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I know that mechanical efficiency is n= power output/power input, but I don't know how to find each of those. I have to find the efficiency at full load running speed.

I was only given mechanical losses (1.04kW), and that's why I'm confused because most equations talk about copper and stator losses and I'm not given them.

If anyone could help explain what equations to use that would be a great help!

Here's an update of other values I was given:

Input voltage= 400V

No. of phases= 3

Frequency= 50Hz

Running speed= 11.9(rev/s)

Total number of poles= 24

Rotor R per phase= 0.66(Ohms)

Rotor X per phase= 5.3(Ohms)

Effective rotor-stator turns ratio= 0.86:1

Mechanical losses= 1.04(kW)

% of running load at start-up= 83%

I don't know if you needed all of that but was not sure what you need to work out efficiency.

Here's an update of what I know so far:

Efficiency (n)= Pout/Pin

Efficiency= (Protor - Presistive rotor - Pmech losses)/(Protor - Presistive rotor)

And to find Protor (which is power flow in rotor)= P= VI. I got my current by doing Ir=s(N2/N1)E1/sqrt(r2^2+(sX2)^2) as it wasn't given.

To find Presistive rotor would I do Prr= ((3Protor)*(2I))-r.

So Effiecincy= ((Protor - Presistive rotor - Pmech losses)/(Protor - Presistive rotor))*100

Are these equations right?

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    \$\begingroup\$ It's 'kW' - 'K' is the SI symbol for the kelvin unit. 'w' is not an SI unit. You must have been given some further information to solve the problem. Please add it into your question using the edit link below it. \$\endgroup\$ – Transistor Jun 8 '19 at 12:16
  • \$\begingroup\$ It's a trick question about equation for "mechanical efficiency" only \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 8 '19 at 12:22
  • \$\begingroup\$ If you were told it is a 10 kW motor, or a motor described in question 2 as a 10 kW motor, that would mean the rated mechanical output is 10 kW and would tell you that the rated (full-load) output power is 10 kW and you could easily determine the mechanical efficiency - making this a "trick" question. More likely, but peculiarly, you need to work out the mechanical power developed assuming that the rotor voltage is input voltage x effective turns ratio. Another peculiar aspect is that you need to divide the total number of poles by the number of phases to get poles in the usual sense. \$\endgroup\$ – Charles Cowie Jun 8 '19 at 13:19
  • \$\begingroup\$ So are you saying to to work out mechanical power I just need to do 400V*0.86=344W for mechanical input? Where do the poles come into it? \$\endgroup\$ – Marie87 Jun 8 '19 at 13:34
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    \$\begingroup\$ I'm voting to close this question as off-topic because the question is a homework or study question without a demonstration of substantial effort to solve. \$\endgroup\$ – Charles Cowie Jun 8 '19 at 13:48
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Mechanical efficiency is the real output power over the developed output power. Thus, mechanical losses are basically friction losses. Keeping this in mind, you have to calculate the power flow in the rotor. Once you have the power in the rotor you should subtract the power lost due to resistance of the motor's winding. Your mechanical efficiency is then simply: (P_rotor-P_resistive_rotor-P_mech_losses)/(P_rotor-P_resistive_rotor).

Hint for calculating P_rotor. Draw the per phase equivalent circuit of the induction motor (using the transformer model). From the circuit and the given parameters you should be able to calculate P_rotor. If not then you can refer to this page: https://myelectrical.com/notes/entryid/251/induction-motor-equivalent-circuit to gain a better understanding of modelling induction motors.

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