Time to Fill an Elevated Storage Tank

by Amit Gulati

Submitted : Spring 2012

I took the two equations for (H) which were given to me and set them equal to each other. I then proceeded to solve for Q² by isolating it on one side and came up with Q² [(M – (1/2201)] = h+50-b. The next step was to solve for Q so I took the square root of each side and formulated an equation for Q (once I’d solved for Q, the resulting equation was multiplied by (-1) so as to avoid taking the square root of a negative answer once certain values were plugged in). I then used the equation I had just found for Q and substituted it into the equation A dh/dt = Q/7.5 and arrived at A dh/dt = (1/7.5) [(b-h-50)/(1/2201) – M]^1/2. Then, I solved for dt and came up with  dt = [7.5 A [(1/2201) – M]^1/2 dh]/(b-h-50)^1/2. In order to get an equation for t so that the time needed to fill the water tank could be determined, I took the integral of the equation for dt and arrived at t= 7.5A [(1/2201)- M]^1/2 [-2(b-h-50)^1/2] and when the equation was simplified and evaluated from 0 to 50, t = 15A [(1/2201)- M]^1/2 [(b-50)^1/2 – (b-100)^1/2. The values for b and M were 162.42 and -.0003 respectively and were found by making a graph of the data provided in Microsoft Excel of Q² vs. H. A linear trend-line was then added which yielded the equation  y = -.0003x + 162.42 where b is -.0003 and M is 162.42. I got 786.907 minutes once I plugged those values into the equation for t, and upon converting that number to hours, I determined that it would take approximately 13.1 hours to fill the tank.

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