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### Mechs questn 8 dec2014 chEng sqa

Posted: Sun Dec 04, 2016 11:36 am
Been trying to understand how to assess the forces and resultant centre of pressure for fluids held in a tank...problem is the fluids are IG gas and oil.

Q8...an oil tank is 4m high and of square cross section with vertical sides 3m wide. The tank contains oil to a depth of 2.5m and the space above the oil is pressurized with inert gas to 60kN/m2 . The density if the oil is 900kg/m3. Calculate each of the following :
a). The theoretical velocity at which the oil would escape from the base of the tank...... (4marks)
b). The total force on the side of the tank.... (6mrks)
c). The position of the resultant center of pressure on one side of the tank.... (6mrks)

Pleased to get some help with explanations soonest....thank you brilliant people

### Re: Mechs questn 8 dec2014 chEng sqa

Posted: Wed Dec 07, 2016 12:15 am
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.

### Re: Mechs questn 8 dec2014 chEng sqa

Posted: Mon Dec 12, 2016 2:49 am

Do you have any idea on how to work out the centre of pressure? The centres of pressure for the base and top will both be at their Geometric centres, each side will be somewhere on the vertical centreline, but how high up?

Calculate the turning moment of the oil force acting through the centre of pressure of the oil force, acting about a horizontal axis of your choice, do the same calculation for the Gas force, ACTING ABOUT the SAME AXIS.

Add the two turning moments and divide them by the Total forces of the oil and Gas, and the result will be the vertical distance of the combined centre of Pressure from your chosen axis.

I hope you are getting somewhere with the question!!

BP

### Re: Mechs questn 8 dec2014 chEng sqa

Posted: Tue Dec 13, 2016 3:42 am
It's rather sad, that this is the question that baffled me when I was studying. I just could not get my head around it.

### Re: Mechs questn 8 dec2014 chEng sqa

Posted: Tue Dec 13, 2016 9:23 am
Well I guess that either Ch Eng Lamont has completely solved the question or given up and hit the bottle!

BP.