Question

A windmill is operating at a steady wind speed of 20 mph. Determine the wind energy: a) per unit mass b) for a mass of 20 lbs c) for a flow rate of 2500 lb/s

Answer 1

Answer:The kinetic energy of wind per unit mass is given by the formula:

E=21​v2

where:

(E) is the kinetic energy per unit mass (in Joules per kilogram, J/kg)

(v) is the wind speed (in meters per second, m/s)

Let’s convert the wind speed from mph to m/s first. 1 mph is approximately equal to 0.44704 m/s. So, 20 mph is equal to (20 \times 0.44704 = 8.9408) m/s.

Now, we can calculate the wind energy per unit mass:

E=21​(8.9408)2=39.92 J/kg

This is the kinetic energy of the wind per unit mass.

For a mass of 20 lbs, we first need to convert the mass from lbs to kg. 1 lb is approximately equal to 0.453592 kg. So, 20 lbs is equal to (20 \times 0.453592 = 9.07184) kg.

The total energy for a mass of 20 lbs is then:

Etotal​=E×m=39.92×9.07184=362.29 J

For a flow rate of 2500 lb/s, we first convert the flow rate from lb/s to kg/s. So, 2500 lb/s is equal to (2500 \times 0.453592 = 1133.98) kg/s.

The power (energy per unit time) for a flow rate of 2500 lb/s is then:

P=E×m˙=39.92×1133.98=45259.5 W=45.26 kW

So, the wind energy is 39.92 J/kg per unit mass, 362.29 J for a mass of 20 lbs, and the power is 45.26 kW for a flow rate of 2500 lb/s. Please note that these calculations assume that all the kinetic energy of the wind can be captured by the windmill, which is not the case in reality due to various losses. The actual energy captured would be less.

The kinetic energy of wind per unit mass is given by the formula:

E=21​v2

where:

(E) is the kinetic energy per unit mass (in Joules per kilogram, J/kg)

(v) is the wind speed (in meters per second, m/s)

Let’s convert the wind speed from mph to m/s first. 1 mph is approximately equal to 0.44704 m/s. So, 20 mph is equal to (20 \times 0.44704 = 8.9408) m/s.

Now, we can calculate the wind energy per unit mass:

E=21​(8.9408)2=39.92 J/kg

This is the kinetic energy of the wind per unit mass.

For a mass of 20 lbs, we first need to convert the mass from lbs to kg. 1 lb is approximately equal to 0.453592 kg. So, 20 lbs is equal to (20 \times 0.453592 = 9.07184) kg.

The total energy for a mass of 20 lbs is then:

Etotal​=E×m=39.92×9.07184=362.29 J

For a flow rate of 2500 lb/s, we first convert the flow rate from lb/s to kg/s. So, 2500 lb/s is equal to (2500 \times 0.453592 = 1133.98) kg/s.

The power (energy per unit time) for a flow rate of 2500 lb/s is then:

P=E×m˙=39.92×1133.98=45259.5 W=45.26 kW

So, the wind energy is 39.92 J/kg per unit mass, 362.29 J for a mass of 20 lbs, and the power is 45.26 kW for a flow rate of 2500 lb/s. Please note that these calculations assume that all the kinetic energy of the wind can be captured by the windmill, which is not the case in reality due to various losses. The actual energy captured would be less.

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Step-by-step explanation: