Pressure on a lateral side of a tank

[Misr]

Hello there,Could you explain to me how to calculate the pressure on the lateral sides of a tank?
[PLAIN]http://desmond.imageshack.us/Himg42/scaled.php?server=42&filename=unleddrw.jpg&res=medium
according to the textbook
P=L/2(ρg)
but I don't understand where this relation has come from
Could you explain it in a better way?

 

[gsal]

Take a look at the 4th posting in https://www.physicsforums.com/showthread.php?t=307277" thread...it explains the formula...and I think the formula that you present incorrectly has a division by 2...unless, the mean something else...

 

[wololoh]

@gsal; remember, without the factor 1/2 you just get the fluid pressure at the bottom of the tank...

 

[gsal]

Here it is more clearly...I hope.

Let's get the nomenclature straight.

Say we have a tank with a total height of L (and is full of water). Let h be any value between 0 and L and let it start (have its origin, value 0) at the very top of the tank and increase downwards...

Then...

The pressure at any given depth is p = ρgh and it is the pressure on the side of the tank at depth h from the top of the tank.

In other words, the pressure along the wall increases linearly proportional to h from h=0 to h=L and it is p=0 and p=ρgL, respectively.

 

[HallsofIvy]

And the fact that the pressure increases linearly means that the average pressure is just the numerical average of the minimum and maximum values. Here, the minimum pressure is 0 at the top and $\rho gl$ (the weight of the water (density times volume time g) divided by the area it presses on gives density times depth times g) at the bottom so the average pressure is $(1/2)\rho l$.

That is, by the way, the "average pressure". Pressure is a "point" function. It makes no sense to talk about the pressure on the entire side. Instead, the pressure integrated over the side, or, for linear change in pressure, the average pressure times the area, gives the total force on the side.

 

[Misr]

Take a look at the 4th posting in this thread...it explains the formula...and I think the formula that you present incorrectly has a division by 2...unless, the mean something else...

I know how to prove that P=hρg
but I don't understand why it is divided by 2
the division by 2 is correct
but can you tell me what do we mean by" pressure on the lateral side"?and how to calculate it?

I already imagine that the pressure at bottom of the tank is the weight of the water column divided by area

 

[gsal]

I think the division by 2 is simply to calculate the 'average' pressure on the side of the tank. But, if you ask me, that's not very useful if you are going to be doing some kind of calculation on the tank...you'd better be working with the maximum pressure...which happens at the bottom.

The pressure on the side of the tank is exactly the same pressure inside the water at the same depth...fluids exert the same pressure in all directions.

Does that help?

 

[Studiot]

But, if you ask me, that's not very useful if you are going to be doing some kind of calculation on the tank...

What sort of calculation did you have in mind?

This average pressure times the area = the force which acts at the centre of pressure.

 

[gsal]

I had some stress calculation in mind, you know, to find out for example how tall can the tank be or how thin can the wall of the tank be...

...as you build a tank from the bottom up and actually, magically, stays full of water as you build it...how tall can you go? will the tank fail at the top, p=0? in the middle p=ρgh/2? at the bottom, p=ρgh?

 

[256bits]

What sort of calculation did you have in mind?

This average pressure times the area = the force which acts at the centre of pressure.

I am not sure what you mean by centre of pressure, but the disributed force on the wall can be replaced by a single force of the same value 1/3 up from the bottom.

 

[Studiot]

I am not sure what you mean by centre of pressure, but the disributed force on the wall can be replaced by a single force of the same value 1/3 up from the bottom.

You must always beware of measuring up from the bottom since the water surface may not coincide with the top of the tank! (Although in this case Misr has shown the water flush with the top)

It is therefore better to say 2/3 of the distance down from the water surface to the bottom.

 

[Redbelly98]

I'm not sure where this 1/3 of the way up from the bottom (or 2/3 down from the top surface) comes from. The 1/2 in the original formula gives either the pressure 1/2 way up the water column, or the pressure averaged over the entire height of the water.

 

[Studiot]

I'm not sure where this 1/3 of the way up from the bottom (or 2/3 down from the top surface) comes from. The 1/2 in the original formula gives either the pressure 1/2 way up the water column, or the pressure averaged over the entire height of the water.

This part of elementary hydrostatics often confuses.

The average pressure is, by definition of average, the pressure which when multiplied by the area give the total force applied to the sidewall of the tank.

As HOI, and Misr's textbook has observed this average is simply 0.5(surface pressure + bottom pressure).

To be horizontally equivalent this could be applied as a single force at any height (depth) on the sidewall.

However, as 256 bits has observed this has to be applied at a point 2/3 of the water depth to be fully equivalent in moment terms, not the centroid of the sidewall area.

This point is the centroid of the pressure -depth diagram which is triangular and is called the centre of pressure. The position of the centre of pressure is different for other pressure diagrams and in general does not coincide with the centre of area (centroid) of the surface under pressure.

It is important to realize that the force is given by the average pressure on the surface, not the pressure at the depth of centre of pressure, times the surface area.

Whilst the calculation is probably irrelevant for a tank sidewall it is vital for calculating the overturning moment of a dam.

 

[Misr]

I think the division by 2 is simply to calculate the 'average' pressure on the side of the tank. But, if you ask me, that's not very useful if you are going to be doing some kind of calculation on the tank...you'd better be working with the maximum pressure...which happens at the bottom.

The pressure on the side of the tank is exactly the same pressure inside the water at the same depth...fluids exert the same pressure in all directions.

Does that help?

Yeah,That's obvious,I know that points on the same horizontal plane have the same pressure but What if was just asked to calculate the pressure on the lateral side of the tank?Should i calculate the average pressure or the max.pressure as you mentioned?

also how to calculate the total pressure on the lateral side?

 

[gsal]

I think the division by 2 is simply to calculate the 'average' pressure on the side of the tank. But, if you ask me, that's not very useful if you are going to be doing some kind of calculation on the tank...you'd better be working with the maximum pressure...which happens at the bottom.

The pressure on the side of the tank is exactly the same pressure inside the water at the same depth...fluids exert the same pressure in all directions.

Does that help?

Yeah,That's obvious,I know that points on the same horizontal plane have the same pressure but What if was just asked to calculate the pressure on the lateral side of the tank?Should i calculate the average pressure or the max.pressure as you mentioned?

also how to calculate the total pressure on the lateral side?

Well, it all depends what you need it for...maybe, BECAUSE the pressure varies along the wall the answer is not necessarily a single number.

 

[Redbelly98]

However, as 256 bits has observed this has to be applied at a point 2/3 of the water depth to be fully equivalent in moment terms, not the centroid of the sidewall area.

Ah, it makes sense to me now. Thank you.

also how to calculate the total pressure on the lateral side?

That's a nonsensical question. There is no "total pressure on the lateral side". There is an average pressure, and there is also a total force. Perhaps you are mixing up those two concepts.

EDIT: or perhaps you are thinking of the partial pressure vs. total pressure that comes up when discussing gas mixtures?

 

[Misr]

Well, it all depends what you need it for...maybe, BECAUSE the pressure varies along the wall the answer is not necessarily a single number.

yeah,ok

That's a nonsensical question. There is no "total pressure on the lateral side". There is an average pressure, and there is also a total force. Perhaps you are mixing up those two concepts.[/quote
okay,so does the total force on the lateral surface=area of the surface*pressure at the bottom?

 

[gsal]

No, because the pressure varies along the wall, you have to integrate...

but because the change is simply linear and is zero at the top and a maximum value at the bottom, it is like a triangle and so the average value is 1/2 the value at the bottom...

 

[Redbelly98]

Or to put it another way,

Total force on lateral surface = (area of surface) * (average pressure on the surface)​

... and the average pressure is the pressure half way up the side.

 

[Studiot]

See also these threads

https://www.physicsforums.com/showthread.php?t=534667

https://www.physicsforums.com/showthread.php?t=538445

 

[Misr]

Or to put it another way,

Total force on lateral surface = (area of surface) * (average pressure on the surface)

so why?

 

[Misr]

so what is the difference between pressure and force?
pressure is the force per unit area so?

 

[Redbelly98]

so what is the difference between pressure and force?
pressure is the force per unit area so?

Yes. The difference between the two is that pressure is force per unit area. I recommend that you look in your textbook, wherever the book first mentions the concept of pressure, and read it. You can also look at this:

http://www.school-for-champions.com/science/pressure.htm

 

[Misr]

That's good.I read the page you provided ,,it is easy to imagine the difference between force and pressure
but still don't understand this equation you provided before:

Or to put it another way,

Total force on lateral surface = (area of surface) * (average pressure on the surface)

... and the average pressure is the pressure half way up the side.

Acoording to your equation:the pressure applied at any point on the lateral surface equals to the average pressure.and by multiplying pressure-force per unit area-times the whole area,thus we get the total force applied on the lateral wall.
and that's not true beacuse the pressure varies along the lateral surface
Could you realize my problem?
Thanks

 

[Studiot]

Misr, your latest reply is so...ooh much better than this one

so why?

because it allows us to see what your difficulty is.



As you so rightly say the real pressure varies from point to point.

Another way to look at the average pressure is to say it is the pressure that would be exerted if it didn't vary from point to point ie was the same at every point, but to have the same effect as the real pressure system.

Of course as for any average some values that make up the average are higher and some are lower. The average is never the largest or smallest value.

Does this make sense?

 

[Misr]

LOLz..That's because I'm not in a hurry today..

Another way to look at the average pressure is to say it is the pressure that would be exerted if it didn't vary from point to point ie was the same at every point, but to have the same effect as the real pressure system.

Okay,How does the average pressure have the same effect as the real pressure?
Thanks

 

[Studiot]

Take a chess board (64 squares each 10 square centimetres in area).

Take 64 flat coins that fit into the squares and weight 100 grams each.

Lay the chess board out on a table.

Place two coins on every black square and none on any white square.

You have an uneven pressure distribution with zero pressure on the white squares but

$$\frac{{200}}{{10}} = 20$$

grams per square centimeter on the black ones.

So the average pressure on all squares is half of zero plus 20 = 10 grams per square centimeter.

This is the same as the pressure you would get if you placed one coin on every square, black or white.

Alternatively you could stack all the coins on one square but the pressure felt by the table under the chessboard would be the same however you stacked the coins.

This pressure would be the average pressure which is the pressure when the coins are spread out evenly.

 

[Misr]

Yes ..you want to say that the average pressure has the same effect as the real pressure,if the point on the lateral side have equal pressure
This is fine but can't imagine this:

Alternatively you could stack all the coins on one square but the pressure felt by the table under the chessboard would be the same however you stacked the coins.

When we put a coin on each square,the pressure on the table=(64*100)/640=10=average pressure
When we snack all of them on one square
the pressure on the table=(64*100)/10
may be you mean the average pressure on the table,in this case i would have no problem :)

so it would produce much larger pressure
thanks very much

 

[Studiot]

Hello misr, you are still confusing force and pressure, but we are getting there (I hope).

It is vitally important to know when to use force and when to use pressure in a calculation.

Each coin exerts a small force on the top of the chessboard.
This is because each coin is small compared to the size of the chessboard so the force it exerts is local. Local means that it only presses down on the part of the board it is sitting on.

When we add up all the forces and add in the weight of the chessboard we obtain the total force on the table.

You cannot add force and pressure, you must convert to all force or all pressure.
Usually converting to all force is easier because adding forces is easy and direct.
Adding pressures together is very complicated and the answer depends on circumstance so we avoid it.

Unlike the coin, the chessboard is not an insignificant size and the total force it imposes on the table is spread out over its whole area. So we can reasonable talk about the pressure on the table. This is, of course, the total force divided by the area of the chessboard.

 

[sophiecentaur]

That's good.I read the page you provided ,,it is easy to imagine the difference between force and pressure
but still don't understand this equation you provided before:

Acoording to your equation:the pressure applied at any point on the lateral surface equals to the average pressure.and by multiplying pressure-force per unit area-times the whole area,thus we get the total force applied on the lateral wall.
and that's not true beacuse the pressure varies along the lateral surface
Could you realize my problem?
Thanks

Why not sort this out for what happens in a spaceship first - when the gravity can be neglected? The pressure everywhere on all the walls of a container will be the same.

Then add the effect of the additional weight of the column of liquid on top.
The "lateral" pressure is only the same at that depth. The 'average pressure', say on a dam wall, is not a particularly meaningful quantity. We have all seen that dam walls are a lot thicker at the bottom than at the top - because the force / pressure is more at the bottom. But the walls, at a particular level along the dam, are all the same thickness. This is because the pressure, everywhere at a given depth, is the same.

 

[Misr]

You mean that i should first find the force that one coin exerts,then the force that all other coins exert and add them to the force of the chess board :!
so the force on the table=(64*100)+force of the chess board

So pressure on the table=6400+force of the chess board/area(I guess area=640)

 

[Misr]

Well,okay,may be you mean that force is still the same whatever the way we distribute it.