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# Reservoir

31 replies to this topic

### #11 SPENGLER

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Posted 09 September 2007 - 09:54 PM

2 days = 48 hours = x
3 days = 72 hours = y
4 days = 96 hours = z
6 hours = a

(x + y)/2= 60
(60+ z)/2=78
(78+6)/2=42

you've calculated a weighted average ... i think. this is more like finding the factorial of the pipes (wrong), not the sum of the pipes and their "volume/time" flow rate.
by your calculations, it would vary depending on the order calculated:
(a + x) / 2 = (6 + 48) / 2 = 54 / 2 = 27
(27 + y) / 2 = (27 + 72) / 2 = 99 / 2 = 49.5
(49.5 + z) / 2 = (49.5 + 96) / 2 = 145.5 / 2 = 72.75 hours????

step back from the problem and as mentioned earlier; if all four are ON, then they can't be slower than the fastest tap at 6 hours. then reverse determine a flow-rate ... say the reservoir is 1 litre. you should get calculations like:

relative flowrates:
6hpl, 48hpl, 72hpl, 96hpl (hpl = hours per litre)

invert to get lph from hpl (and find lowest common denominator for easy math)
6 hpl ==> 1/6 lph = 16/96 lph = 48/288 lph (lph = litres per hour)
2 days ==> 48 hpl ==> 1/48 lph = 2/96 lph = 6/288 lph
3 days ==> 72 hpl ==> 1/72 lph = 4/288 lph
4 days ==> 96 hpl ==> 1/96 lph = 3/288 lph

combined flowrates
(48+6+4+3)/288 lph = 61/288 lph ===> 288/61 hpl = 4.72 hours per litre
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### #12 bonanova

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Posted 09 September 2007 - 11:31 PM

The answer is precisely 12/61 days.

A = 1/2 (reservoirs/day)
B = 1/3
C = 1/4
D = 4

(A + B + C + D) * Time = 1 (Time is in days. A, B, C, and D are in reservoirs per day.)
(1/2 + 1/3 + 1/4 + 4) * Time = 1
(61 / 12) * Time = 1
Time = 12 / 61 days (~ 4.72 hours)

Lucid's analysis is the clearest.
What you get when you turn on more taps is more flow. Find the flow rates and add them.

The taps flow, respectively at rates of 1/2, 1/3, 1/4 and 4 reservoirs per day.
Flowing together, [adding their flow rates] the rate is 61/12 reservoirs per day.

They will fill 1 reservoir in exactly 12/61 days.

That comes to 4 hours, 43 minutes and .. about 16.7213114 seconds.
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### #13 SnoopyGirl

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Posted 20 September 2007 - 06:17 PM

Tap A=2 days=.5 reservoirs in 1 day
Tap B=3 days=1/3 reservoirs in 1 day
Tap C=4 days=.25 reservoirs in 1 day
Tap D=6 hours=4 reservoirs in 1 day

4+.25+.33+.50=5.08 reservoirs/day with all 4

5.08 Res/1Day = 1 Res/x Days
5.08 Res * x = 1ResDay
x = 1ResDay/5.08 Res
x = .20 days = 4.8 hours = 4 hours 48 min.
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### #14 BoilingOil

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Posted 26 September 2007 - 03:27 AM

The answer is precisely 12/61 days.

A = 1/2 (reservoirs/day)
B = 1/3
C = 1/4
D = 4

(A + B + C + D) * Time = 1 (Time is in days. A, B, C, and D are in reservoirs per day.)
(1/2 + 1/3 + 1/4 + 4) * Time = 1
(61 / 12) * Time = 1
Time = 12 / 61 days (~ 4.72 hours)

Lucid's analysis is the clearest.
What you get when you turn on more taps is more flow. Find the flow rates and add them.

The taps flow, respectively at rates of 1/2, 1/3, 1/4 and 4 reservoirs per day.
Flowing together, [adding their flow rates] the rate is 61/12 reservoirs per day.

They will fill 1 reservoir in exactly 12/61 days.

That comes to 4 hours, 43 minutes and .. about 16.7213114 seconds.

'Nuff said. Before even looking at what anyone was answering, I had already determined that it would be 12/61 (or less than 1/5) of a day. I didn't calculate any further, but since 1/5 of a day would be 4h48m, any answer approximating 4h43m16.72s sounds about right to me...

BoilingOil
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### #15 mcleve

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Posted 13 February 2008 - 06:59 PM

4.7213114754098360655737704918041 Hours
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### #16 Moodster

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Posted 16 February 2008 - 05:50 PM

Flow Rate = 1/Time

1/T = 1/t1 + 1/t2 + 1/ t3 + 1/t4

So,

Total Rate (1/T) = 1/4days + 1/3days + 1/2days + 1/6hours = 5.083333333 Resevoirs/Day (also 61/12)

Since Time = 1/Rate

Then Total Time = 1/5.0833333 (also 12/61) = 0.19672 Days or 4.7213 Hours (also 288/61 or about 4 Hours, 43 minutes and 17 seconds)

It's just like adding electrical resistance in parallel.

Edited by Moodster, 16 February 2008 - 05:53 PM.

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Posted 16 February 2008 - 11:28 PM

fractions are more exact so simply saying 12/61days is the best/easyest answer
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### #18 Sharpshark

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Posted 26 February 2008 - 01:52 AM

You didn't saay the four tap...
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### #19 teej

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Posted 29 February 2008 - 08:38 PM

Seems to me that to reach 288 gallons, for example:
Where A is the 4 day tap, B the 3 day tap, C the2 day tap, and D the wuarter day tap

A= 3 gal/h (96 hrs*3=288 gal)
B= 4 gal/h (72 hrs*4=288 gal)
C= 6 gal/h (48 hrs*6=288 gal)
D= 48 gal/h (6 hrs*48=288 gal)

So 61 gal/h to fill 288 gal= 4.721311 hours.

Edited by teej, 29 February 2008 - 08:43 PM.

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### #20 haresh

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Posted 13 March 2008 - 10:39 AM

What I did was:

a= how much water gets filled up in the tank in 6 hours

1a - first tap
0.125a - second tap
0.0833a - third tap
0.0625a - fourth tap

thus

1+.125+.0833+.0625 = running four taps to fill up the reservoir in 6 hours

6 / 1.2708 = 4.7214 hours to fill up the reservoir
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