Answer to Question #223849 in Physics for Saint

Question

Water enters a pipe of diameter 3.0 cm and at a speed of 40 cm/s at the ground level of an office building. The pipe tapers down to 1.0 cm in diameter at the top floor, 35 cm above, where the absolute pressure in 1.2 atm. What is the pressure at the ground level?

Answer 1

Let's first find the speed of the water at the top floor from the law of continuity:

$$A_1v_1=A_2v_2,$$ $$v_2=\dfrac{A_1v_1}{A_2}=\dfrac{\pi r_1^2v_1}{\pi r_2^2},$$ $$v_2=\dfrac{(0.015\ m)^2\cdot0.4\ \dfrac{m}{s}}{(0.005\ m)^2}=3.6\ \dfrac{m}{s}.$$

We can find the pressure at the ground level from the Bernoulli's equation:

$$\dfrac{1}{2}\rho v_1^2+\rho gh_1+P_1= \dfrac{1}{2}\rho v_2^2+\rho gh_2+P_2,$$ $$\dfrac{1}{2}\rho v_1^2+P_1= \dfrac{1}{2}\rho v_2^2+\rho gh_2+P_2,$$ $$P_1=P_2+\dfrac{1}{2}\rho(v_2^2-v_1^2)+\rho gh_2,$$

$P_1=1.2\ atm\cdot\dfrac{101325\ Pa}{1\ atm}+\dfrac{1}{2}\cdot1000\ \dfrac{kg}{m^3}\cdot((3.6\ \dfrac{m}{s})^2-(0.4\ \dfrac{m}{s})^2)+1000\ \dfrac{kg}{m^3}\cdot9.8\ \dfrac{m}{s^2}\cdot0.35\ m=1.31\cdot10^5\ Pa.$