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AC Motor Control: Contactors, Star–Delta & VFDs All parts · 15 pages 1 · Motors & Low-Voltage Supply 2 · Control Devices & Diagrams 3 · Starter & Control Circuits 4 · Variable Frequency Drives 5 · Design, Installation & Lab
// Chapter 21 · AC Motor Control · electrician → student → engineer
AC Motor Control with Magnetic Contactors and VFDs
Three-phase induction motors use more than half of industrial electricity. This course goes from how the motor turns, through reading the nameplate, connecting star or delta, choosing contactors and overloads and reading drawings, to designing a star–delta starter and setting up a VFD. Every example uses the real 230/400 V 50 Hz supply of Thailand, Laos, Cambodia, Vietnam, Myanmar, Malaysia and Singapore. There are formulas with citable IDs and worked examples.
In 30 seconds
An induction motor turns slightly slower than its field: n s = 120 f / p n_s = 120f/p n s = 120 f / p = 1500 rpm (4-pole, 50 Hz), actual ≈ 1470 rpm
Direct-on-line starting draws 6–8 × rated current, so voltage dip and the starting method matter
On 400 V: a 400/690 V motor runs in Δ and can be star–delta started · a 230/400 V motor runs in Y only
Star–delta cuts current and torque to 1/3 · overload at 0.58 In · needs interlocks and a dead time
A VFD varies f and V together (V/f): soft start, full torque and pump/fan energy savings
Learning path by level
Level 1 · electrician · vocational
Wire it right and safely Nameplates, Y/Δ terminals, panel devices, self-holding circuits, reversing, star–delta
Connections → Devices → DOL → Star–delta
Level 2 · technician · supervisor
Select devices and troubleshoot AC-3 contactor size, overload setting, cable size, voltage drop, VFDs and fault tables
Drawings → VFD setup → Design · faults
Level 3 · engineering student · engineer
Understand it with mathematics Equivalent circuit, slip, torque T ∝ V 2 T \propto V^2 T ∝ V 2 , run-up dynamics J d ω / d t = T m − T L J\,d\omega/dt = T_m - T_L J d ω / d t = T m − T L , V/f and the affinity laws
Induction motor → Starting methods compared → VFD principles
Try it: start a 7.5 kW motor four ways
Pick a tab, press START and watch the plot: blue is line current, green is speed and orange is torque. Start direct-on-line first, then switch to star–delta and notice the starting current falling to about 1/3 and the current peak at the change to delta. Then raise the load until star can no longer accelerate it, and open the full simulator with worksheets .
JavaScript is needed for the simulator
Key equations used throughout
H1 Synchronous speed & slip
n s = 120 f p s = n s − n n s n_s = \frac{120 f}{p} \qquad s = \frac{n_s - n}{n_s} n s = p 120 f s = n s n s − n
H2 Motor rated current
I n = P o u t 3 V L η cos φ I_n = \frac{P_{out}}{\sqrt{3}\, V_L\, \eta\, \cos\varphi} I n = 3 V L η cos φ P o u t
H3 Star–delta reduction
I L , Y = 1 3 I L , Δ T Y = 1 3 T Δ I r , O L = I n 3 ≈ 0.58 I n I_{L,Y} = \tfrac{1}{3} I_{L,\Delta} \qquad T_Y = \tfrac{1}{3} T_\Delta \qquad I_{r,OL} = \frac{I_n}{\sqrt{3}} \approx 0.58\, I_n I L , Y = 3 1 I L , Δ T Y = 3 1 T Δ I r , O L = 3 I n ≈ 0.58 I n
H4 V/f and affinity laws
V f = 400 50 = 8 V/Hz P 2 P 1 = ( n 2 n 1 ) 3 \frac{V}{f} = \frac{400}{50} = 8\ \text{V/Hz} \qquad \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3 f V = 50 400 = 8 V/Hz P 1 P 2 = ( n 1 n 2 ) 3
The supply used in this course: 230/400 V 50 Hz
All seven countries use 50 Hz three-phase four-wire systems with line voltages of 380–415 V, following IEC 60038 (230/400 V). Motor-control gear rated 380–415 V 50 Hz is therefore usable across the region. What differs is the declared site voltage, the earthing system and the utility's connection rules — details in 230/400 V 50 Hz supply .
Country Phase / line voltage found on site Frequency Notes for motor work
Thailand 230 / 400 V (formerly 220/380 V) 50 Hz MEA/PEA · TN-C-S · EIT wiring code
Laos 230 / 400 V (220/380 V also found) 50 Hz Grid interconnected with Thailand, IEC equipment
Cambodia 230 / 400 V (220/380 V also found) 50 Hz Measure the real voltage before setting a VFD
Vietnam 220 / 380 V 50 Hz 400 V motors work; 5 % undervoltage raises current slightly
Myanmar 230 / 400 V 50 Hz Voltage varies widely: use voltage and phase relays
Malaysia 230 / 400 V (formerly 240/415 V) 50 Hz Much older 415 V-rated equipment still in service
Singapore 230 / 400 V 50 Hz Installation to SS 638 (based on IEC 60364)
“The same standard” means the same voltage level and frequency so equipment is interchangeable — not that every site has the same voltage. Measure the line voltage before setting overloads and drives.
How to choose a starting method
Method Starting current Starting torque Speed control Good for
Direct-on-line 6–8 In Highest (1.5–2.5 Tn) No Small motors, stiff supply, heavy starts
Star–delta ≈ 2–2.7 In 1/3 of DOL No Pumps, fans, compressors starting unloaded
Soft starter 2–4 In (set) Falls as (I/Is)² No Pumps needing a soft stop (water hammer), conveyors
VFD ≤ 1–1.5 In Up to 1.5 Tn Yes Energy-saving pumps and fans, conveyors, process machines
Every page in this course
Part 1 Motors & Low-Voltage Supply Three-phase induction motor principles, nameplates, the 230/400 V 50 Hz systems of the region, and star/delta terminal connections
n_s 21.1
Three-phase induction motors for industry: rotating field, slip, torque–speed curve, starting current and the nameplate
Three-phase induction motors for industry
Rotating field n_s = 120f/p, slip, equivalent circuit, torque–speed curve, 6–8× starting current, IE2–IE4 efficiency (IEC 60034-30-1), reading the nameplate and rated current from kW
400 V 21.2
Low-voltage 230/400 V 50 Hz supply in Thailand, Laos, Cambodia, Vietnam, Myanmar, Malaysia and Singapore: line and phase voltage, earthing systems and their effect on motors
Low-voltage 230/400 V 50 Hz supply in Thailand, Laos, Cambodia, Vietnam, Myanmar, Malaysia and Singapore
IEC 60038, declared voltages per country from 220/380 to 230/400 V, ±10 % tolerance, TN-C-S and TT earthing, phase sequence, voltage unbalance and 60 Hz motors on 50 Hz
Y/Δ 21.3
Star and delta motor connections: the U1 V1 W1 / W2 U2 V2 terminal box, 230/400 V vs 400/690 V motors, and reversing
Star and delta motor connections
Voltage per winding, line vs phase current, terminal-box links, choosing the connection from the nameplate, which motors can be star-delta started, checking rotation and the mistakes that burn motors
Part 2 Control Devices & Diagrams AC-3 contactors, overload relays, MPCBs, breakers, timers, push-buttons, IEC 60617 symbols and IEC 81346 device codes
K 21.4
Magnetic contactors, overload relays, motor-protection breakers and control devices: AC-3, trip class, timers, push-buttons and pilot lights
Magnetic contactors, overload relays, motor-protection breakers and control devices
Contactor construction, coil, main and auxiliary contacts, utilization categories AC-1 AC-3 AC-4 (IEC 60947-4-1), class 10/20 overloads, MPCBs, aM fuses, Type 1/2 coordination, timers and IEC 60204-1 colours
⊣⊢ 21.5
Reading motor-control drawings: IEC 60617 vs NEMA symbols, IEC 81346 device codes, terminal markings, wire numbers and cross-references
Reading motor-control drawings
Power vs control circuits, contact, coil and timer symbols, terminal markings 1/L1–2/T1, A1–A2, 13–14, 21–22, device letters -Q -K -F -S -H -M, cross-referencing and European, American and Japanese drawing styles
Part 3 Starter & Control Circuits DOL, reversing, a detailed star-delta starter, soft starters and field-proven applied circuits
DOL 21.6
Direct-on-line starter: power circuit, self-holding control circuit, two-location control, jogging and step-by-step device selection
Direct-on-line starter
Self-holding (latching) contact, why STOP is a normally-closed button in series, multi-location control, jog, pilot lights, and selecting the MPCB, contactor and cable for a 7.5 kW 400 V motor
⇄ 21.7
Three-phase motor reversing circuits: swapping two phases, electrical, mechanical and push-button interlocks, and automatic reversing with limit switches
Three-phase motor reversing circuits
Why swapping L1 and L3 reverses rotation, three layers of interlock against a phase-to-phase short, reversing via STOP vs direct, roller doors, hoists, conveyors and limit switches
Y→Δ 21.8
The star–delta starter in detail: power and control circuits, 0.58·In contactor sizing, overload setting, timer and changeover time
The star–delta starter in detail
Current and torque reduced to 1/3, contactors KM1 KM2 KM3, overload at 0.58·In, timer setting, Y–Δ dead time, transition current peaks, phase order on the delta side, closed-transition starters and six-wire cabling
τm 21.9
Star–Delta Analysis: Timer Setting, Star or Delta Operation, Currents, Energy and Numerical Research
Star–Delta Analysis
Timer from the mechanical time constant τm = J·ωs/Tn, dead time and residual voltage, the lagging KM3 order, closed transition, I²t, rotor heat, voltage dip, running in star at light load, case studies, past, present and future, and a dynamic simulation lab
∠α 21.10
Soft starters, autotransformer starters and a comparison of motor starting methods: current, torque, voltage dip and cost
Soft starters, autotransformer starters and a comparison of motor starting methods
Thyristor phase-angle control, voltage ramp, current limit, torque control, bypass contactor, soft stop against water hammer, autotransformer starting and a starting-method selection table
⚙ 21.11
Applied motor-control circuits from real plants: sequence starting, float-switch pumps, duty/standby alternation, pressure-switch compressors, two-speed motors and phase-failure relays
Applied motor-control circuits from real plants
Timed sequence starting, conveyor chains, float or electrode pump control, duty/standby alternation, Hand-Off-Auto, compressors, Dahlander two-speed motors and phase-loss/phase-reversal relays
Part 4 Variable Frequency Drives V/f and PWM principles, wiring, parameters, pump and fan energy savings, and applications
V/f 21.12
Variable frequency drives for motors: rectifier, DC bus, IGBT PWM, V/f and vector control, and correct wiring
Variable frequency drives for motors
540–565 V DC bus from 400 V, PWM, constant V/f and field weakening, sensorless vector, no contactor or capacitor on the output, shielded cable and EMC, cable length and dv/dt, RCD Type B, braking resistor, and single-phase-in three-phase-out drives
n³ 21.13
Setting up and applying VFDs: basic parameters, multi-speed, 0–10 V / 4–20 mA, constant-pressure PID pumping and energy savings from the affinity laws
Setting up and applying VFDs
A 12-step commissioning procedure, accel/decel times, command source, preset speeds, analog inputs, constant-pressure pumps, conveyors, fans, P ∝ n³, energy-saving and payback calculation, temperature derating and bypass
Part 5 Design, Installation & Lab Device selection, cable sizing, protection coordination, panels, testing, troubleshooting and the starter simulator
✓ 21.14
Designing and installing motor circuits: breaker selection, cable size, starting voltage dip, control panels, megger testing and a troubleshooting table
Designing and installing motor circuits
A 10-step design procedure, cable sizing for current and voltage drop, Thai EIT wiring code and IEC 60364, IEC 61439 panels, IP rating, panel heat, insulation-resistance testing, loop checks and common faults with causes
▶ 21.15
Motor starter simulator: DOL, reversing, star–delta, soft starter and VFD with live current, torque, speed and contacts
Motor starter simulator
Press START and STOP and watch power flow through the control circuit, contactors pull in, the timer run and current, torque and speed plotted in real time; inject faults such as overload and phase loss and compare starting methods
Prefixes: H overview · M motor · S supply · C connections · D devices · G drawings · L DOL · R reversing · Y star–delta · T soft starters · A applied circuits · V VFD · P VFD applications · K design
Find a formula
108 formulas
All pages H · Overview M · Induction motor S · 230/400 V 50 Hz supply C · Star & delta connections D · Contactors & overloads G · Symbols & reading diagrams L · Direct-on-line (DOL) R · Forward–reverse Y · Star–delta starter E · Star–delta analysis · research T · Soft starters & autotransformers A · Applied circuits V · VFD principles P · VFD setup & applications K · Design, installation & troubleshooting
H1 Synchronous speed & slip n s = 120 f p s = n s − n n s n_s = \frac{120 f}{p} \qquad s = \frac{n_s - n}{n_s} n s = p 120 f s = n s n s − n
H2 Motor rated current I n = P o u t 3 V L η cos φ I_n = \frac{P_{out}}{\sqrt{3}\, V_L\, \eta\, \cos\varphi} I n = 3 V L η cos φ P o u t
H3 Star–delta reduction I L , Y = 1 3 I L , Δ T Y = 1 3 T Δ I r , O L = I n 3 ≈ 0.58 I n I_{L,Y} = \tfrac{1}{3} I_{L,\Delta} \qquad T_Y = \tfrac{1}{3} T_\Delta \qquad I_{r,OL} = \frac{I_n}{\sqrt{3}} \approx 0.58\, I_n I L , Y = 3 1 I L , Δ T Y = 3 1 T Δ I r , O L = 3 I n ≈ 0.58 I n
H4 V/f and affinity laws V f = 400 50 = 8 V/Hz P 2 P 1 = ( n 2 n 1 ) 3 \frac{V}{f} = \frac{400}{50} = 8\ \text{V/Hz} \qquad \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3 f V = 50 400 = 8 V/Hz P 1 P 2 = ( n 1 n 2 ) 3
M1 Synchronous speed n s = 120 f p rpm ( p = number of poles ) n_s = \frac{120\, f}{p}\ \ \text{rpm} \qquad (p = \text{number of poles}) n s = p 120 f rpm ( p = number of poles )
M2 Slip s = n s − n n s n = n s ( 1 − s ) s = \frac{n_s - n}{n_s} \qquad n = n_s(1 - s) s = n s n s − n n = n s ( 1 − s )
M3 Rotor frequency
M4 Torque from power and speed T = P ω = 60 P 2 π n ⇒ T [ N ⋅ m ] = 9550 P [ kW ] n [ rpm ] T = \frac{P}{\omega} = \frac{60\,P}{2\pi n} \;\;\Rightarrow\;\; T\,[\text{N}{\cdot}\text{m}] = \frac{9550\, P\,[\text{kW}]}{n\,[\text{rpm}]} T = ω P = 2 πn 60 P ⇒ T [ N ⋅ m ] = n [ rpm ] 9550 P [ kW ]
M5 Power split in the rotor P a g = 3 I 2 ′ 2 R 2 ′ s P c u 2 = s P a g P m e c h = ( 1 − s ) P a g T = P a g ω s P_{ag} = 3 I_2'^2 \frac{R_2'}{s} \qquad P_{cu2} = s\,P_{ag} \qquad P_{mech} = (1 - s)\,P_{ag} \qquad T = \frac{P_{ag}}{\omega_s} P a g = 3 I 2 ′2 s R 2 ′ P c u 2 = s P a g P m ec h = ( 1 − s ) P a g T = ω s P a g
M6 Rated current I n = P o u t 3 V L η cos φ P i n = P o u t η I_n = \frac{P_{out}}{\sqrt{3}\,V_L\,\eta\,\cos\varphi} \qquad P_{in} = \frac{P_{out}}{\eta} I n = 3 V L η cos φ P o u t P in = η P o u t
M7 Starting current I s t = ( I a I n ) I n I a I n ≈ 6 – 8 I_{st} = \left(\frac{I_a}{I_n}\right) I_n \qquad \frac{I_a}{I_n} \approx 6\text{–}8 I s t = ( I n I a ) I n I n I a ≈ 6 – 8
M8 Torque versus voltage T ( s ) ∝ V 2 T 2 T 1 = ( V 2 V 1 ) 2 I ( s ) ∝ V T(s) \propto V^2 \qquad \frac{T_2}{T_1} = \left(\frac{V_2}{V_1}\right)^2 \qquad I(s) \propto V T ( s ) ∝ V 2 T 1 T 2 = ( V 1 V 2 ) 2 I ( s ) ∝ V
S1 Line and phase voltage V L = 3 V p h 230 × 3 = 398 ≈ 400 V V p e a k = 2 V r m s V_L = \sqrt{3}\,V_{ph} \qquad 230 \times \sqrt3 = 398 \approx 400\ \text{V} \qquad V_{peak} = \sqrt2\, V_{rms} V L = 3 V p h 230 × 3 = 398 ≈ 400 V V p e ak = 2 V r m s
S2 Three-phase power P = 3 V L I L cos φ S = 3 V L I L P = \sqrt{3}\,V_L I_L \cos\varphi \qquad S = \sqrt{3}\,V_L I_L P = 3 V L I L cos φ S = 3 V L I L
S3 Reversing rotation ( L 1 , L 2 , L 3 ) → ( L 3 , L 2 , L 1 ) ⇒ n s → − n s (L1, L2, L3) \to (L3, L2, L1) \;\Rightarrow\; n_s \to -n_s ( L 1 , L 2 , L 3 ) → ( L 3 , L 2 , L 1 ) ⇒ n s → − n s
S4 Voltage unbalance (NEMA working definition) % V U = max ∣ V L − V ˉ L ∣ V ˉ L × 100 V ˉ L = V 12 + V 23 + V 31 3 \%VU = \frac{\max |V_{L} - \bar V_L|}{\bar V_L} \times 100 \qquad \bar V_L = \frac{V_{12} + V_{23} + V_{31}}{3} % V U = V ˉ L max ∣ V L − V ˉ L ∣ × 100 V ˉ L = 3 V 12 + V 23 + V 31
S5 Current unbalance estimate % I u n b ≈ I s t I n % V U ≈ ( 6 – 10 ) % V U \%I_{unb} \approx \frac{I_{st}}{I_n}\,\%VU \approx (6\text{–}10)\,\%VU % I u nb ≈ I n I s t % V U ≈ ( 6 – 10 ) % V U
S6 60 Hz motor on 50 Hz Φ ∝ V f V 50 = V 60 ⋅ 50 60 n 50 = 5 6 n 60 P 50 ≈ 5 6 P 60 \Phi \propto \frac{V}{f} \qquad V_{50} = V_{60}\cdot\frac{50}{60} \qquad n_{50} = \frac{5}{6} n_{60} \qquad P_{50} \approx \frac{5}{6} P_{60} Φ ∝ f V V 50 = V 60 ⋅ 60 50 n 50 = 6 5 n 60 P 50 ≈ 6 5 P 60
C1 Star (Y) V p h = V L 3 I L = I p h V_{ph} = \frac{V_L}{\sqrt{3}} \qquad I_L = I_{ph} V p h = 3 V L I L = I p h
C2 Delta (Δ) V p h = V L I L = 3 I p h V_{ph} = V_L \qquad I_L = \sqrt{3}\, I_{ph} V p h = V L I L = 3 I p h
C3 Power in either connection P = 3 V p h I p h cos φ = 3 V L I L cos φ P = 3\,V_{ph}\,I_{ph}\cos\varphi = \sqrt{3}\,V_L\,I_L\cos\varphi P = 3 V p h I p h cos φ = 3 V L I L cos φ
C4 Same windings, same supply: Y vs Δ V p h , Y V p h , Δ = 1 3 ⇒ I L , Y I L , Δ = 1 3 T Y T Δ = 1 3 \frac{V_{ph,Y}}{V_{ph,\Delta}} = \frac{1}{\sqrt3} \;\Rightarrow\; \frac{I_{L,Y}}{I_{L,\Delta}} = \frac{1}{3} \qquad \frac{T_Y}{T_\Delta} = \frac{1}{3} V p h , Δ V p h , Y = 3 1 ⇒ I L , Δ I L , Y = 3 1 T Δ T Y = 3 1
C5 Direction of rotation L 1 → U 1 , L 2 → V 1 , L 3 → W 1 ⇒ CW (viewed from shaft end) swap 2 lines ⇒ CCW L1\!\to\!U1,\ L2\!\to\!V1,\ L3\!\to\!W1 \;\Rightarrow\; \text{CW (viewed from shaft end)} \qquad \text{swap 2 lines} \Rightarrow \text{CCW} L 1 → U 1 , L 2 → V 1 , L 3 → W 1 ⇒ CW (viewed from shaft end) swap 2 lines ⇒ CCW
D1 Coil operating range 0.85 U c ≤ U p i c k - u p ≤ 1.1 U c U d r o p - o u t ≈ ( 0.2 – 0.75 ) U c 0.85\,U_c \le U_{pick\text{-}up} \le 1.1\,U_c \qquad U_{drop\text{-}out} \approx (0.2\text{–}0.75)\,U_c 0.85 U c ≤ U p i c k - u p ≤ 1.1 U c U d ro p - o u t ≈ ( 0.2 – 0.75 ) U c
D2 Contactor selection I e ( AC-3 , 400 V ) ≥ I n for AC-4 duty derate to about 0.3 – 0.5 I e ( AC-3 ) I_e(\text{AC-3},\,400\ \text{V}) \ge I_n \qquad \text{for AC-4 duty derate to about } 0.3\text{–}0.5\,I_e(\text{AC-3}) I e ( AC-3 , 400 V ) ≥ I n for AC-4 duty derate to about 0.3 – 0.5 I e ( AC-3 )
D3 Overload setting I r = I n ( DOL, reversing, soft starter ) I r = I n 3 ≈ 0.58 I n ( star–delta, in the phase-current path ) I_r = I_n \;(\text{DOL, reversing, soft starter}) \qquad I_r = \frac{I_n}{\sqrt3} \approx 0.58\,I_n \;(\text{star–delta, in the phase-current path}) I r = I n ( DOL, reversing, soft starter ) I r = 3 I n ≈ 0.58 I n ( star–delta, in the phase-current path )
D4 Trip class check t t r i p ( 7.2 I r ) ≤ class [s] t s t a r t < 0.8 × t t r i p ( I s t / I r ) t_{trip}(7.2\,I_r) \le \text{class [s]} \qquad t_{start} < 0.8 \times t_{trip}(I_{st}/I_r) t t r i p ( 7.2 I r ) ≤ class [s] t s t a r t < 0.8 × t t r i p ( I s t / I r )
D5 Breaking capacity I c u ≥ I k , m a x at the point of installation I k ≈ S t r 3 V L u k ( at the transformer terminals ) I_{cu} \ge I_{k,max}\ \text{at the point of installation} \qquad I_{k} \approx \frac{S_{tr}}{\sqrt3\,V_L\,u_k} \;(\text{at the transformer terminals}) I c u ≥ I k , ma x at the point of installation I k ≈ 3 V L u k S t r ( at the transformer terminals )
D6 No nuisance trip at start I m > ( 1.5 – 2 ) I s t ≈ ( 10 – 13 ) I n ( allowing for the asymmetric first half-cycle inrush ) I_{m} > (1.5\text{–}2)\,I_{st} \approx (10\text{–}13)\,I_n \qquad (\text{allowing for the asymmetric first half-cycle inrush}) I m > ( 1.5 – 2 ) I s t ≈ ( 10 – 13 ) I n ( allowing for the asymmetric first half-cycle inrush )
D7 Control transformer sizing S t r ≥ ∑ S h o l d + S i n r u s h , m a x check voltage dip at pick-up ≤ 5 – 10 % S_{tr} \ge \sum S_{hold} + S_{inrush,\,max} \qquad \text{check voltage dip at pick-up} \le 5\text{–}10\,\% S t r ≥ ∑ S h o l d + S in r u s h , ma x check voltage dip at pick-up ≤ 5 – 10 %
G1 Series = AND
G2 Parallel = OR
G3 NC contact = NOT
G4 Self-holding (seal-in) K 1 = S 0 ‾ ⋅ F 2 ‾ ⋅ ( S 1 + K 1 ) K1 = \overline{S0}\cdot\overline{F2}\cdot(S1 + K1) K 1 = S 0 ⋅ F 2 ⋅ ( S 1 + K 1 )
L1 DOL starting current I s t = k s t I n k s t = I a / I n ≈ 6 – 8 I_{st} = k_{st}\,I_n \qquad k_{st} = I_a/I_n \approx 6\text{–}8 I s t = k s t I n k s t = I a / I n ≈ 6 – 8
L2 Voltage dip at the transformer Δ U % ≈ S s t S s c + S s t × 100 S s t = 3 V L I s t S s c = S t r u k \Delta U\% \approx \frac{S_{st}}{S_{sc} + S_{st}}\times 100 \qquad S_{st} = \sqrt3\,V_L\,I_{st} \qquad S_{sc} = \frac{S_{tr}}{u_k} Δ U % ≈ S sc + S s t S s t × 100 S s t = 3 V L I s t S sc = u k S t r
L3 Cable voltage drop Δ V = 3 I L ( R cos φ + X sin φ ) cos φ s t a r t ≈ 0.3 \Delta V = \sqrt3\, I\, L\,(R\cos\varphi + X\sin\varphi) \qquad \cos\varphi_{start} \approx 0.3 Δ V = 3 I L ( R cos φ + X sin φ ) cos φ s t a r t ≈ 0.3
L4 Seal-in logic K 1 new = F 2 ‾ ⋅ S 0 ‾ ⋅ ( S 1 + K 1 old ) K1_{\text{new}} = \overline{F2}\cdot\overline{S0}\cdot(S1 + K1_{\text{old}}) K 1 new = F 2 ⋅ S 0 ⋅ ( S 1 + K 1 old )
R1 Phase swap K1: ( L 1 , L 2 , L 3 ) → ( U 1 , V 1 , W 1 ) K2: ( L 3 , L 2 , L 1 ) → ( U 1 , V 1 , W 1 ) \text{K1: } (L1, L2, L3)\to(U1, V1, W1) \qquad \text{K2: } (L3, L2, L1)\to(U1, V1, W1) K1: ( L 1 , L 2 , L 3 ) → ( U 1 , V 1 , W 1 ) K2: ( L 3 , L 2 , L 1 ) → ( U 1 , V 1 , W 1 )
R2 Fault if both close I k 2 = V L 1 L 3 2 Z s r c = 3 2 I k 3 ≈ 0.87 I k 3 I_{k2} = \frac{V_{L1L3}}{2 Z_{src}} = \frac{\sqrt3}{2} I_{k3} \approx 0.87\,I_{k3} I k 2 = 2 Z src V L 1 L 3 = 2 3 I k 3 ≈ 0.87 I k 3
R3 Interlocked logic K 1 = S 0 ‾ F 2 ‾ ( S 1 + K 1 ) S 2 ‾ K 2 ‾ K 2 = S 0 ‾ F 2 ‾ ( S 2 + K 2 ) S 1 ‾ K 1 ‾ K1 = \overline{S0}\,\overline{F2}\,(S1 + K1)\,\overline{S2}\,\overline{K2} \qquad K2 = \overline{S0}\,\overline{F2}\,(S2 + K2)\,\overline{S1}\,\overline{K1} K 1 = S 0 F 2 ( S 1 + K 1 ) S 2 K 2 K 2 = S 0 F 2 ( S 2 + K 2 ) S 1 K 1
R4 Slip during plugging s = − n s − n − n s = 1 + n n s ≈ 2 ( n ≈ n s ) s = \frac{-n_s - n}{-n_s} = 1 + \frac{n}{n_s} \approx 2 \quad (n \approx n_s) s = − n s − n s − n = 1 + n s n ≈ 2 ( n ≈ n s )
R5 Rotor heat per operation (load with no torque) E s t a r t = 1 2 J ω s 2 E p l u g ( ω s → 0 ) = 3 2 J ω s 2 E r e v e r s e ( ω s → − ω s ) = 4 ⋅ 1 2 J ω s 2 E_{start} = \tfrac12 J\omega_s^2 \qquad E_{plug\,(\omega_s \to 0)} = \tfrac32 J\omega_s^2 \qquad E_{reverse\,(\omega_s \to -\omega_s)} = 4\cdot\tfrac12 J\omega_s^2 E s t a r t = 2 1 J ω s 2 E pl ug ( ω s → 0 ) = 2 3 J ω s 2 E re v erse ( ω s → − ω s ) = 4 ⋅ 2 1 J ω s 2
Y1 Starting current in star I L , Y = I L , Δ 3 = k s t I n 3 I_{L,Y} = \frac{I_{L,\Delta}}{3} = \frac{k_{st}\,I_n}{3} I L , Y = 3 I L , Δ = 3 k s t I n
Y2 Starting torque in star T Y ( n ) = T Δ ( n ) 3 T s t , Y = T a / T n 3 T n ≈ 0.7 – 0.8 T n T_Y(n) = \frac{T_\Delta(n)}{3} \qquad T_{st,Y} = \frac{T_a/T_n}{3}\,T_n \approx 0.7\text{–}0.8\,T_n T Y ( n ) = 3 T Δ ( n ) T s t , Y = 3 T a / T n T n ≈ 0.7 – 0.8 T n
Y3 Main and delta contactors I e ( KM1 ) = I e ( KM3 ) ≥ I n 3 ≈ 0.58 I n ( AC-3 ) I_e(\text{KM1}) = I_e(\text{KM3}) \ge \frac{I_n}{\sqrt3} \approx 0.58\,I_n \quad (\text{AC-3}) I e ( KM1 ) = I e ( KM3 ) ≥ 3 I n ≈ 0.58 I n ( AC-3 )
Y4 Star contactor I e ( KM2 ) ≥ I n 3 ≈ 0.33 I n ( AC-3 ) I_e(\text{KM2}) \ge \frac{I_n}{3} \approx 0.33\,I_n \quad (\text{AC-3}) I e ( KM2 ) ≥ 3 I n ≈ 0.33 I n ( AC-3 )
Y5 Overload setting (in the phase-current path) I r = I n 3 ≈ 0.58 I n I_r = \frac{I_n}{\sqrt3} \approx 0.58\,I_n I r = 3 I n ≈ 0.58 I n
Y6 Changeover dead time t d e a d ≈ 40 – 100 ms > t a r c , K M 2 + t c l o s e , K M 3 margin t_{dead} \approx 40\text{–}100\ \text{ms} \;>\; t_{arc,KM2} + t_{close,KM3}\ \text{margin} t d e a d ≈ 40 – 100 ms > t a rc , K M 2 + t c l ose , K M 3 margin
Y7 Acceleration time J d ω d t = T m − T L ⇒ t a c c ≈ J Δ ω T m − T L ‾ J\frac{d\omega}{dt} = T_m - T_L \;\Rightarrow\; t_{acc} \approx \frac{J\,\Delta\omega}{\overline{T_m - T_L}} J d t d ω = T m − T L ⇒ t a cc ≈ T m − T L J Δ ω
Y8 Condition before changing to delta T Y ( n ) > T L ( n ) ∀ n < n c h n c h ≈ ( 0.85 – 0.9 ) n n T_Y(n) > T_L(n)\ \ \forall\, n < n_{ch} \qquad n_{ch} \approx (0.85\text{–}0.9)\,n_n T Y ( n ) > T L ( n ) ∀ n < n c h n c h ≈ ( 0.85 – 0.9 ) n n
Y9 Cable sizing for six leads I z ≥ I n 3 per conductor Δ V from the phase current and the actual length I_z \ge \frac{I_n}{\sqrt3} \ \text{per conductor} \qquad \Delta V \text{ from the phase current and the actual length} I z ≥ 3 I n per conductor Δ V from the phase current and the actual length
E1 Winding voltage in Y and Δ V w , Y = V L 3 V w , Δ = V L I L , Δ = 3 I w , Δ V_{w,Y} = \frac{V_L}{\sqrt3} \qquad V_{w,\Delta} = V_L \qquad I_{L,\Delta} = \sqrt3\,I_{w,\Delta} V w , Y = 3 V L V w , Δ = V L I L , Δ = 3 I w , Δ
E2 Line-current ratio at the same speed I L , Y I L , Δ = V L / ( 3 Z w ) 3 V L / Z w = 1 3 Z w = winding impedance at that slip \frac{I_{L,Y}}{I_{L,\Delta}} = \frac{V_L/(\sqrt3\,Z_w)}{\sqrt3\,V_L/Z_w} = \frac13 \qquad Z_w = \text{winding impedance at that slip} I L , Δ I L , Y = 3 V L / Z w V L / ( 3 Z w ) = 3 1 Z w = winding impedance at that slip
E3 Torque ratio (torque ∝ V²) T Y ( s ) T Δ ( s ) = ( V w , Y V w , Δ ) 2 = 1 3 T k , Y T k = 1 3 , s k unchanged \frac{T_Y(s)}{T_\Delta(s)} = \left(\frac{V_{w,Y}}{V_{w,\Delta}}\right)^{2} = \frac13 \qquad \frac{T_{k,Y}}{T_k} = \frac13,\ \ s_k \text{ unchanged} T Δ ( s ) T Y ( s ) = ( V w , Δ V w , Y ) 2 = 3 1 T k T k , Y = 3 1 , s k unchanged
E4 Input power and starting kVA P i n = 3 V L I L cos φ S s t = 3 V L I s t S s t , Y = 1 3 S s t , Δ P_{in} = \sqrt3\,V_L I_L\cos\varphi \qquad S_{st} = \sqrt3\,V_L\,I_{st} \qquad S_{st,Y} = \tfrac13\,S_{st,\Delta} P in = 3 V L I L cos φ S s t = 3 V L I s t S s t , Y = 3 1 S s t , Δ
E5 Kloss approximation T T k = 2 s / s k + s k / s T k , Y = T k 3 \frac{T}{T_k} = \frac{2}{s/s_k + s_k/s} \qquad T_{k,Y} = \frac{T_k}{3} T k T = s / s k + s k / s 2 T k , Y = 3 T k
E6 Slip in star near rated speed T ≈ 3 V w 2 ω s R 2 ′ s ⇒ s Y ≈ 3 s Δ (same load torque) T \approx \frac{3V_w^2}{\omega_s R_2'}\,s \;\Rightarrow\; s_Y \approx 3\,s_\Delta \ \text{(same load torque)} T ≈ ω s R 2 ′ 3 V w 2 s ⇒ s Y ≈ 3 s Δ (same load torque)
E7 Star equilibrium and feasibility T Y ( n Y ) = T L ( n Y ) , d ( T Y − T L ) d n ∣ n Y < 0 T Y ( 0 ) > T L ( 0 ) , n Y ≥ 0.9 n N T_Y(n_Y) = T_L(n_Y),\ \ \frac{d(T_Y - T_L)}{dn}\Big|_{n_Y} < 0 \qquad T_Y(0) > T_L(0),\ \ n_Y \ge 0.9\,n_N T Y ( n Y ) = T L ( n Y ) , d n d ( T Y − T L ) n Y < 0 T Y ( 0 ) > T L ( 0 ) , n Y ≥ 0.9 n N
E8 Run-up time in star J d ω d t = T Y ( ω ) − T L ( ω ) ⇒ t ( ω 1 ) = J ∫ 0 ω 1 d ω T Y ( ω ) − T L ( ω ) J\frac{d\omega}{dt} = T_Y(\omega) - T_L(\omega) \;\Rightarrow\; t(\omega_1) = J\int_0^{\omega_1}\frac{d\omega}{T_Y(\omega) - T_L(\omega)} J d t d ω = T Y ( ω ) − T L ( ω ) ⇒ t ( ω 1 ) = J ∫ 0 ω 1 T Y ( ω ) − T L ( ω ) d ω
E9 Mechanical time constant τ m = J ω s T n ω s = 2 π n s 60 T n = 9550 P n n N \tau_m = \frac{J\,\omega_s}{T_n} \qquad \omega_s = \frac{2\pi n_s}{60} \qquad T_n = \frac{9550\,P_n}{n_N} τ m = T n J ω s ω s = 60 2 π n s T n = n N 9550 P n
E10 Timer setting rule t s e t = max ( 1.2 t 98 ( V n ) , t 95 ( 0.9 V n ) ) = c τ m t_{set} = \max\!\big(1.2\,t_{98}(V_n),\ t_{95}(0.9V_n)\big) = c\,\tau_m t se t = max ( 1.2 t 98 ( V n ) , t 95 ( 0.9 V n ) ) = c τ m
E11 Average-torque estimate t ≈ J Δ ω T Y − T L ‾ t \approx \frac{J\,\Delta\omega}{\overline{T_Y - T_L}} t ≈ T Y − T L J Δ ω
E12 No-load run-up with Kloss (closed form) t = J ω s T k [ 1 − s e 2 4 s k + s k 2 ln 1 s e ] t Y = 3 t Δ t = \frac{J\omega_s}{T_k}\left[\frac{1-s_e^2}{4s_k} + \frac{s_k}{2}\ln\frac1{s_e}\right] \qquad t_Y = 3\,t_\Delta t = T k J ω s [ 4 s k 1 − s e 2 + 2 s k ln s e 1 ] t Y = 3 t Δ
E13 Residual voltage after opening e ‾ ( t ) ≈ e ‾ 0 e − t / T 0 ′ e j ∫ 0 t ( p ω r − ω s ) d τ T 0 ′ = L r R r \underline e(t) \approx \underline e_0\,e^{-t/T_0'}\,e^{\,j\int_0^t (p\,\omega_r - \omega_s)\,d\tau} \qquad T_0' = \frac{L_r}{R_r} e ( t ) ≈ e 0 e − t / T 0 ′ e j ∫ 0 t ( p ω r − ω s ) d τ T 0 ′ = R r L r
E14 Phase drift during the dead time δ ( t d ) ≈ s c h ω s t d + p T L 2 J t d 2 (electrical rad) \delta(t_d) \approx s_{ch}\,\omega_s\,t_d + \frac{p\,T_L}{2J}\,t_d^{2} \quad \text{(electrical rad)} δ ( t d ) ≈ s c h ω s t d + 2 J p T L t d 2 (electrical rad)
E15 Transient driving voltage ∣ Δ V ‾ ∣ = ∣ V ‾ Δ − e ‾ ∣ = V Δ 2 + e 2 − 2 V Δ e cos θ θ = ± 30 ∘ − φ 0 − δ |\Delta\underline V| = |\underline V_\Delta - \underline e| = \sqrt{V_\Delta^2 + e^2 - 2V_\Delta e\cos\theta} \qquad \theta = \pm30^\circ - \varphi_0 - \delta ∣Δ V ∣ = ∣ V Δ − e ∣ = V Δ 2 + e 2 − 2 V Δ e cos θ θ = ± 3 0 ∘ − φ 0 − δ
E16 Changeover peak estimate I p k ≈ k a ∣ Δ V ‾ ∣ V Δ I L R k a ≈ 1.35 – 1.45 I_{pk} \approx k_a\,\frac{|\Delta\underline V|}{V_\Delta}\,I_{LR} \qquad k_a \approx 1.35\text{–}1.45 I p k ≈ k a V Δ ∣Δ V ∣ I L R k a ≈ 1.35 – 1.45
E17 Transition resistor R t = r Z b Z b = V L I n / 3 r ≈ 0.1 – 0.3 R_t = r\,Z_b \qquad Z_b = \frac{V_L}{I_n/\sqrt3} \qquad r \approx 0.1\text{–}0.3 R t = r Z b Z b = I n / 3 V L r ≈ 0.1 – 0.3
E18 Rotor heat, no load E r = 1 2 J ω s 2 ( s 1 2 − s 2 2 ) ⇒ E r ( 1 → 0 ) = 1 2 J ω s 2 independent of voltage E_r = \tfrac12 J\omega_s^2\,(s_1^2 - s_2^2) \quad \Rightarrow \quad E_r(1\to0) = \tfrac12 J\omega_s^2 \ \text{independent of voltage} E r = 2 1 J ω s 2 ( s 1 2 − s 2 2 ) ⇒ E r ( 1 → 0 ) = 2 1 J ω s 2 independent of voltage
E19 Rotor heat with load E r = J ω s 2 ∫ s 2 s 1 s T m ( s ) T m ( s ) − T L ( s ) d s E_r = J\omega_s^2\int_{s_2}^{s_1}\frac{s\,T_m(s)}{T_m(s) - T_L(s)}\,ds E r = J ω s 2 ∫ s 2 s 1 T m ( s ) − T L ( s ) s T m ( s ) d s
E20 I²t and overload thermal budget Q = ∫ i r m s 2 d t B = ∫ ( I / I r ) 2 d t 7.2 2 t c l a s s Q = \int i_{rms}^2\,dt \qquad B = \frac{\int (I/I_r)^2\,dt}{7.2^2\,t_{class}} Q = ∫ i r m s 2 d t B = 7. 2 2 t c l a ss ∫ ( I / I r ) 2 d t
E21 Voltage dip at the PCC Δ V % ≈ S m S s c + S m × 100 S s c = S T u k or S G x d ′ ′ S m = 3 V L I \Delta V\% \approx \frac{S_m}{S_{sc} + S_m}\times100 \qquad S_{sc} = \frac{S_T}{u_k}\ \ \text{or}\ \ \frac{S_G}{x_d''} \qquad S_m = \sqrt3\,V_L I Δ V % ≈ S sc + S m S m × 100 S sc = u k S T or x d ′′ S G S m = 3 V L I
E22 Losses in Y versus Δ P l o s s = P C u 1 + P C u 2 + P F e + P f w P F e , Y ≈ 1 3 P F e , Δ s Y ≈ 3 s Δ P_{loss} = P_{Cu1} + P_{Cu2} + P_{Fe} + P_{fw} \qquad P_{Fe,Y} \approx \tfrac13 P_{Fe,\Delta} \qquad s_Y \approx 3\,s_\Delta P l oss = P C u 1 + P C u 2 + P F e + P f w P F e , Y ≈ 3 1 P F e , Δ s Y ≈ 3 s Δ
E23 Saving and break-even Δ P = P i n , Δ − P i n , Y Δ P ( x b e ) = 0 E y e a r = Δ P ⋅ h cost = E y e a r × tariff \Delta P = P_{in,\Delta} - P_{in,Y} \qquad \Delta P(x_{be}) = 0 \qquad E_{year} = \Delta P\cdot h \qquad \text{cost} = E_{year}\times\text{tariff} Δ P = P in , Δ − P in , Y Δ P ( x b e ) = 0 E ye a r = Δ P ⋅ h cost = E ye a r × tariff
E24 Torque margin in star T k , Y T L = T k 3 T L ≥ 2.5 ⇒ T L ≤ T k 7.5 ≈ 0.4 T n ( T k = 3 T n ) \frac{T_{k,Y}}{T_L} = \frac{T_k}{3\,T_L} \ge 2.5 \;\Rightarrow\; T_L \le \frac{T_k}{7.5} \approx 0.4\,T_n \ (T_k = 3T_n) T L T k , Y = 3 T L T k ≥ 2.5 ⇒ T L ≤ 7.5 T k ≈ 0.4 T n ( T k = 3 T n )
T1 Reduced-voltage starting V V n = k ⇒ I s t = k I s t , D O L T s t = k 2 T s t , D O L \frac{V}{V_n} = k \;\Rightarrow\; I_{st} = k\,I_{st,DOL} \qquad T_{st} = k^2\,T_{st,DOL} V n V = k ⇒ I s t = k I s t , D O L T s t = k 2 T s t , D O L
T2 Torque at a current limit T s t = T s t , D O L ( I l i m I s t , D O L ) 2 T_{st} = T_{st,DOL}\left(\frac{I_{lim}}{I_{st,DOL}}\right)^2 T s t = T s t , D O L ( I s t , D O L I l im ) 2
T3 RMS voltage vs firing angle (resistive approximation) V r m s V = 1 − α π + sin 2 α 2 π \frac{V_{rms}}{V} = \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}} V V r m s = 1 − π α + 2 π sin 2 α
a a a soft starters & autotransformers i_{motor} = a\,i_{st,dol} qquad i_{line} = a^2\,i_{st,dol} qquad t_{st} = a^2\,t_{st,dol} ">T4 Autotransformer starter at tap a a a I m o t o r = a I s t , D O L I l i n e = a 2 I s t , D O L T s t = a 2 T s t , D O L I_{motor} = a\,I_{st,DOL} \qquad I_{line} = a^2\,I_{st,DOL} \qquad T_{st} = a^2\,T_{st,DOL} I m o t or = a I s t , D O L I l in e = a 2 I s t , D O L T s t = a 2 T s t , D O L
T5 Supply dip for any method Δ U % ≈ 3 V L I l i n e , s t S t r / u k + 3 V L I l i n e , s t × 100 \Delta U\% \approx \frac{\sqrt3\,V_L\,I_{line,st}}{S_{tr}/u_k + \sqrt3\,V_L\,I_{line,st}}\times 100 Δ U % ≈ S t r / u k + 3 V L I l in e , s t 3 V L I l in e , s t × 100
A1 Sequence interlock K 2 = K 1 ⋅ T K 3 T ⋅ S 0 ‾ K 1 = S 0 ‾ ( S 1 + K 1 ) K2 = K1 \cdot T_{K3T} \cdot \overline{S0} \qquad K1 = \overline{S0}\,(S1 + K1) K 2 = K 1 ⋅ T K 3 T ⋅ S 0 K 1 = S 0 ( S 1 + K 1 )
A2 Maximum starts per hour t c y c l e = V Q p − q + V q ≥ 4 V Q p ( q = Q p / 2 ) N m a x = Q p 4 V t_{cycle} = \frac{V}{Q_p - q} + \frac{V}{q} \;\ge\; \frac{4V}{Q_p}\ \ (q = Q_p/2) \qquad N_{max} = \frac{Q_p}{4V} t cyc l e = Q p − q V + q V ≥ Q p 4 V ( q = Q p /2 ) N ma x = 4 V Q p
A3 Equal running hours h 1 ≈ h 2 ≈ h t o t a l 2 the standby pump must run at least every 1–2 weeks h_1 \approx h_2 \approx \frac{h_{total}}{2} \qquad \text{the standby pump must run at least every 1–2 weeks} h 1 ≈ h 2 ≈ 2 h t o t a l the standby pump must run at least every 1–2 weeks
A4 Receiver size for compressor cycling t o f f = V t a n k Δ p p 0 q u s e t o n = V t a n k Δ p p 0 ( Q c − q u s e ) t_{off} = \frac{V_{tank}\,\Delta p}{p_0\,q_{use}} \qquad t_{on} = \frac{V_{tank}\,\Delta p}{p_0\,(Q_c - q_{use})} t o ff = p 0 q u se V t ank Δ p t o n = p 0 ( Q c − q u se ) V t ank Δ p
A5 Pole-changing speeds n s , l o w = 120 f 2 p n s , h i g h = 120 f p = 2 n s , l o w n_{s,low} = \frac{120f}{2p} \qquad n_{s,high} = \frac{120f}{p} = 2\,n_{s,low} n s , l o w = 2 p 120 f n s , hi g h = p 120 f = 2 n s , l o w
V1 DC bus voltage V d c ≈ 1.35 V L = 540 V (loaded) V d c , 0 = 2 V L = 566 V (no load) V_{dc} \approx 1.35\,V_L = 540\ \text{V (loaded)} \qquad V_{dc,0} = \sqrt2\,V_L = 566\ \text{V (no load)} V d c ≈ 1.35 V L = 540 V (loaded) V d c , 0 = 2 V L = 566 V (no load)
V2 Maximum output voltage V L L , r m s = 0.612 m V d c ( SPWM , m ≤ 1 ) V L L , m a x ≈ 0.707 V d c ( SVPWM ) V_{LL,rms} = 0.612\,m\,V_{dc}\ (\text{SPWM}, m \le 1) \qquad V_{LL,max} \approx 0.707\,V_{dc}\ (\text{SVPWM}) V LL , r m s = 0.612 m V d c ( SPWM , m ≤ 1 ) V LL , ma x ≈ 0.707 V d c ( SVPWM )
V3 Constant V/f V ( f ) = V n f f n + V b o o s t ( 1 − f f n ) V n f n = 400 50 = 8 V/Hz V(f) = V_n\frac{f}{f_n} + V_{boost}\left(1 - \frac{f}{f_n}\right) \qquad \frac{V_n}{f_n} = \frac{400}{50} = 8\ \text{V/Hz} V ( f ) = V n f n f + V b oos t ( 1 − f n f ) f n V n = 50 400 = 8 V/Hz
V4 Speed with a VFD n = 120 f p − Δ n Δ n ≈ s n n s , n ( constant at rated torque ) n = \frac{120 f}{p} - \Delta n \qquad \Delta n \approx s_n\,n_{s,n}\ \ (\text{constant at rated torque}) n = p 120 f − Δ n Δ n ≈ s n n s , n ( constant at rated torque )
V5 Field weakening (f > f_n) T c o n t ≈ T n f n f T m a x ≈ T m a x , n ( f n f ) 2 P ≈ P n T_{cont} \approx T_n\,\frac{f_n}{f} \qquad T_{max} \approx T_{max,n}\left(\frac{f_n}{f}\right)^2 \qquad P \approx P_n T co n t ≈ T n f f n T ma x ≈ T ma x , n ( f f n ) 2 P ≈ P n
V6 Braking resistor E = 1 2 J ( ω 1 2 − ω 2 2 ) P p e a k ≈ J ω 1 Δ ω t d e c R ≤ V b r 2 P p e a k , R ≥ R m i n E = \tfrac12 J(\omega_1^2 - \omega_2^2) \qquad P_{peak} \approx \frac{J\,\omega_1\,\Delta\omega}{t_{dec}} \qquad R \le \frac{V_{br}^2}{P_{peak}},\ R \ge R_{min} E = 2 1 J ( ω 1 2 − ω 2 2 ) P p e ak ≈ t d ec J ω 1 Δ ω R ≤ P p e ak V b r 2 , R ≥ R min
P1 Minimum ramp time without tripping t a c c , m i n ≈ J ω n T l i m − T L ‾ T l i m ≈ 1.5 T n ( set by the drive current limit ) t_{acc,min} \approx \frac{J\,\omega_n}{T_{lim} - \overline{T_L}} \qquad T_{lim} \approx 1.5\,T_n\ \ (\text{set by the drive current limit}) t a cc , min ≈ T l im − T L J ω n T l im ≈ 1.5 T n ( set by the drive current limit )
P2 Analog reference scaling f = f m i n + ( f m a x − f m i n ) x − x 0 x 1 − x 0 x : 0 – 10 V or 4 – 20 mA f = f_{min} + (f_{max} - f_{min})\,\frac{x - x_0}{x_1 - x_0} \qquad x: 0\text{–}10\ \text{V or } 4\text{–}20\ \text{mA} f = f min + ( f ma x − f min ) x 1 − x 0 x − x 0 x : 0 – 10 V or 4 – 20 mA
P3 Affinity law — flow Q 2 Q 1 = n 2 n 1 \frac{Q_2}{Q_1} = \frac{n_2}{n_1} Q 1 Q 2 = n 1 n 2
P4 Affinity law — head H 2 H 1 = ( n 2 n 1 ) 2 \frac{H_2}{H_1} = \left(\frac{n_2}{n_1}\right)^2 H 1 H 2 = ( n 1 n 2 ) 2
P5 Affinity law — power P 2 P 1 = ( n 2 n 1 ) 3 \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3 P 1 P 2 = ( n 1 n 2 ) 3
P6 System with static head H s y s = H s t + k Q 2 n n 1 ≈ h 0 + ( 1 − h 0 ) q 2 ( h 0 = H s t / H 1 , q = Q / Q 1 ) H_{sys} = H_{st} + kQ^2 \qquad \frac{n}{n_1} \approx \sqrt{h_0 + (1 - h_0)q^2} \quad (h_0 = H_{st}/H_1,\ q = Q/Q_1) H sys = H s t + k Q 2 n 1 n ≈ h 0 + ( 1 − h 0 ) q 2 ( h 0 = H s t / H 1 , q = Q / Q 1 )
P7 Energy saved per year Δ E = ( P t h r o t t l e − P V F D ) h money = Δ E × tariff per kWh \Delta E = (P_{throttle} - P_{VFD})\,h \qquad \text{money} = \Delta E \times \text{tariff per kWh} Δ E = ( P t h ro ttl e − P V F D ) h money = Δ E × tariff per kWh
P8 Simple payback t p a y b a c k = installed VFD cost savings per year t_{payback} = \frac{\text{installed VFD cost}}{\text{savings per year}} t p a y ba c k = savings per year installed VFD cost
K1 Shaft power of a pump or a drive P p u m p = ρ g Q H η p P = T ω = T n 9550 (kW) P_{pump} = \frac{\rho g Q H}{\eta_p} \qquad P = T\,\omega = \frac{T\,n}{9550}\ \text{(kW)} P p u m p = η p ρ g Q H P = T ω = 9550 T n (kW)
K2 Motor size margin P m o t o r ≥ ( 1.10 – 1.15 ) P l o a d , m a x → next standard size P_{motor} \ge (1.10\text{–}1.15)\,P_{load,max} \;\rightarrow\; \text{next standard size} P m o t or ≥ ( 1.10 – 1.15 ) P l o a d , ma x → next standard size
K3 Cable ampacity after derating I z k t e m p k g r o u p ≥ I n I z ≥ I r ( overload protection setting ) I_z\,k_{temp}\,k_{group} \ge I_n \qquad I_z \ge I_r\ (\text{overload protection setting}) I z k t e m p k g ro u p ≥ I n I z ≥ I r ( overload protection setting )
K4 Running voltage drop Δ V % = 3 I n L ( R cos φ + X sin φ ) V L × 100 ≤ 3 – 5 % R = ρ A , ρ C u , 70 ∘ C ≈ 0.0225 Ω mm 2 / m \Delta V\% = \frac{\sqrt3\,I_n\,L\,(R\cos\varphi + X\sin\varphi)}{V_L}\times100 \le 3\text{–}5\,\% \qquad R = \frac{\rho}{A},\ \rho_{Cu,70^\circ C} \approx 0.0225\ \Omega\,\text{mm}^2/\text{m} Δ V % = V L 3 I n L ( R cos φ + X sin φ ) × 100 ≤ 3 – 5 % R = A ρ , ρ C u , 7 0 ∘ C ≈ 0.0225 Ω mm 2 / m
K5 Starting voltage drop Δ V s t % ≈ Δ V t r % + 3 I s t L ( R ⋅ 0.3 + X ⋅ 0.95 ) V L × 100 ≤ 10 – 15 % \Delta V_{st}\% \approx \Delta V_{tr}\% + \frac{\sqrt3\,I_{st}\,L\,(R\cdot0.3 + X\cdot0.95)}{V_L}\times100 \le 10\text{–}15\,\% Δ V s t % ≈ Δ V t r % + V L 3 I s t L ( R ⋅ 0.3 + X ⋅ 0.95 ) × 100 ≤ 10 – 15 %
K6 Earth fault at the cable end (TN system) I f = U 0 Z s ≥ I a Z s ≈ Z s r c + ( 1 + m ) ρ L A p h ( m = A p h / A P E ) I_{f} = \frac{U_0}{Z_s} \ge I_a \qquad Z_s \approx Z_{src} + (1 + m)\,\frac{\rho L}{A_{ph}}\ \ (m = A_{ph}/A_{PE}) I f = Z s U 0 ≥ I a Z s ≈ Z src + ( 1 + m ) A p h ρ L ( m = A p h / A PE )
K7 Enclosure temperature rise and fan airflow Δ T = P l o s s k A e ( k ≈ 5.5 W/m 2 K painted steel ) V ˙ [ m 3 / h ] ≈ 3.1 P l o s s [ W ] Δ T a l l o w \Delta T = \frac{P_{loss}}{k\,A_e}\ \ (k \approx 5.5\ \text{W/m}^2\text{K painted steel}) \qquad \dot V\,[\text{m}^3/\text{h}] \approx \frac{3.1\,P_{loss}\,[\text{W}]}{\Delta T_{allow}} Δ T = k A e P l oss ( k ≈ 5.5 W/m 2 K painted steel ) V ˙ [ m 3 / h ] ≈ Δ T a ll o w 3.1 P l oss [ W ]
K8 Insulation resistance acceptance R i s o ≥ 1 M Ω @ 500 V DC (IEC 60204-1) R 40 ∘ C ≥ 5 M Ω windings < 1 kV (IEEE 43) P I = R 10 m i n R 1 m i n ≥ 2 R_{iso} \ge 1\ \text{M}\Omega\ @\ 500\ \text{V DC (IEC 60204-1)} \qquad R_{40^\circ C} \ge 5\ \text{M}\Omega\ \text{windings} < 1\ \text{kV (IEEE 43)} \qquad PI = \frac{R_{10\,min}}{R_{1\,min}} \ge 2 R i so ≥ 1 M Ω @ 500 V DC (IEC 60204-1) R 4 0 ∘ C ≥ 5 M Ω windings < 1 kV (IEEE 43) P I = R 1 min R 10 min ≥ 2
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Standards and references
IEC 60034-1 (motor ratings) · IEC 60034-8 (terminal markings and rotation) · IEC 60034-30-1 (IE efficiency classes)
IEC 60947-4-1 (contactors and starters, AC-3, Type 1/2 coordination) · IEC 60947-4-2 (soft starters) · IEC 60947-2 (circuit-breakers)
IEC 60204-1 (electrical equipment of machines) · IEC 60617 (symbols) · IEC 81346 (reference designations)
IEC 61800 (adjustable-speed drives) · IEC 60364 / Thai EIT wiring code · IEC 61439 (low-voltage assemblies)
S. J. Chapman, Electric Machinery Fundamentals · B. K. Bose, Modern Power Electronics and AC Drives · manufacturers' starter selection guides
Frequently asked questions Which motors can be star–delta started? Only motors whose delta voltage on the nameplate equals the supply line voltage. On a 400 V system that is a motor marked 400/690 V Δ/Y (usually 4 kW and above). A 230/400 V Δ/Y motor must run in star on 400 V and cannot be star–delta started.
Can the same equipment be used in Thailand and its neighbours? In practice yes. Every country covered here uses 50 Hz three-phase four-wire systems with line voltages of 380–415 V (IEC 60038 nominal 230/400 V), so equipment rated 380–415 V 50 Hz fits all of them. Always measure the actual site voltage and check the local utility's rules.
What is the overload setting for a star–delta starter? When the overload relay sits after the main contactor in the phase-current path (the most common arrangement) set it to 0.58 × the motor In, e.g. 8.5 A for a 14.6 A motor. If it sits in the line-current path before the split, set it to In.
When should I use a VFD instead of star–delta? When you need variable speed, a soft start with high torque, energy savings on pumps and fans that often run below full flow (P ∝ n³), or when the starting voltage dip is a problem. If you only need to reduce the starting current of a light load at fixed speed, star–delta or a soft starter is cheaper.
Mod
Ajarn Mod · Teerawut Savangboon Lecturer in Electrical Engineering, Dhonburi Rajabhat University, Samut Prakan · Profile