## Saturday, December 9, 2017

### Water fountain physics

I was drinking from a water fountain when I began thinking about how the principles behind how they work were related to the fluid physics we just learned. I decided to think about how much pressure must be generated at the pump for the water to shoot out for us to drink.

First I calculated the speed of the water. I estimated the height that it reached after shooting out from the fountain to be roughly 10cm. So we can calculate the water's velocity using conservation of energy.

mgh=1/2mv^2 (mass of the water doesn't matter here as it cancels out)
gh=1/2v^2
2gh=v^2
v=√2gh
v=√2(9.8m/s^2)(0.1m)
v=1.4m/s

Then I found the speed of the water when it is still inside the pump using the equation of continuity. Here, I estimated the pipe in the fountain to be 1cm in diameter and the diameter of the nozzle to be 0.3cm

A1V1=A2V2
(3.14*0.15^2)(1.4m/s)=(3.14*0.5^2)(V2)
V2=0.126 m/s

Finally, I used Bernoulli's equation to find the pressure generated by the pump. I used the location right after the water exits the nozzle as one reference point and the water at the pump (estimated to be roughly 1 meter below the nozzle).

P1+pgh1+1/2pv1^2=P2+pgh2+1/2pv2^2
(1.013*10^5)+(1000kg/m^3)(9.8)(1m)+1/2(1000kg/m^3)(1.4m/s)^2=(1.013*10^5)+P+1/2(1000kg/m^3)(0.126m/s)^2
P=1.1*10^4pa