[hityo]

Hallo,

I'm having some basics doubts in Fluid Mechanics.

I'll try to explain it by giving a example problem and please let me know your expert views.

Problem: A vertical shaft made of Concrete,having Area= 1m2, length= 30 m and roughness of the wall say 2 mm. In these Shaft there is opening of 0.5 m2 at every 5.5 m shaft length of the shaft like a window. From below i.e 0 m there is constant volume flow rate of air 20000 m3/h. (Isotherm =20 C)

So the exit flow rate from which opening will be more? 1,2,3,4
and please Why?


Thanks alot in advance!!

[sbaffini]

Sounds like a bizarre homework to me... why you think being of concrete has relevance for the shaft and the problem at hand to even mention it?

[sbaffini]

Is it closed at the other end and only has these windows? What's the ambient it is inserted in, just ambient? Why vertical has relevance? Why the roughness at the wall has relevance?

[mczero57]

Hi Hityo,

Interesting question !

The opening closest to the bottom will have the largest flow rate. Why ? Because all openings are the same size, and the opening closest to the bottom offers the path of least resistance to the flow. This is regardless of whether the walls are concrete or rough, or whether the flow is going horizontally or vertically.

Best,
Sam

[hityo]

Hallo Paolo,


Thanks for the replies! Concrete does not have any relevance, just I have taken Concrete(2 mm roughness) in my Simulation that is why I mentioned it.



Yeah it closed from other side. Ambient is nor ideal Air at 20 Celcius as mentioned with Isotherm.



Actually I believe the the bottom window will offer more Volume flow rate compare to the above windows. but I just need a expert opinions if I'm missing some Fluid mechanics laws here.



Thanks

[hityo]

Hallo Sam,


Thnaks for the Reply!



Exactly!! I thought the same. but I was in the argument with Colleauge. He meant there will be a static pressure gain and the flow at the top window will be having more flow rate compare to down.



am I missing here something?

[aerosayan]

Originally Posted by sbaffini

Sounds like a bizarre homework to me... why you think being of concrete has relevance for the shaft and the problem at hand to even mention it?

Maybe to teach about viscous losses?


I haven't run the numbers, but if the viscous loss is high enough the flow through the nearest opening might be the highest.

But in reality, I don't think the viscous losses would be high enough to cause a significant drop in flow rate.

[sbaffini]

Your friend has been probably fooled by the geometry but, if all the windows equally discharge in the same ambient, pressure is required to be larger at inlet to have a flow, and it will drop along the way, but won't ever drop below ambient (or you would have reverse flow), nor magically increase.

Still, while I agree, in general, on the fact that the first gets the most and so on, I feel like the details of the geometry might play a role here. For example, if the last window coincides with one end of the shaft, it might be favored with respect to the others.

[Gerry Kan]

Dear Hiyto:

The bottommost window will have the highest flow rate because, as the flow is introduced from the bottom of the shaft, it has the highest static pressure. In addition, it also has the highest hydrostatic pressure. Although, with air as working fluid it contribution is not expected to be significant ( .

The roughness of the concrete introduces losses that proportional to the square of its flow rate (i.e., Reynolds number). If you really want, you first determine the Reynolds number based on the 20000 cu m / h flow rate with the hydraulic diameter as the characteristic length scale (, which in my books should be close to 92600. Then look up the friction factor on the Moody chart with the corresponding relative roughness (.

Since the flow rate along the shaft is constant, so will the head loss and thus does not alter pressure gradient. The bottom of the shaft remains the point of highest pressure.

Of course, the implicit assumption is that the openings do not further alter the pressure distribution along the shaft and over its cross section. I don't know how valid this assumption is, given that the window size is half of the shaft cross section area, but it also seems that whoever came up with this question did not think about it either, as well as the fact that the answer could be achieved without even considering roughness effects, which could arguably be just a red herring.

Gerry.

[hityo]

Originally Posted by sbaffini

Your friend has been probably fooled by the geometry but, if all the windows equally discharge in the same ambient, pressure is required to be larger at inlet to have a flow, and it will drop along the way, but won't ever drop below ambient (or you would have reverse flow), nor magically increase.

Still, while I agree, in general, on the fact that the first gets the most and so on, I feel like the details of the geometry might play a role here. For example, if the last window coincides with one end of the shaft, it might be favored with respect to the others.

but according to Bernoulli, Static pressure will increase at the end of the shaft as velocity is decreasing and if the Pressure losses is less than the Static Pressure rise than the 4th Window will have the more Volume flow rate as compare to 1st Window?

[hityo]

Originally Posted by Gerry Kan

Dear Hiyto:

The bottommost window will have the highest flow rate because, as the flow is introduced from the bottom of the shaft, it has the highest static pressure. In addition, it also has the highest hydrostatic pressure. Although, with air as working fluid it contribution is not expected to be significant ( .

The roughness of the concrete introduces losses that proportional to the square of its flow rate (i.e., Reynolds number). If you really want, you first determine the Reynolds number based on the 20000 cu m / h flow rate with the hydraulic diameter as the characteristic length scale (, which in my books should be close to 92600. Then look up the friction factor on the Moody chart with the corresponding relative roughness (.

Since the flow rate along the shaft is constant, so will the head loss and thus does not alter pressure gradient. The bottom of the shaft remains the point of highest pressure.

Of course, the implicit assumption is that the openings do not further alter the pressure distribution along the shaft and over its cross section. I don't know how valid this assumption is, given that the window size is half of the shaft cross section area, but it also seems that whoever came up with this question did not think about it either, as well as the fact that the answer could be achieved without even considering roughness effects, which could arguably be just a red herring.

Gerry.

Thanks for the detailed explanation, but I'm here really confused because as I mentioned to Paolo.

but according to Bernoulli, Static pressure will increase at the end of the shaft as velocity is decreasing and if the Pressure losses is less than the Static Pressure rise, it will result into the 4th Window will have the more Volume flow rate as compare to 1st Window?

[Gerry Kan]

Originally Posted by hityo

but according to Bernoulli, Static pressure will increase at the end of the shaft as velocity is decreasing and if the Pressure losses is less than the Static Pressure rise, it will result into the 4th Window will have the more Volume flow rate as compare to 1st Window?

Dear Hityo:

I am sorry I don't quite understand what you wrote. Since your mentioned Bernoulli, and from what has been discussed in the thread, I think whoever thought of this question is simply thinking this as a variation of the classic textbook Bernoulli problem (large tank of water with holes drilled at different heights). So, instead of the usual hydrostatic pressure, a dynamic pressure is introduced from the bottom of the shaft in the form of a volumetric flow rate. So the logic goes, if the top is closed, the static pressure will be higher on top of the shaft, and therefore the topmost window has the highest flow rate.

Since the air is considered incompressible, and that a constant volumetric flow rate is prescribed at the bottom shaft opening, we can conclude this must equal to the flow rate through all four windows. Since the windows "big" in relation to the shaft cross section, there will be a significant amount of air leaving the shaft at each window. So you will see a corresponding drop in static pressure. With all four windows having equal area, one can only conclude that the topmost window will have the lowest flow velocity.

Now, let us assume the drop of static pressure is "small" as it leaves each window. In this case, a constant static pressure can be maintained, and thus each window, again having the same area, will have the same flow velocity. If you take into account friction losses and hydrostatic pressure, albeit small, the static pressure reduces again from the bottom to the top, and thus the topmost window should have a slightly lower velocity as the one in the bottom.

I hope this clarification is something you are looking for.

Gerry.

[hityo]

Originally Posted by Gerry Kan

Dear Hityo:

So, instead of the usual hydrostatic pressure, a dynamic pressure is introduced from the bottom of the shaft in the form of a volumetric flow rate. So the logic goes, if the top is closed, the static pressure will be higher on top of the shaft, and therefore the topmost window has the highest flow rate.

Gerry.

why here the Static pressure is more at top?


and not in the below cases as you mentioned?

"Since the air is considered incompressible, and that a constant volumetric flow rate is prescribed at the bottom shaft opening, we can conclude this must equal to the flow rate through all four windows. Since the windows "big" in relation to the shaft cross section, there will be a significant amount of air leaving the shaft at each window. So you will see a corresponding drop in static pressure. With all four windows having equal area, one can only conclude that the topmost window will have the lowest flow velocity.

Now, let us assume the drop of static pressure is "small" as it leaves each window. In this case, a constant static pressure can be maintained, and thus each window, again having the same area, will have the same flow velocity. If you take into account friction losses and hydrostatic pressure, albeit small, the static pressure reduces again from the bottom to the top, and thus the topmost window should have a slightly lower velocity as the one in the bottom""



Actually I have read this thing in a German text book and that is why I am making it very sure and I will Quote it.

"Bei einem geraden Zuluftkanal von konstantem Querschnitt mit vielen gleichgroßen Luftaustrittöffnungen tritt die Luft durchhaus nicht gleichmäßig aus allen Öffnungen aus, sondern die einzelnen Volumenströme werden zum Ende Kanals hin größer. Das ist darauf zurückzuführen, dass sich hinter einem Luftdurchlass im Hauptkanal die Geschwindigkeit verringert, wodurch nach dem Gesetz von Bernoulli der statische Druck ansteigt. Wenn dieser errechbare Druckanstieg größer ist als der Strömungsverlust, erhöht sich der stätische Druck zum Kanalende, und damit wachsen auch die Abzweigvolumina"

Translation: In the case of a straight supply air duct of constant cross-section with many air outlet openings of the same size, the air does not exit evenly from all openings, but rather the individual volume flows increase towards the end of the duct. This is due to the fact that the speed is reduced behind an air passage in the main duct, which, according to Bernoulli's law, increases the static pressure. If this calculable increase in pressure is greater than the loss of flow, the static pressure at the end of the channel increases and the branch volumes increase with it.


That is why I'm going nuts, beacause all the people in the Forum said the Static pressure will be high at the bottom and it will reduce as it goes upwards but this German text book is saying it will high at the Top.

Thanks in advance!!

[AtoHM]

Actually, I was thinking exactly what is written in the book from reading your description (I do just now see the image in the first post) and was a bit confused as everyone said otherwise. However, with the sketch you provided, I am with everybody else, the flow will more of less go directly to the first window and exit there.
Now what I had in mind prior to seeing your sketch was a situation like this one: https://www.google.com/search?q=divi...AKAkel0PubXBYM
And then, I think this is what the book refers to aswell, the last outlet should receive the highest flow rate.

[hityo]

Originally Posted by AtoHM

Actually, I was thinking exactly what is written in the book from reading your description (I do just now see the image in the first post) and was a bit confused as everyone said otherwise. However, with the sketch you provided, I am with everybody else, the flow will more of less go directly to the first window and exit there.
Now what I had in mind prior to seeing your sketch was a situation like this one: https://www.google.com/search?q=divi...AKAkel0PubXBYM
And then, I think this is what the book refers to aswell, the last outlet should receive the highest flow rate.

just because the first opening is aligned to the Inlet Flow? and how it is different then Text book describtion?

ahh thanks for the link...it is so probe the textbook

[Gerry Kan]

Dear Hityo:

Reading the description of your German text, and here is my interpretation. Air enters the duct with a total pressure of . The air is expected to slow down along the duct, due to air resistance as well as friction losses, and the dynamic pressure reduces, which leads to a corresponding increase in static pressure.

I am starting to see the logic in this statement, but I will need some time to convince myself that this is true.

Gerry.

[Edit Dec 3 2020] P.S. - After some thinking I still stand by my original position. Yes, the static pressure could go up along the duct, but gas is also decelerating, thus creating a back pressure which (could) further encourage the flow to escape via the path of least resistance, i.e., the lowest window.

[AtoHM]

Originally Posted by hityo

just because the first opening is aligned to the Inlet Flow? and how it is different then Text book describtion?

I would expect completely different flow characteristics depending on the direction of inflow and thus I doubt a generic description like in the book will be suitable here. I guess there are some assumptions involved regarding width/length ratios of the domain and yes, even the inflow direction even if not explicitly stated.

[sbaffini]

Honestly, I saw this coming since the first post. At this point, we have no idea of how the textbook setup might actually be, and it is assuming a lot about the geometry of side and end of the shaft. Also, it seems anywhere near the picture you posted later... where I have little doubts about the port 1 having the most flow, but not as much as you would probably think.

Also, let me say that the combined pressure change from losses and bernoulli effects can't produce larger pressures that the upstream ones. I hope this sounds obvious to you.

[LuckyTran]

The question can be really misleading. If you consider this problem like a large water tank with quiescent (i.e. stationary) fluid inside, it's clear the bottom hole has the most flow. But once you consider the fluid to be non-stationary and moving inside the enclosure... anything can happen. It depends on the details of how the flow moves through the shaft and negotiates with each window whether there is a static pressure rise or drop. Even if this duct is perfectly contoured to always result in a static pressure rise, in an infinitely long pipe the static pressure at the end of the pipe increases to a finite maximum value (but the flow reaches zero speed) and nothing flows out the last window. It's not a paradox, it just depends.