Question

Consider a length of pipe bent into a U-shape. The inside diameter of the pipe is 0.5 m. Air enters one leg of the pipe at a mean velocity of 100 m/s and exits the other leg at the same magnitude of velocity, but moving in the opposite direction. The pressure of the flow at the inlet and exit is the ambient pressure of the surroundings. Calculate the magnitude and direction of the force exerted on the pipe by the airflow. The air density is 1.23 kg/m3 .

Answer 1

Answer:

The magnitude of the force exerted on the pipe by the air is 4830 N and it acts horizontally

Explanation:

Given the data in the question;

from the Newton's second law of motion;

F = ma

where m is the mass, a is acceleration and F is the force exerted on the pipe due to the airflow in it

now in terns of mass flow;

F = $m^{"}$V

where $m^{"}$ is the mass flow rate, V is the velocity(

so

$m^{"}$ = pAV

$m^{"}$ = p × ($\frac{\pi }{4}$ d² ) × V

where d is the diameter of the pipe( 0.5 m)

p is the density( 1.23 kg/m³ )

velocity v is 100 m/s

so we substitute

$m^{"}$ = 1.23 × ($\frac{\pi }{4}$ (0.5)² ) × 100

$m^{"}$ = 30.75 × $\frac{\pi }{4}$

$m^{"}$ = 24.15 kg/s

Now lets write the equation for the force exerted on the pipe by the airflow

F = $m^{"}$( V₁ - V₂)

where V₁ is velocity at inlet ( 100 m/s )

V₂ is velocity at exit ( - 100 m/s )

so we substitute

F = 24.15 ( 100 - (-100))

F = 24.15 × 200

F = 4830 N

The pipe is symmetric about horizontal axis so the force should also b acting only in the horizontal direction since any force component in the vertical direction is nullified due to this symmetry

Therefore, The magnitude of the force exerted on the pipe by the air is 4830 N and it acts horizontally