- #1
Kelju Ivan
- 26
- 0
I have a sylindrically shaped canister (open from the top) with a height of 25.0cm and a radius of 5.00cm. At the bottom there is a hole with an area of 1.50cm^2 spraying out water. How long does it take to empty the canister?
Subscripts t and b stand for top and bottom and h is the height.
I know that the flowing mass is the same in both ends of the flow tube so [tex] A_t v_t = A_b v_b [/tex] => [tex] v_t = \frac{A_b}{A_t} v_b [/tex]
There's atmospheric pressure at both ends of the canister.
Using Bernoulli's equation I get [tex] p_a + \rho g h + \frac{1}{2}v_t^2 = p_a + \rho g*0 + \frac{1}{2}v_b^2[/tex]
From this I could solve the flow speed at the hole in the bottom. Before I got that far I was stopped by one thought: the flow speed must change when the surface gets lower, that is, when the hydrostatic pressure at the bottom becomes weaker. I assume this is somehow done with integrals, but I really have no idea how. Please help.
Subscripts t and b stand for top and bottom and h is the height.
I know that the flowing mass is the same in both ends of the flow tube so [tex] A_t v_t = A_b v_b [/tex] => [tex] v_t = \frac{A_b}{A_t} v_b [/tex]
There's atmospheric pressure at both ends of the canister.
Using Bernoulli's equation I get [tex] p_a + \rho g h + \frac{1}{2}v_t^2 = p_a + \rho g*0 + \frac{1}{2}v_b^2[/tex]
From this I could solve the flow speed at the hole in the bottom. Before I got that far I was stopped by one thought: the flow speed must change when the surface gets lower, that is, when the hydrostatic pressure at the bottom becomes weaker. I assume this is somehow done with integrals, but I really have no idea how. Please help.
Last edited: