## Friday, December 1, 2017

### Watering the backyard

We all have the experience of watering our backyards with a hose. When I watered my yard, I always pressed the end of the hose into a smaller opening. This way, the water came out at a faster rate and reached further distances. After learning about water flow and pressure in this class, I wanted to measure how fast the water came out when I pressed onto the end of the hose.

The water flowing through the hose had a laminar flow. I assumed that the water flowed in the hose at a speed of 0.5 m/s through a 3 cm radius hose. When I pressed the opening of the hose, the radius of the hose became 1.5 cm.

Using the equation of continuity, I calculated the flow rate of water out of the hose.
A1v1 = A2v2
A1v1 = π x (0.03m)2 x (0.5 m/s ) = 0.0014 m/s
v2 = π x (0.015m)2 = 0.00071 m2
v2 = A1v1/A2 = 0.0014 m/s / 0.00071 m2
v2 = 2.0 m/s

The new flow rate of water at the end of the hose is 2.0 m/s. This speed is 4 times the original flow rate of water through the hose.

In addition to the flow rate, I also wanted to measure how much pressure that the water exerted at the end of the hose.

To calculate the pressure, I used Bernoulli’s equation: P2 + 1/2ρv22 + ρgy2 = P1 + 1/2ρv12 + ρgy1
Since I was holding up the hose about 1m above the ground, the difference in height was 1m. I also assumed that the water flowing in the hose was under pressure of 1 atm.
P2 + ½ ρv22 + ρgy2 = P1 + ½ ρv12 + ρgy1
P2 = P1 + ρg(y1 - y2) + ½ ρ(v12 - v22)
P2 = (1.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-1.0m) + ½ (1.0 x 103 kg/m3)[(0.5 m/s)2 - (2.0 m/s)2]
P2 = 88325 N/m2 = 0.88 atm

The pressure exerted by water at the end of the hose was 0.88 atm. With decreased area of the hose, the pressure of the water decreased.