Electrical Load Calculations for Roll Forming Lines (With Worked Examples)

VFD undervoltage faults during shear/punch events

Roll Forming Machines + Coil Processing Equipment

Electrical load calculation is where most roll forming installations either succeed cleanly or fail with:

nuisance breaker trips
VFD undervoltage faults during shear/punch events
PLC resets from 24V dips
unstable speed control
overheating cables, transformers, and distribution gear

The core problem is simple:

Nameplate kW is not the same as real operating kW.
And “total connected load” is not the same as “maximum demand.”

This page gives a repeatable calculation method and multiple worked examples you can reuse across your 100+ electrical pages.

Word-Based Wiring Context (Use on Every Load Page)

Main power circuit (typical):
MAIN ISOLATOR → MCCB → BUSBAR → BRANCH PROTECTION (MCB/MPCB) → VFD INPUT → VFD OUTPUT (U/V/W) → MOTOR

Control circuit (typical):
24VDC PSU → E-STOP LOOP → SAFETY RELAY → PLC INPUT → PLC OUTPUT → INTERPOSING RELAY → CONTACTOR COIL / DRIVE ENABLE

Flying shear motion circuit (typical servo):
ENCODER → HIGH-SPEED COUNTER → PLC → SERVO DRIVE → SERVO MOTOR → BRAKE RESISTOR

These “word diagrams” help technicians and search engines understand what your calculations relate to.

1) What You’re Calculating: Four Different “Loads”

A professional load package includes four numbers (not one):

Connected Load (kW / kVA)
Sum of nameplate ratings (everything that could run)
Typical Running Demand (kW / kVA)
What the line draws during steady production for a typical product
Peak Production Demand (kW / kVA)
Worst-case product (thicker, higher yield) + acceleration + shear/punch event
Starting / Inrush Event (Amps)
Highest instantaneous current (DOL motors, transformer inrush, hydraulic starts)

If you only provide “connected load,” you will oversize in some places and undersize in the places that actually trip.

2) Key Electrical Terms (Used in Every Example)

2.1 kW, kVA, Power Factor (PF), Efficiency (η)

kW = real power (usable work)
kVA = apparent power (what supply must deliver)
PF = kW / kVA
η = motor/drive efficiency (typical 0.90–0.95)

Relationship:

kVA = kW / PF
kW = kVA × PF

2.2 3-Phase Current Formula

For 3-phase systems:

kVA = ( √3 × V × I ) / 1000
So:

I (A) = ( kVA × 1000 ) / ( √3 × V )

Quick working constant: √3 ≈ 1.732

3) The Correct Workflow for Load Calculations

Step A — Build the Load Schedule (Everything That Can Draw Power)

Split loads into:

Continuous / frequent

main forming motor (often VFD)
hydraulic pump motor (starter or VFD)
uncoiler/recoiler motors (if powered)
leveler (if separate)
stacker/outfeed

Intermittent peak

punch HPU
servo flying shear axis
coil car/upender motors
scrap chopper / shear auxiliaries

Control/auxiliary

PLC/HMI/network
24VDC power supplies
panel fans/AC/heaters
lights/alarms

Step B — Assign “Demand Factors” (Engineering Reality)

A demand factor is the fraction of nameplate expected during typical production.

Practical starting heuristics (adjust with experience):

main forming drive: 0.70–0.95
hydraulic pump (constant run): 0.60–1.00
uncoiler/recoiler: 0.30–0.80
conveyors/stackers: 0.20–0.60
controls/aux: treat as 1.00 (they’re small but constant)

Step C — Compute Typical Running kW

Running kW = Σ (nameplate kW × demand factor)

Step D — Convert to kVA (choose PF assumptions)

Use conservative PF assumptions when you don’t have test data:

VFD-heavy line: PF 0.90 is a reasonable planning assumption
mixed DOL motors: PF 0.85–0.90

Step E — Convert to Running Amps at Site Voltage

Use the 3-phase current formula.

Step F — Evaluate Peak Events Separately

Peak events commonly come from:

acceleration ramps (speed changes)
punching cycles (HPU load spikes)
shear cycles + hold-down clamp
tension correction bursts on uncoiler/recoiler

Peak is not usually “all motors at 100%,” but it can exceed your running demand briefly.

Step G — Identify Starting/Inrush Drivers

Any large DOL motor: can be 6–8× FLA at start
Star-delta: reduced but still significant
Transformers: magnetizing inrush (short, but can trip sensitive protection)
Large VFD banks: DC bus charging (usually controlled, but still a consideration)

4) Worked Example 1 — Roofing Roll Former (400V, 50Hz, Mostly VFD)

Line description (typical high-speed roofing panel line)

Main forming motor (VFD): 22 kW
Hydraulic pump motor (DOL): 7.5 kW
Uncoiler (powered, VFD): 5.5 kW
Stacker/outfeed: 2.2 kW
Controls/aux: 1.0 kW (PLC/HMI/PSU/fans)

4.1 Connected Load (kW)

Connected kW = 22 + 7.5 + 5.5 + 2.2 + 1.0
Let’s add carefully:

22 + 7.5 = 29.5
29.5 + 5.5 = 35.0
35.0 + 2.2 = 37.2
37.2 + 1.0 = 38.2 kW

Connected load = 38.2 kW

4.2 Typical Running Demand (kW)

Apply demand factors:

Main forming: 22 × 0.85 = 18.7 kW
Hydraulic pump: 7.5 × 0.70 = 5.25 kW
Uncoiler: 5.5 × 0.50 = 2.75 kW
Stacker: 2.2 × 0.40 = 0.88 kW
Controls: 1.0 × 1.00 = 1.00 kW

Now sum:

18.7 + 5.25 = 23.95
23.95 + 2.75 = 26.70
26.70 + 0.88 = 27.58
27.58 + 1.00 = 28.58 kW

Typical running demand ≈ 28.6 kW

4.3 Convert Running kW → Running kVA

Assume PF = 0.90 (VFD-heavy)

kVA = 28.58 / 0.90 = 31.76 kVA (rounded)

4.4 Convert kVA → Running Current at 400V

I = (kVA × 1000) / (√3 × V)

Denominator: 1.732 × 400 = 692.8

I = 31.76 × 1000 / 692.8
31,760 / 692.8 ≈ 45.8 A

Running current ≈ 46 A at 400V

4.5 Peak Event Check (Acceleration + Shear/Pump)

Roofing lines often see short spikes when:

main motor ramps hard
hydraulic demand rises near shear events

A practical planning approach is adding a peak margin:

Peak kW ≈ running kW × 1.20 (for a well-designed VFD line)
Peak kW ≈ 28.6 × 1.20 = 34.3 kW

Peak kVA = 34.3 / 0.90 = 38.1 kVA

Peak I at 400V:
38,100 / 692.8 ≈ 55.0 A

So the feeder should comfortably support ~55 A peak plus any DOL inrush.

4.6 Starting/Inrush Driver: 7.5 kW DOL Hydraulic Pump

DOL start current can be 6× FLA (often a safe planning figure).

First estimate pump FLA.
Using rough motor current formula:
I ≈ (kW × 1000) / (√3 × V × PF × η)

Assume PF 0.85, η 0.90 for a small induction motor.

Denominator: 1.732 × 400 × 0.85 × 0.90
1.732×400 = 692.8
692.8×0.85 = 588.88
588.88×0.90 = 530.0 (approx)

I ≈ 7,500 / 530 ≈ 14.2 A (motor running current estimate)

DOL start (6×): ~ 85 A momentary

Engineering implication: your MCCB and feeder must tolerate brief ~85A inrush while the rest of the line is drawing current.

5) Worked Example 2 — Structural C/Z Purlin Line (400V, 50Hz, Higher Torque + Punch)

Line description (typical structural line)

Main forming motor (VFD): 45 kW
Hydraulic pump motor (DOL or VFD): 22 kW
Punch HPU motor (DOL): 15 kW
Uncoiler (powered, VFD): 11 kW
Servo feed or servo axis: 7.5 kW (servo drive)
Stacker/outfeed: 3 kW
Controls/aux/cooling: 2 kW

5.1 Connected Load (kW)

Add carefully:

45 + 22 = 67
67 + 15 = 82
82 + 11 = 93
93 + 7.5 = 100.5
100.5 + 3 = 103.5
103.5 + 2 = 105.5 kW

Connected load = 105.5 kW

5.2 Typical Running Demand (kW)

Demand factors (structural lines tend to run closer to rated when forming thick/high-yield):

Main forming: 45 × 0.90 = 40.5
Hydraulic pump: 22 × 0.75 = 16.5
Punch HPU: 15 × 0.40 = 6.0 (intermittent average, but peak matters!)
Uncoiler: 11 × 0.60 = 6.6
Servo: 7.5 × 0.50 = 3.75
Stacker: 3 × 0.40 = 1.2
Controls: 2 × 1.00 = 2.0

Sum:

40.5 + 16.5 = 57.0
57.0 + 6.0 = 63.0
63.0 + 6.6 = 69.6
69.6 + 3.75 = 73.35
73.35 + 1.2 = 74.55
74.55 + 2.0 = 76.55 kW

Typical running demand ≈ 76.6 kW

5.3 Convert to kVA

Assume PF = 0.88 (more mixed, higher inductive load + drives)

kVA = 76.55 / 0.88 = 87.0 kVA (approx)

5.4 Running Current at 400V

I = 87,000 / 692.8 ≈ 125.6 A

Running current ≈ 126 A

5.5 Peak Production Demand

Structural lines peak during:

acceleration + heavy gauge
punch hits + hydraulic demand
fast indexing between hole patterns

A practical peak approach:

Main forming may hit 100% briefly: add (45 − 40.5) = +4.5 kW
Hydraulic may rise: add 22×(0.95−0.75)= +4.4 kW
Punch HPU may spike near 100% during cycles: add (15 − 6) = +9 kW
Servo may spike: add 7.5×(0.80−0.50)= +2.25 kW

Peak kW ≈ 76.55 + 4.5 + 4.4 + 9 + 2.25
76.55 + 4.5 = 81.05
81.05 + 4.4 = 85.45
85.45 + 9 = 94.45
94.45 + 2.25 = 96.70 kW

Peak kVA = 96.7 / 0.88 = 109.9 kVA

Peak current:
109,900 / 692.8 ≈ 158.6 A

So, even though average is ~126 A, a realistic peak can be ~160 A.

5.6 Starting/Inrush: Punch HPU (15 kW DOL) + Hydraulic (22 kW DOL)

These two can dominate your breaker selection if they start across-the-line.

Estimate motor running currents (rough):

For 15 kW motor at 400V, PF 0.85, η 0.92
Denominator: 692.8×0.85=588.9; ×0.92=541.8
I ≈ 15,000 / 541.8 ≈ 27.7 A
DOL start (6×) ≈ 166 A momentary

For 22 kW motor at 400V, PF 0.85, η 0.92
I ≈ 22,000 / 541.8 ≈ 40.6 A
DOL start (6×) ≈ 244 A

If both start close together, you can see momentary events approaching 400 A plus whatever the rest of the system is doing.

Engineering takeaway: Structural lines should strongly consider VFD or soft-start on large hydraulic HPUs, or enforce start sequencing interlocks so they never start simultaneously.

6) Worked Example 3 — Coil Slitting Line (480V, 60Hz, Multi-Drive + Tension Control)

Coil lines differ because they often have multiple coordinated drives running continuously.

Line description (slitter + recoiler)

Uncoiler drive: 30 kW (VFD)
Pinch rolls: 11 kW (VFD)
Slitter arbor: 45 kW (VFD)
Recoiler: 37 kW (VFD)
Scrap winders: 2 × 7.5 kW (VFD) = 15 kW
Hydraulics/ancillary: 11 kW (starter or VFD)
Controls/aux: 3 kW

6.1 Connected Load (kW)

Add carefully:

30 + 11 = 41
41 + 45 = 86
86 + 37 = 123
123 + 15 = 138
138 + 11 = 149
149 + 3 = 152 kW

Connected = 152 kW

6.2 Typical Running Demand (kW)

Demand factors (coil lines often run many drives, but not always at rated torque):

Uncoiler: 30 × 0.60 = 18
Pinch: 11 × 0.50 = 5.5
Slitter: 45 × 0.75 = 33.75
Recoiler: 37 × 0.70 = 25.9
Scrap winders: 15 × 0.60 = 9
Hydraulics: 11 × 0.60 = 6.6
Controls: 3 × 1.00 = 3

Sum:

18 + 5.5 = 23.5
23.5 + 33.75 = 57.25
57.25 + 25.9 = 83.15
83.15 + 9 = 92.15
92.15 + 6.6 = 98.75
98.75 + 3 = 101.75 kW

Typical running ≈ 101.8 kW

6.3 Convert to kVA

Assume PF = 0.90 (VFD-heavy)

kVA = 101.75 / 0.90 = 113.1 kVA

6.4 Running Current at 480V

Denominator: √3 × 480 = 1.732 × 480 = 831.36

I = 113,100 / 831.36 ≈ 136.1 A

Running current ≈ 136 A at 480V

6.5 Peak Events (Tension Transients + Acceleration)

Coil lines can peak during:

acceleration from threading to speed
tension corrections
slitter bite changes
recoiler torque spikes when diameter increases

A planning peak factor of 1.15–1.30 is common depending on process aggressiveness.

Peak kW ≈ 101.75 × 1.25 = 127.2 kW
Peak kVA = 127.2 / 0.90 = 141.3 kVA
Peak I = 141,300 / 831.36 ≈ 169.9 A

So you might run ~136 A but design for peaks around ~170 A.

Engineering note: With many VFDs, harmonics and heat in feeders/transformers become important. Even if current looks “fine,” RMS heating can be worse than expected.

7) How to Treat Control Loads (PLC/HMI/24VDC) in Calculations

Controls rarely dominate kW, but they dominate reliability.

If your 24VDC system dips:

PLC resets
safety relay drops
drive enable signals fall out
“random” faults multiply

Include control loads as:

continuous (demand factor = 1.0)
and consider separation of “dirty” 24V loads (solenoids) from “clean” (PLC/encoder)

Word-based control flow reminder:
24VDC PSU → fused distribution → clean loads (PLC/encoder/HMI)
24VDC PSU → separate fused distribution → dirty loads (solenoids/relays)

8) Practical Rules for Peak vs Average (What Trips What)

Cables are mostly sized for continuous current + derating + voltage drop.
Breakers must handle continuous current and not nuisance-trip on peaks/inrush.
VFDs trip on undervoltage/overcurrent during acceleration and load spikes.
Transformers fail from thermal overload and harmonics heating (often not obvious at first).

So always compute:

I_run (typical)
I_peak (production worst case)
I_start (largest start event)

9) What to Put in Your Documentation Pack (So It’s “Buyer-Ready”)

Every machine should have a load sheet that includes:

Connected load (kW)
Typical running demand (kW + kVA)
Peak demand estimate (kW + kVA)
Running current at rated voltage
Peak current estimate
Largest start/inrush event (amps) and start sequencing logic
PF assumptions stated clearly
Any harmonic mitigation requirement noted (line reactors/filters)

This prevents disputes and prevents installers from guessing.

10) Buyer Strategy (30%): How to Ask for Correct Load Data

When buying a roll forming or coil line, demand:

Load schedule by motor and subsystem
Starting method for each motor (VFD / DOL / star-delta)
Total connected load and maximum demand estimate
Recommended incoming breaker size and notes about trip curve sensitivity
Any transformer requirement (480↔400 conversions, control transformer taps)
Commissioning checklist including phase rotation and voltage verification

Red flag: supplier only provides “Total kW” with no breakdown and no starting method notes.

6 Frequently Asked Questions

1) What’s the difference between connected load and maximum demand?

Connected load is everything added up at nameplate. Maximum demand is what the line actually draws at worst-case operation including peaks and start events.

2) Why do I need kVA if I already have kW?

Because your supply (transformers, switchgear) must deliver kVA, and PF varies with drives and loading.

3) How do I choose PF for planning if I don’t have measurements?

Use conservative assumptions: 0.90 for VFD-heavy, 0.85–0.90 for mixed induction loads. Document the assumption.

4) What causes nuisance trips even when my running amps seem OK?

Inrush current (DOL motors), acceleration peaks, hydraulic peaks, voltage drop, and power quality issues (harmonics) are common causes.

5) Do VFDs eliminate starting current issues?

They reduce motor inrush substantially, but you still must plan for acceleration peaks, DC bus charging, and any remaining DOL motors.

6) What’s the single biggest mistake in load calculations for roll forming lines?

Treating all motors as if they run at 100% continuously, and ignoring start/peak events. That leads to wrong breaker behavior and unstable production.

Final Engineering Summary

Electrical load calculations for roll forming and coil processing lines must produce:

Connected load
Typical running demand
Peak demand
Starting/inrush events

Using a structured method:

build a load schedule
apply demand factors
convert kW → kVA using PF
convert kVA → amps at site voltage
separately evaluate peaks and starting currents