Question

A 75 -hp compressor in a facility that operates at full load for \(2500 \mathrm{h}\) a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is \(\$ 0.06 / \mathrm{kWh}\), the annual electricity cost of this compressor is \((a) \$ 7802\) \((b) \$ 9021\) \((c) \$ 12,100\) \((d) \$ 8389\) \((e) \$ 10,460\)

Short answer

Answer: (b) $9021

Step-by-step solution

Convert the compressor's power

First, we need to convert the compressor's power from horsepower (hp) to kilowatts (kW). The conversion factor is 1 hp = 0.7457 kW. So, for a 75-hp compressor, the power in kW is: \(75 \ \mathrm{hp} \times 0.7457 \frac{\mathrm{kW}}{\mathrm{hp}}\approx56.178\ \mathrm{kW}\).

Calculate the input power required

Next, we need to find the input power required by the motor. Considering the motor's efficiency of 93%, the input power is given by: \(\mathrm{Input\ Power\ (kW)} = \frac{\mathrm{Output\ Power\ (kW)}}{\mathrm{Efficiency}} = \frac{56.178\ \mathrm{kW}}{0.93}\approx60.405\ \mathrm{kW}\).

Compute the total consumed energy

Now, we need to multiply the input power by the operating hours to find the total energy consumed in kilowatt-hours (kWh). The compressor operates for 2500 hours a year, so the consumed energy is: \(\mathrm{Energy\ Consumed\ (kWh)} = 60.405\ \mathrm{kW} \times 2500\ \mathrm{h} = 151012.5\ \mathrm{kWh}\).

Calculate the annual electricity cost

Finally, we can compute the annual electricity cost by multiplying the consumed energy by the unit cost of electricity, which is \(0.06/\mathrm{kWh}\): \(\mathrm{Annual\ Electricity\ Cost} = 151012.5\ \mathrm{kWh} \times \$ 0.06\ / \mathrm{kWh} = \$ 9060.75\). Among the given options, the closest value is \((b)\ \$ 9021\).

Key concepts

Compressor Energy Cost Calculation

When confronted with a problem involving the energy cost calculation for a compressor, it's important to approach it methodically. The underlying objective is to determine how much it will cost to run a compressor over a certain period, using specific pricing information for electricity.

Consider the compressor's power rating, usually given in horsepower (hp), which is a unit of power commonly used in the engineering and business world. Knowing the power rating, one can calculate the actual power usage in kilowatts (kW), because electricity costs are typically billed based on kilowatt-hours (kWh). Hence, the first step involves a conversion from horsepower to kilowatts using the equation: \( \text{Power} (\text{kW}) = \text{Power} (\text{hp}) \times 0.7457 \).

Once converted, it's instrumental to consider the device's efficiency, as not all the power consumed is used effectively. The efficiency is a ratio of the output power (the practical work the compressor does) to the input power (the energy required to produce that work), which is provided in the problem. By dividing the compressor's power by the motor efficiency, you can find the total input power required. Multiply this by the number of hours the compressor runs annually, and you will have total energy consumed in kWh. Finally, determine the cost of running the device by multiplying the total energy consumption by the electrical energy cost per kWh.

Electric Motor Efficiency

Electric motor efficiency is a measure of how well a motor converts electrical energy into mechanical energy, and it plays a crucial role in determining the cost-effectiveness of running machinery such as compressors. Typically expressed as a percentage, this efficiency is a key factor when calculating the overall energy required to operate an electric motor-driven device.

An efficiency of 93 percent, for instance, implies that for every 100 kW of electricity consumed, only 93 kW are converted into work while the remaining 7 kW are lost as heat, noise, or vibration. To find how much input power an electric motor needs given its output power and efficiency, the formula used is: \( \text{Input Power (kW)} = \frac{\text{Output Power (kW)}}{\text{Efficiency}} \).

Understanding and applying this concept is critical when one aims to optimize energy consumption as well as to reduce costs. For the best energy-saving results, it is ideal to select motors with higher efficiencies, which can significantly lower energy usage and therefore, operating expenses over the motor's lifetime.

Power Conversion Horsepower to Kilowatts

In many industries, power output and mechanical work are often measured in horsepower, but electricity bills and mechanical power in the scientific realm are commonly calculated in kilowatts. Consequently, for some calculations, it's necessary to convert horsepower to kilowatts to align with the power unit used in electricity cost and consumption calculations.

The conversion factor between these two units is a fixed value, with 1 horsepower being equivalent to approximately 0.7457 kilowatts. The conversion equation is therefore simple: \( \text{Power} (\text{kW}) = \text{Power} (\text{hp}) \times 0.7457 \).

By applying this conversion, users can transition smoothly between the mechanical and electrical aspects of power measurement, allowing for precise cost estimations and better planning regarding energy use. This simple yet powerful conversion process is a cornerstone for various applications within thermodynamics and energy management.