## 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**= 5.70 – 5.00/3.30 – 2.90 A_{n}

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)

**The voltage ratio is determined by the following equations:**

U_{Δ} = (U_{A} – U_{min}) / (U_{max} – U_{min})

which is in this case: **U _{Δ} =** (254 – 220) / (227 – 220) =

**0.6**

U_{Y} = (U_{A} – U_{min}) / (U_{max} – U_{min})

which is in this case: **U _{Y} =** (440-380) / (480-380) =

**0.6**

So, **U _{Δ} = U_{Y}**

#### 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.

**Rule-of-thumb:**The external motor overload relay is always set to the nominal current shown on the nameplate.

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

Shafi

Good explanation . Need specific value of current at which relay should trip the motor with 440V Y and 3.1A setting.

Also plz explain if overload setting is made at nominal current of motor, mentioned on nameplate . Then what is the use of 1.15 or 1.25 times rated current setting. Thanks

SLTAN TEKLAY

I WANT TO DOWNLOAD ELECTRICAL VIDEOS

SAYAK DAS

THANK YOU SIR FOR THE EXPLANATION .I WANT TO KNOW IF THE FREQUENCY IS 50 HZ THEN CAN I USE THIS FORMULA? PLEASE GIVE ME THE ANSWER

THANK YOU

Umar

I need more diagrams about motor connections

Glen Sullivan

Thank you for a good explanation. I have a couple of people I work with that believe if you have a fast acting overtemp relay, you dont need overloads. If it gets hot, it’s overloaded. Not very fast enough for lost phases or locked rotors. I am not sure if it’s even legal as the standard calls for overload fitted to motors over 370w. Has anyone here got a correct answer?

Samir Ismail

Dear Sullivan

Normally the Fast Acting O.T.R. have an embedded sensor(s) in the motor armature coils, suppose for some reasons these sensors are no more working, then your motor is not protected at all.

It’s preferable to use the usual Over Load protection mean as the “Main Protection One” and the F.A.O.T.R. as a “Backup”, in this case you realize a complete protection of your motor against the overload events.