3-ph power basics
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3 Phase Power BasicsThomas GreerExecutive Director – Engineering ServicesTLG Services
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AgendaTerminologyBasic Electrical CircuitsBasic Power Calculations
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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!
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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
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What You Will Take HomeUnderstand basic terminology in electrical circuits and power systemsAble to perform basic power calculations
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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”.
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Direct CurrentDirect Current (DC) - Current flows in one directionCommon DC source - battery
DC Current
Time
Current
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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
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VoltageIs the electrical potential or force that causes current to flow in a circuit.Unit measure is the volt, abbreviated “V”.
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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”
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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
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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
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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
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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 = ?
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Are You Still There?Any Questions?
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AC Waveform - 3 Phase
90
150
210
270
330
120 180
240
300
360
A
C
BOne Cycle
Frequency# Cycles Per SecondHertz
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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
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Distorted Sinewave
Voltage Waveform with distortion caused by load withswitching SCR’s
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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.
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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
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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
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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.
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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
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CapacitorElectrical device that stores electrical energy.Does not allow instantaneous voltage changeCapacitance - storage capability of capacitor– Measured in “farads”
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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
-
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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”
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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
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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
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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
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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
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Still With Me?
Any Questions?
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Single Phase Systems
220/230/240V - 50 Hz 110/115/120V - 60 Hz load voltages may be obtained from these systems
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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
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Three Phase Systems
380/400/415
480V220/
N 480V (Line-to-Line)
Delta Connected SystemNo NeutralLine-To-Line Voltages Only
480V
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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
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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
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Three Phase SystemsWorldwide Voltages available– 60Hz
• 600/346V (Canada)• 480/277V• 208/120V• 220/127V (Mexico)
– 50Hz• 380/220V• 400/230V• 415/240V
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38
POWER CALCULATIONS
Putting it All Together
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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
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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
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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
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Determining kVA From Power Profile of Equipment
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
√
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Determining kVA From Power Profile of Equipment
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)
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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%
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Determining kVA from Power Profile of Equipment
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
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Determining kVA from Power Profile of Equipment
Equipment Voltage Load Phase “A” Phase “B” Phase “C”
Terminal #2 120v / 1 Phase 4.0A 4.0
Terminal #3 120v / 1 Phase 4.0A 4.0
Terminal #4 120v / 1 Phase 4.0A 4.0
Terminal #5 120v / 1 Phase 4.0A 4.0
Terminal #6 120v / 1 Phase 4.0A 4.0
Total Phase Load 68.5 69.5 71.5
Load Calculations by Phase (continue)
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Determining kVA from Power Profile of Equipment
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
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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
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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
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The End