the required documents is 3000
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e

r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer:
P > 142.5 N (→)
the motion sliding
Explanation:
Given
W = 959 N
μs = 0.3
If we apply
∑ Fy = 0 (+↑)
Ay + By = W
If Ay = By
2*By = W
By = W / 2
By = 950 N / 2
By = 475 N (↑)
Then we can get F (the force of friction) as follows
F = μs*N = μs*By
F = 0.3*475 N
F = 142.5 N (←)
we can apply
P - F > 0
P > 142.5 N (→)
the motion sliding
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius,
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax 
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
Answer:
a. 0.28
Explanation:
Given that
porosity =30%
hydraulic gradient = 0.0014
hydraulic conductivity = 6.9 x 10⁻4 m/s
We know that average linear velocity given as



The velocity in m/d ( 1 m/s =86400 m/d)
v= 0.27 m/d
So the nearest answer is 'a'.
a. 0.28