Answer:
a. 4
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m
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