Pressure on a lateral side of a tank


by Misr
Tags: lateral, pressure, tank
Misr
Misr is offline
#1
Oct5-11, 07:21 PM
P: 391
Hello there,Could you explain to me how to calculate the pressure on the lateral sides of a tank?

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?
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gsal
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#2
Oct5-11, 11:51 PM
P: 838
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...
wololoh
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#3
Oct6-11, 03:13 AM
P: 5
@gsal; remember, without the factor 1/2 you just get the fluid pressure at the bottom of the tank...

gsal
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#4
Oct6-11, 08:35 AM
P: 838

Pressure on a lateral side of a tank


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
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#5
Oct6-11, 08:57 AM
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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 [itex]\rho gl[/itex] (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 [itex](1/2)\rho l[/itex].

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
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#6
Oct6-11, 05:09 PM
P: 391
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
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#7
Oct7-11, 04:41 AM
P: 838
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
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#8
Oct7-11, 05:52 AM
P: 5,462
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
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#9
Oct7-11, 08:35 AM
P: 838
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
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#10
Oct8-11, 02:50 PM
P: 1,260
Quote Quote by Studiot View Post
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
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#11
Oct8-11, 03:24 PM
P: 5,462
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
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#12
Oct8-11, 05:58 PM
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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
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#13
Oct8-11, 06:18 PM
P: 5,462
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 realise 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
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#14
Oct8-11, 06:53 PM
P: 391
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
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#15
Oct8-11, 07:06 PM
P: 838
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
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#16
Oct8-11, 07:11 PM
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Quote Quote by Studiot View Post
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.

Quote Quote by Misr View Post
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
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#17
Oct9-11, 03:16 PM
P: 391
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
[quote]
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
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#18
Oct9-11, 04:38 PM
P: 838
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...


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