Subtopic: Sources of power on the farm – human, animal, mechanical, electrical, wind, solar and biomass
Question Statement:
A pair of Haryana bullocks having an average body weight of $W_b = 450\text{ kg}$ per bullock is used for ploughing. Each bullock can exert a sustained continuous tractive pull equal to $10\%$ of its body weight over an 8-hour workday. The bullocks walk at an average continuous working speed of $v = 3.6\text{ km/h}$. Taking $g = 9.81\text{ m/s}^2$, the sustained power developed by the pair of bullocks in $\text{kW}$ is ________ (round off to 2 decimal places).
Answer Key:
0.88
Step-by-Step Solution & Explanation
```
Bullock Power Output:
Pair of bullocks: 2 animals
Weight per bullock: W_b = 450 kg
Sustained pull per bullock: 10% of body weight
Speed: v = 3.6 km/h = 1.0 m/s
```
**Step-by-step Calculation:**
1. Sustained draft pull exerted by each bullock ($P_1$):
$P_1 = 0.10 \times (m \times g) = 0.10 \times (450\text{ kg} \times 9.81\text{ m/s}^2) = 441.45\text{ N}$
2. Total draft pull exerted by the pair of bullocks ($P_{\text{pair}}$):
$P_{\text{pair}} = 2 \times P_1 = 2 \times 441.45\text{ N} = 882.90\text{ N}$
3. Continuous working speed in $\text{m/s}$:
$v = \frac{3.6 \times 1000}{3600} = 1.0\text{ m/s}$
4. Power developed by the pair of bullocks ($P$):
$P = P_{\text{pair}} \times v = 882.90\text{ N} \times 1.0\text{ m/s} = 882.90\text{ W} = 0.8829\text{ kW} \approx 0.88\text{ kW}$
This matches the standard rule of thumb that a pair of medium Indian draft bullocks develops approximately $1\text{ hp} \approx 0.75\text{--}0.90\text{ kW}$.