An incompressible fluid is flowing through a vertical pipe with a constriction. The wide section is 2.00 cm in diameter and is at the top of the pipe. The pressure of the fluid in the wide section at the top is 200 kPa. The velocity of the fluid in the wide section is 4.00 m/s. The narrow section is located 4.00 m below the wide section. What is the diameter of the narrow section for the pressure of the fluid in the narrow section to equal the pressure in the wide section? (density of the fluid is 1,000 kg/m³) A) 1.50 cm B) 1.36 cm C) 1.28 cm D) 1.12 cm E) 1.00 cm

Question

Quizplus · Accepted Answer

Using Bernoulli's equation, we can relate the pressure and velocity at two different points in a fluid flowing through a pipe. Since the fluid is incompressible, the volume flow rate (Q) must be constant throughout the pipe.

Bernoulli's equation:

P1 + 1/2 * rho * v1^2 = P2 + 1/2 * rho * v2^2

where 
P1 = pressure at point 1 (wide section) 
v1 = velocity at point 1 
P2 = pressure at point 2 (narrow section) 
v2 = velocity at point 2 
rho = density of fluid

We want to find the diameter of the narrow section (d2) that makes P2 = P1. We are given P1 = 200 kPa and v1 = 4.00 m/s. We know that the volume flow rate is constant at any point in the pipe, so we can relate the velocities and diameters using the equation:

Q = A1 * v1 = A2 * v2

where A1 and A2 are the cross-sectional areas of the wide and narrow sections, respectively. Since the pipe is vertical and the fluid is incompressible, the cross-sectional areas are related to the diameters as:

A = pi * d^2 / 4

Combining these equations gives:

d2 = sqrt((4*A1)/(pi*v1)) = 1.28 cm

Therefore, the correct answer is C.

An Incompressible Fluid Is Flowing Through a Vertical Pipe with a Constriction