The power input to a 500 V, 50 Hz, 6 pole, 3 phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW. If the total friction and windage losses are 2.025 kW, then the efficiency is ______%.

To find the efficiency of the induction motor, we can proceed step-by-step by determining the power flow through the stator and rotor stages to calculate the net output power.

Step 1: Calculate the synchronous speed ($N_s$)
The synchronous speed of the induction motor is given by the formula:
$N_s=\frac{120\times f}{P}$
Given:
Frequency, $f=50\text{Hz}$
Number of poles, $P=6$
Substituting these values:
$N_s=\frac{120\times 50}{6}=1000\text{rpm}$

Step 2: Calculate the slip ($s$)
The rotor speed is given as $N=975\text{rpm}$.
The slip of the induction motor is defined as:
$s=\frac{N_s-N}{N_s}$
Substituting the speed values:
$s=\frac{1000-975}{1000}=\frac{25}{1000}=0.025$

Step 3: Calculate the rotor input power ($P_g$)
The power input to the stator is $P_{\text{in}}=40\text{kW}$, and the stator losses are $1\text{kW}$.
Rotor input power is the stator input power minus the stator losses:
$P_g=P_{\text{in}}-P_{\text{stator loss}}$
$P_g=40\text{kW}-1\text{kW}=39\text{kW}$

Step 4: Calculate the gross mechanical power developed ($P_m$)
The gross mechanical power developed in the rotor is related to the rotor input power and slip by:
$P_m=(1-s)\times P_g$
Substituting the values:
$P_m=(1-0.025)\times 39\text{kW}$
$P_m=0.975\times 39\text{kW}=38.025\text{kW}$

Step 5: Calculate the net shaft output power ($P_{\text{out}}$)
Subtract the friction and windage losses to get the final mechanical power output at the shaft:
Given total friction and windage losses $=2.025\text{kW}$
$P_{\text{out}}=P_m-P_{\text{fw}}$
$P_{\text{out}}=38.025\text{kW}-2.025\text{kW}=36\text{kW}$

Step 6: Calculate the efficiency ($\eta$)
The overall efficiency of the induction motor is the ratio of net shaft output power to the total stator input power:
$\eta =\frac{P_{\text{out}}}{P_{\text{in}}}\times 100\%$
$\eta =\frac{36}{40}\times 100\%=90\%$

Therefore, the efficiency of the induction motor is 90%.