General help on a single phase motor | on ElectriciansForums

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M

markhughes

I have a single phase motor showing the following information

P2 0.18Kw 50Hz
230 Volts
1.70Amps
Cos 0.97

How does the 0.18Kw work out?

My working is P=I*V*PF - P=1.70*230*0.97
P=379.27w not 1.8Kw

Help please!
 
Your calculation gives the input electrical power, but P2 is the output mechanical power of 180W or approx. 1/4 hp, so you also need to multiply by the efficiency to get this figure.
 
A common missing factor in motor calculations is efficiency and it seems this hasn't been included in yours..

KW= I x V x PF x Eff

or

I = KW / (V x PF x Eff)

As you don't have all the info here then you equation will be out, by what margin depends on the motor and its normal losses.
 
My working is P=I*V*PF .
You forgot efficiency P=I*V*PF*Efficiency. I also think they're being very optimistic with the PF as 0.97 as well. Finally they stated the output as 0.18kW, not 1.8kW that you stated after your calculation.

**edit** Oops, only saw the first reply before I posted.
 
Last edited:
The efficiency calc is correct but your units are suspect.

Nm would usually be used to express Neuton Meters which is a measure of torque and not a unit of efficiency.

η Is the lower case Greek letter eta, not the normal English letter n. η[SUB]m[/SUB] is used to denote efficiency and isn't a unit of it....if that makes sense.

Efficiency is a ratio and it doesn't have any units.
 
Last edited:
The efficiency calc is correct but your units are suspect.

Nm would usually be used to express Neuton Meters which is a measure of torque and not a unit of efficiency.

η Is the lower case Greek letter eta, not the normal English letter n. η[SUB]m[/SUB] is used to denote efficiency and isn't a unit of it....if that makes sense.

Efficiency is a ratio and it doesn't have any units.

Thanks for that explanation.

So if I understand this correctly, would I be correct in saying :-

This motor is rated as having an output of 0.18kW, however due to it's efficiency (or inefficiency) uses 0.379kW of power to work.
 
Thanks for that explanation.

So if I understand this correctly, would I be correct in saying :-

This motor is rated as having an output of 0.18kW, however due to it's efficiency (or inefficiency) uses 0.379kW of power to work.

I think your confusion here is you trying to match 2 different meanings... A 0.18KW motor refers to the mechanical energy output available at the shaft which is sometimes given in horse power (which is less confusing), the KW rating of a motor is not the electrical wattage that is used to power the motor so a 1kw motor will not use 1kw of electrical energy at full load.
 
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