energy management of electrical equipment

7/31/2019 Energy Management of Electrical Equipment

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By

Kiran.R

M.Tech (CEM)


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Understanding how your facilitys power factoraffects your electricity cost is an important part ofan energy management program.

Low power factor can be expensive and inefficient. Many utility companies charge an additional fee if

a facilitys power factor is less than an establishedvalue.

Low power factor also can reduce the electricalsystems distribution capacity by increasingcurrent flow and causing voltage drops


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o Power factor involves the relationship between twotypes of power: Working Power and Reactive Power.

o Most loads in electrical distribution systems areinductive, which means that they require anelectromagnetic field to operate.

o Inductive loads require two kinds of current:

Working Power performs actual work of creating

heat, light, motion, etc. Reactive Power sustains the electromagnetic field.

PF measures how effectively electrical power is beingused.


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Working power is expressed in kilowatts orkW (one kilowatt = 1,000 watts). This registersas kilowatt-hours on your electric meter. The

total capacity required is expressed in kilovolt-amperes, or kVA (Electrical equipment such asgenerators and transformers is rated in kVA).

Power factor is the ratio of the working power

to the total capacity required to provide thispower or kW/kVA


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5

Examples of Electric Equipment and Their Power Factor

Name of Equipment Power Factor Percent

Lightly loaded inductionmotor

.20

Loaded induction motor .80

Incandescent lamps 1

Neon-lighting equipment .30 - .70

All types of resistance heatingdevices (e.g. toaster, spaceheater)

1

Different types of electric equipment have different Power Factors and

consequently different efficiencies and current requirements:


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6

1. Real Power or Working Power (kW) - Measured

2. Reactive Power (kVAR) - Measured

3. Apparent Power (kVA) - Calculated

Power Triangle

Working Power (kW)

Apparent Power (kVA)

ReactivePower (kVAR)

kVA = kW + kVAR2 22

Pythagorean Theorem: c = a + b2 22

c

ab


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Calculating Power Factor

kVAh = kWh

2

+ kVARh

2

Metered kW x 0.95

PF

Power Factor = kWhkVAh

Average PF over month

= kW billed


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Example 1: January 2006 Billing

Monthly meter readings: 1,625 kW, 762,600 kWh,

846,600 kVARh

762,600 kWh2 + 846,600 kVARh2 = kVAh

kVAh = 1,139,425

762,600 kWh

PF = 1,139,425 kVAh = 66.9 %

1,625 KW x 0.95

.669 = 2,308 KW billed


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Power factor can impact your energy costs.Improving the power factor can improveefficiency, often resulting in significant economicsavings.

Benefits of Improving Power Factor Reduced energy costs

Lower transmission and distribution losses in yourelectrical system

Higher and more quality voltage regulation Increased capacity available to serve actual working

power requirements

Reduced non-productive loading on the system


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Places the cost burden more on those customers

causing the problem rather than all rate payers.

Improves system efficiency by reducing losses

Enhances system operation and reliability.

Increases system capacity capabilities.


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Minimize operation of idling or lightly loadedmotors.

Replace standard motors as they burnout withenergy efficient motors sized correctly.

Install capacitors in the circuit .


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Passive and Active power factor correction

Passive and Active filtering of network

Accepting non-sinusoidal voltage / current inthe system.


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Work with a qualified electrical contractor orengineer to study your application.

Different strategies of correction, static versusbulk.

Inverters, variable speed drives, and solid statesoft starters will affect design.

Power factor can be affected by harmonics,which can only be addressed with filters


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energy management of electrical equipment

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