Because the density of the IG is negligible it can be ignored.
Calculate all the pressures due to the Liquid in the tank, and add on the Gas pressure.
For part a, pressure at the base of the tank is Ro g h + gas pressure. i.e ( 0.9 x 9.81 x 2.5) + 60 kN/m2
Assuming Atmospheric pressure outside the tank that would be the pressure accelerating the oil out of the base, but I cant remember the formula for that off hand, but it should be easy to find.
For part b there are various ways of breaking up the calculation, one way would be to calculate the Force of the Gas pressure and then calculate the Force created by the water pressure and add them.
Total Area of the side of the tank, on which gas pressure is acting is 4 x 3 = 12. Force = Pressure x Area so Force of Gas is 12 x 60 = 720 kN.
Area of the side of the tank, on which oil pressure is acting is 2.5 (depth of oil) x 3 (width of side) = 7.5 m2
Please note the above only applies to rectangular tanks, for "ship shaped " tanks you would have to use Calculus and or Simpsons rule to get the results.
Pressure due to head of oil at the oil surface = 0
Pressure due to Head of oil at base of tank (from Part a) =0.9 x 9.81 x 2.5 = 22.075
Mean Pressure is half the above. (0 + 22.075)/2
Force on the side of the tank, due to oil pressure is the area calculated above x mean pressure calculated above. 7.5 x 22.075/2 = 82.78125 kN.
To calculate the Centre of Pressure you have to calculate the moments exerted by all the forces acting on all the areas about the top or bottom corner of the tank.
Having done that, the centre of pressure is the single point at which, if the total force on the side of the tank as calculated above was acting, it would produce the same turning moment.
How do you think you would do that??
Hope this helps.
Last edited by Big Pete
on Mon Dec 12, 2016 3:02 am, edited 2 times in total.
It is always better to ask a stupid question than to do a stupid thing.