I Explain Bernoulli at the molecular level?

[A.T.]

But in the second case, a rocket accelerating through the air, it is not the air that is accelerating but your frame of reference (and the rocket of course). For this case no pressure gradient needs to exist in the air to facilitate the acceleration. This is because the air is not accelerating with respect to an inertial frame of reference.

So if the cart here is accelerating to the right in still air, the force from the "airflow" on the cart at a particular instant is not different than the force from the air on a cart as if it were moving at a constant velocity? Here the purple arrows are area Normals, not velocity.

What does the force on your air accelerating cart, have to do, with the pressure gradient in non-accelerated air, that @Arjan82 talks about?

 

[erobz]

What does the force on your air accelerating cart, have to do, with the pressure gradient in non-accelerated air, that @Arjan82 talks about?

I'm trying to see why when I solve the equations for an accelerating cart vs a a cart moving at constant velocity, I get different forces acting on the cart from the air. My interpretation of the statement is the force (from the air) shouldn’t be different just because the cart is accelerating. So I want to see where I go wrong in these solutions. Where I am not being careful enough in the reduction of the equations. The pressure gradient in the direction of the flow should integrate over the frontal area of the cart to contribute to this force. I get a force that is dependent on velocity $v$ in both cases, and in the accelerating case $v$ is time dependent. I'm not seeing the disconnection between applied force from the air mass flow and pressure in the mass flow. It is feeling like a free lunch. I'll share my analysis tomorrow.

 

[Arjan82]

In case you're wondering about wings, Bernoulli is violated since a wing adds energy to the air. This is easier to visualize with a propeller, but it is the same principle with a wing. Just behind the wing the work done is mostly related to a pressure jump as the air flows downwards behind the wing. The affected air is slowed down and pressure reduced due to the surrounding air performing "negative" work on the affected air, and the velocity of the affected air when its pressure returns to ambient is called the exit velocity, and reflects the work performed by a wing.

This is actually not true. If you analyze a (non-accelerating) wing from a wing-attached reference frame Bernoulli holds just fine (if you ignore viscosity, so you are looking at potential flow). Take XFoil, which is a code that can compute pressures around airfoils (2D section of a wing). This code compute the airflow around an airfoil using the potential flow equations and then relies on Bernoulli to compute the pressure. This works just fine. Also for a propeller. And, in fact, most (all?) potential flow codes rely on Bernoulli to compute pressure.

So, how does this work since a propeller clearly adds energy to the flow, which would violate Bernoulli. The trick is to analyze the flow in the wing-attached reference frame (or propeller-blade attached reference frame, but that's non-inertial, so be careful...). From that reference frame's perspective, there is no energy added to the flow.

Also, at least for subsonic flow, there is no pressure jump anywhere, the pressure is smooth. Are you perhaps referring to the jump in the potential behind the wing, which you need to apply the Kutta condition on a wing? Potential flow is rotation-free, but a wing adds rotation to the flow, so to analyze a wing with potential flow you need a discontinuity (i.e. a surface over which a discontinuity can exist) in your solution. This surface is located behind the wing, attached to the trailing edge, over which the potential ($\phi$) has a discontinuity (a jump). This is often referred to as the wake model of a wing (or propeller). So this introduces a jump in the potential, and thus also in flow direction over that surface.

This is how a wake model for a wing looks:

Or, including roll-up of the tip vortices:

And this for a propeller (the blue plane). This is a (computational) surface that models the wake of the propeller.

 

[Arjan82]

I'm trying to see why when I solve the equations for an accelerating cart vs a a cart moving at constant velocity, I get different forces acting on the cart from the air. My interpretation of the statement is the force (from the air) shouldn’t be different just because the cart is accelerating. So I want to see where I go wrong in these solutions. Where I am not being careful enough in the reduction of the equations. The pressure gradient in the direction of the flow should integrate over the frontal area of the cart to contribute to this force. I get a force that is dependent on velocity $v$ in both cases, and in the accelerating case $v$ is time dependent. I'm not seeing the disconnection between applied force from the air mass flow and pressure in the mass flow. It is feeling like a free lunch. I'll share my analysis tomorrow.

I'm wondering which equations you are solving, since they look different in a non-inertial frame of reference.

 

[user079622]

The stagnation pressure however does feel the acceleration. It will indicate a somewhat higher pressure, until the velocity is constant again.

Are you sure in this?
If ball hit the wall with constant speed of 20km/h, it is the same as if ball accelerate and in time t1 hit the wall with 20km/h. Isn't this same case for air particle in fluid?

For ease of analysis we usually attach the frame of reference to the object. We also do this trick when we do measurements in windtunnels. In that case you are comparing:
1) a stationary rocket sitting in a windtunnel that is generating a certain airspeed, say V,
2) a rocket moving at that same speed V through stationary air.
This works fine because in both cases the frame of reference is inertial.

If you now add acceleration, in the first case, in the windtunnel the air actually needs to be accelerated, generating a pressure gradient. But in the second case, a rocket accelerating through the air, it is not the air that is accelerating but your frame of reference (and the rocket of course). For this case no pressure gradient needs to exist in the air to facilitate the acceleration. This is because the air is not accelerating with respect to an inertial frame of reference.

What is that diagram supposed to show? Pressure gradient (which is frame invariant) is associated with frame invariant proper acceleration of the fluid, not with its frame dependent coordinate acceleration in some non-inertial frame (like the rest frame of the rocket).

1. Accelerated object in stationary air is not equivalent to a stationary object in accelerated airflow.
2. Object with const. speed in stationary air is equivalent to a stationary object in const. speed airflow.

You want to say that this?

@A.T. Diagram show missile acceleration to the left in stationary air and pressure gradient that "appear" to exist(but gradient don't exist).
If point 1. is correct, that mean force/acceleration/pressure gradient is frame dependent?
Proper acceleration of fluid is acceleration relative to which frame? Any inertial frame?

 

[Arjan82]

Are you sure in this?
If ball hit the wall with constant speed of 20km/h, it is the same as if ball accelerate and in time t1 hit the wall with 20km/h. Isn't this same case for air particle in fluid?

I think for the ball this is only true in the case of a perfect collision which takes 0 time. A real collision takes a little bit of time, in which the ball is still accelerating and thus more energy is added to the system. So the impulse is increased.

In the case of a fluid, you've already showed the potential flow equations:

If $f(t)=0$ (no forcing term) then if there is no acceleration $p/\rho + 1/2 | \nabla\phi |^2=0$, which is Bernoulli. But if there is acceleration then the $\partial \phi / \partial t$ term is non-zero, adding to the stagnation pressure. Basically the assumption that the stagnation pressure is equal to the dynamic pressure (which you need to compute $V$) does not hold anymore during acceleration.

The effect is tiny though, so in practice it has no relevance.

1. Accelerated object in stationary air is not equivalent to a stationary object in accelerated airflow.
2. Object with const. speed in stationary air is equivalent to a stationary object in const. speed airflow.

You want to say that this?

Exactly right.

@A.T. Diagram show missile acceleration to the left in stationary air and pressure gradient that "appear" to exist(but gradient don't exist).
If point 1. is correct, that mean force/acceleration/pressure gradient is frame dependent?
Proper acceleration of fluid is acceleration relative to which frame? Any inertial frame?

Acceleration is frame in-dependent, as long as the frame is inertial. Force and pressure gradient is therefore also frame independent.

However, for point 1 we are comparing two different situations, not the same situation from two reference frames.

 

[user079622]

I think for the ball this is only true in the case of a perfect collision which takes 0 time. A real collision takes a little bit of time, in which the ball is still accelerating and thus more energy is added to the system. So the impulse is increased.

But in both case ball touch wall with speed of 20km/h, then start to decelerate to zero.
So F= m x (20km/h-0) / time of deceleration from 20km/h to 0
I don't see where difference comes in math

Acceleration is frame in-dependent, as long as the frame is inertial. Force and pressure gradient is therefore also frame independent.

However, for point 1 we are comparing two different situations, not the same situation from two reference frames.

Yes, in case accelerated rocket in stationary air, rocket is non-inertial frame.

So if the cart here is accelerating to the right in still air, the force from the "airflow" on the cart at a particular instant is not different than the force from the air on a cart as if it were moving at a constant velocity?

I am interested in this calculation.

 

[user079622]

@Arjan82
So if rocket is stationary and airflow first flow with constant speed,then accelerate. During acceleration we have pressure gradient, what will static port show now compare when speed was constant?

 

[Arjan82]

So, we're back at the start...

During acceleration we have pressure gradient

No, we do not... not due to the acceleration anyway...

The pressure gradient is a field function. It has a value (a vector) at each point in space. So all but the most simple flows (no flow, or same velocity everywhere) have a pressure gradient somewhere. So both constant speed flow around a rocket as the flow around an accelerating rocket has a pressure gradient somewhere.

The discussion is about the pressure gradient which would be due to the acceleration of the rocket, not the gradient that is due to the presence of the rocket. If the flow is accelerating, not the rocket, there would be a gradient in the flow everywhere, also far away from the rocket. But if the flow is not accelerating, but the rocket is, then the only gradients in the flow you'll find are close to the rocket, which are caused by the presence of the rocket. The gradient far away from the rocket is zero. Also, the rocket has a large parallel mid-section. There, the pressure gradient is also (almost) zero.

 

[user079622]

So, we're back at the start...



No, we do not... not due to the acceleration anyway...

Didn't we say if airflow is accelerated, that we have pressure gradient?

 

[Arjan82]

Nope, we didn't say that... read back..

 

[user079622]

Nope, we didn't say that... read back..

So if accelerated airflow flow parallel over stationary static port, he will show same static pressure as if airflow flows with constant speed or zero speed?

 

[Arjan82]

So if accelerated airflow flow parallel over stationary static port, he will show same static pressure as if airflow flows with constant speed or zero speed?

Correct indeed.

 

[user079622]

Correct indeed.

What is difference in; object accelerate in stationary air vs stationary object in accelerated airflow?

 

[Arjan82]

What is difference in; object accelerate in stationary air vs stationary object in accelerated airflow?

In the second case the bulk flow (i.e. the flow around the rocket, not necessarily close to the rocket) is actually accelerating, which requires a pressure gradient to do.

In the first case the bulk flow is not accelerating, but the frame of reference is accelerating.

 

[erobz]

I'm wondering which equations you are solving, since they look different in a non-inertial frame of reference.

Let me just start with Mass - Continuity for the cart, incompressible flow: The control volume is moving to the right with $v \boldsymbol{i}$ at any particular instsant and the cart is accelerating to the right:

So, continuity for the flow inside/passing through the control volume is given by:

$$ 0 = \frac{d}{dt} \int_{cv} \rho d V\llap{-} + \int_{cs} \rho ( \boldsymbol{V} \cdot d \boldsymbol{A} ) $$

Density $\rho$ constant in throughout flow spatially and in time. As well as control volume not deforming in time.

$$ \implies 0 = \cancel{ \frac{d}{dt} \int_{cv} \rho d V\llap{-}}^{0} + \int_{cs} \rho ( \boldsymbol{V} \cdot d \boldsymbol{A} ) $$

$\boldsymbol V$ is the velocity of the flow relative to the control surface at stations 1 and 2 , with uniform velocity distribution:

$$ 0 = (\boldsymbol {V_1} \cdot d {A_1}) + (\boldsymbol {V_2} \cdot d {A_2}) $$

$$ 0= \langle -V_1\boldsymbol {i} \rangle \cdot \langle A_1 \boldsymbol {i} \rangle + \langle -V_2 \cos \theta \boldsymbol {i} + V_2 \sin \theta \boldsymbol {j} \rangle \cdot \langle A_2 \boldsymbol {j} \rangle $$

$$ V_1 A_1 = V_2 A_2 \sin \theta $$

My first concern.

Is it ok to assume $A_1$ and $A_2$ are related geometrically:

Hence $V_2 = V_1 \tan \theta$

or are the required to be related through The First Law? I want to see if I have this step under control before moving forward. Thanks for any help.

 

[user079622]

In the second case the bulk flow (i.e. the flow around the rocket, not necessarily close to the rocket) is actually accelerating, which requires a pressure gradient to do.

But that I say in post #158 and #160. So what I said wrong?

 

[Arjan82]

$$ 0= \langle -V_1\boldsymbol {i} \rangle \cdot \langle A_1 \boldsymbol {i} \rangle + \langle -V_2 \cos \theta \boldsymbol {i} + V_2 \sin \theta \boldsymbol {j} \rangle \cdot \langle A_2 \boldsymbol {j} \rangle $$

This is incorrect.

First of all, there is a difference between $-V_1$ and $V_1 < 0$. The first means that somewhere in your analysis you need the negative $V_1$, the second means that $V_1$ has a negative value. However, in the analysis, it doesn't yet matter which values, positive or negative, the symbols have. That only comes when you are computing the numerical values, at the end of the analysis.

You are also confusing the vector $\boldsymbol{V_1}$ with the component of the vector $V_1$. So, both vectors $\boldsymbol{V_1}$ and $\boldsymbol{V_2}$ have two components. I presume that with $V_1$ you mean the first component of the vector $\boldsymbol{V_1}$, but what is then the first component of $\boldsymbol{V_2}$ called? It is probably better to adopt a notation like $\boldsymbol{V_1} = [v_{1i}, v_{1j}]$ and $\boldsymbol{V_2} = [v_{2i}, v_{2j}]$.

Third, $\boldsymbol V_2$ nor $\boldsymbol A_2$ have anything to do with $\theta$. The surface $A_2$ has no angle $\theta$ in it.

So the equation should be:
$$
\langle0\rangle = \langle v_{1i} \text{, } v_{1j}\rangle \cdot \langle a_{1i} \text{, } a_{1j}\rangle + \langle v_{2i} \text{, } v_{2j}\rangle \cdot \langle a_{2i} \text{, } a_{2j}\rangle
$$

$$ V_1 A_1 = V_2 A_2 \sin \theta $$

This is also wrong. It will become
$$
v_{1i} a_{1i} = v_{2j} a_{2j}
$$
This makes use of the fact that $v_{1j}$, $v_{2i}$, $a_{1j}$ and $a_{2i}$ are all zero (the way you've drawn it).

Is it ok to assume $A_1$ and $A_2$ are related geometrically:
View attachment 361178
or are the required to be related through The First Law? I want to see if I have this step under control before moving forward. Thanks for any help.

Is the drawing meant to be 3D? Otherwise I don't understand it. But you can define the control volume anyway you want, they're not bounded by any law other than that the geometry has to make sense (not self intersecting and all that). In the way you draw them they are indeed geometrically related.

 

[A.T.]

1. Accelerated object in stationary air is not equivalent to a stationary object in accelerated airflow.
2. Object with const. speed in stationary air is equivalent to a stationary object in const. speed airflow.

"Stationary" is ambiguous. There is no absolute rest.
"Accelerated" is ambiguous. You have to differentiate proper acceleration and coordinate acceleration (see below).

If point 1. is correct, that mean force/acceleration/pressure gradient is frame dependent?

Pressure forces, pressure gradient and proper acceleration are all frame invariant.

Proper acceleration of fluid is acceleration relative to which frame? Any inertial frame?

Proper acceleration is what an accelerometer measures, and every frame must agree what a given accelerometer will measure, so proper acceleration is frame invariant.

Coordinate acceleration is dv/dt relative to some frame, so it's frame dependent. Coordinate acceleration equals proper acceleration in free falling frames.

 

[Arjan82]

But that I say in post #158 and #160. So what I said wrong?

What you say in post #158 and #160 is indeed wrong, it is not equal to what I said... Unless you are trying to analyze a rocket that is not accelerating but the air around it is accelerating.

 

[erobz]

This is incorrect.

First of all, there is a difference between $-V_1$ and $V_1 < 0$. The first means that somewhere in your analysis you need the negative $V_1$, the second means that $V_1$ has a negative value. However, in the analysis, it doesn't yet matter which values, positive or negative, the symbols have. That only comes when you are computing the numerical values, at the end of the analysis.

You are also confusing the vector $\boldsymbol{V_1}$ with the component of the vector $V_1$. So, both vectors $\boldsymbol{V_1}$ and $\boldsymbol{V_2}$ have two components. I presume that with $V_1$ you mean the first component of the vector $\boldsymbol{V_1}$, but what is then the first component of $\boldsymbol{V_2}$ called? It is probably better to adopt a notation like $\boldsymbol{V_1} = [v_{1i}, v_{1j}]$ and $\boldsymbol{V_2} = [v_{2i}, v_{2j}]$.

$$ \boldsymbol{V_1} = V_1 \langle -1\boldsymbol{i} + 0\boldsymbol{j}\rangle $$

The un-bolded $V_1$ is a magnitude. I omitted the "0j" in the vector $\boldsymbol{V_1}$, because in the dot product it vanishes anyhow.

Same thing with the areas bolded is vector, unbolded magnitude $d \boldsymbol{A_1} = dA_1 \langle 1\boldsymbol{i} + 0\boldsymbol{j} \rangle$


As far as $\boldsymbol{V_2}$ goes, I try to assume that the streamlines exiting the control volume at 2 are parallel to the face of the cart. hense:

$\boldsymbol{V_2} = V_2 \langle - \cos \theta\boldsymbol{i} + \sin \theta \boldsymbol{j}\rangle$

Likewise $d \boldsymbol{A_2} = dA_2 \langle 0\boldsymbol{i} + 1\boldsymbol{j} \rangle$

Again, omitting components that vanish in the dot product. It was very half arsed, sorry if that was confusing.

Is the drawing meant to be 3D? Otherwise I don't understand it. But you can define the control volume anyway you want, they're not bounded by any law other than that the geometry has to make sense (not self intersecting and all that). In the way you draw them they are indeed geometrically related.

I'm trying to represent that the "flow" over the inclined surface is uniformly distributed along an axis coming into/out of the page.

 

[A.T.]

I'm trying to see why when I solve the equations for an accelerating cart vs a a cart moving at constant velocity, I get different forces acting on the cart from the air. My interpretation of the statement is the force (from the air) shouldn’t be different just because the cart is accelerating.

I have no idea how you arrived at this interpretation of @Arjan82's statement. And I see no reason to assume the forces necessarily must be equal. If the cart accelerates, then you don't have a steady flow in the rest frame of the cart.

 

[user079622]

What you say in post #158 and #160 is indeed wrong, it is not equal to what I said... Unless you are trying to analyze a rocket that is not accelerating but the air around it is accelerating.

We have two cases:
case 1. object accelerate in stationary air
case 2. stationary object in accelerated airflow.

Here I refer to stationary rocket in accelerated airflow, so flow accelerate in every point around rocket, upstream,downstream, up, down etc. I don't talk about local accelerations due to rocket body influence on flow, in front of nose high pressure, at back wake etc..
Static port must be placed at position with zero "local acceleration",
otherwise it will show wrong results, so these local accelerations are automatically out of the discussion.

Pressure forces, pressure gradient and proper acceleration are all frame invariant.

Proper acceleration is what an accelerometer measures, and every frame must agree what a given accelerometer will measure, so proper acceleration is frame invariant.

Coordinate acceleration is dv/dt relative to some frame. Coordinate acceleration equals proper acceleration in free falling frames.

Proper acceleration is same in relation to any non inertial frame?

I have no idea how you arrived at this interpretation of @Arjan82's statement. And I see no reason to assume the forces necessarily must be equal. If the cart accelerates, then you don't have a steady flow in the rest frame of the cart.

If ball hit the wall with constant speed of 20km/h, it is the same as if ball accelerate and in time t1 hit the wall with 20km/h. isnt it?

 

[erobz]

If ball hit the wall with constant speed of 20km/h, it is the same as if ball accelerate and in time t1 hit the wall with 20km/h. isnt it?

Yeah, I want to second this. This is how I interpret what was said.

If this was happening in a near vacuum, and there were a line of very small particles (atoms?) sitting there at rest and the cart with a vertical face is coming along impacting them sequentially, conservation of momentum an energy apply here and depend on what the velocity of the cart is when it strikes any particular atom (whether or not energy is conserved in the collision).

If the cart is accelerating while it's hitting these particles seems to be immaterial to the outcome for the particle. Its behavior depends on the relative mass between the particle and cart, and velocity of the cart, not the carts acceleration.

So I feel like I better derive the same force from the air on the cart as a function of the carts velocity whether the cart is accelerating or moving at constant velocity under the action of the for $F$ pushing it along.

 

[jbriggs444]

Proper acceleration is same in relation to any non inertial frame?

Proper acceleration is an invariant. That means that taking it "in relation to" some particular frame, inertial or otherwise is nonsense.

Yes, it is a fixed quantity regardless of the frame one has chosen to adopt.

 

[Arjan82]

Yeah, I want to second this. This is how I interpret what was said.

If this was happening in a near vacuum, and there were a line of very small particles (atoms?) sitting there at rest and the cart with a vertical face is coming along impacting them sequentially, conservation of momentum an energy apply here and depend on what the velocity of the cart is when it strikes any particular atom (whether or not energy is conserved in the collision).

If the cart is accelerating while it's hitting these particles seems to be immaterial to the outcome for the particle. Its behavior depends on the relative mass between the particle and cart, and velocity of the cart, not the carts acceleration.

So I feel like I better derive the same force from the air on the cart as a function of the carts velocity whether the cart is accelerating or moving at constant velocity under the action of the for $F$ pushing it along.

I've explained in post #156 why this is wrong. Do you think that explanation was wrong?

 

[erobz]

I've explained in post #156 why this is wrong. Do you think that explanation was wrong?

No, I think I missed it.

So the velocity of the particle becomes unclear as the body of the particle deforms. Lets assume rigid wall, elastic ball. On the tiny timescales of the collision duration I would agree that a greater net impulse should be delivered to the ball from the wall in the case where the wall is accelerating (toward the ball of course). You did say more or less the difference is practically negligible though.

I think that unless I consider this highly detailed model, the continuum mechanics (greatly simplified- ignorantly) I'm trying to correctly apply should be ignorant of this difference in force between an accelerated cart and an inertial cart.

 

[A.T.]

Here I refer to stationary rocket in accelerated airflow, ...

Read this again:

"Stationary" is ambiguous. There is no absolute rest.
"Accelerated" is ambiguous. You have to differentiate proper acceleration and coordinate acceleration.

Proper acceleration is same in relation to any non inertial frame?

Proper acceleration is the same in any frame, so it's not "in relation to any frame"

If ball hit the wall with constant speed of 20km/h, it is the same as if ball accelerate and in time t1 hit the wall with 20km/h. isnt it?

"Constant speed" is ambiguous. There is no absolute speed.
"Accelerate" is ambiguous. You have to differentiate proper acceleration and coordinate acceleration.

 

[rcgldr]

Are you perhaps referring to the jump in the potential behind the wing, which you need to apply the Kutta condition on a wing?

Yes in my prior post, I mentioned that the pressure jump occurs behind the wing. In level flight, the weight of the aircraft is transmitted to the air via acceleration (work is done), and the air eventually transfers that weight to the surface of the earth. In closed system, consisting of air and aircraft, such as a model plane inside a large sealed box, the weight of the system is the same if the aircraft is resting at the bottom or in the air with no vertical component of acceleration. Lift involves downwash as seen in this video of an owl flying through a wall of bubbles, generating vortices that move downwards:

 

[Arjan82]

We have two cases:
case 1. object accelerate in stationary air
case 2. stationary object in accelerated airflow.

I'm not sure if you understand that these are two physically different cases, this is not just a change in reference frame. Case 1 is a rocket accelerating through the air, case 2 is a wind tunnel test where the rocket is bolted to the ground and the air speed in the tunnel is speeding up.

Here I refer to stationary rocket in accelerated airflow, so flow accelerate in every point around rocket, upstream,downstream, up, down etc. I don't talk about local accelerations due to rocket body influence on flow, in front of nose high pressure, at back wake etc..
Static port must be placed at position with zero "local acceleration",
otherwise it will show wrong results, so these local accelerations are automatically out of the discussion.

Ok, you are analyzing the windtunnel case here. Is that what you intent to do?

 

[Arjan82]

Yes in my prior post, I mentioned that the pressure jump occurs behind the wing.

A jump implies a change in pressure over zero distance, so a discontinuity in the pressure field. In that sense there is no jump in the pressure. Actually, the pressure distribution is quite smooth around a wing. Also, there is only a small pressure gradient behind the wing at all. The pressure difference that causes lift is between the bottom and top of the wing. This is what the pressure distribution around a wing (airfoil in this case, but close enough) looks like:

In level flight, the weight of the aircraft is transmitted to the air via acceleration (work is done),

Let me repeat myself: if you analyze the flow in a wing-fixed reference frame there is no work being done to the fluid in the flow domain.

and the air eventually transfers that weight to the surface of the earth. In closed system, consisting of air and aircraft, such as a model plane inside a large sealed box, the weight of the system is the same if the aircraft is resting at the bottom or in the air with no vertical component of acceleration. Lift involves downwash as seen in this video of an owl flying through a wall of bubbles, generating vortices that move downwards:

True, but I'm not sure how that relates to what you previously said.

 

[rcgldr]

A jump implies a change in pressure over zero distance ...

"So there is an abrupt change in pressure across the propeller disk". Other articles use the term pressure jump to describe this. The distance is not zero, but it is small.

https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/propth.html

if you analyze the flow in a wing-fixed reference frame there is no work being done to the fluid in the flow domain.

The air ahead of a wing has no downwards component of flow, while the air behind a wing does have a downwards component of flow, and this component is greater in magnitude than the relative reduction in backwards flow related to drag (if lift to drag ratio is high enough, and relative velocity is not very high), so a net energy increase of the affected air, so work is done. Simplified models ignore the induced downwash, but more realistic models include induced downwash.

 

[Arjan82]

"So there is an abrupt change in pressure across the propeller disk". Other articles use the term pressure jump to describe this. The distance is not zero, but it is small.

https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/propth.html

This is an actuator disk model. That is not the same as 'pressure over a wing'... The actuator disk model models the effect of a propeller, not the propeller itself. It is also not in the 'blade-fixed reference frame' as I noted a few times, since it does not include the propeller blade at all.

The air ahead of a wing has no downwards component of flow, while the air behind a wing does have a downwards component of flow, and this component is greater in magnitude than the relative reduction in backwards flow related to drag (if lift to drag ratio is high enough, and relative velocity is not very high), so a net energy increase of the affected air, so work is done. Simplified models ignore the induced downwash, but more realistic models include induced downwash.

Nope, this is just simply not true. There is no model that I know that ignores downwash. Downwash is directly related to the 'lift induced drag' which is an essential part of the drag of a wing or propeller blade (usually that's around 80% or 90% of the total drag!). So a model without downwash is useless.

Also, the air is effectively rotated, not accelerated. You increase the vertical velocity of the flow but you reduce the horizontal velocity. The earliest useful model of a wing was called 'the lifting line model' This model is just a vortex in space. A vortex does not add energy, it adds rotation to the flow (and it does model downwash!). However, this was a huge step forward in computations around wings.

 

[erobz]

@Arjan82 Just to be clear. I assume $\boldsymbol {V_2}$ ( the outflow velocity vector ) is parallel to the incline in the frame of reference of the cart:

$$ 0 = V_1\langle -1 i + 0j \rangle \cdot A_1 \langle 1i+0j \rangle + V_2\langle - \cos \theta i + \sin \theta j \rangle \cdot A_2 \langle 0i+1j \rangle $$

Giving scalar equation:

$$ V_1 A _1 = V_2 A_2 \sin \theta$$

Is this incorrect?

 

[Arjan82]

Ah... you had assumed the direction of V2, that was not clear to me. Also V1 and V2 are the magnitude of the vectors V1 and V2, that was also not clear to me... But then your equation is correct indeed.

 

[erobz]

Ah... you had assumed the direction of V2, that was not clear to me. Also V1 and V2 are the magnitude of the vectors V1 and V2, that was also not clear to me... But then your equation is correct indeed.

Thank You, yeah I did it half assed- I deserved a misread.

Is it ok to assume that the outflow vector is parallel ( more or less) to the slope in the carts reference frame, or is there a strong reason that it shouldn't be that will haunt me later? Assuming incompressible, inviscid flow.

 

[rcgldr]

increase the vertical velocity of the flow but reduce the horizontal velocity.

True but those velocities normally do not cancel out (more on this in the last paragraph).

Switch this to using the air as a frame of reference, with initial velocity 0 before a wing passes through a volume of air, and mostly downwards (lift) and somewhat forwards (drag) velocity after a wing passes through that volume of air. In this frame of reference, the wing increases the kinetic and pressure energy of the air immediately behind the wing. The average velocity where the affected air's pressure returns to ambient is called the exit velocity. This also explains how the air eventually transmits the weight of an aircraft onto the surface of the earth.

That pressure jump can be explained as the air streams above and below a wing converging (colliding) behind the wing, with a net downwards change of angle in the converged streams.

Another way to consider this is a glider in a steady (non-accelerating) descent, gravitational potential energy decreases, energy of the air increases, conservation of energy. From the air's frame of reference, most of that energy increase is related to downwash (lift), a small amount of it to forward wash (drag), and smaller still amounts related to sound and temperature. From the aircraft | wing's frame of reference, there is still energy conservation, so the energy of the air is still increased, but a small component of drag is in the direction of gravity. In the case of Nimbus 4 glider, with a 60:1 glide ratio at 110 kps, I doubt that small component of drag in the direction of gravity accounts for much.

 

[user079622]

I'm not sure if you understand that these are two physically different cases, this is not just a change in reference frame. Case 1 is a rocket accelerating through the air, case 2 is a wind tunnel test where the rocket is bolted to the ground and the air speed in the tunnel is speeding up.



Ok, you are analyzing the windtunnel case here. Is that what you intent to do?

Yes it can be in wind tunnel or in open wind tunnel or theoretical model where all atmosphere is accelerate.

 

[A.T.]

"Constant speed" is ambiguous. There is no absolute speed.
"Accelerate" is ambiguous. You have to differentiate proper acceleration and coordinate acceleration.

So force that ball exert on wall is not equal in both case?

That's not what I wrote. I wrote that your cases are not well defined, because you use ambiguous terms.

 

[Arjan82]

Yes it can be in wind tunnel or in open wind tunnel or theoretical model where all atmosphere is accelerate.

There is a difference in 'can be' and 'is'... You cannot compare that case with a rocket in free flight if the rocket is accelerating. It doesn't seem that you are getting that point.

 

[user079622]

@Arjan82

your wrote: "case 2 is a wind tunnel test where the rocket is bolted to the ground and the air speed in the tunnel is speeding up."

I respond to that: "Yes it can be in wind tunnel or in open wind tunnel or theoretical model where all atmosphere is accelerate."

what I write wrong?

 

[Arjan82]

True but those velocities normally do not cancel out (more on this in the last paragraph).

Momentum needs to cancel out, not velocity. Can you further substantiate this claim? How do you relate that to the fact that a wing can literally be modelled as a vortex, as already been shown by Prandtl in 1918 I believe?

Switch this to using the air as a frame of reference, with initial velocity 0 before a wing passes through a volume of air, and mostly downwards (lift) and somewhat forwards (drag) velocity after a wing passes through that volume of air.

Ok, I don't know how to make it more clear than I already did, what I'm saying is that there is no energy added to the flow, and therefore that Bernoulli is valid if and only if you look at the problem from a reference frame that is attached to the wing!

So if you use 'air as reference frame' (let's not use fluid as a reference frame, it is earth-fixed with zero wind or something) then you can indeed not use Bernoulli. But Bernoulli is used in calculations over a wing or a propeller all the time. In fact, I'm making a living by doing that...

In this frame of reference, the wing increases the kinetic and pressure energy of the air immediately behind the wing. The average velocity where the affected air's pressure returns to ambient is called the exit velocity. This also explains how the air eventually transmits the weight of an aircraft onto the surface of the earth.

You're mixing up your models sir... 'exit velocity' is not used in flow over a wing or over a propeller. An actuator disk has an 'exit velocity', or the flow through a nozzle or shroud. But those are not the cases we are talking about now.

That pressure jump can be explained as the air streams above and below a wing converging (colliding) behind the wing, with a net downwards change of angle in the converged streams.

That's just BS. A fluid cannot sustain a pressure jump unless you are going supersonic. Talking about a pressure jump you were referring to the actuator disk model. The actuator disk is a model where you apply a pressure jump as a means to model the effect of a propeller, not the propeller itself!

Another way to consider this is a glider in a steady (non-accelerating) descent, gravitational potential energy decreases, energy of the air increases, conservation of energy.

From the earth's frame of reference yes, from the glider's frame of reference no, because from the glider's frame of reference there is no altitude loss...

From the air's frame of reference, most of that energy increase is related to downwash (lift), a small amount of it to forward wash (drag), and smaller still amounts related to sound and temperature.

True, although changes in energy due to sound and temperature are many orders of magnitude lower than due to drag, viscosity, turbulence...

From the aircraft | wing's frame of reference, there is still energy conservation,

True

so the energy of the air is still increased,

Wrong. This is where you are mistaken. From the wing's frame of reference there is no energy added. For energy to be added to the system you need you need a moving boundary. But since you are looking at the problem from the wing's perspective, there is no moving boundary!

but a small component of drag is in the direction of gravity.

?? I don't know what you mean here.

In the case of Nimbus 4 glider, with a 60:1 glide ratio at 110 kps, I doubt that small component of drag in the direction of gravity accounts for much.

 

[A.T.]

There is a difference in 'can be' and 'is'... You cannot compare that case with a rocket in free flight if the rocket is accelerating. It doesn't seem that you are getting that point.

At least it’s not clear what "all atmosphere is accelerate" is supposed to mean.

If you could apply uniform homogeneous proper acceleration to a an entire airmass by some force field (not by pushing the air mass at one end), that would be equivalent to the non-inertial frame of the rocket, where the airmass is subject to the uniform inertial force -ma. In neither case you have a pressure gradient in the free-stream, because the acceleration of an air packet is not due to the force from neighboring packets.

But in a wind tunnel you cannot keep uniform velocity across the test section, while changing that velocity over time. The air that is currently in the test section, must be properly accelerating to achieve that change over time, which implies a pressure gradient.

 

[A.T.]

From the aircraft | wing's frame of reference, there is still energy conservation, so the energy of the air is still increased,

What?

In the inertial rest frame of the wing, there is no work done by/on the wing. So, except of heat and sound, all the energy that the air has, stays in the air, none is added or removed. Ideally the air is just changing direction, so its macroscpic kinetic energy stays constant. But realistically it is slowed down so some macroscpic kinetic energy is converted to heat and sound.

 

[user079622]

Yeah, I want to second this. This is how I interpret what was said.

Key question is, can force which is causing the object to accelerate diminishes instantaneously with its contact with the wall? I think in real life not.

 

[Arjan82]

Key question is, can force which is causing the object to accelerate diminishes instantaneously with its contact with the wall? I think in real life not.

In real life nothing is instantaneous.

 

[A.T.]

If this was happening in a near vacuum, and there were a line of very small particles (atoms?) sitting there at rest and the cart with a vertical face is coming along impacting them sequentially, conservation of momentum an energy apply here and depend on what the velocity of the cart is when it strikes any particular atom (whether or not energy is conserved in the collision).

If the cart is accelerating while it's hitting these particles seems to be immaterial to the outcome for the particle. Its behavior depends on the relative mass between the particle and cart, and velocity of the cart, not the carts acceleration.

So I feel like I better derive the same force from the air on the cart as a function of the carts velocity whether the cart is accelerating or moving at constant velocity under the action of the for $F$ pushing it along.

Yes, in rare gas, where the interaction of the particles with each other can be ignored, everything is much simpler, and the acceleration of the cart plays no role, just it's instantaneous velocity. So, if your derivation is based on this assumption, you should get the same force.

But in a dense gas / liquid I would not rely on this for arbitrarily large accelerations.

 

[Arjan82]

@Arjan82

your wrote: "case 2 is a wind tunnel test where the rocket is bolted to the ground and the air speed in the tunnel is speeding up."

I respond to that: "Yes it can be in wind tunnel or in open wind tunnel or theoretical model where all atmosphere is accelerate."

what I write wrong?

Since your original question was about what a well-placed static port would measure on an accelerating rocket, I would assume you mean a rocket in free flight. So now you are saying you are talking about a rocket that is bolted on the ground. So I'm lost to what your question now actually is.

Also, you said 'can be'. That implies it als 'can be' something else. So it might be that you are thinking, 'well you can talk about a rocket bolted to the ground, but you don't have to. It might also just be in free flight'. Which would be the wrong conclusion.

So, I'm trying to probe whether you got the message that when the rocket is accelerating, you cannot^*) switch between case 1 and 2, these are physically different cases.

Last: I don't know what you mean with 'an open wind tunnel'.

And just to state what I think should now be obvious: if a rocket is bolted to the ground and the air around it is accelerated, the static probe will not measure the ambient pressure. What it will measure highly depends on the actual configuration.

*) not in the same way as case 1 I mean. You need different equations to describe the situation.

 

[rcgldr]

From the earth's frame of reference yes, from the glider's frame of reference no, because from the glider's frame of reference there is no altitude loss...

In the inertial rest frame of the wing, there is no work done by/on the wing.

Consider earth, air, and glider as a closed system, and the earth as a frame of reference. Gravitational potential energy is a function of distance between earth and glider. regardless of what inertial frame of reference is used. From the glider's frame of reference, the earth is moving closer to the glider, gravitational potential energy decreases, energy of the air increases, energy is conserved.

 

[Arjan82]

Is it ok to assume that the outflow vector is parallel ( more or less) to the slope in the carts reference frame, or is there a strong reason that it shouldn't be that will haunt me later? Assuming incompressible, inviscid flow.

Depends on what you want to do with the results. It is not really correct. It depends on the next steps.