Question

A fan is to accelerate quiescent air to a velocity to \(12 \mathrm{m} / \mathrm{s}\) at a rate of \(3 \mathrm{m}^{3} / \mathrm{s}\). If the density of air is \(1.15 \mathrm{kg} / \mathrm{m}^{3}\) the minimum power that must be supplied to the fan is \((a) 248 \mathrm{W}\) \((b) 72 \mathrm{W}\) \((c) 497 \mathrm{W}\) \((d) 216 \mathrm{W}\) \((e) 162 \mathrm{W}\)

Short answer

a) 248 W b) 330 W c) 410 W d) 500 W Answer: a) 248 W

Step-by-step solution

Find the mass flow rate

To calculate the mass flow rate, multiply the flow rate by the air density: mass flow rate = flow rate × air density We have the flow rate (\(3 \,\mathrm{m^3/s}\)) and air density (\(1.15 \,\mathrm{kg/m^3}\)) given in the problem statement. Plug in the values to get the mass flow rate.

Apply the kinetic energy equation

The kinetic energy equation is: $$ power = \frac{1}{2} \times \text{mass flow rate} \times \text{final velocity}^2 $$ where 'power' is the required minimum power, 'mass flow rate' is calculated in Step 1, and 'final velocity' is the desired velocity (\(12 \,\mathrm{m/s}\)).

Calculate the power

Plug in the mass flow rate and final velocity values into the kinetic energy equation from Step 2: $$ power = \frac{1}{2} \times (3 \,\mathrm{m^3/s} \times 1.15 \,\mathrm{kg/m^3}) \times (12 \,\mathrm{m/s})^2 $$

Evaluate the result and choose the correct option

Calculate the power from Step 3 and compare the result to the given answer options: (power equals the calculated value rounded to the nearest whole number) In our case, power = 248 W, which matches option (a). So, the correct answer is \((a) 248 \,\mathrm{W}\).

Key concepts

Mass Flow Rate

Understanding the mass flow rate is crucial for various applications in thermodynamics and fluid mechanics. The mass flow rate, typically denoted by the symbol \( \dot{m} \), is a measure of the mass of a substance that passes through a given surface per unit time. It plays a pivotal role in analyzing systems like fans, pumps, and turbines where continuous flow is involved.

To calculate the mass flow rate, you need the volume flow rate and the density of the fluid. The formula is given by \( \dot{m} = \text{flow rate} \times \text{density} \). In the context of the exercise, the flow rate of air through the fan is \(3 \,\mathrm{m^3/s}\) and the density of air is \(1.15 \,\mathrm{kg/m^3}\), therefore the mass flow rate is the product of these two values. The mass flow rate in this scenario represents the amount of air mass the fan moves per second, which directly affects the fan power requirement.

Kinetic Energy Equation

The kinetic energy equation in the context of power calculation represents the work done to accelerate the fluid to a certain speed. Here, power is the rate of doing work or the rate at which energy is transferred. This concept is derived from the kinetic energy (KE) principle where \(KE = \frac{1}{2} m v^2\), with 'm' being the mass and 'v' the velocity.

In our case, to find the minimum power supplied to the fan, we apply a modified kinetic energy equation which is \(power = \frac{1}{2} \times \text{mass flow rate} \times \text{velocity}^2\). By calculating both the mass flow rate and multiplying it by the square of the final velocity, we identify the power the fan needs to impart the desired velocity to the airflow. This equation shows the linear relationship with the mass flow rate and the quadratic relationship with velocity, emphasizing why an increase in either significantly affects power.

Velocity of Air

In problems like this, the velocity of air is a determining factor for the kinetic energy the fan needs to impart to the air to reach the desired speed. Velocity is a vector quantity that describes both the speed and direction of a moving object, and in the case of the exercise, it is the speed at which the air is propelled by the fan.\

The given velocity, \(12 \,\mathrm{m/s}\), represents how fast the air must move after the fan accelerates it from rest. An important point to note is that the kinetic energy, and thus the power requirement, increases with the square of the velocity. This non-linear relationship means that even small increases in the target velocity can lead to significant increases in the required power—a key reason why accurate calculation of velocity is essential in designing and operating fan systems.