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An exercise room has 6 weight-lifting machines that have no motors and 7 treadmills each equipped with a 2.5 -hp (shaft output) motor. The motors operate at an average load factor of \(0.7,\) at which their efficiency is \(0.77 .\) During peak evening hours, all 12 pieces of exercising equipment are used continuously, and there are also two people doing light exercises while waiting in line for one piece of the equipment. Assuming the average rate of heat dissipation from people in an exercise room is \(600 \mathrm{W}\), determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions.

Short Answer

Expert verified
Based on the given conditions and calculations, the total heat gain in the exercise room from people and equipment at peak load conditions is approximately 10,584.74 watts.

Step by step solution

01

Convert motor output power to watts

We need to convert motor output power from horsepower (hp) to watts (W): 1 hp = 746 W Motor output power (P_output) = 2.5 hp × 746 W/hp = 1865 W
02

Calculate total input power to motors

Now we need to find the total input power (P_input) to the motors: P_input = P_output / efficiency P_input = 1865 W / 0.77 ≈ 2422.08 W
03

Calculate heat dissipation from motors

The heat dissipated (Q_motors) from all 7 treadmills can be calculated as: Q_motors = (P_input - P_output) * 7 * load_factor Q_motors = (2422.08 W - 1865 W) * 7 * 0.7 ≈ 2184.74 W
04

Calculate heat dissipated by people

For all 12 people exercising and the 2 people waiting in line, we can calculate the heat dissipated (Q_people) as: Q_people = (12 + 2) * 600 W = 14 * 600 W = 8400 W
05

Determine total heat gain in room

Now, we can calculate the total rate of heat gain in the room (Q_total) by summing up the heat dissipated from the motors and the people: Q_total = Q_motors + Q_people Q_total = 2184.74 W + 8400 W ≈ 10584.74 W The rate of heat gain of the exercise room from people and equipment at peak load conditions is approximately 10584.74 W.

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Key Concepts

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

Heat Dissipation
Understanding heat dissipation is crucial when evaluating the temperature changes within a given environment, such as an exercise room. Heat dissipation refers to the process by which heat is released or spread out from a source into the surrounding area. In the context of our exercise, heat is dissipated from both the exercising equipment, like treadmills, and the individuals using them.

Each motor within the exercising equipment generates heat as a by-product of its operation. This heat is not completely captured by the equipment's mechanical parts and hence is released into the environment, contributing to the overall heat gain of the room. Similarly, humans also dissipate heat due to metabolic processes, especially when exercising. In fact, the average person dissipates approximately 600W of heat during moderate activity. When managing the temperature of a facility, it is important to account for all heat sources to ensure a comfortable environment is maintained.
Motor Efficiency
Motor efficiency is a measure of how well a motor converts electrical power into mechanical power, which is the useful work. It's defined as the ratio of the output power to the input power and is usually expressed as a percentage. In our exercise scenario, the treadmills' motors have an efficiency of 77% at a load factor of 0.7. This means that 77% of the electrical energy is converted into mechanical energy, while the remaining 23% is lost, primarily as heat.

Understanding motor efficiency is vital for calculating the total heat gain in the exercise room. The higher the efficiency, the less heat is generated for a given amount of mechanical work. It is worth noting that efficiency can vary depending on the load on the motor. To ensure accuracy in your calculations, always use the efficiency figure that matches the motor’s operating conditions. Considering motor efficiency allows us to determine the electrical input required, which in turn helps us calculate the heat dissipated from the motors into the room.
Power Conversion
Power conversion is the process of changing one form of power into another. In the exercise room scenario, the electrical power supplied to the treadmill motors must be converted into mechanical power to operate the treadmills. The measure of power, especially in relation to motors, is often given in horsepower (hp) but can be easily converted to watts (W), a more commonly used unit of power in physics. Specifically, 1 hp is equivalent to 746 W.

In our exercise, we convert the motors' power output from horsepower to watts to align with the standard power units used in thermodynamics and to facilitate the calculation of the heat gain in the room. Proper power conversion is key to ensuring accurate measurements and calculations are made. When completed correctly, these conversions and calculations provide a comprehensive view of the energy flow within a system and are essential for applications in engineering and environmental control.

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Most popular questions from this chapter

A university campus has 200 classrooms and 400 faculty offices. The classrooms are equipped with 12 fluorescent tubes, each consuming \(110 \mathrm{W}\), including the electricity used by the ballasts. The faculty offices, on average, have half as many tubes. The campus is open 240 days a year. The classrooms and faculty offices are not occupied an average of \(4 \mathrm{h}\) a day, but the lights are kept on. If the unit cost of electricity is \(\$ 0.11 / \mathrm{kWh}\), determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.

The inner and outer surfaces of a 0.5 -cm-thick \(2-\mathrm{m} \times\) \(2-m\) window glass in winter are \(15^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C},\) respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\), determine the amount of heat loss, in \(\mathrm{kJ}\), through the glass over a period of \(10 \mathrm{h}\). What would your answer be if the glass were \(1-\mathrm{cm}\) thick?

A \(2-k W\) electric resistance heater in a room is turned on and kept on for 50 min. The amount of energy transferred to the room by the heater is \((a) 2 \mathrm{kJ}\) \((b) 100 \mathrm{kJ}\) \((c) 3000 \mathrm{kJ}\) \((d) 6000 \mathrm{kJ}\) \((e) 12,000 \mathrm{kJ}\)

On a hot summer day, the air in a well-sealed room is circulated by a 0.50 -hp fan driven by a 65 percent efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan.) The rate of energy supply from the fanmotor assembly to the room is \((a) 0.769 \mathrm{kJ} / \mathrm{s}\) \((b) 0.325 \mathrm{kJ} / \mathrm{s}\) \((c) 0.574 \mathrm{kJ} / \mathrm{s}\) \((d) 0.373 \mathrm{kJ} / \mathrm{s}\) \((e) 0.242 \mathrm{kJ} / \mathrm{s}\)

A room is heated by an iron that is left plugged in. Is this a heat or work interaction? Take the entire room, including the iron, as the system.

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