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
