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A room is cooled by circulating chilled water through a heat exchanger located in a room. The air is circulated through the heat exchanger by a 0.25 -hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25 -hp equipment is 54 percent. Determine the rate of heat supply by the fan- motor assembly to the room.

Short Answer

Expert verified
Answer: To determine the rate of heat supply by the fan-motor assembly to the room, follow these steps: 1. Convert the 0.25 hp output to watts: Power (watts) = 0.25 hp * 746 watts/hp 2. Calculate the input power to the electric motor: Input Power = Power (watts) / 0.54 3. Determine the power converted to heat: Power converted to Heat = Input Power - Output Power 4. The power converted to heat in Step 3 is the rate of heat supply by the fan-motor assembly to the room.

Step by step solution

01

Convert the horsepower rating to watts

We have the power output of the fan in horsepower (0.25 hp) and need to convert it to watts. We can use the conversion factor: 1 hp = 746 watts Now, multiply the horsepower by the conversion factor to obtain the power in watts. Power (watts) = 0.25 hp * 746 watts/hp
02

Calculate the input power to the electric motor

To determine the input power to the motor, we need to use the motor's efficiency (54% or 0.54). We can use the formula: Input Power = Output Power / Efficiency Using the power (watts) calculated in Step 1 as the output power, calculate the input power. Input Power = Power (watts) / 0.54
03

Determine the power converted to heat

Finally, we can determine the power that is converted to heat. The power that is not used as output power (i.e., shaft output) will be the power converted to heat. We can use the following formula: Power converted to Heat = Input Power - Output Power Using the input power calculated in Step 2 and the output power (in watts) from Step 1, calculate the power that is converted to heat.
04

Calculate the rate of heat supply by the fan-motor assembly

The power converted to heat that we calculated in Step 3 is the rate of heat supply by the fan-motor assembly to the room.

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Key Concepts

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

Energy Conversion in Thermodynamics
Understanding the concept of energy conversion in thermodynamics is vital for comprehending various processes that involve heat and work. Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In thermodynamics, we often discuss the transformation of energy from one form to another.

For instance, in the textbook exercise provided, a fan driven by an electric motor circulates chilled water through a heat exchanger to cool a room. Here, electrical energy is being converted into mechanical energy by the motor to drive the fan. This conversion isn't 100% efficient, as no process in the real world is, which leads us to understand that some of the electrical energy is lost in the form of heat.

In a broader perspective, within power systems, generators convert mechanical energy to electrical energy, while motors do the reverse. The efficiency of these devices is a crucial factor, as it determines how much input energy is successfully converted to the desired output, and how much is inadvertently turned into waste heat.
Electrical-to-Mechanical Energy Efficiency
The efficiency of an electric motor is a measure of how well it converts electrical energy into mechanical energy, which is the useful output we need for applications like fans, pumps, and machines. In the provided exercise, the fan's motor has an efficiency of 54 percent. This means that for the electric energy the motor consumes, only 54 percent is transformed into mechanical energy.

The efficiency can be represented mathematically by the formula: \[\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%\]
Here, the output power is the mechanical power delivered by the motor's shaft to drive the fan. The input power is the electrical power consumed. By understanding this, students can gauge the performance of the motor and calculate the inevitable heat generation, which is the amount of input energy that doesn't contribute to the mechanical output. The lower the efficiency, the higher the heat generated, which in some scenarios could be detrimental to the system's overall functionality.
Power and Heat Relationship
The relationship between power and heat in an electrical system, such as the fan-motor assembly in our exercise, reveals the intrinsic connection between energy usage and heat production. Power (P) is the rate at which work is done or energy is transferred over time. When energy is converted from one form to another (in this case, from electrical to mechanical), the inefficiency of the process often leads to heat generation.

The heat generated (Q) can be calculated by subtracting the useful mechanical power output (P_output) from the total power input (P_input) to the system. Mathematically, it is expressed as: \[Q = P_{\text{input}} - P_{\text{output}}\]
In this context, all the energy that isn't converted into mechanical energy for the fan ends up as heat within the room. If we treat the fan-motor assembly as a heat source, being able to calculate this heat supply is crucial in understanding and managing the thermal load within the room. For energy management and system design, this relationship emphasizes the importance of selecting efficient equipment to minimize unnecessary heat gain, especially in temperature-critical environments.

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Most popular questions from this chapter

The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price. Suppose a utility company is selling electric power for \(\$ 0.05 / \mathrm{kWh}\) at night and is willing to pay \(\$ 0.12 / \mathrm{kWh}\) for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir \(40 \mathrm{m}\) above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump-motor operates as a turbine- generator during reverse flow. Preliminary analysis shows that a water flow rate of \(2 \mathrm{m}^{3} / \mathrm{s}\) can be used in either direction. The combined pump-motor and turbine-generator efficiencies are expected to be 75 percent each. Disregarding the frictional losses in piping and assuming the system operates for \(10 \mathrm{h}\) each in the pump and turbine modes during a typical day, determine the potential revenue this pump-turbine system can generate per year.

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