How to know if you set the correct current on a motor thermal overload relay

Start with full-load current…

The full-load current at a given voltage indicated on the nameplate is normative for setting the overload relay. Because of the variable voltages around the world, motors for pumps are made to be used at both 50 Hz and 60 Hz in a wide voltage range.

Therefore, a current range is indicated on the motor’s nameplate. The exact current capacity can be calculated when we know the voltage.

Calculation example

When we know the precise voltage for the installation, the full-load current can be calculated at 254 Δ/440 Y V, 60 Hz. The data is indicated on the nameplate as shown on the illustration on below:

• f = 60 Hz
• U = 220-277 ∆/380 – 480 Y V
• I_n = 5.70 – 5.00/3.30 – 2.90 A

60 Hz data calculation:

• U_a = actual voltage 254 ∆/440 Y V (actual voltage)
• U_min = 220 ∆/380 Y V (Minimum values in the voltage range)
• U_max = 277 ∆/480 Y V (Maximum values in the voltage range)

Calculation of the actual full-load current (I)

I_min = 570/3.30 A
(Current values for Delta and Star at minimum voltages)

I_max = 500/2.90 A
(Current values for Delta and Star at maximum voltages)

Now, it is possible to calculate the full-load current by means of the first formula:

• I for Delta values: 5.70 + (5.00 – 5.70) × 0.6 = 5.28 = 5.30 A
• I for Star values: 3.30 + (2.90 – 3.30) × 0.6 = 3.06 = 3.10 A

The values for the full-load current correspond to the permissible full-load current of the motor at 254 ∆/440 Y V, 60 Hz.

However if motors are designed with a service factor, which is then shown on the nameplate eg. 1.15, the set current for the overload relay can be raised by 15% compared to full-load current or to the service factor amps (SFA) which is normally indicated on the nameplate.

If the motor is connected in star = 440 V 60 Hz the overload relay then has to be set to 3.1 A.

Reference // The motor book by Grundfos