Question

a gardener wishes the water from his hose was able to reach flowers 3m away when he holds the nozzle horizontally at a height of 1m. He finds out that the water has a speed of 6.64 m/s when it leaves his hose. The nozzle of his hose has an inner diameter of 1.5 cm. The other end is connected to a pipe with an inner diameter of 2 cm. What must the pressure in the exit pipe be for the water from his hose to reach his flowers?

Answer 1

Answer on Question#39741, Physics, Other

A gardener wishes the water from his hose was able to reach flowers 3m away when he holds the nozzle horizontally at a height of 1m. He finds out that the water has a speed of 6.64 m/s when it leaves his hose. The nozzle of his hose has an inner diameter of 1.5 cm. The other end is connected to a pipe with an inner diameter of 2 cm. What must the pressure in the exit pipe be for the water from his hose to reach his flowers?

Solution

According to the conservation of mass law:

$$v_1 A_1 = v_2 A_2,$$

where $v_1$ - speed of water in the pipe, $v_2$ - speed of water in the hose, $A_1$ - inner area of pipe, $A_2$ - inner area of hose.

According to Bernoulli law:

$$P + \frac{\rho v_1^2}{2} = \frac{\rho v_2^2}{2},$$

where $P$ - the pressure in the exit pipe, $\rho$ - density of water.

$$P = \frac{\rho v_2^2}{2} - \frac{\rho v_1^2}{2} = \frac{\rho}{2} (v_2^2 - v_1^2) = \frac{\rho v_2^2}{2} \left(1 - \left(\frac{A_2}{A_1}\right)^2\right) = \frac{\rho v_2^2}{2} \left(1 - \left(\frac{d_2}{d_1}\right)^4\right).$$ $$P = \frac{1000 \cdot 6.64^2}{2} \left(1 - \left(\frac{1.5}{2}\right)^4\right) = 15 \, \text{kPa}.$$

Answer: $15 \, \text{kPa}$.