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 [tex]= \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F = \sqrt{2400^2 + 1385.6^2} = 2771.26 \ N[/tex]
The force that must be applied to maintain the vane speed constant is 2771.26 N
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
