$2.)$ The power supplied to a three$-$phase induction motor is $50$ kW and the stator losses are $2kW.$ If the slip is $4\%,$ determine:

$($a$)$ the rotor copper loss,

Input power to rotor $=$ stator input power $-$ stator losses

Slip $=\frac{\text{r}\text{o}\text{t}\text{o}\text{r}\text{c}\text{o}\text{p}\text{p}\text{e}\text{r}\text{l}\text{o}\text{s}\text{s}}{\text{r}\text{o}\text{t}\text{o}\text{r}\text{i}\text{n}\text{p}\text{u}\text{t}}$

rotor copper loss $=$

$($b$)$ the mechanical power developed by the rotor,

Total mechanical power developed by the rotor $=$ rotor input power $-$ rotor losses

$($c$)$ the output power of the motor if friction and windage losses are $1$ kW $,$ and

Output power of the motor $=$ power developed by the rotor $-$ friction and windage losses

$($d$)$ the efficiency of the motor, neglecting rotor iron losses.

Efficiency, $=(\frac{\text{o}\text{u}\text{t}\text{p}\text{u}\text{t}\text{p}\text{o}\text{w}\text{e}\text{r}}{\text{i}\text{n}\text{p}\text{u}\text{t}\text{p}\text{o}\text{w}\text{e}\text{r}})100\%=$

$3.)$ A $10$ kVA transformer steps up the voltage from $240$ V to $2400$ V at $60$ Hz and has the following resistance values of its windings. A $10$ kVA transformer steps up the voltage from $240$ V to $2400$ V at $60Hz$ and has the following resistance values of its windings:

$x_{11}=20,x_{22}=2000,r_1=0,01,x_1=0,06,x_2=6,r_2=1$

Determine, using the equivalent circuit coupled through mutual inductances, the primary voltage that is necessary to apply to the primary terminals to have nominal voltage in the secondary$,$ when the transformer delivers full load, at a power factor of $0\mathrm{.}9$ lagging.

Solution

The equivalent circuit is the following: