# Bernoulli equation related to minimum pressure inside train

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1. Jun 23, 2016

### Soren4

1. The problem statement, all variables and given/known data
The train is travelling in a gallery at $v=70 km/h$. Outside air is not moving and pressure is $p_0=101325 Pa$ and temperature $T=283 K$. The area of the face of the train is $A=9 m^2$. Determine the minimum area of the section of the gallery such that the pressure inside the train does not go down $P=90000Pa$

2. Relevant equations
Bernoulli equation

3. The attempt at a solution
I cannot really relate the Surface of the sections of train or the gallery with the pressure difference, what is the relation that one should use?

2. Jun 24, 2016

### andrevdh

This looks very much like a fluid that moves through a constriction, that is the air in the tunnel (gallery) has to flow through a smaller area when the train passes, which would reduce its pressure as the train passes.

3. Jun 24, 2016

### Soren4

Thanks for the reply! Didn't think about that! But in the problem is the pressure inside the train that goes down.. I'm quite confused about which of the two masses of air is moving (the one in the train or the one in the gallery). In the text it is said that the air outside the train is steady..

4. Jun 24, 2016

### andrevdh

Surely it will move when the train passes! It has to flow over the oncoming train to make room for it. The same process happens when an aeroplane moves through the air. The air flows over the plane and especially it's wings. What is moving the air or the aeroplane? A passenger will say that the air is moving past the aeroplane, while an air molecule will say that the aeroplane is moving past it. All motion is relative. The pressure of the moving air will decrease and thereby decrease the pressure in the train, unless it is sealed perfectly like an aeroplane (in which the pressure is regulated). The same happens in a car as anyone with sinus problems (like me ) will tell you. Long trips can really sometimes reak havoc and cause problems due to the reduced pressure and my system not being able to equalize.

Last edited: Jun 24, 2016
5. Jun 24, 2016

### Soren4

Thanks a lot for this answer! That's a point that confuses me in general: I'm totally ok with Galilean relativity but when using bernoulli equation pressure is linked to the velocity and the pressure cannot be "relative". I mean if the pressure is less in the train than outside, it cannot be the reverse, even if we have the two possible cases: the train is moving and the air is steady and the reverse. But that cannot be right. Is there something I'm missing?

6. Jun 24, 2016

### haruspex

I am not following your reasoning there, so do not see the difficulty. As andrevdh posted, you can consider it in the reference frame of the train. (The only doubt is that the walls of the tunnel are also moving relative to the train.)
It does not change which will be the high pressures and which will be low.

7. Jun 24, 2016

### rcgldr

The stated premise is that outside air is not moving with respect to the ground (gallery), which would eliminate issues related to drag (which would violate Bernoulli). The implied premise is that the outside air is not moving (with respect to the ground) must be referring only to the air in front of and behind the train (zero drag). If the outside air was not moving along the sides of the train, then there's no change in speed, just compression of the air. Another implied premise is that the pressure of the air inside the train is the same as the pressure of the air along the sides of the train, even though the air inside the train moves at a different velocity than the air outside the train, (this would be possible if the venting of the train involved the equivalent of static ports).

As already posted above, the desired approach is to use the trains frame of reference and consider the air to speed up along the sides of the train in the same manner as a constricted flow. From a ground based frame, the air would travel backwards along the sides of the train.

Last edited: Jun 24, 2016
8. Jun 24, 2016

### Soren4

Thanks a lot for the reply to my question! I thought about it again but still I do not see how can be no difference in the pressure from the frame of reference chosen. I considered the two possible frame of reference:
• Frame of reference of the train: $v_{air \, inside \, the \, train}=0 km/h$ and $v_{air \, outside \, the \, train}=70 km/h$ $\implies$ $p_{\, outside \, the \, train}=p_{atm}-\frac{1}{2} \rho v_{air \, outside \, the \, train}^2$ and $p_{\, inside \, the \, train}=p_{atm}$
• Frame of reference of the gallery: $v_{air \, inside \, the \, train}=70 km/h$ and$v_{air \, outside \, the \, train}=0 km/h$ $\implies$ $p_{\, inside \, the \, train}=p_{atm}-\frac{1}{2} \rho v_{air \, inside \, the \, train}^2$ and $p_{\, outside \, the \, train}=p_{atm}$
That's how the velocity changes the pressure, as far as I could understand about Bernoulli equation. Did I get it completely wrong?

9. Jun 24, 2016

### Nidum

The question is basically twaddle but here is one possible way to answer it :

Air at some distance in front of and behind the train is static . Train in passing through the gallery is displacing air and the displaced air flows backwards around the train in the clearance gap between train and gallery walls .

This displaced air will be moving with a velocity which can be calculated . This air velocity will be a function of the train velocity . The actual velocity of the train is only relevant as far as it is needed to calculate the displaced air velocity .

Last edited: Jun 24, 2016
10. Jun 24, 2016

### rcgldr

It appears that you're supposed to assume the static pressure of the air inside the train is the same as the static pressure of the air outside the side of the train (not the front or the back), despite the differences in velocity. This could be possible if the trains ventilation was the effective equivalent of static ports, which are not sensitive to differences between internal and external velocities (as long as the speeds are reasonably below mach 1).

Say the cross sectional area of the gallery (tunnel) is 18 m2, then from the trains frame of reference the air's velocity increases from 70 km/h to 140 km/h as it moves around the train. From the ground frame of reference, I would assume that the air travels backwards at 70 km/h (same 70 km/h difference in speed), however the calculated pressure of the air along the sides of the train will differ, depending on the frame of reference. The issue here is that the velocity related component of energy or in this case, the energy per unit volume of air relative to velocity or a change in velocity is frame dependent, and it's not clear which frame of reference that the problem wants you to use.

In the real world, the static pressure of the air around the train would be independent of frame, but the problem ignores factors like drag, which violates Bernoulli. If drag is included, then from both frames of reference, the air is accelerated forwards, increasing its velocity (from zero to non-zero) from a ground frame of reference, and decreasing it's velocity (from 70 km/h to something less) from the trains frame of reference.

Last edited: Jun 24, 2016
11. Jun 24, 2016

### TSny

Bernoulli's equation assumes that you are using a reference frame in which the fluid flow is steady (i.e., velocity field is time independent). See for example
http://www.loreto.unican.es/Carpeta2012/EJP%28Mungan%29Bernoulli%282011%29.pdf [Broken]

So, which frame sees the flow as steady? The ground frame or the train frame?

Also, you have to be careful when comparing pressure at two points that are not on the same streamline. Thus, a point well inside the train and a point just outside an open window of the train are not on the same streamline. You cannot apply Bernoulli's equation directly to these two points.

Last edited by a moderator: May 8, 2017
12. Jun 24, 2016

### rcgldr

or the frame invariant version of Bernoulli's equation from that article could be used:

$p1 - p2 = \frac{\rho}{2} \ (v1-v2)^2 \ ( \frac{a1 + a2}{a1-a2})$

Although this is the correct answer, I'm not sure if this is the answer that the problem is asking for.

Last edited by a moderator: May 8, 2017
13. Jun 24, 2016

### TSny

Yes, this formula will give the answer. I'm with you, I don't think the problem expects the student to be familiar with this. Instead, it is easier to just use the usual form of Bernoulli's equation in the frame in which the flow is steady.

14. Jun 26, 2016

### Soren4

Thanks for the answer and the useful link! The flow is steady where the velocity is not time dependent $\frac{\partial v}{\partial t}=0$ . I would say that the frame of reference of the train is the one where the flow is steady, because in the ground frame the velocity of air in a fixed point changes in time because of the passage of the train.

About the problem of streamlines: as far as I understood I can use Bernoulli between different stream lines if flow is irrotational, anyway that's a bit confusing, because I read some (I think quite classical) problems where there is an house and a storm with strong wind outside and it is asked to find the force on the wedge of the house (inside of which the wind is not moving).

Also in that case one can use Bernoulli equation between a point outside and a point inside the house or is that the wrong way to do such exercise (from some point of view similar to the one in question)?

Last edited by a moderator: May 8, 2017
15. Jun 26, 2016

### rcgldr

That should be where the velocity field is not time dependent; the velocity itself does change. An alternative perspective on this is the frame where the velocity is inversely proportional to the cross sectional area it flows through. Using the frame invariant version of Bernoulli can be used to verify or eliminate having to choose the appropriate frame.

Only if the stream lines are parallel and connected by the equivalent of a static port, which "hides" beneath a boundary layer so that the static pressure on both sides of the static port is the same regardless of speed differences in the streamlines (within reason, below mach 1 speeds).

This depends on how the house is vented. Say it's sealed everywhere except on the upwind face, so that the vents are similar to pitot tubes that stop the streamlines that correspond to those vents. The house pressure will be the [[Stagnation_pressure]] of the wind.

16. Jun 27, 2016

### andrevdh

I would thought that there is a transitional region in the air in front of the train where it starts to flow to go around the train into the constricted area and thus speed up from zero. From then onwards the air is moving relative to the train at and speed equal to the train's speed and thus create a region of reduced pressure around it.

17. Jun 27, 2016

### rcgldr

The problem states that from a ground based frame of reference the air has zero velocity, referring to the air in front and behind the train (zero drag). The speed of the air flowing backwards around the train depends the speed of the train and the cross sectional area of the train versus the cross sectional area of the gallery (tunnel).

18. Jun 27, 2016

### andrevdh

How would the speed of the train determine or influence the speed at which the air flows past it? I thought it is a relative speed so that the speed of the air past the train is also the same speed as the train past the air?

19. Jun 27, 2016

### rcgldr

Since the problem states that the air in front and behind the train has zero velocity, then the speed of the train is the speed of displacement of air. If train travels faster, the air moves backwards around the train faster. The cross-sectional area of the gallery (tunnel) also affects the speed of the air traveling backwards around the train, the smaller the cross-sectional area of the gallery, the faster the air moves backwards around the train.