A 3-bottom moldboard plow of width $30\text{ cm}$ per bottom is operating at a depth of $15\text{ cm}$ with a forward speed of $4.5\text{ km/h}$. If the specific draft of the soil is $0.6\text{ kg/cm}^2$, calculate the drawbar power required by the plow in $\text{kW}$. (Take acceleration due to gravity $g = 9.81\text{ m/s}^2$)
Answer Key:
9.93
Step-by-Step Solution & Explanation
Given:
- Number of bottoms $n = 3$
- Width per bottom $w = 30\text{ cm}$
- Total width of cut $W = 3 \times 30 = 90\text{ cm}$
- Depth of cut $d = 15\text{ cm}$
- Cross-sectional area of furrow slice $A = W \times d = 90 \times 15 = 1350\text{ cm}^2$
- Specific draft $D_s = 0.6\text{ kg/cm}^2$
Total draft in $\text{kgf}$:
$D = A \times D_s = 1350 \times 0.6 = 810\text{ kgf}$
In Newtons:
$D = 810 \times 9.81 = 7946.1\text{ N}$
Forward speed $v$:
$v = 4.5\text{ km/h} = \frac{4.5 \times 1000}{3600} = 1.25\text{ m/s}$
Drawbar power required:
$P = \frac{D \times v}{1000} = \frac{7946.1 \times 1.25}{1000} \approx 9.93\text{ kW}$