The Curious Case of Pascal Barrels

[Usamah Jundi]

Hello! I've been confused about fluid statics, to be more specific, about the hydrostatic pressure. supposedly i have a cylindrical barrel of height h and radius of r:

(a). if i were to calculate the total Force inside the barrel, do i use the (Force on the lid + Force on the Sides + Force on the bottom) while the pressure on the sides is defined as an integration of the dF = P * dA ?

(b).how about the pressure?

and

(c) i suppose you guys have seen the Pascal's Experiment on Bursting the barrel with a long tube on top of the barrel filled with water. is the pressure inside the barrel only the hydrostatic pressure of the water inside the tube plus the average pressure inside the barrel ( i suppose the atmospheric pressure cancels out because it's applied to the tube and the barrel) , or i can just calculate the force by using the hydrostatic pressure of the water + the integration of the dF on the sides of the barrel, and the Force on the lid and bottom?

if i have to use the average pressure, why?thanks!

forever in curiousity,

Usamah Jundi Abdillah.

 

[BvU]

Hello Usamah,

supposedly i have a cylindrical barrel of height h and radius of r

Is not enough to completely describe the situation. For completeness, consider adding the following:

• filled with a liquid, e.g. water
• plug closed without changing the pressure from atmospheric and temperature unchanged
• vertical on a flat floor, in air (not 100 m under water).

(a)
In that case there still is no point asking for the force inside the barrel. What you probably mean is the absolute value of force difference between the inside and outside integrated over the entire area of the barrel material. Which, as you say, can be obtained by integrating $\ P \cdot dA \$.

In the situation I described the contribution from the force on the lid is zero (atmospheric inside and outside). The bottom also (normal force offsets weight of barrel plus liquid) .

(b)
what about it ? Everywhere where there is liquid you expect $P = \rho g h$ + 1 atm, where $h$ is the height of the liquid above the point where you measure.

(c)
In this experiment the 1 atm at the fluid level is replaced by $\rho g h_2$ + 1 atm, where $h_2$ is the height of the liquid in the tube.
And no, I didn't know this experiment had a name and I sure never saw it demonstrated
You can still do the calculation, just with a different P.
And the barrel will burst sideways, is my bet -- but then again, you want a cylindrical barrel and there the lid/side transition may well be the weakest point. Pascal didn't have oil drums; he had real tuns (barrel-shaped barrels with staves and a lid that tightens itself when pressurized).

$\rho$ is the density of the liquid, about 1000 kg/m³ for water.

--

 

[Usamah Jundi]

Hello Usamah,

Is not enough to completely describe the situation. For completeness, consider adding the following:

• filled with a liquid, e.g. water
• plug closed without changing the pressure from atmospheric and temperature unchanged
• vertical on a flat floor, in air (not 100 m under water).

(a)
In that case there still is no point asking for the force inside the barrel. What you probably mean is the absolute value of force difference between the inside and outside integrated over the entire area of the barrel material. Which, as you say, can be obtained by integrating $\ P \cdot dA \$.

In the situation I described the contribution from the force on the lid is zero (atmospheric inside and outside). The bottom also (normal force offsets weight of barrel plus liquid) .

(b)
what about it ? Everywhere where there is liquid you expect $P = \rho g h$ + 1 atm, where $h$ is the height of the liquid above the point where you measure.

(c)
In this experiment the 1 atm at the fluid level is replaced by $\rho g h_2$ + 1 atm, where $h_2$ is the height of the liquid in the tube.
And no, I didn't know this experiment had a name and I sure never saw it demonstrated
You can still do the calculation, just with a different P.
And the barrel will burst sideways, is my bet -- but then again, you want a cylindrical barrel and there the lid/side transition may well be the weakest point. Pascal didn't have oil drums; he had real tuns (barrel-shaped barrels with staves and a lid that tightens itself when pressurized).

$\rho$ is the density of the liquid, about 1000 kg/m³ for water.

--

Ah, i forgot to mention that the barrel is filled with water on a flat foor on sea level, and it is closed.

 

[BvU]

Well, that is what I made of it. But here at PF you want to be complete in the 'problem description' -- lots of lengthy threads result because that isn't always the case.

Anyway, does my answer make sense for you ?

 

[Usamah Jundi]

Well, that is what I made of it. But here at PF you want to be complete in the 'problem description' -- lots of lengthy threads result because that isn't always the case.

Anyway, does my answer make sense for you ?

The only thing i am confused now is that isn't the water inside the barrel also contributes to the force exerted to the sides of the barrel?

 

[BvU]

It sure does. "... 1 atm at the fluid level is replaced by ... " so that $\ P = \rho g (h +h_2 )$ + 1 atm.

$\$

 

[Usamah Jundi]

It sure does. "... 1 atm at the fluid level is replaced by ... " so that $\ P = \rho g (h +h_2 )$ + 1 atm.

$\$

now one last question ( sorry for asking too much)

why did he used the midpoint of the barrels height to calculate the "average pressure" ? shouldn't he just integrated it from the ∫ρgh*2πr*dh all the wayy to the bottom? or would it just yield the same result?

 

[BvU]

The latter: the pressure varies linearly with height, so the integral is the same as taking height times pressure half way.

 

[Usamah Jundi]

The latter: the pressure varies linearly with height, so the integral is the same as taking height times pressure half way.

thanks so much! i also discovered that the total force after integration will eventually be F = ½ρgh²C

thanks!