Engineering >> Civil & Environmental EngineeringTime to Fill an Elevated Storage Tankby 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 Q^{2} by isolating it on one side and came up with Q^{2} [(M – (1/2201)] = h+50b. 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) [(bh50)/(1/2201) – M]^{1/2}. Then, I solved for dt and came up with dt = [7.5 A [(1/2201) – M]^{1/2 }dh]/(bh50)^{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(bh50)^{1/2}] and when the equation was simplified and evaluated from 0 to 50, t = 15A [(1/2201) M]^{1/2} [(b50)^{1/2} – (b100)^{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 Q^{2} vs. H. A linear trendline 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.
