A cylindrical tank with 4 square feet cross section has h feet of water in it. Water is pumped into the tank a?

Question

A cylindrical tank with 4 square feet cross section has h feet of water in it. Water is pumped into the tank at .08 cu ft /sec. Water is also drained from the bottom of the tank at .12h cu ft / sec. Find time it takes to increase the height from 0.1 ft to 0.5 ft.

Answer 1

Water is drained from the bottom of the tank faster than pumping into the tank, so there is no question of height increase.

Answer 2

I've drilled shallow wells in my subject part time and set up many pumps. You could have a number of problems right here that may or won't affect you. A pump, of any sort, can and will push water to any length, that's now not the difficulty. It can be getting the pump to the water. The second is, pumping the water from the good. Your at a superior distance and you've got a variance in the water height for some motive, i might say the underground go with the flow or aquifer. This is what i'd endorse: Put in a Wayne Submersible pump, half of horse. It'll provide you with all the water you wish to have at any distance just about eternally and, they're not high priced. When you have a concern with an submersible, then i would propose a Wayne one 1/3 horse with a tank and a bladder expansion tank on the apartment for pressure. This is going to fee a little. Your going to have problems with the pump pulling the water at this distance and retaining up, the well. The variable water top is telling me the aquifer is not that first-class. These shallow wells glide according to the weather, the moist weather has a better give. Your in a shallow aquifer. Go for the submersible.

Answer 3

First, to maximise the chance of detecting crass blunders, make an estimate of the MINIMUM time.



If there were NO drainage from the bottom, the time would be 4 * (0.5 - 0.1) / 0.08. This is 20 seconds.



The inflow is 0.08 - 0.12h, so the rate of increase of h is (0.08 - 0.12h) / 4.



dh/dt = 0.02 - 0.03h



To improve on the estimate of minimum time above, we can think about the rate at which the height is increasing at the beginning and end of the filling.



When h is 0.1 dh/dt = 0.02 - 0.03 * 0.1 = 0.017



At this value of dh/dt, the time taken would be 0.4 / 0.017 = 23.5 seconds



When h = 0.5, dh/dt = 0.02 - 0.03 * 0.5 = 0.005



At this value of dh/dt, the time taken would be 0.4 / 0.005 = 80 seconds



So we know that the answer we are looking for is /somewhere/ near the middle of the range from 23 to 80.



To find the actual answer:



dh/dt = 0.02 - 0.03h



so



dt = dh/(0.02 - 0.03h)



The integral of 1 / (a + b*x) is 1/b * ln(a + b*x)



[Note: ln is Napierian logarithm, that is, to the base e]



so the integral of 1 / (0.02 - 0.03h) is -1/0.03 * ln(0.02 - 0.03h)



so t2 - t1 = -1/0.03 * ( ln(0.02 - 0.03*h2) - ln(0.02 - 0.03*h1) )



so that



t2 - t1 = 1/0.03 * ( ln(0.02 - 0.03*h1) - ln(0.02 - 0.03*h2) )



so that



t2 - t1 = 1/0.03 * ln(0.017/0.005)



and so



t2 - t1 = 40.8 seconds

Answer 4

dh/dt = 0.12h - 0.02

The general solution of the equation is of the form

h = A + Ce^(-0.12t),

The drainage rate will equal the input rate,

and dh/dt will be 0, when

0.08 = 0.12 h => h = (2/3),

and the integration constant C may be evaluated by setting h=0.1 ft when t=0;

0.1 ft = (2/3) ft - C

C = 2/3 - 1/10 = 17/30 ft



h = (2/3) ft - (17/30) ft * e^(-0.12 t)

0.5 ft = (2/3) ft - (17/30 ft * e^(-0.12 t)

(17/30) e^(-0.12t) = 1/6

e^(-0.12 t) = 5/17

0.12 t = ln(17/5) = 1.224

t is a little more than 10 seconds.