Question

Calculate the horse power developed by a pair of bullocks in pulling U.P. No-2 plough at forward speed of 3 km/h, when the average draft at dynamometer indicated is 85 kgf.

• A. 0.94 hp
• B. \(9.4 \times 10^{-4}\) hp
• C. 3.4 hp
• D. \(3.4 \times 10^3\) hp

Hint:

To perform this calculation quickly during exams, use the combined formula:
\[ \text{Horsepower (hp)} = \frac{\text{Draft (kgf)} \times \text{Speed (km/h)}}{270} \]
(Since \(75 \times 3.6 = 270\) in the denominator).
\[ \text{hp} = \frac{85 \times 3}{270} = \frac{255}{270} \approx 0.94\text{ hp} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Power is defined as the rate of doing work.
In farm power calculations, the power developed by draft animals depends on the draft force they exert and their forward speed.
Key Formula or Approach:
1. Work done per second (kgf\(\cdot\)m/s):
\[ \text{Work done/sec} = \text{Draft (kgf)} \times \text{Speed (m/s)} \]
2. Speed conversion from km/h to m/s:
\[ v \text{ (m/s)} = \text{Speed (km/h)} \times \frac{1000}{3600} \]
3. Metric Horsepower (hp) conversion:
\[ 1 \text{ hp} = 75 \text{ kgf}\cdot\text{m/s} \]
\[ \text{Horsepower (hp)} = \frac{\text{Work done/sec (kgf}\cdot\text{m/s)}}{75} \]

Step 2: Detailed Explanation:
Let us identify the given values from the problem statement:
- Draft force, \(F = 85\text{ kgf}\)
- Forward speed, \(v = 3\text{ km/h}\)
First, convert the speed from km/h to m/s:
\[ v = 3 \times \frac{1000}{3600} = \frac{3000}{3600} = \frac{5}{6} \approx 0.833\text{ m/s} \]
Next, calculate the rate of work done (Power in kgf\(\cdot\)m/s):
\[ \text{Work done/sec} = F \times v = 85\text{ kgf} \times 0.8333\text{ m/s} \approx 70.833\text{ kgf}\cdot\text{m/s} \]
Now, convert this value into metric horsepower:
\[ \text{Horsepower (hp)} = \frac{70.833}{75} \approx 0.944\text{ hp} \]
Therefore, the horsepower developed by the pair of bullocks is approximately 0.94 hp.
This matches Option (A).

Step 3: Final Answer:
The correct option is (A).