We know that:
Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely.
Therefore, the power potential of the wind is its kinetic energy, which is V² /2 per unit mass, and mV² / 2 for a given mass flow rate:
e (mech) = k e = V² /2
e (mech) = (10 m/s)² / 2
To convert the units from m/s to kJ/kg,
e (mech) = 50 m² /s² x ( 1 kJ/kg / 1000 m² /s²)
e (mech) = 0.050 kJ/kg
Now,
We also know the relation
m = ρVA
Also A = π D²/4, so above equation becomes:
m = ρV (π D²/4)
m = (1.25 kg/ m³) ( 10 m/s) x ( 3.14 x 80 m x 80 m)/ 4
m = 62,800 kg/s
So,
E (mech) = m x e (mech)
E (mech) = 62,800 kg /s x 0.050 kJ/kg
E (mech) = 3,140 kW
Therefore, the 3140 kW of actual power can be generated by this wind turbine at the given conditions.
Answer:
158
Explanation:
The input is 6.185/10.000 = 0.6185 of the full-scale of the A/D converter. If a 10V input is represented by a register value of 255, then the expected register value for this input is about ...
0.6185·255 = 157.7175 ≈ 158
_____
In hexadecimal, that is 9E.