I Explain Bernoulli at the molecular level?

[erobz]

Cart that accelerate, in front produce force on air particle, particle with same force push back on the cart.
This additional force dont exist in constant velocity case.
Isnt it?

Yeah, I agree (I think we all do now) that is the reality of an accelerating cart. If a wall is moving at constant velocity it will have displacement $\delta x \approx v \delta t$. If the wall is accelerating at constant $a$, then it has displacement $\delta x \approx v \delta t + \frac{1}{2}a (\delta t)^2$. So, presumably more work will be done on the particle per unit $\delta t$ in the case where the wall is accelerating assuming $\delta t$ doesn't strongly depend on $a$.

The devil here is in the details of the unsteady terms I crossed out in post no 225. You can work on the equations I provided (with now some hopefully sensible examples), make some different assumptions and pull the levers. That will be the same thing I am doing. Perhaps start with a constant velocity control volume and try to make some assumptions about the flow-mass acceleration inside it and see what happens to the equations.

 

[A.T.]

What is your actual point about the energy in rest frame of the glider in steady decent? That the stationary glider is adding energy to the air?

Simplified velocity vectors for a wing in level flight (v is smaller than actual).

What do you mean by "(v is smaller than actual)". Is v the freestream velocity, or is it already reduced due to interaction with the wing?

The incoming air is deflected mostly downwards (lift), and somewhat forwards (drag). The resulting vector is greater in magnitude, reflecting an increase in energy.

Greater in magnitude, than the freestream velocity?

That would violate energy conservation: You could put a bunch of those airfoils in a circular pipe, and they would keep speeding up the flow.

A static object cannot do positive work on anything, and thus cannot transfer energy to the flow. Just like the static road cannot do positive work on the wheel, as was explained to you ad nauseam in this previous thread:

Post in thread 'Is there any work done by static friction when accelerating a car?'

Jan 27, 2020

the response I received is similar to what I've seen elsewhere:

"The road is, in fact, doing work on the car to propel it forward."

The response is incorrect. If the road is doing work on the car then why doesn’t the energy increase?

It is pretty funny that you have scores of correct responses here, but because that one incorrect response is what you wanted to hear you immediately grab onto it. You appear to be immune to any logical arguments and simply set on sticking with your incorrect notion regardless.

• Dale

• 

Look at the definition of mechanical power:
P = F dot v
https://en.wikipedia.org/wiki/Power_(physics)#Definition
Since the angle between relative flow velocity (freestream) and force by the air on the wing is always less than 90°, the angle between relative flow velocity (freestream) and force by the wing on the air is always greater than 90°. So the dot product of F_on_air and v_air is negative, meaning that no energy is added to the air by the wing in the rest frame of the wing, and the air must actually slow down on average in that frame.

 

[rcgldr]

A static object cannot do positive work on anything, and thus cannot transfer energy to the flow.

You're correct. I deleted that post (brain fade on my part). I don't recall where I got those images from. There would be a decrease in energy: Δ v would be angled more forwards, and | v + Δ v | < | v | (using | ... | for magnitude).

Getting back to the glider in a steady descent, using the glider as frame of reference, the energy of the air affected by the glider is decreasing. Energy is conserved, so some form of energy is increasing. As noted in that answer from Stack Exchange, since the potential well is moving with respect to the glider's frame of reference, "the concept of potential energy no longer offers a quick and easy way to keep track of the energy movements."

There have already been posts that agree that if using the earth | air as a frame of reference, gravitational potential energy decreases and energy of the air increases.

Getting back on topic, I previously posted about the OP question about Bernoulli at the molecular level, would apply in the case of an idealized Venturi tube with no change in total energy, but in the case of an aircraft the total energy increases as the speed of the aircraft increases, and that static pressure remains constant in that case.

 

[A.T.]

Getting back to the glider in a steady descent, using the glider as frame of reference, the energy of the air affected by the glider is decreasing.

Yes, locally around the glider the kinetic energy of the air is reduced in the rest frame of the glider.

But the kinetic energy of the atmosphere as a whole cannot be continuously decreasing in the rest frame of the glider, because here the Earth is moving at constant speed, and the atmosphere must move along with the Earth. So, the moving Earth must transfer energy to the now slower moving atmosphere to compensate for the local KE reduction by the static glider. This is where the gravitational potential energy of the Earth-glider-system is going to.

 

[rcgldr]

1) I think this discussion about energy balance of the entire Earth glider system should be a separate thread, because this one is already too long and too confused.

I agree with this, if someone is willing to take the time to do this. I wouldn't mind a restart on the earth glider system, especially for a glider based frame of reference versus gravitational potential energy.

I'm also thinking that the best example of Bernoulli at the molecular level would be a idealized Venturi tube and a fluid or gas with zero viscosity, and no change in total energy.

Wings complicate things because viscosity is needed to cause a gas to follow a convex surface.

For some of the other cases mentioned, static pressure remains constant while total energy changes as relative speed changes.

 

[A.T.]

I wouldn't mind a restart on the earth glider system, especially for a glider based frame of reference versus gravitational potential energy.

If my post #224 doesn't sufficently answer this, then you can start a new thread. Consider replacing the glider with something falling in the atmosphere vertically at constant terminal velocity. Then you have a simpler 1D case that captures the same issues regarding energy, for which the horizontal motion of the glider is not key.

 

[rcgldr]

If my post #224 doesn't sufficiently answer this, then you can start a new thread.

I'm still wondering about the horizontal component, especially if you consider the earth's surface to be frictionless, in which case, there could be a continuous change in the horizontal component of air velocity until the glider lands. If the earth's surface is not frictionless, then a horizontal component of flow would apply a toque onto the earth. However, I'm not looking for answer to this now, as I need to do more research. It only took me about 3 minutes to figure out DDWFTTW was possible, but I'm still having issues with the earth + air + glider situation.

 

[A.T.]

... but I'm still having issues with the earth + air + glider situation.

That's no reason for hijacking someone else's thread. Start your own.

 

[rcgldr]

That's no reason for hijacking someone else's thread. Start your own.

I agree. I'm waiting to see if someone wants to move all the glider related posts to another thread. If they just want to delete them, I wouldn't have an issue with that either, but don't know about the others involved. I did go through and did strike-through on my prior posts.

 

[rcgldr]

A static object cannot do positive work on anything, and thus cannot transfer energy to the flow.

Getting back on topic, in the case of a static Venturi, as the tube diameter decreases, pressure energy decreases while kinetic energy increases. In a real world situation, the total energy decreases due to losses, but kinetic energy can be increased. Indy and Formula 1 race cars use Venturi principle to reduce pressure underneath a car which also speeds up the relative (to the car) air flow underneath the car. Diffusers at the back of the car smooth out the transition at the rear to reduce losses (drag).

 

[erobz]

Lets assume that the flow in the control volume is uniformly distributed, and experiences a constant acceleration along the circular arc $s$.

In other words $\ddot s = k$ implies:

$$ \frac{d \dot s }{d \theta} \frac{d \theta }{dt} = k $$

Since $\dot s = r \dot \theta$:

$$ \frac{d \dot s }{d \theta} \frac{\dot s }{r} = k $$

Separate variables, integrate this over $\theta$, with $\dot s = V$ implies that the scalar velocity of the flow along at any point along the arc is given by:

$$ V = \sqrt{2kr \theta + V_1^2} $$

Does anyone have complaints about the "lever pulling" I'm about to do for fear of basic representation of the unsteady terms in the Mass Continuity/ Energy equations before I go further?

I believe I can get the pressure $P_2$ at the outlet as function of $V_1, \dot V_1, k$ etc... (the velocity of the cart, cart acceleration, etc... ). Explicitly I'm not sure, but at least implicitly it looks like I have it.

 

[Lnewqban]

I agree. I'm waiting to see if someone wants to move all the glider related posts to another thread. If they just want to delete them, I wouldn't have an issue with that either, but don't know about the others involved. I did go through and did strike-through on my prior posts.

Paging @berkeman and @russ_watters.

 

[sophiecentaur]

While I agree with @jbriggs444 that this is a potentially unnecessary and unhelpful complication,

We are initially taught a lot of Physics with the particle model because it's the least abstract way into things like the gas laws and EE topics. Kinetic Theory starts with with 3D boxes and molecules hitting the walls. So I'd bet that everyone posting here (with Physics knowledge) started that way.

But some things are better left behind when the Science gets advanced. You can't always look back to 'check' the new new stuff.