Question

The compressed-air requirements of a plant are being met by a 100 -hp screw compressor that runs at full load during 40 percent of the time and idles the rest of the time during operating hours. The compressor consumes 35 percent of the rated power when idling and 90 percent of the power when compressing air. The annual operating hours of the facility are \(3800 \mathrm{h}\), and the unit cost of electricity is \(\$ 0.115 / \mathrm{kWh}\). It is determined that the compressed-air requirements of the facility during 60 percent of the time can be met by a 25 -hp reciprocating compressor that consumes 95 percent of the rated power when compressing air and no power when not compressing air. It is estimated that the 25 -hp compressor runs 85 percent of the time. The efficiencies of the motors of the large and the small compressors at or near full load are 0.90 and \(0.88,\) respectively. The efficiency of the large motor at 35 percent load is \(0.82 .\) Determine the amount of energy and money saved as a result of switching to the 25 -hp compressor during 60 percent of the time.

Short answer

Answer: The amount of money saved can be calculated using the following steps: 1. Calculate the energy consumption of the 100-hp compressor during full load and idle time. 2. Calculate the energy consumption of the 25-hp compressor during air compression. 3. Calculate the difference in energy consumption between both compressors. 4. Convert the energy difference from hp-hours to kWh. 5. Calculate the cost savings using the energy difference in kWh and the unit cost of electricity. The resulting value will represent the amount of money saved as a result of switching to the 25-hp compressor during 60 percent of the time.

Step-by-step solution

Calculate energy consumption of 100-hp compressor

First, we need to find out the amount of energy consumed by the 100-hp compressor during full load and idle time. The energy consumption during full load (when compressing air) is equal to the rated power times the full-load power percentage and the operating hours during full-load: $$Energy_{full\_load} = 100\,\text{hp} \times 0.9 \times (0.4 \times 3800\,\text{h})$$ During idle time, the energy consumption is equal to the rated power times the idle percentage and the operating hours when idling: $$Energy_{idle} = 100\,\text{hp} \times 0.35 \times (0.6 \times 3800\,\text{h})$$ The total energy consumed by the 100-hp compressor is the sum of the energy consumed during full-load and idle: $$Energy_{100\_hp} = Energy_{full\_load} + Energy_{idle}$$

Calculate energy consumption of 25-hp compressor

Next, we need to find out the amount of energy consumed by the 25-hp compressor during air compression. This is equal to the rated power times the full-load power percentage and the operating hours during compression: $$Energy_{25\_hp\_comp} = 25\,\text{hp} \times 0.95 \times (0.6 \times 3800\,\text{h} \times 0.85)$$ Note that the 25-hp compressor uses no power when not compressing air, so we do not need to calculate its energy consumption during idle time.

Calculate the difference in energy consumption

Now that we have the energy consumption of both compressors, we can calculate the difference in energy consumption as follows: $$Energy\_Difference = Energy_{100\_hp} - Energy_{25\_hp\_comp}$$

Convert energy difference to kWh

We need to convert the energy difference from hp-hours to kWh to be able to calculate the cost savings. To do this, we use the conversion factor 0.746 kW/hp: $$Energy\_difference\_kWh = Energy\_Difference \times 0.746$$

Calculate the cost savings

We can now calculate the cost savings resulting from the difference in energy consumption. This is equal to the energy difference in kWh multiplied by the unit cost of electricity: $$Cost\_Savings = Energy\_difference\_kWh \times \$0.115 \,/\,\text{kWh}$$ The resulting value will be the amount of money saved as a result of switching to the 25-hp compressor during 60 percent of the time.

Key concepts

Understanding Thermodynamics in Air Compression

Thermodynamics plays a crucial role in understanding energy conversion and efficiency in air compressors. In simple terms, thermodynamics is the study of how energy is transformed, particularly from electrical energy into mechanical work and heat during the compression process.

When air is compressed, the motor of the compressor converts electrical energy into mechanical energy. However, not all of this energy is effectively turned into the compressed air; some is lost as waste heat, a classic example of the second law of thermodynamics in action, which involves inevitable inefficiency in energy conversions. Therefore, knowing the thermodynamic properties of the system is vital when calculating total energy consumption. For students grappling with textbook exercises, appreciating this concept is essential for understanding why and how a more efficient compressor, like the 25-hp reciprocating type, can lead to significant energy and cost savings.

Maximizing Compressor Efficiency

Compressor efficiency is essentially the ratio of the useful work provided by the compressor to the electrical energy consumed. It's influenced by factors like design, motor efficiency, full-load and idling power consumption percentages, and operating conditions.

In the exercise we're examining, there are two types of compressors: a screw compressor and a reciprocating compressor. The screw compressor, which is larger and stays on even when not actively compressing air, consumes more energy overall due to its idling power usage. The reciprocating compressor operates much more efficiently, primarily because it uses no power when not in use. When targeting content for students, explaining these differences in running and idle power consumptions and their effect on overall efficiency is key. Understanding that the efficiency varies at different loads—as indicated by the motor efficiency data—will enable learners to grasp why switching compressors can result in substantial energy savings.

Calculating Electricity Cost Savings

Electricity cost savings are an important consideration for any industrial process, as they directly impact the bottom line. When dealing with air compressors, as in our exercise, calculating potential cost savings involves determining the energy used by both compressors, converting this to kWh (since electrical energy is billed per kWh), and then applying the cost of electricity to find out how much money can be saved.

By identifying and comparing the power consumption of both the 100-hp and 25-hp compressors under different operating conditions, students can uncover potential savings. It is also critical to explain energy unit conversions, which often involve the conversion of horsepower-hours to kilowatt-hours using the factor 0.746 kW/hp. Ultimately, the exercise underscores the real-world value of analyzing the efficiency of machines and optimizing operations to reduce costs—a tangible application of textbook principles to everyday decision-making in industry.