Question Bank • QB_FM_062 NAT • 1 Mark

Section 2: Farm Machinery

Farm Machinery

Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses

Question Statement:

A $2.2\text{ m}$ wide seed drill operates at a forward speed of $5.4\text{ km/h}$. The field efficiency of the operation is $75\%$. Calculate the effective field capacity in $\text{ha/h}$.

Answer Key:

0.89

Step-by-Step Solution & Explanation

Theoretical Field Capacity ($TFC$): $TFC = \frac{W \times S}{10} = \frac{2.2 \times 5.4}{10} = 1.188\text{ ha/h}$ Effective Field Capacity ($EFC$): $EFC = TFC \times \frac{\eta}{100} = 1.188 \times 0.75 = 0.891\text{ ha/h}$
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