An incompressible fluid is flowing through a pipe with a constriction. The pipe is on an incline with an angle of 30.0 degrees. The narrow section is $10.0 \mathrm{~m}$ from the wide section and the narrow section is lower than the wide section. The velocity of the fluid in the wide section of the pipe is $4.00 \mathrm{~m} / \mathrm{s}$. The diameter of the wide section is $12.00 \mathrm{~cm}$ and the diameter of the narrow section is $9.00 \mathrm{~cm}$. The pressure of the fluid in the wide section is $250 \mathrm{kPa}$. What is the pressure in the narrow section of the pipe? (density of the fluid is 1,000 $\left.\mathrm{kg} / \mathrm{m}^{3}\right)$
A) $275 \mathrm{kPa}$
B) $295 \mathrm{kPa}$
C) $240 \mathrm{kPa}$
D) $252 \mathrm{kPa}$
E) $282 \mathrm{kPa}$

Question

Quizplus · Accepted Answer

To solve this, we can use the Bernoulli's equation, which in this case also accounts for the change in height (gravitational potential energy) due to the incline. The equation is $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$, where $P$ is pressure, $\rho$ is the density of the fluid, $v$ is the velocity of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height. The velocity in the narrow section ($v_2$) can be found using the continuity equation $A_1v_1 = A_2v_2$, where $A$ is the cross-sectional area of the pipe. Given the diameters, we can find the areas ($A_1 = \pi(0.06)^2$ and $A_2 = \pi(0.045)^2$), and thus find $v_2$. With $v_2$, we can then solve for $P_2$ using Bernoulli's equation, including the height difference ($h_2 - h_1 = 10\sin(30^\circ)$). The correct answer accounts for the increase in kinetic energy and decrease in potential energy as the fluid moves down the incline, leading to a higher pressure in the narrow section.

An Incompressible Fluid Is Flowing Through a Pipe with a Constriction