Question

: A pneumatic "cannon" is a device that launches a low mass projectile from a cylindrical tube using pressurized air stored upstream of the projectile. The projectile is held in place while the pressurized air is introduced to the cylinder. After the trigger is released, the projectile accelerates down the tube and the pressure upstream of the projectile drops according to the relation.

p = π [ 1 − (x/d)²] /25

where π is the initial (high) pressure in the tube before the trigger is released, x is the distance traveled by the projectile, and d is the diameter of the tube. If we have a 5.0 cm diameter tube, with a projectile initially located 5.0 tube diameters upstream of the tube exit, and the initial pressure upstream of the projectile is 12 bar. Determine the work done by the air on the projectile during the time it is in the tube. Assume that there is no friction between the tube and the projectile.

Answer 1

Answer:

Work done = 125π J

Explanation:

Given:

P = P_i * ( 1 - (x/d)^2 / 25)

d = 5.0 cm

x = 5 * d cm = 25d

Pi = 12 bar

Work done = integral ( F . dx )

F (x) = P(x) * A

F (x) =  (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)

Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx

For Limits 0 < x < 5d

Work done = (πd^2 / 4) * P_i  integral ( (1 - (x/d)^2) / 25)) . dx

Integrate the function wrt x

Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )  

Evaluate Limits 0 < x < 5d :

Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)

Work done = (πd^2 / 4) * P_i * (10*d / 3)

Work done = (5 π / 6)d^3 * P_i

Input the values:

Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)

Work done = 125π J