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Nairobi PowerENGINEERS · EST. 1984

SOLUTIONS / APPLICATION GUIDE

Relays, monitoring & protection

Identify the protection function and control circuit before choosing a relay. Convert CT currents and assemble the information needed for a coordination review.

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Illustrative relays, monitoring & protection equipment
AI-generated equipment illustration.

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CHOOSE YOUR APPROACH

Start with the application.

Control and interface

Confirm coil voltage, contact arrangement, switching category and load inrush.

Monitoring

Define the voltage, phase, current or process condition and required response.

Protection

Use a fault study and device curves to coordinate protection; do not guess pickup or time delay.

BUILD A USEFUL BRIEF

Bring these details.

  • Relay model, function, supply and wiring diagram
  • CT ratio, secondary rating, class and burden
  • Protected equipment ratings and fault-level study
  • Existing settings, trip circuit and test records
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Does the CT calculator provide relay settings?

No. It converts ideal primary and secondary current. It does not evaluate CT saturation, burden, class, pickup or time coordination. Never open-circuit an energised CT secondary.

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COMPONENTS THAT COMPLETE THE APPLICATION

Specify the whole working system.

Start with the equipment duty. Add the control, protection and measurement functions that your project needs; these are options to review, not a pre-approved assembly.

Images are AI-generated illustrations of equipment categories, not LOVATO model photographs. Confirm dimensions and ratings from the selected model datasheet.

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Electronic Protection Relays ↗Panel Instruments & Metering ↗PLC, SCADA & BMS Automation Panels ↗

METHODS & WORKED EXAMPLES

Understand the estimate.

These examples use the default inputs. Change the live controls above for your own scenario; the reference examples below remain fixed.

CT current conversion — worked example

Ideal secondary current = primary current × rated secondary / rated primary. This does not evaluate CT burden, accuracy or saturation.

Example inputs

  • CT rated primary: 400 A
  • CT rated secondary: 5 A
  • Primary current: 240 A

Example result

  • CT ratio: 80:1
  • Ideal secondary current: 3 A
  • Primary rating utilisation: 60 %

Assumptions & limits

  • Isecondary = Iprimary × rated secondary / rated primary.
  • Ideal ratio only; no allowance for saturation, phase error, accuracy class or burden.
  • Never open-circuit an energised CT secondary. Protection settings require a coordination study and approved testing.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Technical reference: Schneider Electric: current transformer guidance. This reference does not certify the website calculation or an NPE assembly.

Short-circuit current — worked example

Example inputs

  • Transformer rating: 1000 kVA
  • Impedance Z: 5 %
  • Secondary voltage: 415 V — industrial
  • Phase: 3Φ (three phase)
  • Cable attenuation: 10 %

Example result

  • Transformer FLC: 1,391 A
  • Prospective fault: 25 kA
  • Breaking capacity: 36 kA

Assumptions & limits

  • Cable attenuation applied: 10 %
  • Infinite upstream source assumed — utility contribution ignored
  • Confirm the utility fault level with KPLC before finalising ratings
  • Preliminary planning only. Final selection needs project-specific engineering verification.
kVA ↔ amps — worked example

Example inputs

  • Value: 500
  • Conversion: kVA → amps
  • Phase: 3Φ (three phase)
  • System voltage: 415 V — industrial
  • Power factor: 0.8

Example result

  • Line current: 696 A
  • Apparent power: 500 kVA
  • Active power: 400 kW

Assumptions & limits

  • Phase system: 3Φ balanced at 415 V
  • Multiplier used: √3 × V
  • Assumes a balanced, linear load profile
  • Preliminary planning only. Final selection needs project-specific engineering verification.

METHODS & WORKED EXAMPLES

Understand the estimate.

These examples use the default inputs. Change the live controls above for your own scenario; the reference examples below remain fixed.

DC control power budget — worked example

Example inputs

  • DC output voltage: 24 V DC
  • PLC and other continuous load: 25 W
  • Simultaneously held DC coils: 6
  • Holding power per DC coil: 3 W
  • Simultaneous sensors: 10
  • Power per sensor: 0.5 W
  • Reserve allowance: 25 %

Example result

  • Connected continuous load: 48 W
  • Budget with reserve: 60 W
  • DC output current budget: 2.5 A

Assumptions & limits

  • Pload = PLC / other W + coil count × holding W + sensor count × sensor W.
  • Pbudget = Pload × (1 + reserve / 100); Ibudget = Pbudget / DC voltage.
  • Reserve is a planning assumption, not a manufacturer requirement.
  • All listed loads must match the selected DC voltage. AC coil VA and DC coil watts are not interchangeable.
  • Check coil pull-in peaks, simultaneous starting, temperature and altitude derating, overload behaviour and branch protection separately. This tool does not select a supply model.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Method reference: MEAN WELL — Power supply selection FAQ

Three-phase voltage unbalance — worked example

Example inputs

  • L1–L2 voltage: 415 V
  • L2–L3 voltage: 410 V
  • L3–L1 voltage: 420 V

Example result

  • Average line voltage: 415 V
  • Maximum deviation: 5 V
  • Voltage unbalance: 1.2 %

Assumptions & limits

  • Unbalance (%) = 100 × maximum absolute deviation from the average / average.
  • Use line-to-line readings taken under the same operating condition.
  • No pass/fail limit or trip delay is selected. Check equipment limits and investigate supply conditions with a qualified engineer.
  • This method uses voltage magnitudes only; it is not an IEC negative-sequence calculation.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Method reference: US DOE — Energy Management for Motor-Driven Systems