense to me; why would the pressure be greater when assuming that all of the dynamic pressure becomes static versus when the flow still continues to move. I have definitely done something wrong here.Attached is my work.

You don’t use Bernoulli or Reynolds transport. They are used together for the angled lid analysis. Bernoulli is conservation of energy along a streamline, Reynolds is momentum rate( i.e. Newtons Second Law) for a control volume.Question 1:
What is the "Area" of the outlet?Question 2:
If mass is conserved and we are neglecting the effects of compressibility, what is the outlet velocity of the flow in the angled lid scenario? Assume a uniform velocity distribution across the area of the outlet.

Question 3:
Where is the pressure force acting on the inlet side of the control volume?

It would be best if you presented your analysis in the math formatting used on the site. see: LaTeX Guide

erobz said:

You don’t use Bernoulli or Reynolds transport. They are used together for the angled lid analysis. Bernoulli is conservation of energy along a streamline, Reynolds is momentum rate( i.e. Newtons Second Law) for a control volume.Question 1:
What is the "Area" of the outlet?Question 2:
If mass is conserved and we are neglecting the effects of compressibility, what is the outlet velocity of the flow in the angled lid scenario? Assume a uniform velocity distribution across the area of the outlet.

Question 3:
Where is the pressure force acting on the inlet side of the control volume?

It would be best if you presented your analysis in the math formatting used on the site. see: LaTeX Guide

Hi erobz,

I understand that Bernoulli is conservation of energy while RTT is conservation of momentum, but I am not sure how I can use these at the same time to come up with an equation for the force on the lid.

Addressing your questions:
1: Realistically, this area would be something of an irregular shape, but for my analysis, I simply assumed that the flow would maintain its shape deflecting off of the lid - you can picture the flow as a jet that maintains its cross-section from when it was in the tube. I should have mentioned that the pipe is of a known diameter/area.

2: Given my 'jet' assumption of a constant cross section, the magnitude of velocity of the fluid stays the same across its interaction with the lid - the speed and shape of the air stay the same.

3: For the angled scenario, the interaction between the airflow and lid is just past the outlet of the pipe, so its within atmospheric conditions. For the closed scenario, I'm considering the pressure build up (dynamic to static) just before the lid opens for the flow to exit. I'm not sure if I'm understanding your question correctly here.

I'm not sure if I'm approaching this problem correctly. My end goal here is to have the force of the airflow against the lid as a function of the lid's angle. I just ran into this discrepancy when considering these separate scenarios of when the raincap is closed, versus when it is angled, which made me question whether or not I'm doing this correctly.

I apologize for posting with just handwriting, I am new here and not used to LaTeX and was also hoping that the diagram would be helpful. I will try to re-post/comment in LaTeX soon.

Magic865 said:

Hi erobz,

I understand that Bernoulli is conservation of energy while RTT is conservation of momentum, but I am not sure how I can use these at the same time to come up with an equation for the force on the lid.

For starters I think you should do a force/moment balance directly on the lid. Its weight, and pressure force acting at an unknown position (normal to the lid) such that static equilibrium is satisfied.

Magic865 said:

Addressing your questions:
1: Realistically, this area would be something of an irregular shape, but for my analysis, I simply assumed that the flow would maintain its shape deflecting off of the lid - you can picture the flow as a jet that maintains its cross-section from when it was in the tube. I should have mentioned that the pipe is of a known diameter/area.

Yeah, its not straightforward in the least bit. You are going to need to do some significant simplification and hope it all holds together under scrutiny.

For an approximation I would try ##A_o = 2r ( 2r \sin \theta ) \cos \theta = 2r^2 \sin (2\theta) ##. We are looking for the area normal to the velocity as a rectangle of width ( into the page ) of ##2r##

Let me know if you would like further clarification on that. Having the flow not change shape seems ( to me ) to be an excessive oversimplification that it will be of little to no utility in the problem you wish to solve.

Magic865 said:

2: Given my 'jet' assumption of a constant cross section, the magnitude of velocity of the fluid stays the same across its interaction with the lid - the speed and shape of the air stay the same.

The velocity is found via continuity ( under the incompressible flow approximation )

##A_i v_i = A_o v_o##

Magic865 said:

3: For the angled scenario, the interaction between the airflow and lid is just past the outlet of the pipe, so its within atmospheric conditions. For the closed scenario, I'm considering the pressure build up (dynamic to static) just before the lid opens for the flow to exit. I'm not sure if I'm understanding your question correctly here.

There is a (gauge) pressure acting over the lower control surface that you have not factored in. The control volume is an isolated system. That pressure in the surrounding flow is an external force acting on the control volume. The outlet we will take to be atmospheric pressure.

Magic865 said:

I'm not sure if I'm approaching this problem correctly. My end goal here is to have the force of the airflow against the lid as a function of the lid's angle. I just ran into this discrepancy when considering these separate scenarios of when the raincap is closed, versus when it is angled, which made me question whether or not I'm doing this correctly.

Hopefully we can get there checking any misunderstandings (and assumptions) along the way without too much torment.

Magic865 said:

I apologize for posting with just handwriting, I am new here and not used to LaTeX and was also hoping that the diagram would be helpful. I will try to re-post/comment in LaTeX soon.

Thank You!

https://en.m.wikipedia.org/wiki/Stagnation_pressure

Just to clarify a bit more my intentions:



You are missing the external force ##P_iA_i## acting over control surface inlet. ##P_i## will be found from Bernoulli's Principle as a function of the inlet velocity ##v_i##.

The outlet velocity as a function of ##v_i## is found from conservation of mass ## A_iv_i = A_ov_o## for "incompressible" flow (this is a simplifying assumption - the flow is compressible).

Your force balance in the vertical direction will be:

$$P_i A_i - R_y = \dot m v_i \left( \frac{A_i}{A_o} \sin \theta - 1 \right)$$

Then you will apply Bernoulli's ( inlet to outlet) to eliminate ##P_i## ( with ##P_o= 0 \rm{guage}##)

Repeat analysis for horizontal direction, and ##R_x,R_y## can be put in terms of the mass flow rate ##\dot m##

Then for some angle ##\theta## you balance the moments about the hinge of the lid as a function of ##\dot m##. ##R_{\perp}## at distance ##l## (along the lid) must balance the torque from the weight of the lid for there to be a solution.

That is the approach I was planning to take. It would be fun to see where (if) it makes sense?

erobz said:

Just to clarify a bit more my intentions:

View attachment 329595

You are missing the external force ##P_iA_i## acting over control surface inlet. ##P_i## will be found from Bernoulli's Principle as a function of the inlet velocity ##v_i##.

The outlet velocity as a function of ##v_i## is found from conservation of mass ## A_iv_i = A_ov_o## for "incompressible" flow (this is a simplifying assumption - the flow is compressible).

Your force balance in the vertical direction will be:

$$P_i A_i - R_y = \dot m v_i \left( \frac{A_i}{A_o} \sin \theta - 1 \right)$$

Then you will apply Bernoulli's ( inlet to outlet) to eliminate ##P_i## ( with ##P_o= 0 \rm{guage}##)

Repeat analysis for horizontal direction, and ##R_x,R_y## can be put in terms of the mass flow rate ##\dot m##

Then for some angle ##\theta## you balance the moments about the hinge of the lid as a function of ##\dot m##. ##R_{\perp}## at distance ##l## (along the lid) must balance the torque from the weight of the lid for there to be a solution.

That is the approach I was planning to take. It would be fun to see where (if) it makes sense?

Hi erobz,

Sorry for the delay. This is my first time really using LaTeX and it has been pretty painful starting off.
As promised, below is my original hand calc that I scanned, but correctly put in LaTeX now:

Scenario 1 in my first post was when the lid is closed, and the Bernoulli Principle is applied:

$$ P_1 + \frac 1 2 \rho v^2_1 + \rho g h_1 = P_2 + \frac 1 2 \rho v^2_2 + \rho g h_2 $$

$$ \text{where:}$$

$$\rho g \left( h_2 - h_1 \right) \approx 0 $$

$$P_2 - P_1 \approx 0$$

$$v_2 = 0$$

$$ \text{such that:}$$

$$\frac 1 2 \rho v^2_1 = P_2$$

$$\frac 1 2 \rho \left( \frac {\dot m} {A \rho} \right)^2 = P_2$$

$$ \text{Then the force, using F = PA, would be:}$$

$$F = \frac {\dot m^2} {2 A \rho} $$Scenario 2 in my original post was considering an angled opening of the lid, this angle being ## \theta##, where I applied the Reynold Transport Theorem (RTT):$$\frac d {dt} m \vec V = \frac d {dt} \int_{CV} \vec V \, dm + \sum_{Outlets} \vec {V}_o \dot m_o - \sum_{Inlets} \vec V_i \dot m_i$$
$$ \text{Where: } \frac d {dt} \int_{CV} \vec V \, dm = 0 \text{ due to steady state}$$

This leaves us with the simplified RTT equation:

$$\sum F = \sum_{Outlets} \vec V_o \dot m_o - \sum_{Inlets} \vec V_i \dot m_i$$

Neglecting Losses:
$$|\vec V_o| = |\vec V_i| = V$$
$$\dot m_o = \dot m_i = \dot m = \rho A V$$

$$ \text{Breaking up the } \vec V_o \text{ and } \vec V_i \text{ vectors into components, we get:} $$
$$\vec V_i = \vec u_i + \vec v_i \text{ where: } \vec u_i = 0$$
$$\vec V_o = \vec u_o + \vec v_o$$
$$= V \cos \theta + V \sin \theta$$

Looking at the sum of forces in the y-direction that act on the flow, there is only ##R_y##, the reaction force of the lid acting in the negative y-direction:
$$-R_y = \dot m V \sin \theta - \dot m V$$
$$= \dot m V \left( \sin \theta - 1 \right)$$
$$R_y = \dot m V \left(1 - \sin \theta \right)$$
$$ = \frac {\dot m^2} {A \rho} \left(1 - \sin \theta \right)$$
$$ \text{Such that when } \theta \text{ approaches } 0, R_y \text{ approaches: }$$
$$F = \frac {\dot m^2} {A \rho}$$
In contrast to the first equation:
$$F = \frac {\dot m^2} {2 A \rho} $$

Now addressing all your points in the order that you stated them.
I have started a moment/force balance on the lid. The gravitational forces are relatively simple, I ran into an issue with this airflow force and isolated only this part of the problem when making my post. I just was not sure if I am accounting for it correctly.

Your approximated outlet area is very vital input. Thank you for that! I think this is the first place where I went off course. In my assumptions, I stated:
$$\dot m_o = \dot m_i = \dot m = \rho A V$$
but really, I should have stated:
$$\dot m_o = \dot m_i = \dot m = \rho A_i V_i = \rho A_o V_o$$
I treated the 2 areas the same and I factored them out as if they were the same value.

I do have a question about your approach however; where does the ##\cos \theta## come from?
Shouldn't it just be ##2r \sin \theta## for the vertical height of the area?

For the pressure acting on the control volume, it definitely exists, but it is very small. My anticipated mass flow rates are tiny for the area that they are going through - I'm expecting a static pressure of about 0.1kPa this close to the exit. This is why I stated that ## P_2 - P_1 \approx 0 ## - but yes, it currently is missing.

Thank you for your help so far - it has been very valuable, but I am still confused about my underlying problem of Bernoulli and Reynold. If you can consider the 2 below cases to know what I mean.For case 1, imagine that there is a circular cup that is fixed, as pictured below. It is exposed to a uniform airflow upward. Using my approach with Bernoulli's equation, all of the dynamic pressure/energy of the flow is converted to static pressure (similar to a pitot tube analysis) and the final force acting on the cup would be ##F = \frac {\dot m^2} {2 A \rho}## where ##\dot m## would be the mass airflow hitting the bottom of the cup and ##A## would be that same area

Now consider the same thing happening, but a cylinder of airflow matching the shape and mass airflow of the previous scenario, but hitting a flat plate perfectly perpendicular, such that the mass flowrate splits evenly after and half goes one way and the other half goes the other way.

In this case, looking at the vertical components, we will be left with only ##F = \frac {\dot m^2} {A \rho}##

First question here is: are these equations correct? Am I misinterpreting something?
Second question is: why are the final force values so different? I would have expected them to either be much closer, or for the force in the first scenario to be larger given that all the kinematic energy goes away in that case.

Magic865 said:

I do have a question about your approach however; where does the ##\cos \theta## come from?
Shouldn't it just be ##2r \sin \theta## for the vertical height of the area?

It's really just a simplification. I want the velocity to be normal to the outlet, parallel with the plate. The further out we go the less I think that is a good approximation. The air is diffusing is 3d space, I think we want to avoid using the fundamental form the last term where we are taking the dot prouct:

$$\sum \boldsymbol F = \frac{d}{dt}\int_{cv} \boldsymbol v \rho ~d V\llap{-} + \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A $$

Magic865 said:

For the pressure acting on the control volume, it definitely exists, but it is very small. My anticipated mass flow rates are tiny for the area that they are going through - I'm expecting a static pressure of about 0.1kPa this close to the exit. This is why I stated that ## P_2 - P_1 \approx 0 ## - but yes, it currently is missing.

We have to include it otherwise the application of Bernoulli's would be inconsistent. We are already neglected elevation head, if you neglect pressure there is simply no reservoir for kinetic energy to flow into. The pressure is the crucial element in the computation. It's necessarily greater than the ambient pressure, or else the analysis attempt falls flat on its face.

Magic865 said:

Thank you for your help so far - it has been very valuable, but I am still confused about my underlying problem of Bernoulli and Reynold. If you can consider the 2 below cases to know what I mean.For case 1, imagine that there is a circular cup that is fixed, as pictured below. It is exposed to a uniform airflow upward. Using my approach with Bernoulli's equation, all of the dynamic pressure/energy of the flow is converted to static pressure (similar to a pitot tube analysis) and the final force acting on the cup would be ##F = \frac {\dot m^2} {2 A \rho}## where ##\dot m## would be the mass airflow hitting the bottom of the cup and ##A## would be that same area
View attachment 329606

Where is the outflow in this? If nothing is coming out, then momentum is accumulating ( i.e its rate of change is non-nero - negative) in the control volume, so you have:

$$\sum \boldsymbol F = \overbrace{\frac{d}{dt}\int_{cv} \boldsymbol v \rho ~d V\llap{-}}^{\text{<0}} -\dot m v_i $$

So there is necessarily some force acting on the mass entering the control volume that is taking its velocity from ##v_i## to ##0##. So its the same force you have in the second scenario plus something else. The force will be greater.

EDIT: also you are flipping the sign on the forces. They are the forces that either accelerate the control volume (velocity of the control surface always must be referenced from inertial frame), or they are the forces keeping the control volume inertial ( not accelerating ). So in the both cases you present, the net external force acting on the control volume is opposite the direction you have shown. The forces are restraining forces responsible for holding the control volume still (as it is). I don't want there to be confusion about negative signs.

erobz said:

The air is diffusing is 3d space, I think we want to avoid using the fundamental form the last term where we are taking the dot prouct:

∑F=ddt∫cvvρ dV−+∫csvρV⋅dA

Ok, I agree, but the angle of the dot product here is the angle between the control surface normal vector and the flow vector. We should either not call this variable ##\theta## as this is the lid angle (lets say this is ##\alpha##), or we take our area with the assumption that all the outgoing flow is perpendicular to this area, in which case ##\alpha = 90°, \cos \alpha = 1##

erobz said:

We have to include it otherwise the application of Bernoulli's would be inconsistent.

Ok, lets leave this in

erobz said:

Where is the outflow in this?

There is no outflow. Consider this as the same thing that happens with a pitot tube; initially it does accumulate the fluid, and the pressure rises, but at a certain point, it stop accumulating fluid and stays at a pressure proportional to the flow rate hitting the cup - the stagnation pressure. At this point, the flow is actually going around the cup. Have a look at these solved problems to see what I mean:
Pitot Tube in Water
Manometer Pitot Tube in Wind Tunnel
I guess the question is; can I apply the logic in these videos to a fixed cup receiving a flow?

Magic865 said:

There is no outflow. Consider this as the same thing that happens with a pitot tube; initially it does accumulate the fluid, and the pressure rises, but at a certain point, it stop accumulating fluid and stays at a pressure proportional to the flow rate hitting the cup - the stagnation pressure. At this point, the flow is actually going around the cup. Have a look at these solved problems to see what I mean:
Pitot Tube in Water
Manometer Pitot Tube in Wind Tunnel
I guess the question is; can I apply the logic in these videos to a fixed cup receiving a flow?

The cup doesn't know how big it is. By applying RTT here you are telling the equation that mass is entering the cup at some velocity, and it is being accelerated to rest inside the cup, nothing is coming out (or around it), only going in. The accumulation term is for all time non-zero negative in the model. To calculate the force (restraining the cup) you have to say something concrete about this acceleration i.e. it needs characterization. Bernoulli's only applies to steady flow. The flow is distinctly non-steady inside the cup, so it's probably not so straightforward to sort out the meshing of the theories.

Anyhow, the idea I have is that Bernoulli's is used with RTT (Momentum) when the flow is steady and being directed by the lid:

$$P_i = \frac{1}{2} \rho v_i^2 \left( \left( \frac{A_i}{A_o} \right)^2 - 1 \right)$$

So sub that into the relation I gave in #6. You can then get ##R_y(\dot m, \theta ), R_x( \dot m, \theta) ##. You'll have the magnitude of ##R## from that.

Now, there is something bothering me... I'm not finding enough equations to fix the point of application of the force on the lid. To me it seems like for a given lid weight, the angle is fixed by the mass flowrate ##\dot m##. I can't see a light at the end of the tunnel anymore...

hmmm...

Perhaps there is something not in plain sight in my assumptions that demands a uniform pressure distribution on the wetted surface that I'm not picking up on. That would fix the angle as a function of the mass flowrate like I expect. Better think about whether or not that is a reasonable assertion.

I forgot about this ( its been a while and you have probably solved it), but just in case...

I believe we need to make some simplifying assumption that we have a uniform pressure distribution acting on over the lid, applied to the center of the lid. Since we are ignoring viscosity the magnitude ##\vec{R}## must be normal to the lid. ( i.e. ##\vec R## acting on the control volume is just third law manifestation of the pressure distribution acting on the lid). So the final equation comes from the sum of torques on the lid in steady state.

Let ##M## be the mass of the lid.

$$\sum \tau = \sqrt{ R_y^2 + R_x^2 } \cancel{r} - M g \cos \theta \cancel{r} = 0 $$

So putting everything in terms of ##Q## via continuity (the volumetric flowrate in the pipe) the final system:

$$ P_i A_i - R_y = \frac{\rho Q^2}{A_i}\left( \frac{A_i}{A_o} \sin \theta - 1 \right) \tag{1 - RTT}$$

$$ R_x = \frac{\rho Q^2}{A_o} \cos \theta \tag{2 - RTT} $$

$$ P_i = \frac{\rho Q^2}{2} \left( \frac{1}{A_o^2} - \frac{1}{A_i^2} \right) \tag{3 - Bernoulli} $$

$$ \sqrt{ R_y^2 + R_x^2 } = M g \cos \theta \tag{4 - moment} $$

So after combining the system you pick a ##\theta##( lid angle) and solve for ##Q##.

Lastly I realized I didn't explain where the ##\cos \theta## factor came from. I believe it comes from the uniform velocity distribution in the continuity equation ##Q = vA##. I think implicitly we are saying that the velocity is normal to the area in that equation, but I could be mistaken.

In my opinion we either want the component of the velocity parallel to the area normal, or the area normal to the velocity. I'm up for discussion on this though, because it is not 100% clear to me now that a have another go at it.