A three-phase,50 Hz, 4- pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to5% . The % speed regulation of the motor (rounded off to two decimal places) is ______

The correct answer is 4.21 (or 4.21%).

Step 1: Understand Speed Regulation in Induction Motors

Percent speed regulation of a motor is defined as the change in rotor speed from no-load to full-load, expressed as a percentage of the full-load speed:

$\text{\% Speed Regulation}=\frac{N_{\text{nl}}-N_{\text{fl}}}{N_{\text{fl}}}\times 100$

where:
- $N_{\text{nl}}$ is the no-load rotor speed.
- $N_{\text{fl}}$ is the full-load rotor speed.

Step 2: Calculate Synchronous Speed ($N_s$)

Given data:
- Supply frequency ($f$) = 50 Hz
- Number of poles ($P$) = 4

Using the formula for synchronous speed:

$N_s=\frac{120\times f}{P}$

$N_s=\frac{120\times 50}{4}=1500\text{rpm}$

Step 3: Calculate No-Load Rotor Speed ($N_{\text{nl}}$)

Given no-load slip ($s_{\text{nl}}$) = 1% = 0.01.

$N_{\text{nl}}=N_s(1-s_{\text{nl}})=1500\times (1-0.01)=1500\times 0.99=1485\text{rpm}$

Step 4: Calculate Full-Load Rotor Speed ($N_{\text{fl}}$)

Given full-load slip ($s_{\text{fl}}$) = 5% = 0.05.

$N_{\text{fl}}=N_s(1-s_{\text{fl}})=1500\times (1-0.05)=1500\times 0.95=1425\text{rpm}$

Step 5: Compute Percentage Speed Regulation

Substitute the speeds into the speed regulation formula:

$\text{\% Speed Regulation}=\frac{1485-1425}{1425}\times 100$

$\text{\% Speed Regulation}=\frac{60}{1425}\times 100\approx 4.21\text{\%}$

Thus, the percent speed regulation of the motor rounded off to two decimal places is 4.21.