Question

An electric motor operates at full-load of 100 KW for 10 min, at 1 / 2 load for next 20 min, no-load for the next 20 min and this cycle repeats continuously. Find the continuous rating of the suitable motor.
1. \(\sqrt {1000} \;kW\)
2. \(\sqrt {300} \;kW\)
3. \(\sqrt {3000} \;kW\)
4. 75 kW

Answer 1

Correct Answer - Option 3 : \(\sqrt {3000} \;kW\)

Concept:

• Since the primary limitation for the operation of an electric motor is its temperature rise, hence motor rating is calculated on the basis of its average temperature rise.
• The average temperature rise depends on the average heating which itself is proportional to the square of the current and the time for which the load persists.

 

For example, if a motor carries a load L₁ for time t₁ and load L2 for time t₂ and so on, then

Average heating ∝ L₁² t₁ + L₂² t₂ + ..........+ L_n² t_n, and the rating of the motor is given as

∴ size of the motor \(= \sqrt {\frac{{L_1^2\;{t_1} + L_2^2\;{t_2} + \ldots + L_n^2\;{t_n}}}{{{t_1} + {t_2} + \ldots + {t_n}}}} \)

Calculation:

Let the full load L₁ = 100 kW operated for time t₁ = 10 min

Half-full load L₂ = L₁ / 2 = 50 kW operated for time t₂ = 20 min

No-load L₃ = 0 KW operated for t₃ = 20 min

∴ Rating of the motor\( = \sqrt {\frac{{{{100}^2} \times 10 + {{50}^2} \times 20}}{{10 + 20 + 20}}} \)

∴ Rating of the suitable motor is = \(\sqrt {3000} \;kW\)