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
Explanation:
Diesel cycle:
All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows
1. Adiabatic reversible compression
2.Heat addition at constant pressure
3.Adiabatic reversible expansion
4.Constant volume heat rejection
In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.
Applications of diesel cycle:
Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.
I won leader solution contain 0.46 mL of hydronic I said of 0.3 potassium
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = 
d = 1.5
= 1.767 inches²
<h3>Solve for inertia</h3>
= 0.2485inches⁴
Solve for the tensile force from here
30x10³ = 
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
Read more on tensile force here: brainly.com/question/25748369
