Question Bank • QB_FP_105 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust

Question Statement:

A centrifugal governor on an agricultural engine operates at $1800\text{ rpm}$ at full load and rises to $1900\text{ rpm}$ at no load. Calculate the governor speed regulation (droop) in percentage.

Answer Key:

5.41

Step-by-Step Solution & Explanation

Governor speed regulation (droop): $R = \frac{N_1 - N_2}{N_{\text{mean}}} \times 100$ Where: - No load speed $N_1 = 1900\text{ rpm}$ - Full load speed $N_2 = 1800\text{ rpm}$ - Mean speed $N_{\text{mean}} = \frac{1900 + 1800}{2} = 1850\text{ rpm}$ $R = \frac{1900 - 1800}{1850} \times 100 = \frac{100}{1850} \times 100 \approx 5.405\%$ (Or based on rated speed $100/1800 = 5.56\%$).
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