Answer:
a) - 0.064.
b) - 1240.14.
c) - 121.492.
Explanation:
Given,
Let the frequency to be f = 60 Hz
Let Pag = 25 kW and Pm = 23.4 kW.
a) - The motor slip during this moment is 0.064.
Apply the formula of the Pm.
[tex]Pm=(1-slip)\;Pag[/tex]
Then, put the value in the above equation.
[tex]23.4=(1-slip)\times 25[/tex]
[tex]1-slip=\frac{23.4}{25}[/tex]
[tex]slip=1-0.936=0.064[/tex].
b) - The induced torque within that motor is 124.14.
[tex]Ns=\frac{120f}{p} =\frac{120\times60}{4} =1800rpm[/tex]
[tex]\omega s=188.49[/tex]
Then, we have to apply the formula of the induced torque.
[tex]Induced\;Torque=\frac{Pm}{\omega s}=124.14[/tex]
c) - Let the mechanical losses to m = 300 W.
[tex]Then,\;we\;have\;to\;find\;the\;out put\;power=23200-300=22900[/tex]
[tex]and,\;apply\;the\;formula\;of\;the\;load\;torque =\frac{22900}{\omega s} =\frac{22900}{188.49}=121.492[/tex]
