Question

The compressed-air requirements of a plant at sea level are being met by a 90 -hp compressor that takes in air at the local atmospheric pressure of \(101.3 \mathrm{kPa}\) and the average temperature of \(15^{\circ} \mathrm{C}\) and compresses it to \(1100 \mathrm{kPa}\). An investigation of the compressed-air system and the equipment using the compressed air reveals that compressing the air to \(750 \mathrm{kPa}\) is sufficient for this plant. The compressor operates \(3500 \mathrm{h} / \mathrm{yr}\) at 75 percent of the rated load and is driven by an electric motor that has an efficiency of 94 percent. Taking the price of electricity to be \(\$ 0.105 / \mathrm{kWh},\) determine the amount of energy and money saved as a result of reducing the pressure of the compressed air.

Short answer

Answer: The amount of energy saved as a result of reducing the pressure of the compressed air is 0.53 × 10^8 Wh, and the amount of money saved is approximately $55,515.

Step-by-step solution

Analyze the compressor operation

Given: The compressor operates 3500 hours per year at a 75% rated load and is driven by a 90 hp electric motor that has an efficiency of 94%. The price of electricity is \(\$0.105 / \mathrm{kWh}\). The compressor takes in air at \(101.3 \mathrm{kPa}\) and \(15^\circ \mathrm{C}\) and compresses it to the initial pressure of \(1100 \mathrm{kPa}\) and the new pressure of \(750 \mathrm{kPa}\).

Calculate the initial energy consumption

To find the energy consumption, we first need to find the work done by the compressor. The work done by the compressor can be expressed as: \(W_{compressor} = Power \times Load \times Efficiency\) We are given the power of the electric motor (\(90 \, \text{hp}\)) and its efficiency (94%). We first need to convert the power in horsepower to watts. \(Power_{watts} = Power_{hp} \times 746\) Substituting the values: \(Power_{watts} = 90 \, \text{hp} \times 746 = 67260 \, \text{W}\) Now, we can find the work done by the compressor at the initial pressure: \(W_{initial} = Power_{watts} \times Load \times Efficiency = 67260 \, \text{W} \times 0.75 \times 0.94\) \(W_{initial} = 47410.65 \, \text{W}\) Finally, to calculate the energy consumption, we multiply the work done by the number of hours the compressor operates per year: \(E_{initial} = W_{initial} \times 3500 \, \text{h} = 47410.65 \, \text{W} \times 3500 \, \text{h} = 1.66 \times 10^8 \, \text{Wh}\)

Calculate the new energy consumption

We repeat the calculation for the new energy consumption, using the new pressure of \(750 \mathrm{kPa}\) , after which we will find the difference between initial energy consumption and new energy consumption. Since the problem does not provide any direct information about how the compressor work changes with pressure, we will assume that the compressor work is proportional to the pressure ratio. Pressure ratio = \(\frac{P_{new}}{P_{initial}}\) Pressure ratio = \(\frac{750}{1100}\) = 0.6818 The new work done by the compressor (\(W_{new}\)) can be calculated using the pressure ratio: \(W_{new} = W_{initial} \times \text{Pressure ratio} = 47410.65 \, \text{W} \times 0.6818 = 32322.94 \, \text{W}\) Now, we can find the new energy consumption: \(E_{new} = W_{new} \times 3500 \, \text{h} = 32322.94 \, \text{W} \times 3500 \, \text{h} = 1.13 \times 10^8 \, \text{Wh}\)

Calculate the energy saved

We can now find the energy saved by subtracting the new energy consumption from the initial energy consumption: Energy saved = \(E_{initial} - E_{new} = 1.66 \times 10^8 \, \text{Wh} - 1.13 \times 10^8 \, \text{Wh} = 0.53 \times 10^8 \, \text{Wh}\)

Calculate the money saved

Finally, we can calculate the money saved using the price of electricity: Money saved = Energy saved × Price per kWh = \(0.53 \times 10^8 \, \text{Wh} \times \frac{\$0.105}{1 \, \text{kWh}}\) Money saved = \(0.53 \times 10^8 \, \text{Wh} \times \frac{\$0.105}{1000 \, \text{Wh}} = \$55515\) So the amount of energy saved as a result of reducing the pressure of the compressed air is \(0.53 \times 10^8 \, \text{Wh}\) and the amount of money saved is approximately $55,515.