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#21
As far as it goes, I made that problem up on the spot based on something from my calculus book. I haven't the faintest clue what the real answer would be.
As far as it goes, I made that problem up on the spot based on something from my calculus book. I haven't the faintest clue what the real answer would be.
Actually that's not quite true although you've got the "tube" bit right -- the volume of this is essentially the volume of cylinder - so it's relatively simple maths to work out the cylinder dimensions to solve the problem. (Again assuming an incompressible liquid - not a gas - and assume normal atmospheric pressure). There's no EXACT answer as some calculus will be required but my answer is a good practical correct one.
Imagine an infinitely large bucket that already contains a huge volume of water. A 100 gallon one is a good example here.
At 2 gals a minute it should take around around 50 mins to empty the bucket with no water coming in -- we can ignore the effects of the water pressure here as it will be relatively constant over the lifetime of this problem.
You could have larger bucket which is supplied at > 2 gals a min ( or < 2 gals a min).
The outflow doesn't depend on the rate of inflow (until the bucket is empty of course).
Actually to SUPPLY the bucket at 2 gallons a minute you've got a similar problem - the answer is the same as the pipe needs to be 1/2 inch approx.
(Ok you could have a 6 inch diameter pipe or anything > 1/2 inch SUPPLYING the water but I assume the problem meant what's the SMALLEST size that would do the job and we are considering the OUTFLOW).
a 1/2 inch hole will do the trick -- try it and see with say a 5 litre gas can - drill a 1/2 inch hole in it - fill it up with water and see how long it takes to empty. Convert US gals to litres and you should come up with the magic answer 2.2 gals / min requires a 1/2 inch diameter pipe. (Of course if you increase the pressure significantly you might get a bit more but I assume here that the water is flowing at approx 1 Bar (1 Atm) pressure.
(And this is AFTER I've just come back from the pub - now having a look again at Man Utd's win today -- back at the top of the league).
Cheers
jimbo
Ya can't do that - you're starting with an arbitrary number, 10 in this case.
You could start with any arbitrary number to count down, say 25. Then the countdown would be:
25, 24, 23, 22, 21.
Of course we can't add 5 more fingers of the other hand to the 21, to yield 26...
The solution to get the correct amount of digits would be:
(n2 - n1) + 1 = amount of digits.
(25 - 21) + 1 = 5, just as the other (10) countdown:
(10 - 6) + 1 = 5...
Then and only then, we can add the 5 more to get the ? how-ever many...
I've seen snags like this with 'buffer-sizes', where the +1 wasn't done:
(Last_index - First_index) is always short by one,
even when using the 'c' base index of [0]:
(4 - 0) = 4, but to get all the items, we need the + 1...