20.2 Industrial Motor Control, Logic Circuits & PLC Automation

20.2 Industrial Motor Control, Logic Circuits & PLC Automation

Industrial motor control systems, magnetic contactors, reduced-voltage starting topologies, and Programmable Logic Controllers (PLCs) form vital practical components of the PRC Registered Electrical Engineer (REE) Licensure Examination. Three-phase induction motors drive the vast majority of industrial loads (pumps, compressors, conveyors, fans). Electrical engineers must design control circuits that safely start, stop, reverse, speed-control, and protect electric motors while complying with Philippine Electrical Code (PEC) guidelines.

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1. Industrial Motor Starting Techniques & Performance Comparisons

Direct-On-Line (DOL) starting of large 3-phase induction motors causes massive inrush current ($500\% \text{ to } 800\%$ of full-load current), creating severe line voltage dips that disturb neighboring facility loads. Reduced-voltage starting limits starting current to acceptable utility limits.

Comparison of Motor Starting Methods

Mathematical Formulations

Wye-Delta ($\,\text{Y}\text{--}\Delta$) Starting:

When connected in Delta (run mode), phase voltage equals line voltage ($V_{\text{phase},\Delta} = V_L$) and line current is $I_{\text{line},\Delta} = \sqrt{3} I_{\text{phase},\Delta} = \frac{\sqrt{3} V_L}{Z_{\text{st}}}$. During Wye starting, phase voltage is $V_{\text{phase},Y} = \frac{V_L}{\sqrt{3}}$, so phase current is $I_{\text{phase},Y} = \frac{V_L}{\sqrt{3} Z_{\text{st}}}$. Line current equals phase current:

$$I_{\text{line},Y} = \frac{V_L}{\sqrt{3} Z_{\text{st}}} = \frac{1}{3} \left( \frac{\sqrt{3} V_L}{Z_{\text{st}}} \right) = \frac{1}{3} I_{\text{line},\Delta}$$ $$\text{Starting Torque: } T_{\text{start}} \propto V_{\text{phase}}^2 \implies T_{\text{start},Y} = \left( \frac{1}{\sqrt{3}} \right)^2 T_{\text{start},\Delta} = \frac{1}{3} T_{\text{start},\Delta}$$

Autotransformer (Korndorfer) Starting:

For an autotransformer tap ratio $x$ (where $x = 0.50, 0.65, 0.80$):

• Motor terminal voltage: $V_m = x V_L$
• Motor starting current: $I_m = x I_{\text{DOL}}$
• Transformer primary line current (by power equality $V_L I_{\text{line}} = V_m I_m$): $$I_{\text{line}} = x \left( \frac{V_m}{V_L} \right) I_{\text{DOL}} = x^2 I_{\text{DOL}}$$
• Motor starting torque: $$T_{\text{start}} = x^2 T_{\text{DOL}}$$

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2. Hardwired Relay Logic & Motor Control Circuits

Elementary ladder diagrams represent control circuits using standard IEEE/NEMA symbols.

Standard Forward-Reverse Motor Starter Circuit

                  FORWARD / REVERSE MOTOR CONTROL LADDER DIAGRAM

   L1 |---[ STOP ]---+-+---[ F-START ]---+---[ R-NC ]---( F COIL )---+---[ TOR ]---| N
      |              | |                 |                            |
      |              | +---[ F-NO  ]-----+                            |
      |              |                                                |
      |              +-----[ R-START ]---+---[ F-NC ]---( R COIL )----+
      |                |                 |
      |                +---[ R-NO  ]-----+

Essential Control Elements:

1. Stop Pushbutton: Normally Closed (NC) contact connected in series with the entire control circuit.
2. Start Pushbutton: Normally Open (NO) momentary contact.
3. Holding (Seal-In) Contact: NO auxiliary contact of contactor ($F\text{-NO}$ or $R\text{-NO}$) connected in parallel across the Start button to maintain coil energization after button release.
4. Electrical Interlocking: NC contact of Forward contactor ($F\text{-NC}$) wired in series with Reverse coil ($R$), and vice versa. Prevents shorting two line phases if both pushbuttons are pressed simultaneously.
5. Thermal Overload Relay (TOR): NC contact ($95\text{--}96$) placed in series with contactor coils. Opens automatically when sustained motor overcurrent causes bimetallic strip bending.

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3. Programmable Logic Controllers (PLCs)

A Programmable Logic Controller (PLC) is a ruggedized industrial digital computer that continuously monitors input signals from sensors/switches, executes a stored control program, and actuates output devices (contactors, solenoids, indicator lamps).

PLC Hardware Architecture

• CPU (Central Processing Unit): Microprocessor executing system firmware and user ladder logic.
• Input Module: Converts field AC/DC signals ($24\ \text{V DC}, 120\ \text{V AC}$) into internal logic voltages using optocouplers for electrical noise isolation.
• Output Module: Drives field loads via relays, triacs, or solid-state transistors.
• Power Supply: Converts plant AC voltage into regulated $24\ \text{V DC}$ bus power.

The Four-Stage PLC Scan Cycle

                             PLC SCAN CYCLE REPEAT LOOP

       +-------------------------------------------------------------------+
       |                                                                   |
       v                                                                   |
  +----------+         +---------------+         +-----------+       +-----------+
  | 1. Input | ------> | 2. Logic Exec | ------> | 3. Output | ----> | 4. System |
  |   Scan   |         |   (Program)   |         |   Scan    |       | Overhead  |
  +----------+         +---------------+         +-----------+       +-----------+

1. Input Scan: CPU reads state of physical input terminals and updates the Input Image Table memory.
2. Logic Execution Scan: CPU executes ladder rungs sequentially from top to bottom, evaluating conditions using Input Image Table values and updating the Output Image Table.
3. Output Scan: CPU writes Output Image Table states simultaneously to physical output module terminals.
4. System Diagnostics / Overhead: Performs self-testing, communication servicing, and watchdog timer reset.

PLC Scan Time ($T_{\text{scan}}$): Total duration required to complete one full cycle (typically $1 \text{ to } 20\ \text{milliseconds}$). High-speed events occurring faster than $T_{\text{scan}}$ require hardware interrupt inputs.

IEC 61131-3 Programming Languages

• Ladder Diagram (LD): Graphical language mimicking hardwired relay logic rungs.
• Function Block Diagram (FBD): Graphical blocks representing logic functions (AND, OR, TON, PID).
• Structured Text (ST): High-level block-structured text language (similar to Pascal/C).
• Instruction List (IL): Low-level assembly-like language.
• Sequential Function Chart (SFC): Flowchart-based state machine programming.

Standard PLC Timers & Counters

• Timer On-Delay (TON): Output turns ON after input rung remains continuous for Preset Time ($PT$).
• Timer Off-Delay (TOF): Output turns OFF after input rung goes false for Preset Time ($PT$).
• Retentive Timer (RTO): Retains accumulated time ($ACC$) even if input rung turns false; requires explicit Reset ($RST$) instruction.
• Count Up (CTU) / Count Down (CTD): Increments/decrements accumulated count on false-to-true rung transitions.

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Solved Board Exam Examples

Example 1: Wye-Delta Starter Currents & Torque Calculation

Problem: A $3$-phase, $460\ \text{V}$, $60\ \text{Hz}$, $50\ \text{HP}$ squirrel-cage induction motor has a rated full-load current of $65\ \text{A}$ and a Direct-On-Line (DOL) starting current of $420\ \text{A}$ with a starting torque of $280\ \text{N}\cdot\text{m}$. Calculate: (a) Line starting current under Wye-Delta starting, (b) Motor starting torque under Wye-Delta starting, and (c) Percentage reduction in starting line current compared to DOL.

Solution:

1. Calculate line starting current under Wye-Delta starting ($I_{\text{start, Y}}$): $$I_{\text{start, Y}} = \frac{1}{3} I_{\text{DOL}} = \frac{420\ \text{A}}{3} = 140.0\ \text{A}$$
2. Calculate motor starting torque under Wye-Delta starting ($T_{\text{start, Y}}$): $$T_{\text{start, Y}} = \frac{1}{3} T_{\text{DOL}} = \frac{280\ \text{N}\cdot\text{m}}{3} = 93.33\ \text{N}\cdot\text{m}$$
3. Calculate percentage reduction in line current: $$\% \text{Reduction} = \left( 1 - \frac{1}{3} \right) \times 100\% = 66.67\%$$

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Example 2: Autotransformer Starter Tap Calculation

Problem: A $2300\ \text{V}$, $3$-phase induction motor draws a DOL starting current of $1,200\ \text{A}$ from the line. The facility distribution system limits starting line current to a maximum of $500\ \text{A}$. Determine: (a) The maximum allowable autotransformer voltage tap $x$, and (b) The resulting starting torque as a percentage of DOL starting torque.

Solution:

1. Use autotransformer line current equation ($I_{\text{line}} = x^2 I_{\text{DOL}}$): $$x^2 I_{\text{DOL}} \le 500\ \text{A} \implies x^2 (1200) = 500$$ $$x^2 = \frac{500}{1200} = 0.41667 \implies x = \sqrt{0.41667} = 0.6455 \quad (64.55\%$$
2. Select nearest standard lower autotransformer tap: $$x_{\text{standard}} = 0.65 \quad (65\%\text{ tap yields } I_{\text{line}} = (0.65)^2 \times 1200 = 507\ \text{A})$$ To strictly respect the $500\ \text{A}$ limit, select the $50\%$ tap ($x = 0.50$): $$I_{\text{line, 50\%}} = (0.50)^2 \times 1200 = 300\ \text{A}$$
3. Calculate starting torque percentage for $50\%$ tap: $$T_{\text{start}} = x^2 T_{\text{DOL}} = (0.50)^2 T_{\text{DOL}} = 0.25 T_{\text{DOL}} \quad (25.0\%\text{ of DOL torque})$$