A water tank 3m in diameter and 6m high is made from a steel taving a thickness of 12mm. When tank is filled w/water determine the circumference stress. Det. the tank stressL at the buttom of the tank when it is filled w/water. if the circumference stress is limited to 5Mpa what is the max. height of the water?
Volume of tank = ? @patbigFat
not stated sir ..
nono... can u work it out please?
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V=Pi x r ^2 x h ?
yup
is it 169.6464m?
no
what are u using for r?
raduis sir?
yeah... what number are u plugging into ur equation tho... for r?
3 sir?
no... 3 = diameter radius is half of diameter
so give it another shot with r = 1.5 so V=??
56.54 m3
i got 42.41... how are u calculating this? 3.14159 x 1.5^2 x 6 =42.4115
my mistake sorry ..
all good dude now... surface area of inside of tank? = ??
i tot when i sqr 1.5 = 3 so i did a manual calculating .... my bad....
SA = 2 × pi × r2 + 2 × pi × r × h ?
70.686 m^2 ?
guyz help?
hoop stress \(\large \sigma_{\theta} = \frac{P \ D}{2t}\) D = 3, t = 12mm get pressure from bernoulli \(P = \rho g h\) h = 6 plug and play [if you know some calculus, the formula is deriveable; but i think it is also pretty standard for thin walled pressure vessels. been ages since i did this stuff.] you can get steel's material properties online, for the end bit you will need its tensile strength and the hoop stress is measured around the circumference , ie tangentially to the circle
actually you don't need to look at steel at all you are told "the circumference stress is limited to 5Mpa" so you are now solving h for a given \(\sigma_{\theta}\)
Uhmm.m P=pgh by the way what's "pg" ?
\(\rho \equiv\) density of fluid, here water g \(\equiv\) gravity = \(9.8 m/s^2\)
sorry guys, fell asleep last night
howd u go with this?
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