3-PH Power Basics

3 Phase Power BasicsThomas GreerExecutive Director – Engineering ServicesTLG Services

Emerson Confidential

AgendaTerminologyBasic Electrical CircuitsBasic Power Calculations


Why This Electricity Stuff?

–To Become an Electrical Engineer?–So We Won’t Have to Call Our AE?–To Moonlight Teaching at the University?

I Don’t Think So!


Why This Electricity Stuff?Able to talk the talkFundamental language with customers, consultants, and contractors in this industryImproved technical skills help you to meet and exceed the expectations of your customers


What You Will Take HomeUnderstand basic terminology in electrical circuits and power systemsAble to perform basic power calculations


CurrentThe movement of electrons in a circuit. It is the flow of electricity.Unit of measure is the ampere abbreviated “AMP” or ‘A”.Represented in equations by the letter “I”.


Direct CurrentDirect Current (DC) - Current flows in one directionCommon DC source - battery

DC Current

Time

Current


Alternating CurrentAlternating Current (AC) - Current flows first in one direction and then the other, reversing direction periodicallyCommon AC source - Commercial Power (AC Generator)

+

-

AC CurrentCurrent

Time


VoltageIs the electrical potential or force that causes current to flow in a circuit.Unit measure is the volt, abbreviated “V”.


ImpedanceImpedance is the total opposition a circuit offers to the flow of electric current – DC circuit impedance include resistance only– AC circuit impedance includes resistance and reactance

• Reactance comes from inductors and capacitors

Measured in ohms (Ω)Represented in equations by the letter “Z”


Electric CircuitRoute in which current flows from a power source to a load and back to the power source.

V ZAC Power Source

Switch

Load


Hydraulic and Electric Circuit Analogy

+-

Battery developingelectrical pressure

Direction of currentflow

Resistance(electrical load)

Wire conductingcurrent flow

Pipe conductingwater flow

Pump generating mechanical pressure

Mechanical Load

Electric Circuit


Ohm’s Law

I = VZ

I = Current (Amps)

V = Voltage (Volts)

Z = Impedance (Ohms)

V = IZSolving for Voltage or Impedance

VI

Ohm’s Law - The current in an electric circuit is directly proportional to the applied voltage and inversely proportional to the circuit impedance.

Z =

or


Applying Ohm’s Law

Z=10ΩV = 120VAC

Example: AC circuit with resistive electric heater load of 10 ohms.

I = V/ZI = 120/10I = 12A

I = ?


Are You Still There?Any Questions?


AC Waveform - 3 Phase

90

150

210

270

330

120 180

240

300

360

A

C

BOne Cycle

Frequency# Cycles Per SecondHertz


Peak and RMS Values

RMS value of an AC current is equal to the DC current which will produce the same average heating effect in a given resistance

1.0 Peak 1.0 (170V)

0.7

0.90.8

RMS 0.707 (120V)0.60.50.40.30.2

00.1

-0.1-0.2-0.3-0.4-0.5-0.6-0.7-0.8-0.9-1.0

Irms = .707 · Ipeak

Ip = √2 · Irms

For Sinewave


Distorted Sinewave

Voltage Waveform with distortion caused by load withswitching SCR’s


HarmonicsUsed as Building Blocks to Define a non Sinusoidal Waveform.– Periodic Sinusoidal Components– Multiples of Fundamental

• 3rd Harmonic of 60Hz Sinewave is 180Hz

Harmonic Distortion - A current or voltagewaveform includes includes non 60Hz components. Therefore, it is a distorted sinewave. Most real world 3 phase loads include harmonic distortion.


PowerRate of Doing Work

P=V * I

P = Power (Volt Amperes or Watts)V = Voltage (Volts)I = Current (Amperes)Z = Impedance (Ohms)

Since, V = I * Z , Power can also be expressed as follows:

P = V2/Z and P = I2Z


AC PowerApparent Power– Total power measured in Volt-Amperes or VA.

Obtained from the measured current and voltage.

– KVA (Single Phase) = (V * A) / 1000– KVA (Three Phase) = (VLN * A * 3) / 1000 or– KVA (Three Phase) = (VLL * A * √3) / 1000

Where √3 = 1.732


AC PowerReal Power– Power which is actually available to do work.

• Total power (KVA) includes reactive components due to inductance and capacitance. Power useful for work is resistive component only.

• Measured in KW (kilowatts)• Must be obtained by measurement with a Wattmeter or

calculated.


Power FactorRatio of Real Power to Apparent Power

PF = KW / KVA– Power Factor is described as leading or lagging

based on whether the current leads or lags the voltage

– For a sinusoidal current and voltage the power factor equals the cosine of the phase angle between the current and voltage


CapacitorElectrical device that stores electrical energy.Does not allow instantaneous voltage changeCapacitance - storage capability of capacitor– Measured in “farads”


Capacitor

The capacitor current is out of phase with the generated voltage, and leads the voltage by 90 degrees.

Capacitor voltage and current+

Voltage

Current

0° 90° 180° 270° 360°Time

0

-


Inductor

Device which stores electrical energy.Impedes instantaneous change in current.Inductance - measure of the amount of interaction between alternating current and resultant changing electromagnetic fields in a device.Unit of measure is “henry”


The inductor current is out of phase with the generated voltage, and lags the voltage by 90 degrees.

Voltage

Current

0° 90° 180° 270° 360°Time

0

-

+

Inductor

Inductor voltage and current


Lead and Lag Power Factor Components

Single - PhaseTransformer

Three - PhaseTransformer

Choke

Induction Motor

Lagging Power Factor Leading Power Factor

Capacitor

Filter

Unity Power Factor

• Incandescent Lamps• Heaters• PFC Power Supplies• Synchronous Motors


Efficiency

Efficiency = = Power out Kw outPower in Kw in

110 kVA 100 kva load

Input and output PF must be known as efficiency is a ratio of Kw’s

EX: PF in = PF out (this case only) = 0.8Find efficiency.

Efficiency = = .91•(100) = 91%100 (.8)110 (.8)

Ratio of useful output energy to total useful input energy


System Efficiencies

BuildingXformer

99%

UPS90%

Load PS80%

StepdownXformer

98%

Overall Efficiency = (.99 * .98 * .9 * .8) = 70%

Sample System

Overall system efficiency is obtained by multiplying efficiencies of series components


Still With Me?

Any Questions?


Single Phase Systems

220/230/240V - 50 Hz 110/115/120V - 60 Hz load voltages may be obtained from these systems


Single Phase Systems

Neutral

Three load voltages may be obtained from this system

1. 120 volt single phase, two wire2. 240 volt single phase, two wire3. 120/240 volt sing;e phase, three wire

120V

240V


Three Phase Systems

380/400/415

480V220/

N 480V (Line-to-Line)

Delta Connected SystemNo NeutralLine-To-Line Voltages Only

480V


Three Phase Systems

380/400/415

480V

220/230/240N

277V (Line-to-Neutral)480V (Line-to-Line)

277V

Wye Connected System

Load voltages obtained from 480V systems

1. 277 volt single phase, two wire (L-N)2. 480 volt single phase, two wire3, 480 volt three phase, three wire4. 480/277 volt three phase, four wire


Three Phase Systems

380/400/415

208V

220/230/240N

To find the line-to-neutral voltage if the line-to-linevoltage is 208V

V 2081.73 1.73 120V

120V (Line-to-Neutral)208V (Line-to-Line)

120V


Three Phase SystemsWorldwide Voltages available– 60Hz

• 600/346V (Canada)• 480/277V• 208/120V• 220/127V (Mexico)

– 50Hz• 380/220V• 400/230V• 415/240V

38

POWER CALCULATIONS

Putting it All Together


Determining kVA of Power Feeder Service (Single Phase)

KVA = V • A

Assume a single phase 120 entrance service specified at 20 A.

KVA = = 2.4

1000

120 • 201000


Determining kVA of Power Feeder Service (Three Phase)

KVA = V • A 3 1000 √

EXAMPLE 2: Assume a 3 phase 208/120 entrance service specified at 200A.

KVA = = 72

75kVA UPS should be selected.

208 • 200 • 1.7321000


Determining kVA From Power Profile of Equipment

Simple Addition of KVA Values

EQUIPMENT VOLTAGE /

PHASE LOAD 1 CPU 208 / 3 Phase .11 KVA1 Controller 208 / 3 Phase 12 Amps4 Disc 208 / 1 Phase 6 Amps Each1 Printer 208 / 1 Phase 5 Amps6 Terminal 120 / 1 Phase 4 Amps Each



EXAMPLE (cont)EQUIPMENT CALCULATION INDIVIDUALCPU None Required KVA = .11

Controller KVA = KVA = 4.3

Disc KVA = KVA = 1.25

Printer KVA = KVA = 1.0

Terminal KVA = KVA = 0.48

P

V •A • 31000

V •A1000

V •A1000

V •A1000

√



EQUIPMENT KVA EACHTOTAL

KVA LOAD1 CPU @ .11 0.111 Controller @ 4.3 4.34 Disc @ 1.25 5.01 Printer @ 1.0 1.06 Terminal @ 0.48 2.9

Total KVA 24.20

EXAMPLE (cont)


Determining kVA from Power Profile of Equipment

A 30kVA UPS could be selected as a minimum

To allow for growth a larger unit should be selected. This should be discussed with your customer to determine what size is needed.– Rule of thumb is 20% - 30%



Equipment Voltage Load Phase “A” Phase “B” Phase “C”

CPU 208v / 3 Phase 30.5 30.5 30.5 30.5

Controller 208v / 3 Phase 12.0A 12.0 12.0 12.0

Disc #1 208v / 1 Phase 6.0A 6.0 6.0

Disc#2 208v / 1 Phase 6.0A 6.0 6.0

Disc #3 208v / 1 Phase 6.0A 6.0 6.0

Disc #4 208v / 1 Phase 6.0A 6.0 6.0

Printer 208v / 1 Phase 5.0A 5.0 5.0

Terminal #1 120v / 1 Phase 4.0A 4.0

Load Calculations by Phase



Equipment Voltage Load Phase “A” Phase “B” Phase “C”






Total Phase Load 68.5 69.5 71.5

Load Calculations by Phase (continue)



Calculating the kVA from the most heavily-loaded phase (phase C):

kVA = 208V • 71.5A • 3 kVA = 25.81000

√

Load Calculations by Phase (continued)

A 30kVA UPS could be selected as a minimum


Something to take home

Single phase capacity– V x A = VA

– 120 x 100 = 12 Kva

Three phase capacity– V x A x 1.73 = VA

– 208 x 100 x 1.73 = 36 Kva


Something more to take home

Power factor = Kw / Kva– Kva = Kw / Pf

Must know kVA and kW to properly select UPS size– kW can be determined from PF and kVA

Maximum UPS output at rated power factor– 100Kva/80kW unit can be fully loaded at 80Kva if load PF is 1.0

The End

3-PH Power Basics

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