Consider the flow of air through a wind turbine whose blades sweep an area of diameter D (in m). The average air velocity through the swept area is V (in m/s). On the bases of the units of the quantities involved, show that the mass flow rate of air (in kg/s) through the swept area is proportional to the air density, the wind velocity, and the square of the diameter of the swept area.
There's no diagram or image, btw.
not very complicated
velocity v (do not use V, that is usually volume)
Area that air flows though= pi R^2 = pi D^2/4
so
Volume of air through/second = v (pi D^2/4)
but mass = density times volume so mass /second
= rho v D^2 (pi/4)
To show that the mass flow rate of air through the swept area is proportional to the air density, the wind velocity, and the square of the diameter of the swept area, we can analyze the units of the quantities involved.
The mass flow rate of air is the amount of air passing through a given area per unit of time. It is typically represented by the symbol Ẇ (pronounced "dot m"), and its unit is in kilograms per second (kg/s).
The air density is defined as the mass of air per unit volume. It is typically represented by the symbol ρ (pronounced "rho"), and its unit is in kilograms per cubic meter (kg/m³).
The wind velocity is the speed at which the air is moving. It is represented by the symbol V, and its unit is in meters per second (m/s).
The diameter of the swept area is represented by the symbol D, and its unit is in meters (m).
To determine the relationship between these quantities, we can consider the dimensions of each variable involved.
First, let's analyze the units of the term ρVD², which combines air density, wind velocity, and the square of the diameter:
- ρ has units of kg/m³.
- V has units of m/s.
- D² has units of m².
Multiplying these quantities together, we get:
ρVD² = kg/m³ * m/s * m² = kg * m / (m³ * s) = kg/s,
which corresponds to the unit of mass flow rate (Ẇ).
Therefore, we can conclude that the mass flow rate of air through the swept area is proportional to the air density, the wind velocity, and the square of the diameter of the swept area.
To determine the mass flow rate of air through the swept area of a wind turbine, we need to examine the units of the quantities involved: air density (ρ), wind velocity (V), and the diameter of the swept area (D).
1. First, let's consider the units of air density (ρ). Air density is typically measured in kilograms per cubic meter (kg/m³), which implies that it represents the mass of air per unit volume.
2. Next, let's examine the units of wind velocity (V). Wind velocity is measured in meters per second (m/s), which represents the rate of displacement of air molecules over a specific distance in a given time. Hence, it measures the speed at which air is flowing.
3. Finally, let's consider the units of the diameter of the swept area (D). Diameter is typically measured in meters (m) and represents a linear dimension.
To determine the units of the mass flow rate, we need to combine these quantities in a way that ensures the resulting units are in kilograms per second (kg/s). We can achieve this by considering the relationship between mass, volume, and time:
Mass = Density × Volume
Volume = Area × Distance
Distance = Velocity × Time
Combining these relationships, we find:
Mass = Density × (Area × Distance)
= Density × (Area × (Velocity × Time))
= Density × (Area × Velocity) × Time
As we want the mass flow rate per unit time (kg/s), we can rewrite this as:
Mass Flow Rate = Density × (Area × Velocity)
Now, let's analyze the units of this expression:
- Density (ρ): kg/m³
- Area (πD² / 4, where D is diameter): m²
- Velocity (V): m/s
Multiplying these together, we get:
Mass Flow Rate = (kg/m³) × (m²) × (m/s)
= kg × (m³/m²) × (m/s)
= kg/s
Thus, we have shown that the units of the mass flow rate of air through the swept area are proportional to the air density, the wind velocity, and the square of the diameter of the swept area.