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Energy savings

March 1, 2011
- by
Fred Fairchild
When looking at ways to conserve electrical energy in operating a feed mill, one must first understand what the energy is costing. The cost of electrical energy is normally the highest expense in the operations of the mill.

**THE ELECTRIC BILL **

Each plant receives a monthly bill for its electrical use. The different parts of the bill may be presented in various ways depending on the power company’s billing format and the specific contact with the mill facility. In most cases, the electric bill may be divided into three billable components, the sum of which determines the total cost of the bill. Those three parts are: the total amount of energy used, the rate of use (demand) and how efficiently the power is used (Power Factor). In simple form, a monthly electric bill may look like this:

- Energy Use: 1454400 KWH /KVA @ $0.2704 = $39,326.98
- Rate of Use (Demand): 3229 KVA @ $7.30 per month = $23,571.70
- Demand Adjustment of 4% (Power Factor): $23571.70 x .04* = $942.87
- TOTAL CHARGES = $63,841.55

*The power factor for the billing period was 0.86. The contract requires that the power factor be 0.9 or greater to avoid a power factor penalty (added charge) on the bill. Based on the bill, we will take a look at how the costs are

determined.

**POWER FACTOR **

Power factor is based on the relationship of the two types of currents — reactive and active — required in the electrical

system.

Reactive current is energy that is needed to energize the field of the motor windings. It is energy that is not used up,

just transferred back and forth between the motor and the power system each half cycle. This current does not turn the

watt-hour meter and therefore is not shown on the amount of energy used. However, the power company lines and equipment must be sized to handle this current load.

Active current is the energy used to actually do the work. This current is recorded on the watt-hour meter and is seen

on the energy bill as the amount of energy used in the billing cycle. Likewise, the power company lines and equipment

must be sized to also handle this current load.

The power company does not like to deal with reactive current because they have to generate and transport it, but

don’t get paid for it. Due to the characteristics of electricity and its use, a small portion of the reactive current is lost and

not returned to the power company. The power company expects the power system in the plant to function well enough to maintain the power factor at 0.9 or better. If less of the reactive power is returned, the power company may initiate a penalty factor called a power factor penalty, and it is added to the billed cost.

Power factor is the cosine value of the angle incorporated in the triangle shown in fi gure 1. If there is no loss of Reactive Current (VARS), the Power Factor Angle would be 0. The cosine of 0 is 1.00. Likewise, if the Reactive Current is equal to the Active Current, the Power Factor Angle would be 45 degrees, making the Power Factor (cosine of 45 degrees) 0.707. If the power company requires a cosine of 0.90 minimum, the Power Factor Angle will be about 26 degrees.

Power Factor is dependent on the capacitance characteristics of the electrical system and its branches. The Power Factor

may be controlled and sometimes improved by using energy-efficient motors, capacitors, AC variable frequency drives,

solid state motor starters, and keeping motors sized to run at loads near their capacity.

**DEMAND **

Demand is the rate at which the power is used as determined by the total amount of energy used over a specific amount time. This is usually a 15- or 30-minute window that is constantlypects the power system in the plant to function well enough to maintain the power factor at 0.9 or better. If less of the reactive power is returned, the power company may initiate a penalty factor called a power factor penalty, and it is added to the billed cost.

Power factor is the cosine value of the angle incorporated in the triangle shown in figure 1. If there is no loss of Reactive Current (VARS), the Power Factor Angle would be 0. The cosine of 0 is 1.00. Likewise, if the Reactive Current is equal to the Active Current, the Power Factor Angle would be 45 degrees, making the Power Factor (cosine of 45 degrees) 0.707. If the power company requires a cosine of 0.90 minimum, the Power Factor Angle will be about 26 degrees.

Power Factor is dependent on the capacitance characteristics of the electrical system and its branches. The Power Factor

may be controlled and sometimes improved by using energy-efficient motors, capacitors, AC variable frequency drives,

solid state motor starters, and keeping motors sized to run at loads near their capacity.

**DEMAND **

Demand is the rate at which the power is used as determined by the total amount of energy used over a specific amount time. This is usually a 15- or 30-minute window that is constantly moving forward as time passes. For each window of time, the total amount of power used is determined to find the window in which the greatest amount of energy is used.

The maximum demand calculated for the billing cycle determines the demand rate for all of the energy used over the entire billing cycle. The power company has to have a supply system capable of delivering power at the highest demand rate. If the demand exceeds the contract limit for power supplied, the power company may add an additional penalty as a multiplier on the electric bill. This charge is usually applied to all the energy used within the billing cycle, even if the highest demand was for only one 15-minute window during the whole billing cycle.

Let’s use an automobile speedometer as an analogy to more clearly explain demand. We will take an extended trip and look at four different 15-minute segments during the trip. Using the odometer, we will measure the highest amount of miles traveled in each segment. This is analogous to the demand rate for each time period. The speedometer needle will be used to indicate instantaneous speeds during the time period and would be analogous to the instantaneous use of power.

In the fi rst segment, the automobile is driven through a residential section of town. The highest speed attained during

this 15-minute segment is 30 miles per hour, and a total distance of five miles is driven. Even though a speed of 30

miles per hour was reached for a short period, the average speed is only 20 miles per hour and the demand is five miles.

In the second segment, the automobile is driven through an industrial area. The highest speed attained is 45 miles per hour, and a distance of nine miles is driven. The average speed for this segment is 36 miles per hour and the demand is nine miles.

In the third segment, the automobile is now outside the city limits on a two-lane highway. The highest speed attained is 75 miles per hour, and the distance traveled is 12 miles. The average speed for this segment is 48 miles per hour and the demand is 12 miles.

In the fourth segment, the automobile is on a freeway and the cruise control is set at 65 miles per hour. A total of 16.25 miles is driven in the 15-minute time period. The average speed is 65 miles per hour and the demand is 16.25 miles.

Analyzing the four segments and using the odometer to measure demand, the highest demand was the fourth segment, 16.25 miles, even though the third segment had the highest instantaneous speed of 75 miles per hour. Starting a large motor is like passing a car: the instantaneous use of power is high for a few seconds, but the “miles driven” does not change that much because the time duration of the high use is very short.

Production scheduling can have a major affect on a mill’s demand rate. If you must produce a given tonnage of feed

per day, you have several choices in relationship to the power demand created by the production schedule. Assume the

amount of power required to make the feed is the same regardless of how long you take to make it.

Consider the following analogy: You must drive 1,200 miles in 24 hours (one day). You drive 120 miles per hour for

10 hours, or drive 50 miles per hour for 24 hours. In either case you will get the 1,200 miles covered. In the first scenario, the demand is 30 miles (miles driven in 15 minutes), while the demand for the second scenario is 12.5 miles. Assume it takes the same amount of total electrical energy to drive the 1,200 miles regardless of how long it takes to make the drive. The demand at 120 miles per hour is 2.4 times the demand at 50 miles per hour. Thus a lower overall electric bill may be possible by keeping the demand rate lower.

Another way to minimize demand is to stagger the number of machines in operation at one time. One method could be load shedding. This is a procedure whereby certain non-critical pieces of machinery can be shut down to avoid going over a set demand limit. The machinery that is shed (shut down) can be restarted when demand on the system has been lowered due to no longer using some of the other machinery. This is an important part of production scheduling.

Keeping control of both power factor and demand is critical in managing the electrical costs for the plant. It is also important that you understand the power company bill and what can be done to minimize electric power cost.