1.
Recommend the KW-rating of a standard 4-wheel tractor that when attached to the proper
implement could perform any of the operation shown below under the given conditions:
OPERATION
IMPLEMENT
Plowing
4-30 cm disk plow at
15 cm plowing depth
3-m width spike tooth
Harrowing
harrow
Cultivation
4-row cultivator
Furrowing
3-row furrower
DRAFT REQ'T
10 N/sq. cm.
SPEED
5 kph
1.75 KN/m
1.70 KN/cultivator
1.60 KN/furrower
bottom
8 kph
7.5 kph
8 kph
Assume that the tractor would operate at 75% of its rated Drawbar Power
Sol’n:
Plowing:
𝐷𝐵𝑃 = 𝑆𝑑𝐴 →
1000
(5𝑘𝑝ℎ × (3600)) (10𝑁/𝑠𝑞. 𝑐𝑚)(4 × 30𝑐𝑚 × 15𝑐𝑚)
1000
= 25𝐾𝑊
Harrowing:
𝐷𝐵𝑃 = 𝑆𝑑𝐴 → [8𝑘𝑝ℎ × (
1000
)] (3𝑚)(1.75𝐾𝑁/𝑚) = 11.67𝐾𝑊
3600
Cultivation:
𝐷𝐵𝑃 = 𝑆𝑑𝐴 → [7.5𝑘𝑝ℎ × (
1000
)] (4 − 𝑟𝑜𝑤 𝑐𝑢𝑙𝑡)(1.70𝐾𝑁/𝑐𝑢𝑙𝑡) = 14.17𝐾𝑊
3600
Furrowing:
𝐷𝐵𝑃 = 𝑆𝑑𝐴 → [8𝑘𝑝ℎ × (
1000
)] (3 − 𝑟𝑜𝑤 𝑓𝑢𝑟)(1.60𝐾𝑁/𝑓𝑢𝑟) = 10.67𝐾𝑊
3600
Since Plowing is the highest calculated DBP, choose it because it satisfies the power
requirement of other farm operation.
For the tractor to operate 75% only of its rated DBP,
𝑇𝑟𝑎𝑐𝑡𝑜𝑟 𝐷𝐵𝑃 =
25𝐾𝑊
= 33.33𝐾𝑊
0.75
Hence, you may recommend 35KW Tractor.