Reduced electrical costs can be achieved directly by effective utilization and improved efficiency of electrical facilities and by optimizing the mechanical and process systems requiring electrical power. Substantial electrical savings may be possible through shutdown of non-essential electrical equipment.
Saving Energy in Electrical System
Consider the following specific opportunities which could result in further minimization of electrical expense.
Power Factor Control
Alternating current is composed of two components: active and reactive. The active component is measured in kilowatts (KW) and the reactive component is measured in kilovars (KVAR). When current flows through an electrical circuit, only the part of it that is in phase with the voltage, the active component, delivers any useful power. The reactive component delivers no useful work, although additional generation capacity is required to produce it. For this reason, many utilities make up for the expense of producing reactive power by including power factor provisions in their rate structures.
Power factor is the ratio of active power (KW) to apparent power (KVA) or,
Typically, power factor lower than 0.90 or higher than 1.0 are subjected to a cost penalty, therefore, maintaining the power factor between 0.90 and 0.95 is advisable.
There are several methods of power factor correction. These include:
- Installation of capacitors.
- Properly sizing or rerating of motors and driven equipment.
- Use of synchronous motors instead of induction motors, if applicable.
Improving Power Factor with Capacitors
Generally, inexpensive power factor correction can be made on existing equipment by adding capacitors to the electrical feeders. Capacitors can either be placed on the high voltage side of the main plant power lines at a point closest to the electrical power generator, or they may be placed at individual motors. Individual motor capacitors provide the maximum benefit throughout the electrical system. However, power factor correction by this method is much more expensive than placing capacitors on the high voltage lines. Each option has its own merits:
HIGH VOLTAGE FEEDER
– Least expensive.
– Decreases plant electrical cost.
ON INDIVIDUAL MOTORS
– More expensive.
– Decreases plant electrical cost.
– Improve existing line capacity.
– Deceases line and transformer loss.
– Improves voltage regulation.
Review of the purchased power rate schedule and monitoring the power factor penalty portion of the electrical bill will determine whether or not additional power factor correction is needed.
The demand charge portion of an electric bill essentially reimburses the utility company for the cost of installing and maintaining generating facilities capable of supplying the peak kilowatt load required by the plant. This charge is based on maximum kilowatt usage over a short period of time usually 15 or 30 minutes. Review of the electrical bill will quantify the demand charge. Since the maximum short-time demand recorded is used to calculate the demand charge for the remaining term of the contract period, it is important to minimize peak electrical usage.
The load management technique commonly used to decrease the demand charge consists of the shutting down or rescheduling of non-essential loads to reduce demand during peak demand period. The first step is to examine demand records to determine when peak demands typically occur. Operation must then be observed to determine the equipment primarily responsible for the peaks.
Finally, it must be determined if this equipment can be shut down during the peak demand period or rescheduled for a period with lower demand. All light and/or intermittent electrical loads should be rescheduled if possible. The savings can be significant and requires no capital expense. Some operations have profited by installation of peak demand alarms for use in manual load shedding. System can be used to shed unnecessary loads automatically during peak demand periods. Automatic load shedding systems also have been installed to sequentially shed electrical loads on a predetermined priority basis. At a minimum, daily usage of purchased electrical power should be reviewed for management of peak demand as described above.
Control of the starting of electric motors is another method of lessening demand charges. Starting motors require high current loads which during peak load periods should be avoided. If the field has a high number of wells on timers, care should be exercised to ensure that the starts are staggered.
High Efficiency Motors
In general, electrical motors are very efficient when compare to other source of mechanical energy.
An engineer has the choice between a 100 HP electrical motor rated at 90% efficiency and a high efficiency motor rated at 96%. What is the savings if the motor runs 8320 hours per year?
Savings = 100 HP x 8320 HRS/YR x 0.746 KWH/HP-HR x (0.96 – 0.90) x $ 0.05 KWH
Savings = $1862.00 per year
Variable Speed Motor Control
Variable speed motors should be considered where speed reduction of driven equipment would result in significant energy savings. For example, power required for a centrifugal pump operated at half speed is only ⅛ of the power required at full speed. Pumping energy lost across flow control valves for less than rated operation would be substantially conserved through speed reduction. Variable speed motors may be justifiable for continuous operating equipment, but whose load varies over a wide range.
Purchased Power vs. Onsite Generation
In light of increased fuel and maintenance costs for aging internal combustion engines associated with the generation of electricity, a review should be made of the cost per KWH of onsite electrical generation. Of course, there are essential electrical loads which must never suffer an electrical power outage. However, it should be possible to supply a significant portion of the plants electrical power requirements with purchased electric power, should it be less costly. The following example compares the option of purchased power to that of onsite generation.
A power plant supplies a normal load of 500 KW with three 2-cycle gas engines operating at 75% of full load. What is the cost per KWH for onsite generation? Will purchasing electric power at $ 0.05 per KWH provide a cost savings? If so, what is the annual savings?
Total cost = fuel cost + maintenance cost + oil cost
Total cost/day = $ 689 + ($ 30000/365) +$ 30 = $ 800
Cost/KWH = ($ 800) / (500 KW x 24 HR/DAY) = $ 0.0667
Savings = ($0.0667/KWH – $0.05/KWH) x (500 KW) x (8760 HRS/YR) = $ 73150/YR savings
The above example indicates it is more economical to purchase power from an utility rather than to generate the power onsite.
Energy Conservation Guidelines
– Maintain purchased electrical power factor above the level which would invoke a billing penalty, usually greater than 0.9 P.F.
– Review operations and implement, if justifiable, closer demand control and load management.
– Consider variable speed controllers for continuous operating motor driven equipment if operating load varies significantly, especially if low load operation is of long duration.
– Evaluate the use of high efficiency electric motors whenever a purchase or repair of an electric motor is necessary.
– Insure equipment is properly sized for current loading.