Electric Drives in Utilization Of Electrical Energy : Practice MCQ Questions & Answers with Explanations

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 +..........+ Ln2 tn, and the rating of the motor is given as ∴ size of the motor $= \sqrt {\frac{{L1^2\;{t1} + L2^2\;{t2} + \ldots + Ln^2\;{tn}}}{{{t1} + {t2} + \ldots + {tn}}}}$ Calculation: Let the full load L1 = 200 kW operated for time t1 = 10 min Half-full load L2 = L1 / 2 = 100 kW operated for time t2 = 20 min From above concept, ∴ Rating of the motor$= \sqrt {\frac{{{{200}^2} \times 10 + {{100}^2} \times 20}}{{10 + 20}}} =\sqrt{\frac{60\times 10^4}{30}}=100\sqrt {2}$ kW