Question

Calculate the total time required to harvest 2.50 ha of forage crop with a mower having 2.0 m long cutter bar. The speed of operation is \(5 \text{ km}\cdot\text{h}^{-1}\) and the machine efficiency is 75 per cent.

• A. 1.55 h
• B. 3.33 h
• C. 2.22 h
• D. 4.66 h

Hint:

Remember the metric field capacity formula:
\[ \text{Field Capacity (ha/h)} = \frac{\text{Width (m)} \times \text{Speed (km/h)} \times \text{Efficiency (\%)}}{1000} \] This makes calculations direct and less prone to unit conversion errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The rate of field coverage by agricultural machinery is expressed as field capacity.
Theoretical field capacity assumes continuous, $100\%$ efficient operation.
Actual field capacity accounts for time losses during turning, clogging, and unloading, which are represented by the machine's field efficiency.
Key Formula or Approach:
The theoretical field capacity ($TFC$, in ha/h) is:
\[ TFC = \frac{W \cdot S}{10} \] The actual field capacity ($AFC$, in ha/h) is:
\[ AFC = TFC \cdot \eta \] The total operating time ($T$, in hours) is:
\[ T = \frac{\text{Area } (A)}{AFC} \] where:
$W = 2.0\text{ m}$ (width of cutter bar)
$S = 5\text{ km/h}$ (operating speed)
$\eta = 0.75$ ($75\%$ machine efficiency)
$A = 2.50\text{ ha}$ (total crop area)

Step 2: Detailed Explanation:
First, calculate the theoretical field capacity:
\[ TFC = \frac{2.0 \text{ m} \times 5 \text{ km/h}}{10} = 1.0 \text{ ha/h} \] Next, calculate the actual field capacity by applying the machine's efficiency:
\[ AFC = 1.0 \text{ ha/h} \times 0.75 = 0.75 \text{ ha/h} \] Finally, determine the total time required to harvest the field:
\[ T = \frac{2.50 \text{ ha}}{0.75 \text{ ha/h}} = 3.33 \text{ hours} \]

Step 3: Final Answer:
The total time required to harvest $2.50\text{ ha}$ is $3.33\text{ hours}$.