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Compressed air is one of the key utilities in manufacturing facilities, and the total installed power of compressed-air systems in the United States is estimated to be about 20 million horsepower. Assuming the compressors to operate at full load during one-third of the time on average and the average motor efficiency to be 90 percent, determine how much energy and money will be saved per year if the energy consumed by compressors is reduced by 5 percent as a result of implementing some conservation measures. Take the unit cost of electricity to be \(\$ 0.11 / \mathrm{kWh}\).

Short Answer

Expert verified
Answer: The annual energy saved with a 5% reduction in energy consumption is 1,929,866,000 kWh and the annual money saved is $212,285,260.

Step by step solution

01

Convert the total installed power to kilowatts

We are given the total installed power in the United States as 20 million horsepower. We need to convert this to kilowatts for our calculations. 1 HP (horsepower) = 0.7355 kW (kilowatts) Total installed power (kW) = 20,000,000 * 0.7355 = 14,710,000 kW
02

Calculate the annual energy consumption of the compressors

To find the annual energy consumption, we need to take into account that the compressors operate at full load during one-third of the time on average and the average motor efficiency is 90%. Energy consumption per hour (kWh) = Total installed power (kW) * Operating hours * Motor efficiency Energy consumption per hour (kWh) = 14,710,000 * (1/3) * 0.9 = 4,407,000 kWh Since there are 8760 hours in a year, the annual energy consumption is: Annual energy consumption (kWh) = 4,407,000 * 8760 = 38,597,320,000 kWh
03

Find the annual energy consumption reduced by 5%

After implementing the conservation measures, the energy consumed by the compressors is reduced by 5%. Reduced energy consumption (kWh) = Annual energy consumption * 5% = 38,597,320,000 * 0.05 = 1,929,866,000 kWh
04

Calculate the annual money saved with 5% reduced energy consumption

We are given the unit cost of electricity as \(\$0.11 / \mathrm{kWh}\). Using this, we can calculate the annual money saved. Annual money saved (\() = Reduced energy consumption (kWh) * Unit cost of electricity (\)/kWh) = 1,929,866,000 * \(0.11 = \)212,285,260 So the annual energy and money saved with a 5% reduction in energy consumption are 1,929,866,000 kWh and $212,285,260, respectively.

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Key Concepts

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

Compressed Air System Efficiency
Compressed air systems are common in industrial settings, where they are used to operate tools, equipment, and processes. Improving the efficiency of these systems can lead to significant energy and cost savings.

Efficiency in compressed air systems can be measured by how much air is produced per unit of electricity consumed. Factors affecting efficiency include the condition of the equipment, the design of the system, and how well it is maintained. Leaks, for instance, can waste a significant amount of compressed air and reduce system efficiency. Measures such as regular maintenance, repairing leaks, and optimizing the system design can lead to improved efficiency and reduced energy consumption.

Implementing energy conservation measures can result in lower operating costs and also contribute to sustainability efforts by reducing the carbon footprint associated with energy use.
Electricity Consumption Calculation
Understanding how to calculate electricity consumption is key to identifying potential savings in a compressed air system. The exercise provided walks through the conversion of total installed power from horsepower to kilowatts and the calculation of the energy consumption of compressors.

To calculate the annual energy consumption, you need to know the total installed power, how long the system operates under full load, and the motor efficiency. You use these values to calculate the energy consumption per hour, and then multiply by the number of hours in a year. Once you have this annual energy consumption, you can calculate the energy reduction from implemented conservation measures and the associated cost savings.

This process of calculation can guide you to make informed decisions on where to focus energy-saving efforts for the highest financial return. It's important to use accurate and current data for these calculations to ensure precise results.
Energy Savings Measures
Taking energy savings measures is vital for both environmental conservation and operational cost reduction. In the context of compressed air systems, such measures might include optimizing systems operation, proactive maintenance, and updating equipment.

For instance, introducing variable speed drives (VSDs) allows compressors to run at speeds matching the demand, reducing electricity use. Another measure is to ensure that the system is free from leaks, which can waste a significant percentage of compressed air. Properly adjusting the pressure to the minimum necessary and using high-efficiency equipment are other practical steps.

By implementing measures such as these, industries can significantly lower their energy consumption, as reflected in our exercise, which calculated a substantial energy and monetary saving of 5%. This directly exemplifies how energy-efficient practices yield both ecological and economical benefits.

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Most popular questions from this chapter

Argon gas expands in an adiabatic turbine steadily from \(600^{\circ} \mathrm{C}\) and \(800 \mathrm{kPa}\) to \(80 \mathrm{kPa}\) at a rate of \(2.5 \mathrm{kg} / \mathrm{s} .\) For isentropic efficiency of 88 percent, the power produced by the turbine is \((a) 240 \mathrm{kW}\) \((b) 361 \mathrm{kW}\) \((c) 414 \mathrm{kW}\) \((d) 602 \mathrm{kW}\) \((e) 777 \mathrm{kW}\)

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Steam enters an adiabatic turbine steadily at \(400^{\circ} \mathrm{C}\) and \(5 \mathrm{MPa}\), and leaves at \(20 \mathrm{kPa}\). The highest possible percentage of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is \((a) 4 \%\) \((b) 8 \%\) \((c) 12 \%\) \((d) 18 \%\) \((e) 0 \%\)

Air is compressed steadily and adiabatically from \(17^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\) to \(200^{\circ} \mathrm{C}\) and \(400 \mathrm{kPa} .\) Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is \((a) 0.76\) \((b) 0.94\) \((c) 0.86\) \((d) 0.84\) \((e) 1.00\)

The temperature of an ideal gas having constant specific heats is given as a function of specific entropy and specific volume as \(T(s, v)=A v^{1-k} \exp \left(s / c_{v}\right)\) where \(A\) is a constant. For a reversible, constant volume process, find the expression for heat transfer per unit mass as a function of \(c_{v}\) and \(T\) using \(Q=\int T d S .\) Compare this result with that obtained by applying the first law to a closed system undergoing a constant volume process.

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