Question

A garden hose with a diameter of 0.70 in. has water flowing in it with a speed of 0.63 m/s and a pressure of 1.3 atmospheres. At the end of the hose is a nozzle with a diameter of 0.22 in.
Find the speed of water in the nozzle.
Find the pressure in the nozzle.

Answer 1

Answer on Question 75823, Physics, Mechanics, Relativity

Question:

A garden hose with a diameter of $0.70 \, in$ has water flowing in it with a speed of $0.63 \, m/s$ and a pressure of 1.3 atmospheres. At the end of the hose is a nozzle with a diameter of $0.22 \, in$.

(a) Find the speed of water in the nozzle.

(b) Find the pressure in the nozzle.

Solution:

(a) We can find the speed of water in the nozzle from the Law of Continuity:

$$A_1 v_1 = A_2 v_2,$$

here, $A_1$, $A_2$ are the cross-section areas of the hose and the nozzle, respectively; $v_1, v_2$ are the speeds of the water flowing through the hose and the nozzle, respectively.

Then, from this formula we can find the speed of water in the nozzle:

$$v_2 = v_1 \frac{A_1}{A_2} = v_1 \frac{\pi d_1^2}{\pi d_2^2} = 0.63 \, \frac{m}{s} \cdot \frac{\pi \cdot (0.70 \, in)^2}{\pi \cdot (0.22 \, in)^2} = 6.38 \, \frac{m}{s}.$$

(b) We can find the pressure in the nozzle from the Bernoulli’s equation:

$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2,$$

here, $P_1$, $P_2$ are the pressures in the hose and nozzle, respectively; $v_1$, $v_2$ are the speeds of the water flowing through the hose and the nozzle, respectively; $\rho = 1000 \, kg/m^3$ is the density of the water.

From this equation we can find the pressure in the nozzle:

$$\begin{array}{l} P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) = \\ = 1.3 \, atm \cdot \frac{101325 \, Pa}{1 \, atm} + \frac{1}{2} \cdot 1000 \, \frac{kg}{m^3} \cdot \left( \left(0.63 \, \frac{m}{s}\right)^2 - \left(6.38 \, \frac{m}{s}\right)^2 \right) \\ = 1.1 \cdot 10^5 \, Pa = 1.1 \, atm. \end{array}$$

Answer:

(a) $v_{2} = 6.38 \frac{m}{s}$.

(b) $P_{2} = 1.1 \cdot 10^{5} \, Pa = 1.1 \, atm$.

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