Back to pressure. Bernoulli’s principle stats that the faster a fluid moves, the lower its pressure. This is the basis for the operation of a python. They water flowing out of the faucet is moving and thus at a lower pressure than the air in the hose. Since pressure tries to equalize, the air in the hose lowers as air runs out of the python outlet with the water. This causes the water to be “sucked” out of the tank. Saying it sucks isn’t quite right. When you suck on a straw you don’t really suck, you cause the pressure in the atmosphere to push the liquid down and it goes up through the straw since the pressure is lower. Here is a picture to help show this.
Bernoulli’s equation is:
P1 + (½ x D x V1^2) + DxGxH1 = P2 + (½ x D x V2^2) + DxGxH2. P is atmospheric pressure, D is density of the fluid, V is the velocity of the fluid, G is the rate of gravity, and H is the depth of the
You can use this to find a flow rate based on pressure. The problem with finding this using a python is that I don’t know what the water pressure coming out of your faucet is (and you can vary it). However, with a regular gravel vac, you can calculate this quite easily. The side with the 1’s is going to be the hose output, and the side with the 2’s is going to be the tank. Since the tank pressure is basically the same as the atmospheric pressure the P’s cancel. The tank water is not moving (for practical purposes), so (½ x D x V2^2)=0. We will let the hose output be zero, so DxGxH1 = 0. This leaves us with:
½ x D x V1^2= D x G x H2
H2 is the height from the hose outlet to the top of the water in the tank. The D’s cancel and multiplying both sides by 2 we get.
V1^2=2xGxH2 or V1=sqrt(2xG x H) sqrt=square root.
Lets just say that the height is about 1.5 feet from the top of the tank water, to the hose out put. Gravity is 32 feet/sec/sec, so
V1= Sqrt ( 2x32x 1.5) =9.8 feet/ sec = 6.7 mph
If the radius of the hose is r= 1/8 of an inch, or 1/96 of a foot then the area of its opening is A=Pi x r^2 = 3.14 x (1/48)^2=.00034 square feet
The flow rate in volume is then, the speed times the area of the opening.
VxA= 8.9 f/s x .00034 square feet = .003 cubic feet per second. Since 1 gallon is .133 cubic feet, then (.003 cf/s) / (.133 cf/gal)= .023 gal/sec.
To get sec/gal, take the inverse of .023, which is 1/.023= 44 sec/gal.
To do a 50% water change in a ten gallon, it takes about 220 seconds, or 3 minutes 40 seconds(5x44). I haven’t timed it or measured the actual height from the hose opening to the water surface, but it does sound about right.
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