Question

Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the pipe is 1.8 cm, and the radius of the nozzle is 0.49 cm. The speed of the water in the pipe is 0.56 m/s. Treat the water as an ideal fluid, and determine the absolute pressure of the water in the pipe. (The atmosphere pressure is P = 1.01 105 Pa, and the density of water is rho = 1.00 103 kg/m3.)

Answer 1

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$, and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

$v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s$

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

$P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa$