Understanding Demand
Most customers pay for the energy they use (measured in kilowatt-hours, abbreviated kWh). Larger users of electricity are also charged for something called demand (measured in kilowatts, abbreviated kW)
- What is “Demand" – A measure of the number of energy-consuming units, or the amount of service or output, for which energy inputs are required.
- What is “kW” – Rate of using Electricity (Demand). Example, ten 100-watt lamps consume electricity at the rate of 1,000 watts, or 1 (one) kWh (kilowatt per hour), electrical energy actually used.
- What is 1 (one) “Horse Power” – Under full load, consumes power at the rate of 1 (one) kilowatt (kW) hour per hour.
To understand Demand, think about your trip driving to work. As you entered the highway, you may have started at 65 mph. Due to traffic you only drove 45 mph for the majority of the trip. Although your engine is capable of exceeding 100 mph Your speedometer never reached that speed.
Demand works the same way for an electric customer. Just like your cars engine has a maximum speed that could be reached, a business has a maximum peak DEMAND that can be reached if they turned on every piece of equipment at the same time.
So to compare to the speedometer example, what was the max speed you were able to reach on trip to you way to work for one month? If you were able to drive 60 mph for 15 minutes and never exceeded the speed. Then this would be your peak demand…even if you drove 55 mph for the rest of the month.
So to compare to the speedometer example, what was the max speed you were able to reach on trip to you way to work for one month? If you were able to drive 60 mph for 15 minutes and never exceeded the speed. Then this would be your peak demand…even if you drove 55 mph for the rest of the month.
How does the demand affect the customer’s electric bill? The demand charge will be a large part of the bill if the customer uses a lot of power over a short period of time, and a smaller part of the bill if the customer uses power at a more or less constant rate throughout the month. Let’s look at two examples using an irrigation pump scenario:
1. A customer runs a 50 horsepower (hp) irrigation pump for only five hours during July1
Demand Charge = 50 hp x .746 kW/hp x $8.03/kW = $299.52
Energy Charge = 50 hp x .746 kW/hp x 5 Hr x $0.034/kWh = $6.34
2. The same customer runs a 50 hp irrigation pump constantly through the entire month of July
Demand Charge = 50 hp x .746 kW/hp x $8.03/KW = $299.52
Energy Charge = 50 hp x .746 kW/hp x 744 Hr x $0.034/kWh = $943.54
As you can see, the demand charge portion of the customer’s power bill does not change, whether the pump runs fifteen minutes or all month. However, the energy charge portion of the power bill does depend on the amount of time the pump runs. So a customer who is careful and doesn’t run the pump more hours than necessary will save money on the energy bill.
1. A customer runs a 50 horsepower (hp) irrigation pump for only five hours during July1
Demand Charge = 50 hp x .746 kW/hp x $8.03/kW = $299.52
Energy Charge = 50 hp x .746 kW/hp x 5 Hr x $0.034/kWh = $6.34
2. The same customer runs a 50 hp irrigation pump constantly through the entire month of July
Demand Charge = 50 hp x .746 kW/hp x $8.03/KW = $299.52
Energy Charge = 50 hp x .746 kW/hp x 744 Hr x $0.034/kWh = $943.54
As you can see, the demand charge portion of the customer’s power bill does not change, whether the pump runs fifteen minutes or all month. However, the energy charge portion of the power bill does depend on the amount of time the pump runs. So a customer who is careful and doesn’t run the pump more hours than necessary will save money on the energy bill.
Taking the above scenario into consideration, how can that particular business save money on their demand charge?
- Make sure the motor is the correct size for the pump. An oversized motor could be increasing the demand and cost business money. It’s also possible that a newer, more efficient pump and motor combination may be available that would save on demand and energy.
- Make sure the pump and motor combination is the correct size for the system. If the business is using a 75 hp pump when a 50 hp pump would do the job, they are wasting 25 hp or 18.7 kW of demand, (25 hp x .746 kW/hp = 18.7 kW).
- Make sure that worn pumps, motors, nozzles, and leaks are not increasing flow to the point where demand has increased. A business can get an idea of what the demand should be by multiplying the total connected horsepower on the meter by .746 kW/hp. This should be approximately the demand amount on their utility bill. Be aware that any additional machinery connected to the meter could increase the demand.
- Know the date the meter is read each month. If the business only runs the pump for a day or two during a billing period (for example, at the beginning or end of the year), the demand cost could far exceed the energy cost.
