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?
Best Answer
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.