In places such as central Long Island, many properties do not allow the space for in-ground pools, and so many families choose to install above-ground units to help cool off in the summertime. Come Fall, one will often see streams of chlorinated water running along curbs as families drain their pools to prevent leaves from clogging the pump systems. One day, we needed to siphon the water from our own pool, however my dad and I argued between which length size of siphon tube to purchase, as the longer the tube, the greater the cost. The leaves were already starting to come down, and I argued that a longer tube would allow a greater exit flow rate so that the pool could be drained before the leaves started to accumulate. It was the last weekend of the summer, so we had only 48 hours at most to drain the water before the pump was at stake of clogging. In the diagram below, A is the height of the water, at around 5 feet deep or about 1.5 meters. There is a 2 meter gate above the surface of the water through which the tube cannot fit, and so must be placed over. The pool rests on stilts, and so the tube can end at some point beneath the body of water height wise. I needed to come up with a length of tube that would allow the entire body of water to drain within 48 hours. Initially we thought of a 7 meter long tube, whose end would be at approximately the bottom of the pool, creating C=0.
Using Torcelli's theorum, we would find that the exit speed of the water would be equal to √(2g(y2-y1). Because y2-y1 is equivalent to the height difference between the hose ending and the water, we substitute it with h, or 3.5-2=1.5 meters. Plugging this in, we have
√(2g(y2-y1) -> √(2(9.8)(1.5)= 5.4 m/s
With the diameter of the tube being about 4 centimeters, we can now find the flow rate of the exiting water.
Cross sectional area=(.02)^2(pi)=.00125 meters^2
Flow rate=(.00125)meters^2(5.4)meters^3/second=0.00675 meters^3/second
We assume the pool to be 11 meters in diameter. Recalling the water height is 5 meters tall, we can model the volume of the pool as a 3-dimensional cylinder.
Volume of a cylinder= (pi)(5.5)^2(5)=475.155 meters^3 of water.
Time for water to completely drain would then be= (475.155)m^3/(.00675)m^3/second= approximately 70,394 seconds.
With a 7 meter long tube, it would take approximately 70,394 seconds, or 19 hours. So, saving both time and money, a 7 meter long tube would be sufficient to drain the pool before the leaves would start to fall.