This question asks us to find the value of the slip for a 3-phase, 4-pole, 50 Hz induction motor running at a specific speed. Understanding slip is crucial for analysing the performance of an induction motor.
Understanding Induction Motor Slip
An induction motor operates on the principle of electromagnetic induction. The rotating magnetic field created by the stator winding induces current in the rotor, causing it to rotate. However, for induction to occur, the rotor speed must always be less than the speed of the rotating magnetic field. This difference in speed is known as slip.
Slip is typically expressed as a fraction or a percentage of the synchronous speed. The synchronous speed is the speed of the rotating magnetic field.
Calculating Synchronous Speed ($N_s$)
The synchronous speed of the magnetic field in an AC motor is determined by the frequency of the power supply and the number of poles in the motor's stator winding. The formula for synchronous speed ($N_s$) in revolutions per minute (r.p.m.) is:
$$N_s = \frac{120f}{P}$$
Where:
- $f$ is the supply frequency in Hertz (Hz)
- $P$ is the number of poles
Given in the question:
- Frequency ($f$) = 50 Hz
- Number of poles ($P$) = 4
Let's calculate the synchronous speed:
$$N_s = \frac{120 \times 50}{4}$$
$$N_s = \frac{6000}{4}$$
$$N_s = 1500 \text{ r.p.m.}$$
So, the synchronous speed of the rotating magnetic field is 1500 r.p.m.
Calculating Motor Slip (s)
Slip ($s$) is defined as the difference between the synchronous speed ($N_s$) and the rotor speed ($N_r$), divided by the synchronous speed. The formula for slip is:
$$s = \frac{N_s - N_r}{N_s}$$
Given in the question:
- Synchronous speed ($N_s$) = 1500 r.p.m. (calculated above)
- Rotor speed ($N_r$) = 1455 r.p.m.
Let's calculate the slip as a fraction:
$$s = \frac{1500 - 1455}{1500}$$
$$s = \frac{45}{1500}$$
Now, we can simplify the fraction:
$$s = \frac{45 \div 15}{1500 \div 15} = \frac{3}{100}$$
So, the slip as a fraction is 0.03.
Converting Slip to Percentage
To express slip as a percentage, we multiply the fractional slip by 100%:
$$\text{Slip percentage} = s \times 100\%$$
$$\text{Slip percentage} = 0.03 \times 100\%$$
$$\text{Slip percentage} = 3\%$$
The value of the slip for the given 3-phase induction motor is 3%.
Let's compare this result with the given options:
| Option | Value |
|---|---|
| 1 | 2% |
| 2 | 3% |
| 3 | 4% |
| 4 | 5% |
Our calculated slip value of 3% matches Option 2.
Revision Table: Induction Motor Fundamentals
| Term | Definition | Formula (where applicable) |
|---|---|---|
| Synchronous Speed ($N_s$) | Speed of the rotating magnetic field in the stator. | $N_s = \frac{120f}{P}$ (r.p.m.) |
| Rotor Speed ($N_r$) | Actual mechanical speed of the rotor shaft. | Always less than $N_s$ during normal motor operation. |
| Slip ($s$) | Relative speed difference between $N_s$ and $N_r$, expressed as a fraction or percentage of $N_s$. | $s = \frac{N_s - N_r}{N_s}$ |
| Frequency ($f$) | Frequency of the AC power supply. | - |
| Poles ($P$) | Number of magnetic poles in the stator winding. | - |
Additional Information on Induction Motor Slip
Slip is a critical parameter in understanding induction motor operation:
- Starting: At the moment of starting, the rotor is stationary ($N_r = 0$). Therefore, the slip is $s = \frac{N_s - 0}{N_s} = 1$ or 100%. High slip at start leads to high induced voltage and current in the rotor, producing high starting torque.
- Running under Load: As the motor takes on load, the rotor speed ($N_r$) decreases slightly. This increases the slip ($s$). An increase in slip leads to a stronger induced current in the rotor, which in turn generates more torque to meet the load demand.
- No-load Running: When the motor runs with no mechanical load, the rotor speed ($N_r$) is very close to the synchronous speed ($N_s$). The slip is very small (typically 1% to 2%).
- Efficiency: Slip represents the speed difference that causes rotor losses (mostly rotor I²R losses). A higher slip generally means higher rotor losses and lower efficiency for a given torque.
- Frequency of Rotor Current: The frequency of the current induced in the rotor is directly proportional to the slip frequency ($f_r = s \times f$). At synchronous speed ($s=0$), the rotor frequency is zero, meaning no current is induced, and thus no torque is produced. This is why the rotor must run slower than synchronous speed.
The slip value is a direct indicator of how much the rotor is lagging behind the rotating magnetic field, and consequently, the torque being produced by the motor.
Important Questions from Miscellaneous-Engineering
- Which one of the following lists identifies the wastes from common manufacturing and industrial process, such as solvents that have been used in cleaning or degreasing operations?
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly
- Which one of the following is a time-dependent reversible process in which materials under constant composition and volume soften when remolded?
- The compressibility of a saturated, clay-water system is determined by means of the apparatus devised by Terzaghi known as
