Question

A fan is to accelerate quiescent air to a velocity of 12 m/s at a rate of 3 m³/s. If the density of air is 1.15 kg/m³, what is the minimum power that must be supplied to the fan?

Answer 1

The minimum power that must be supplied to the fan to accelerate quiescent air to a velocity of 12 m/s at a rate of 3 m³/s, given the density of air as 1.15 kg/m³, can be calculated using the formula for power. The result is 41.4 Watts.

Step-by-step explanation:

To determine the minimum power required by the fan, we utilize the formula for power, which is defined as the rate at which work is done or energy is transferred. Power
`(\(P\))` is given by the formula
`\(P = (W)/(t)\),`where
`\(W\)` is the work done and
`\(t\)` is the time taken. In this scenario, the work done by the fan is equivalent to the change in kinetic energy of the air passing through it.

The kinetic energy of a flowing fluid can be expressed as
`\(KE = (1)/(2) m v^2\),` where
`\(m\)` is the mass of the fluid passing through the fan per unit time and
`\(v\)` is the velocity of the fluid. Given the mass flow rate
`(\(m = \rho * V\)`, where
`\(\rho\)` is the density and
`\(V\)` is the volume flow rate), and the density of air as 1.15 kg/m³, and velocity as 12 m/s, we can calculate the mass passing through the fan per unit time.

Substituting the values into the kinetic energy formula and considering the rate of change of kinetic energy as the power supplied by the fan, we get
`\(P = (1)/(2) * \rho * V * v^2\)`. Upon substitution of the given values, we find the minimum power required by the fan to be 41.4 Watts. This power output is necessary to accelerate the air to the specified velocity at the given rate.