[ASK] Is This Debit Question Solvable?

Monox D. I-Fly · 2017-03-29 15:28:39

A water container whose volume is 400 liters has 3 vertical holes. The first hole is located at the height which shows 300 liters. The second hole is located at the height which shows 150 liters. The third hole is parallel to the bottom. If the water in the container is drained by opening the holes one by one from the top to the bottom (second hole isn't opened until the first hole can no longer drain the water, third hole also isn't opened until the first and second holes can no longer drain the water), and the water container is empty in 10 hours, what are the debit (speed of draining water) of each holes?

Bob · 2017-03-29 19:36:19

hi Monox D. I-Fly

Unless all holes have the same rate, I cannot see how this can be solved.

Bob

iamaditya · 2017-03-29 20:03:08

bob bundy wrote:

Unless all holes have the same rate, I cannot see how this can be solved.

All the holes cannot have the same rate since the lateral pressure(pressure exerted on the side walls of the container) of a fluid inside a container id directly proportional to the depth. So the pressure at the hole parallel to the bottom will have the highest pressure, the hole marked 150ml will have less high pressure, and the top hole will have the least pressure. So, if all the holes are opened simultaneously, water will flow out from the third hole at the fastest rate, the 2nd the next fastest and the third with the slowest rate. Now, it is said that the holes are opened one by one. So, you need to know the lateral pressure at that hole. Now, if we know the pressure, and the area of each hole then we will know the force as Pressure*Area. After working out the force we can get the time required for draining that volume of water. Then dividing the total time by the total volume of water drained we can get the rate of debit of each holes.

Bob · 2017-03-31 20:26:22

hi iamaditya

Thanks for the suggestion. It sounds good, but, as I understand it, the velocity of 'ejection' is dependent on the height above the hole. So, even if we assume all the holes are equal circles, the velocity will diminish as the water level drops. So you'd have to integrate wrt the height with upper and lower limits for each hole. Not impossible but horribly complicated. I shall ponder this further. Maybe I can find a way to do this.

Bob

iamaditya · 2017-03-31 23:04:50

Hi bob,

Yes the pressure will be constantly changing(more correctly decreasing) since the water in the tank constantly decreases and hence the force will be decreasing too.

So you'd have to integrate wrt the height with upper and lower limits for each hole.

Well, if it can be solved, then its good but as you know I don't know calculus I hope you can solve it.

I shall ponder this further. Maybe I can find a way to do this.

Yes, you can. Try a bit and you will find it (I hope).

Bob · 2017-03-31 23:08:40

Here's my attempt, based on iamaditya's pressure idea.

I am assuming that the container has a uniform cross section ( meaning the volume is proportional to the height) and that the holes are small and all the same.

Torricelli's law states:

where V is the volume, t the time, g the acceleration due to gravity, and h the height of the water above the hole.

I'll get a general formula when the height drops from h to zero.

The height drops 100, then 150, then 150 'height' units in times t1, t2, t3 so

Adding

So

No idea if this is correct.

Bob

iamaditya · 2017-03-31 23:10:38

The question says that the total time for draining the whole water is 10 hrs. Just add all the 3 times and then check whether it is coming 10 hrs or not.

Bob · 2017-03-31 23:13:07

These times do add up to 10 (checked before I posted) but that doesn't prove the analysis is correct.

Bob

iamaditya · 2017-03-31 23:14:22

Hmm, but..then how can you check it?

Bob · 2017-03-31 23:18:17

You get a container and make three holes ........

Bob

iamaditya · 2017-03-31 23:28:07

You mean to say that we can check it by doing the actual practical. Hmm, well I'll see when I get time and will do it. I love doing science experiments at home, and I will try this out.

P.S. You should have Rationalized the denominator in each case; then the fractions would be much more simpler.

Bob · 2017-04-01 00:28:49

Yes, you can do that. Back in pre-calculator days it made it a lot easier to evaluate as a decimal. I had to memorise the square roots of 2, 3 and 5 so I could do such things.

Nowadays it doesn't make much difference so I've stopped bothering.

But I'm glad you did because it highlighted an error I made when transcribing from paper to the post. I should have typed:

I've edited the earlier post and the rationalise versions should be

and

My apologies for that.

Bob

iamaditya · 2017-04-01 19:57:33

Ok, dont worry it happens. Anyways I'm taking the decimal approximation of each of the expressions:

And together they add upto 10 Hrs.

Bob · 2017-04-01 21:18:01

Thanks. If you look again at this line

That is the point where I use the value of 10 to replace t1+t2+t3. So it's unlikely my three answers won't add up to 10 (as there's not much working after that point) even if the rest of the analysis is rubbish.

Bob

Math Is Fun Forum

#1 2017-03-29 15:28:39

[ASK] Is This Debit Question Solvable?

#2 2017-03-29 19:36:19

Re: [ASK] Is This Debit Question Solvable?

#3 2017-03-29 20:03:08

Re: [ASK] Is This Debit Question Solvable?

#4 2017-03-31 20:26:22

Re: [ASK] Is This Debit Question Solvable?

#5 2017-03-31 23:04:50

Re: [ASK] Is This Debit Question Solvable?

#6 2017-03-31 23:08:40

Re: [ASK] Is This Debit Question Solvable?

#7 2017-03-31 23:10:38

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#8 2017-03-31 23:13:07

Re: [ASK] Is This Debit Question Solvable?

#9 2017-03-31 23:14:22

Re: [ASK] Is This Debit Question Solvable?

#10 2017-03-31 23:18:17

Re: [ASK] Is This Debit Question Solvable?

#11 2017-03-31 23:28:07

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#12 2017-04-01 00:28:49

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#13 2017-04-01 19:57:33

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#14 2017-04-01 21:18:01

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