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:
Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3
Explanation:
The step by step and appropriate substitution is as shown in the attachment.
From number of moles = Concentration x volume
number of moles = number of particles/ Avogadro's number
Volume = mass/density, the appropriate derivation to get the number of moles of atoms
Answer:

Explanation:
Given data:
Diffusion constant for nitrogen is 
Diffusion flux 
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law

Take C_A as point on which nitrogen concentration is 2 kg/m^3

Assume X_A is zero at the surface


Answer:
Explanation: Here it is: 67 Hope that helps! :)