Waterfall & Motor Overload!!

[OnTheTools]

As part of a large waterfall installation I'm installing a single phase 0.75kW motor with time clock and need advice on the overload setting please! I was told anything above 0.5hp/0.37kW required a contactor with thermal overload?!

I'm aware the 3 phase calculation would be...
kW/ Vl x sqrt 3 x p.f but I'm unsure of what to do on a single phase installation.

Does it even require a contactor would you guys just wire it straight up from the time clock? Help and thanks!

 

[GMES]

The op's profile states he is a Trainee, go easy on him please chaps.

 

[Sparkie30]

As part of a large waterfall installation I'm installing a single phase 0.75kW motor with time clock and need advice on the overload setting please! I was told anything above 0.5hp/0.37kW required a contactor with thermal overload?!

I'm aware the 3 phase calculation would be...
kW/ Vl x sqrt 3 x p.f but I'm unsure of what to do on a single phase installation.

Does it even require a contactor would you guys just wire it straight up from the time clock? Help and thanks!

Tel got it bang on as usual.

 

[davesparks]

Unless the rules have changed since I trained under the 16th then any motor of 0.75KW or more requires a motor starter (not necessarily a contactor and thermal overload relay, other options are available)

 

[telectrix]

I =P/V. i.e. 750/240 = 3.125A.

 

[telectrix]

read this:

Fractional / Integral Horsepower Manual Motor Starters

 

[Marvo]

I =P/V. i.e. 750/240 = 3.125A.

The wattage or kilowatt rating of a motor refers to it's shaft power and not the electrical consumption. It is possible to work backwards from the kilowatt rating to get the electrical consumption but you need to know lots of info like efficiency and PF etc and many of these are variable depending on the mechanical load.

Take the maximum run current from the info plate on the motor itself or for the manufacturers data sheet and set your thermal overload at that figure.

 

[OnTheTools]

I =P/V. i.e. 750/240 = 3.125A.

Yes please go easy on me! I haven't been long out of college long.. Now I feel a fool that was a lot simpler than expected haha. Appreciated though!

 

[Marvo]

A 0.75kW (1HP) single phase motor should have a maximum run current in the region of between 6 and 7 Amps unless it's a newer generation high efficiency thing. As per previous post the kW rating refers to mechanical output so don't use it to calc the run current.

It was far less confusing when motors were sized in Horsepower rather than kW.

 

[pennychew]

Tel got it bang on as usual.

How did you predict Tel would get it spot on!!

(Although to be fair he always does!)

 

[OnTheTools]

Please explain what your on about?

 

[Sparkie30]

How did you predict Tel would get it spot on!!

(Although to be fair he always does!)

because I got wrong 1st and edited my post mate.

 

[pennychew]

Ahh right!

 

[Marvo]

**sigh**

Maybe I'm not explaining this very well.

The fact it's called a 0.75 kW motor means it's power output is 750 watts or basically 1 horsepower.

The 0.75kW does not refer to the electrical input power requirement of the motor.

You can't use this figure in the I=P/V formula for the same reason you can't specify a run current for a horse that's lifting a 75Kg weight one meter in height in one second.

 

[rapparee]

In other words a motor is not 100% efficient