# Thermofluid mechanics, calculating load on a steel water tank

An open cylindrical water tank 2.4m diameter rests on three legs evenly positioned around the periphery of its circular base. The base and sides of the tank are made from steel with relative density 7.8 and thickness 20mm. If the tank is 2.1m high determine the load on each leg when the tank is empty, and full The volume of the steel in the sides of the tank can be approximated by V=dth

i tried calculating it and i got an answer however i didnt even use the volume equation, tho why would they give it in the question if im not going to use it? that's why i'm asking for your help, i used $F=((\rho*g*h)+101\mathrm{kPa})$*area

• what do you mean by density and area? Maybe the density of steel and the area of the circle with 2.4m diameter? If so, then... since the tank is not filled with stell, this would be wrong – Ilja Apr 24 '16 at 0:20
• yes, density of steel and area of steel, it doesnt work out, could you tell me what formula i would need to use please – Mark Apr 24 '16 at 10:03
• if you took the area of steel (i.e. the thin ring of diameter 2.4m and width 20mm), then it's right. Well, still not quite, since the 101kPa dont belong there (they act from below too), and you forgot the bottom plate :) Why don't you just calculate the mass – Ilja Apr 24 '16 at 10:20
• if i calculate the mass, than multiply it by gravity, to get the weight, it says open tank so i can't add the atm pressure to force, i have to calculate in regard to pressure – Mark Apr 24 '16 at 10:45
• the atmospheric pressure does nothing, it's there everywhere and cancels itself out. Or to be more precise, it's very slightly decreasing with height - if you consider this deviation you get just the buoyant force from air on the tank. So this is equivalent to replacing $\rho_\mathrm{steel}$ by $\rho_\mathrm{steel}-\rho_\mathrm{air}$ – Ilja Apr 24 '16 at 10:49

## 1 Answer

Yes, the solution you posted in your last comment is good.

With the sole criticism, that it is a little too precise. If the density of steel has only two digits, then the result should not have 4 as in 10.37kN
But this is very advanced criticism :)

Also minor points, to be even more precise: for the circumference volume of the mantle you could equally well insert $2.36$m instead of $2.4$m - it's something in between, actually.
Also, the radius of the bottom should be only $2.36$m, since the you counted the part outside in the mantle already.
What I want to say: don't try to make it too precise, since some of the values (density in this case) don't allow this anyway. Your water volume is exact, but not substracting the thickness would be sufficient also. But it's good.

• oh alright, thank you a lot ! at least someone helped me with feedback, my professor doesnt :) – Mark Apr 27 '16 at 0:21