How long does it take for the canister to empty?

[Kelju Ivan]

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 $$A_t v_t = A_b v_b$$ => $$v_t = \frac{A_b}{A_t} v_b$$

There's atmospheric pressure at both ends of the canister.

Using Bernoulli's equation I get $$p_a + \rho g h + \frac{1}{2}v_t^2 = p_a + \rho g*0 + \frac{1}{2}v_b^2$$

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.

 

[Tide]

The flow speed does decrease - because h decreases!

 

[Kelju Ivan]

Yes, I understand that, but I have know idea how to apply integrals to problems like these. I have other similar problems which I can't solve because of this.

One of them is about a pool. At the side of a pool there's a hatch (dimensions a and b), which is hinged from the bottom and which's top is just at the surface of the water. Now I need to find the net force caused on the hatch by the pressure and the torque caused by the same force.

The deeper in the water, the stronger the force caused by the pressure. Torque on the other hand has a larger effect near the top, where the lever is longer. How do I go about with applying these integrals?

 

[Kelju Ivan]

Okay, I ignored it and got a time of 5.91s. Seems kinda fast, but the volymetric flow looks to be consistent with another problem with the same canister. I guess that's it then, thanks.

How about that pool with the hatch? In my book (Young and Freedman's University Physics) there's a similar problem and it's got a hint telling to calculate the torque of a thin horizontal strip at depth h and integrate that over the hatch.

How does that work exactly?

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The answer I was replying to seems to have been deleted while I wrote this. Hmmh...

 

[Clausius2]

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 $$A_t v_t = A_b v_b$$ => $$v_t = \frac{A_b}{A_t} v_b$$

There's atmospheric pressure at both ends of the canister.

Using Bernoulli's equation I get $$p_a + \rho g h + \frac{1}{2}v_t^2 = p_a + \rho g*0 + \frac{1}{2}v_b^2$$

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.
.....
Okay, I ignored it and got a time of 5.91s. Seems kinda fast, but the volymetric flow looks to be consistent with another problem with the same canister. I guess that's it then, thanks.

Hi, Kelju Ivan. What have you ignored? You mean, the variations of v due to variations of h?.

We will consider steady Bernoulli equation. I mean, just the fluid movement is started, there is a period of time in which the flow is accelerated downwards. Thus, the process is unsteady, and you would have to add another term to this form of Bernoulli equation. But we suppose (if you don't want say it to me) that time to be smaller enough compared with characteristic time of downwards flow. Then:

$$p_a + \rho g h + \frac{1}{2}v_t^2 = p_a + \rho g*0 + \frac{1}{2}v_b^2$$

Continuity: $$A_t v_t = A_b v_b$$

And now Continuity again : $$v_t=-\frac{dh}{dt}$$ where h(t) is the height of the fluid column. If we insert this result in Bernoulli equation:

$$v_t=-\frac{dh}{dt} =\sqrt{\frac{2gh}{(A_t/A_b)^2-1}}$$

In this form you only have approached about steadiness of the flow. You could simplify this by saying that "vt" is much smaller than "vb" due to At/Ab>>1. It is the same thing that writting "vt=0". In fact, if Ab/At<<1, the time variations of h(t) are negligible compared with the residence time of a fluid particle inside the vessel:

residence time: $$t_r \approx \frac{D_b}{U_o}$$

characteristic time of h variations: $$t_o \approx \frac{h_o}{U_o}$$

where Db=bottom's hole diameter; ho=initial height; Uo=characteristic velocity.

You can realize that tr<<to; so that you could linearize the differential equation around "ho" and solve it with the appropiate initial conditions.

I hope it could help you a bit.

Regards.

 

[Kelju Ivan]

Yes, I ignored the variations in v, because some other user advised me to do so (but then deleted his post).

I must say that I didn't understand where you were getting at with those characteristic times and velocities. What are they? I've never seen those equations you presented, I don't think they're necessary for this task.

I'm starting to trust the thought that this exercise should be thought in the simpler way of having a constant speed. I'll find out on Thursday.

Currently I'm more interested in the hatch problem with the pool. With some help from a friend I integrated to find the force affecting the hatch. I got F=½rho*g*b^2*a

Should I now just calculate F*b (the height of the hatch) or do I need more integration?