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### 3-phase WYE example

Course: EE 369, Fall 2010
School: Mohawk
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Word Count: 139

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Industrial EE369 Automation 3-phase Wye System basics THREE PHASE BALANCED Y-CIRCUIT Given loads: Phase (A)=25 0 Phase (B)=25 0 Phase (C)=25 0 Line voltage = 600 Volts Phase sequence: ABC Using VBC as the reference, calculate all the voltages and currents. VAB=600 +120 Volts VBC=600 0 Volts VCA=600 -120 Volts VAN=347 90 Volts VBN=347 -30 Volts VCN=347 -150 Volts ILA=347 90 V 25 0 = 13.88 90 A ILB=347 -30 V...

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Industrial EE369 Automation 3-phase Wye System basics THREE PHASE BALANCED Y-CIRCUIT Given loads: Phase (A)=25 0 Phase (B)=25 0 Phase (C)=25 0 Line voltage = 600 Volts Phase sequence: ABC Using VBC as the reference, calculate all the voltages and currents. VAB=600 +120 Volts VBC=600 0 Volts VCA=600 -120 Volts VAN=347 90 Volts VBN=347 -30 Volts VCN=347 -150 Volts ILA=347 90 V 25 0 = 13.88 90 A ILB=347 -30 V 25 0 = 13.88 -30 A ILC=347 -150 V 25 0 = 13.88 -150 A IN=ILA+ILB+ILC=0 0 A THREE UNBALANCED PHASE Y-CIRCUIT Given loads: Phase (A)=15 0 Phase (B)=62 -88 Phase (C)=50 78 VL=208 Volts Phase sequence: ABC Using VAN as the reference, calculate all the voltages and currents. VAN=120 0 Volts VBN=120 -120 Volts VCN=120 -240 Volts VAB=208 30 Volts VBC=208 -90 Volts VCA=208 -210 Volts 1 EE369 Industrial Automation 3-phase Wye System basics ILA=120 0 V 15 0 = 8.0 0 A ILB=120 -120 V 62 -88 = 1.935 -32.0 A ILC=120 -240 V 50 78 = 2.4 42 A IN=ILA+ILB+ILC=11.4393 2.91 A 2
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Mohawk - EE - 369
EE369 Industrial Automation 3-Wire Edison Distribution Transformer Example Refer to the following diagram and calculate the currents I1, I2, I3 and In. Also determine the total secondary side (load) power.Solution: Vs1 = (208/2) / 4 = 26V Vs2 = (208/2) /
Mohawk - EE - 369
Mohawk - EE - 369
EE369 Industrial Automation Counters Practice Problems 1. Write a PLC program that controls three lights as follows: Pressing a N.O. pushbutton PB1 once turns L1 on and keeps it on Pressing PB1 twice turns L2 on and keeps it on Pressing PB1 three times tu
Mohawk - EE - 369
Mohawk College of Applied Arts and Technology Electrical and Computer Engineering Technology Department EE369 / 370 Industrial Automation T01 (Mon, Tue, Fri) This document shows the updated schedule for the lectures and the labs. The instructors shall att
Mohawk - EE - 369
EE369 Midterm 1 Practice Problem Solutions1. a) b)ZT = 220.2 -74.4 Assuming source voltage is 120V: Vr = 27.2 74.4 V Vc = 144.5 -15.6 V VL=29.4 154.4 V The voltage phasor diagram should be drawn as a closed loop to indicate KVLholds. 2. a) b) c) Ratio
Mohawk - EE - 369
EE369 Industrial Automation Quiz 11 PLC Applications Name: _ Date: _Develop a program that will display a 1 minute count-down sequence on a 2-digit 7segment LED display that is wired to an 8-point output card (O:003) when a N.O. pushbutton `Start' is pre
Mohawk - EE - 369
Mohawk - EE - 369
EE370 Industrial Automation Lab PLC Lab #2 This is a modification to the &quot;Boiler System&quot; problem in PLC lab #2 in the lab manual. Please follow this procedure instead of the one in the lab manual. 1 BOILER SYSTEM Hot water boiler systems commonly have saf
Mohawk - EE - 369
Create ladder logic to implement this project on the PLC5. A machine is used in an automotive plant to install tires onto rims to make finished wheels. The operation is as described below: An operator ensures that the correct style of rim and tire are in
Mohawk - EE - 369
Mohawk - EE - 369
Using the PLC5 system at your station, design the logic to solve the following problem: Access to a parking lot is to be controlled by a PLC. The parking lot can only hold 50 cars maximum. Agate at the entrance and another gate at the exit control the acc
Mohawk - EE - 369
Mohawk - EE - 369
Allen-Bradley1785 PLC-5 Programmable ControllersAddressing Reference ManualImportant User InformationBecause of the variety of uses for the products described in this publication, those responsible for the application and use of this control equipment
Mohawk - EE - 369
Allen BradleyClassic 1785 PLC 5 Family Programmable Controllers(Cat. No. 1785 series)Hardware Installation ManualImportant User InformationBecause of the variety of uses for the products described in this publication, those responsible for the applic
Mohawk - EE - 369
1785 PLC-5 Programmable Controllers Quick ReferenceStatus Bits. 3-1 Relay. 3-2 Timer . 3-5 Counter . 3-7 Compare . 3-9 Compute . 3-12 Logical . 3-22 Conversion . 3-24 Bit Modify and Move. 3-26 File . 3-28 Diagnostic . 3-30 Shift Register. 3-32 Sequencer
Mohawk - EE - 369
page 0FS = first scanT1 = ST2 AST1A T1C*B T3 T4 ST2 C+BST3T2 = ST1 B T3 = ST3 ( C B ) T4 = ST2 ( C + B ) ST1 = ( ST1 + T1 ) T2 + FS ST2 = ( ST2 + T2 + T3 ) T1 T4 ST3 = ( ST3 + T4 T1 ) T3BT2ST2A T1ST1Automating Manufacturing Systems T2C BBS
Mohawk - EE - 369
Mohawk - EE - 369
EE369 Industrial Automation PLC Practice Problems1. Develop a 24-hour digital clock that keeps track of hours, minutes and seconds. (Hint. Increment a counter every 60 seconds and another counter every minute.)2. When a motor is turned off because of an
Mohawk - EE - 369
EE369 Industrial Automation PLC references1. PLC-5 hardware installation manual:-Ch 1 pgs 1-1 to 1-6 Ch 8 pgs 8-1 to 8-5 Ch9 Ch 10 pgs 10-1 to 10-3 Ch 11 - troubleshooting2. PLC-5 quick instruction manual3. PLC-5 addressing reference manual:-pg 2 m
Mohawk - EE - 369
EE369 Industrial Automation PLC Sequential Practice Problem (Container example) A container is loaded with 50 items at a loading dock. The loading chute only opens if the container is at the loading dock. Once full, the container is moved forward by a con
Mohawk - EE - 369
POWER FACTOR CORRECTIONExisting installation208 Volts 60 Hz75 HP Motor PF=0.75P=75 HP x 746 W = 55.950 kW S=55.950 kW/0.75 = 74.600 kVA Q=74.600 kVA x sin 41.41 deg. = 49.3426 kVAR'sAfter PF correction What value of capacitor is needed to increase th
Mohawk - EE - 369
EE369 Industrial Automation Module 1 Single Phase Review Problems1. For figure 1, find the currents through the inductor and thecapacitor.Figure 1 2. For figure 2, find the current through the resistor.Figure 23. For figure 3, find the currents thoug
Mohawk - EE - 369
EE369 Industrial Automation Combo RLC Problem: A series-parallel circuit is shown below. Calculate the total circuit impedance Zt and the voltage across the resistor, VR. E = 90V @ 60Hz R = 600 C1 = 470F C2 = 200FSolution: Zt = 18.9 - 89.1 VVR = 63.13
Mohawk - EE - 369
EE369 Industrial Automation Complex numbers Problems: 1. Convert into rectangular coordinates:a) b)220 240120 552. Convert into polar coordinates: a) 50-15j b) -25+25j 3. Solve the following:a) b)35 - 90 + 25 - 90 70 80 / 100 - 30Solutions: 1. a) -
Mohawk - EE - 369
EE369 Industrial Automation Parallel RLC Problem: A parallel circuit is shown below. Calculate the total circuit impedance Zt and the current through the inductor, IL. E = 120V @ 60Hz R = 40 L = 100mH C = 200FSolution: Zt = 18.22 IL = 3.18 - 62.91 - 90
Mohawk - EE - 369
EE369 Industrial Automation Series RLC Problem: A series circuit is shown below. Calculate the total circuit impedance Zt and the voltage across the capacitor, Vc. E = 28V @ 60Hz R = 40 L = 10mH C = 20FSolution: Zt = 134.9 Vc = 27.52 - 72.75 - 17.25 V
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
Mohawk - EE - 369
EE369 Industrial Automation Relay Ladder Logic Practice ProblemsUse the Common Electronic Symbols chart and the notes form the lecture to draw the ladder logic diagram (control circuit) for the following problems.1. A three-phase motor is controlled usi
Mohawk - EE - 369
EE369 Relay Ladder Logic Practice problems 1. Motor M1 is controlled by a standard Start-Stop-Seal circuit as follows: when the start pushbutton is pressed, a red light is turned on and 10 seconds later, M1 starts to run and seals itself. The light stays
Mohawk - EE - 369
Design PLC5 ladder logic to simulate the following temperature data from a temperature sensor. Conditions: At start up the temperature is 20 C. After 5 seconds the temperature rises 10 Degrees. When the temperature reaches 30 C. it remains constant for 15
Mohawk - EE - 369
Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMP - EE357
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Mohawk - COMM - 1000345
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Mohawk - COMM - 1000345
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