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Chapter 4: Problem 107

A \(0.8-\mathrm{m}\) -diameter fan is driven by a \(1-\mathrm{kW}\) motor. The fan pulls in air from a large room, and air leaves the fan at \(10 \mathrm{~m} / \mathrm{s}\). Estimate the efficiency of the fan in converting electrical energy to the energy added to the air by the fan (i.e., the efficiency of the fan-motor system). Assume standard air.

Short Answer

Expert verified
The efficiency of the fan-motor system is approximately 30.75%.

Step by step solution

01

Find the Mass Flow Rate

To find the mass flow rate, we start by determining the volumetric flow rate using the formula \( Q = A \times v \), where \( A \) is the cross-sectional area of the fan and \( v \) is the velocity of air. For a circular area, \( A = \pi \times \left(\frac{d}{2}\right)^2 \). Here, \( d = 0.8 \) meters, so \( A = \pi \times (0.4)^2 = 0.5024 \) m虏. The volumetric flow rate \( Q = 0.5024 \times 10 = 5.024 \) m鲁/s. The mass flow rate \( \dot{m} \) is given by \( \dot{m} = \rho \times Q \), where \( \rho \approx 1.225 \) kg/m鲁 for standard air, yielding \( \dot{m} = 1.225 \times 5.024 = 6.15 \) kg/s.
02

Calculate the Kinetic Energy Added to the Air

The kinetic energy added to the air can be found using the formula \( \dot{E}_{\text{kinetic}} = \frac{1}{2} \dot{m} v^2 \), where \( \dot{m} = 6.15 \) kg/s and \( v = 10 \) m/s. Substituting the values, we find \( \dot{E}_{\text{kinetic}} = \frac{1}{2} \times 6.15 \times 10^2 = 307.5 \) Watts.
03

Compute the Efficiency of the Fan-Motor System

The efficiency \( \eta \) of the fan-motor system is defined as the ratio of the useful power output to the power input, \( \eta = \frac{\dot{E}_{\text{kinetic}}}{P_{\text{motor}}} \). Here \( \dot{E}_{\text{kinetic}} = 307.5 \) Watts and \( P_{\text{motor}} = 1000 \) Watts. Therefore, the efficiency \( \eta = \frac{307.5}{1000} = 0.3075 \) or 30.75%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Flow Rate
Understanding mass flow rate is key when analyzing systems involving fluids like air or water. It measures the amount of mass passing through a surface per unit of time. In the context of a fan system, it tells us how much air the fan is moving.
We calculate the mass flow rate using the relation \( \dot{m} = \rho \times Q \), where \( \dot{m} \) represents mass flow rate, \( \rho \) is the air density, and \( Q \) is the volumetric flow rate.
To get the volumetric flow rate \( Q \), we use the formula \( Q = A \times v \), with \( A \) being the cross-sectional area of the fan and \( v \) the velocity of the air leaving the fan.
  • Cross-sectional area (A): For circular fans, calculate \( A \) using \( A = \pi \times (\frac{d}{2})^2 \).
  • Volumetric flow rate (Q): Multiply the area by the air velocity.
  • Density (\( \rho \)): For air at normal conditions, it is approximately 1.225 kg/m鲁.
This way, you can calculate the mass flow rate, which gives a clearer picture of the fan's performance.
Kinetic Energy
Kinetic energy is the energy of motion. When air is pushed by the fan, it gains speed and thus gains kinetic energy. Understanding the kinetic energy helps us quantify how much energy the fan transfers to the air.
The formula to calculate kinetic energy added to the air by the fan is \( \dot{E}_{\text{kinetic}} = \frac{1}{2} \dot{m} v^2 \). This measures the power output in terms of kinetic energy that the moving air possesses.
  • Mass flow rate (\( \dot{m} \)): As calculated above, this is a crucial factor.
  • Velocity (v): The speed of the air leaving the fan also influences the kinetic energy.
With these figures, you can determine how effectively the fan is propelling air, converting electrical energy into motion. This is an essential step to evaluate the energy transfer involved in the fan operation.
Energy Conversion Efficiency
Energy conversion efficiency indicates how well a device converts one form of energy to another. In the context of a fan-motor system, it describes how efficiently electrical energy is turned into the kinetic energy of the air.
Efficiency \( \eta \) is calculated using the equation \( \eta = \frac{\dot{E}_{\text{kinetic}}}{P_{\text{motor}}} \), where \( \dot{E}_{\text{kinetic}} \) represents the power in kinetic energy and \( P_{\text{motor}} \) the electrical power supplied to the motor.
  • Useful power output (\( \dot{E}_{\text{kinetic}} \)): This indicates the effective energy converted to motion.
  • Power input (\( P_{\text{motor}} \)): The actual electric power supplied to the motor.
A higher efficiency percentage signifies a better conversion rate of electrical energy into air motion. Therefore, knowing the efficiency helps in understanding the performance and energy consumption of the system, essential for evaluating device efficacy and potential energy savings.

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Most popular questions from this chapter

A blade attached to a turbine rotor is driven by a stream of water that has a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The blade moves with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and deflects the stream of water through an angle of \(80^{\circ} ;\) the entrance and exit flow areas are each equal to \(2 \mathrm{~m}^{2}\). Estimate the force on the moving blade and the power transferred to the turbine rotor.

Water at \(20^{\circ} \mathrm{C}\) is contained in a pressurized tank that is supported by a cable as shown in Figure 4.71 . The (gauge) air pressure above the water in the tank is \(600 \mathrm{kPa}\), the depth of water in the tank is \(1 \mathrm{~m}\), the diameter of the tank is \(1 \mathrm{~m}\), the weight of the tank is \(1 \mathrm{kN},\) and there is a \(50-\mathrm{mm}\) -diameter orifice in the bottom of the tank. The discharge coefficient of the orifice is 0.8 . Estimate the tension in the cable supporting the tank.

A room has dimensions \(10 \mathrm{~m} \times 10 \mathrm{~m} \times 3 \mathrm{~m},\) and air from the room is vented outside through a 100 -mm- diameter vent. The head loss in this type of vent can be estimated as \(h_{\ell}=0.1 V^{2} / 2 g\), where \(V\) is the flow velocity through the vent. If the pressure inside the room is \(2 \mathrm{kPa}\) higher than the outside pressure, estimate the volume flow rate of air through the vent and the time it takes to fully exchange the air in the room. Assume standard air.

Consider a 50-mm-diameter thin-plate orifice centrally located in a \(110 \mathrm{~mm}\) -diameter pipe as shown in Figure 4.67 . When water at \(20^{\circ} \mathrm{C}\) flows at a rate of \(0.04 \mathrm{~m}^{3} / \mathrm{s}\) through the pipe, a pressure drop of \(150 \mathrm{kPa}\) between sections 1 and 2 is measured. If the wall friction is negligible, estimate the force of the water on the orifice plate.

Angel Falls in Venezuela is the second-highest waterfall in the world, with a vertical drop of approximately \(979 \mathrm{~m}\). If adiabatic conditions were to exist along the entire length of the waterfall, estimate the difference in temperature in an element of water in the upstream pool and this same element of water in the downstream pool after the kinetic energy acquired during the fall has been dissipated.
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