Question

Speed of a 3-\(\phi\), 4-pole, 60 Hz induction motor at 75% of full-load is 1700 rpm. The speed at full-load can be

• A. \( 1750\text{ rpm} \)
• B. \( 1800\text{ rpm} \)
• C. \( 1700\text{ rpm} \)
• D. \( 1600\text{ rpm} \)

Hint:

An induction motor always slows down as mechanical load is added. $$\text{No-Load Speed} > \text{Partial-Load Speed} > \text{Full-Load Speed}$$ Since the speed at $75%$ load is $1700\text{ rpm}$, the full-load speed must be less than $1700\text{ rpm}$. This eliminates options (1), (2), and (3) without requiring complex slip calculations.

Answer 1

The Correct Option is D

Concept: The synchronous speed ($N_s$) of a three-phase induction motor depends on the supply frequency ($f$) and the number of stator poles ($P$): $$N_s = \frac{120 \cdot f}{P}$$ The actual rotor operating speed ($N$) of the induction motor is less than the synchronous speed due to slip ($s$), expressed as: $$N = N_s(1 - s)$$ As the mechanical load on the induction motor increases from a partial load (e.g., $75%$) to full-load ($100%$):
• The motor must produce more torque to balance the load, which requires a larger rotor current.
• To induce a larger current, the rotor must slow down slightly relative to the rotating magnetic field, increasing the slip ($s$).
• Therefore, as load increases, the actual rotor speed $N$ decreases.

Step 1: Calculate the synchronous speed of the machine.
Given values:
• Number of poles, $P = 4$
• Frequency, $f = 60\text{ Hz}$ $$N_s = \frac{120 \cdot 60}{4} = \frac{7200}{4} = 1800\text{ rpm}$$

Step 2: Compare the partial-load speed with the synchronous speed.
At $75%$ of full-load, the speed is given as $1700\text{ rpm}$. This is less than $1800\text{ rpm}$, which is correct for normal motor operation.

Step 3: Deduce the speed behavior at full-load.
When the load increases from $75%$ to $100%$ full-load: $$\text{Load} \uparrow \quad \Rightarrow \quad \text{Slip } s \uparrow \quad \Rightarrow \quad \text{Rotor Speed } N \downarrow$$ This means the rotor speed at full-load must be strictly less than the partial-load speed of $1700\text{ rpm}$: $$N_{\text{full-load}} < 1700\text{ rpm}$$ Let us evaluate the available choices based on this constraint:
• (1) $1750\text{ rpm}$ (Incorrect, higher than $1700\text{ rpm}$)
• (2) $1800\text{ rpm}$ (Incorrect, equal to synchronous speed)
• (3) $1700\text{ rpm}$ (Incorrect, equal to $75%$ load speed)
• (4) $1600\text{ rpm}$ (Correct, lower than $1700\text{ rpm}$) The only physically valid speed option below $1700\text{ rpm}$ is $1600\text{ rpm}$. Therefore, option (4) is correct.