Question

A horizontal bend in a pipeline conveying 1 m3/s of water gradually reduces from 600 mm to 300 mm in diameter and deflects the flow through an angle of 60°. At the larger end the pressure is 170 KN/m2. Determine the magnitude and direction of the force exerted on the bend.

Answer 1

Answer:

The force excerted on the bend = 45.3 kN

This happens under an angle of tan^-1(1.7/4.2) = 22° to the x-direction

Explanation:

Step 1: Data given

A horizontal bend in a pipeline conveying 1m³/s of water

Diameter reduces from 600 mm to 300 mm

angle = 60°

At the larger end, the pressure = 170 KN/m²

Step 2:

1 m³/s = A1 *V1 = A2*V2

⇒ with A1 = the area at side 1

⇒ with A2 = the area at side 2

V1 = 1/((π/4)(0.6²)) = 3.537 m/s

V2 =  1/((π/4)(0.3²)) = 14.147 m/s

p1/p*g + (V1)²/2g = p2/p*g + (V2)²/2g

⇒ with p1 =170 *10³ N/m²

⇒ with p = 10³

⇒ with g = 9.81 m/s²

⇒ with p2 = TO BE DETERMINED

⇒ with V1 = 3.537 m/s

⇒ with V2 = 14.147 m/s

p2/fg = (170*10³)/(10³*9.81) + (3.537²)/(2*9.81) - (14.147²)/(2*9.81)

p2/fg = 7.767

p2 = 7.767 * 9810

p2 = 7.62*10^4 N/m²

Gravity forces are 0 along the horizontal plane. The only forces acting on the fluid mass = pressure and momentum forces.

Let's consider Fx and Fy as 2 components of total force F excerted by the bent boundary surface on the fluid mass.  

In x-direction we have: p1*A1 + Fx - p2A2cos∅ = pQ(V2cos∅ - V1)

In y-direction we have: 0 + Fy - p2A2sin∅ = pQ(V2sin∅-0)

Fx = 10³*(14.147cos60° - 3.537) + 7.62 * 10^4 * π/4 *(0.3²) * cos60° - 170*10^3 * π/4 *(0.6²)

Fx = -4.2 *10^4 N  (The negative sign shows the direction to the left)

Fy = 10³*(14.147sin60°) + 7.62 *10^4 * π/4 *(0.3²) *sin 60°

Fy = 1.7*10^4 N ( The positive sign shows the direction is upwards)

The law of motion says: the forces Rx and Ry excerted by the fluid on the bend will be equal and opposite to Fx and Fy:

Rx = -Fx = 4.2 *10^4 N ( positive sign means direction to the right)

Ry = -Fy = -1.7 *10^4 N  (Negative sign means direction downwards)

The resultant force on the bend:

R = √((Rx)² + (Ry)²

R = √((4.2 * 10^4)² + (-1.7*10^4)²)

R = 45310 N = 45.3 * 10³ N = 45.3 kN

The force excerted on the bend = 45.3 kN

This happens under an angle of tan^-1(1.7/4.2) = 22° to the x-direction