0
$\begingroup$

A closed tank rectangular in plane with vertical sides is 1.8m deep and contains water to a depth of 1.2m.

Air is pumped into the space above the water until the air pressure is 35kN/m2.

If the length of one wall of the tank is 3m, determine the resultant force on this wall and the height of the centre of pressure above the base.

So far my calculations have been to calculate the pressure due to the water which is: rho(g)(h) = 1000(9.81)(1.2) = 11,772 N/m^2. I then multiply this by the arc of the side of the tank covered by water which is: 1.2(3) = 3.6m^2.

Multiplying these gives 42379.2N.

I am not entirely sure what to do with the air pressure. Whether to multiply it again by 3.6 (the area of the side of the tank covered by water) or to multiply it by the area of the top of the water but I would not have enough details to calculate this.

$\endgroup$
  • $\begingroup$ Is the air pumped into the tank resulting in pressure that is above atmospheric pressure? There are two ways to specify pressure: 1) absolute pressure; 2) gauge pressure, which is everything above or below ambient pressure. $\endgroup$ – David White Feb 20 '17 at 2:57
2
$\begingroup$

Above the water line, the pressure on the wall is that of air pressure, so you can calculate the area of the wall above the water line and multiply it by the air pressure to get the force on that section. Below the water line, the pressure varies with depth as $$P=\rho g h +P_0$$ where $P_0$ is the air pressure above the water and $h$ is the distance below the water. To get the force below the water line you need to integrate the pressure. For each section of the wall with height $dh$, the force is given by $$dF = (\rho g h +P_0)Ldh$$ You then need to integrate this force to the bottom of the tank and add it to the force from the top of the tank.

Sorry for any typos, phone posting.

$\endgroup$
0
$\begingroup$

First, the pressure on the wall will be a function of depth- so near the top the water pressure is very low, while lower down it is higher. Averaged across the contact surface it will be equivalent to the pressure at the mid point.

Now you increase the pressure in the entire vessel by pressurizing the air space above it. This adds the same pressure everywhere. You can therefore take the air pressure and multiply it by the entire area of the side wall, and add that to the water pressure at a depth of 0.6m multiplied by the area wetted by the water.

I hope this is sufficient help...

$\endgroup$
  • $\begingroup$ Would I be correct so in saying that the final resultant force would be 1000(9.81)(0.6)(1.2)(3) + 35000(1.8)(3)? This being the pressure at depth 0.6 multiplied by the area covered by water plus the air pressure multiplied by the whole area of the wall. $\endgroup$ – Cian Dec 12 '15 at 21:51
  • $\begingroup$ @Cian looks right to me $\endgroup$ – Floris Dec 12 '15 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.