Question

Water from a stationary nozzle impinges on a moving vane with turning angle θ = 120. The vane moves away from the nozzle with constant speed, U = 10 m=s, and receives a jet that leaves the nozzle with speed V = 30 m=s. The nozzle has an exit area of 0.004 m 2. Find the force that must be applied to maintain the vane speed constant.

Answer 1

Answer:

The force that must be applied to maintain the vane speed constant is 2771.26 N

Explanation:

Given;

turning angle of the vane = 120°

control volume velocity, u = 10 m/s

absolute velocity, v = 30 m/s

nozzle area = 0.004 m²

The force acting on the vane has horizontal and vertical components:

Based on Reynolds general control volume system;

The horizontal force component of the system, ∑Fₓ = ρW²A(1-cosθ)

where;

ρ is the density of water = 1000 kg/m³

W is the relative velocity = Absolute velocity - control volume velocity

W = v - u

    = 30 - 10 = 20m/s

∑Fₓ = ρW²A(1-cosθ) = 1000 x 20² x 0.004 (1 - cos 120) = 2400 N

The vertical force component of the system, ∑Fy = ρW²A(sinθ)

∑Fy = ρW²A(sinθ) = 1000 x 20² x 0.004 x sin(120) = 1385.6 N

The magnitude of the force applied $= \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{2400^2 + 1385.6^2} = 2771.26 \ N$

The force that must be applied to maintain the vane speed constant is 2771.26 N

Answer 2

Answer:

The force that must be applied to maintain the vane speed constant 1599.96N

Explanation:

Turning angle of the vane = 120°

Vane's velocity, u = 10 m/s

Jet's velocity, v = 30 m/s

nozzle exit area = 0.004 m²

The force is in Vertical and Horizontal components,

from Reynolds general control volume system;

The horizontal force component of the system, ∑Fₓ = ρW²A(cosθ)

where;

ρ is the density of water = 1000 kg/m³

W is the relative velocity = jet's velocity - vane's velocity

That is, W = v - u = 30 - 10 = 20m/s

∑Fₓ = ρW²A(cosθ) = 1000 x 20² x 0.004 (cos 120) = 800N(In negative horizontal direction)

The vertical force component of the system, ∑Fy = ρW²A(sinθ)

∑Fy = ρW²A(sinθ) = 1000 x 20² x 0.004 x sin(120) = 1385.6 N

The magnitude of the force applied is the resultant of vertical and horizontal components.

By Pythagoras's theorem,

F^2 = (Vertical component of the force)^2 + (Horizontal component of the force)^2

F = sqrt(1385.6^2 + 800^2)

F = 1599.96N