ABE TUTORIALS

AGRICULTURAL AND BIOSYSTEMS ENGINEERING

PROBLEM:
HOW LONG (IN MIN) WILL IT TAKE A FARMER TO PLOW 60 HECTARES WITH FIVE-DISK PLOW PULLED BY A TRACTOR AT 6 KILOMETERS PER HOUR? THE EFFECTIVE WIDTH OF CUT IS 35.56 CM PER DISK AND EFFICIENCY IS 80%.

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a. 5.85 | b. 70.3 | c. 1.2 | d. 140.6 | e. nota

View Solution

Step 1: Total Width of Cut (W)
W = 5 disks × 0.3556 m/disk = 1.778 m

Step 2: Effective Field Capacity (EFC)
EFC = (W × S × Ef) / 10 = (1.778 × 6 × 0.80) / 10 = 0.85344 ha/hr

Step 3: Time Required
Time (hr) = 60 ha / 0.85344 ha/hr ≈ 70.3 hr
Time (min) = 70.3 hr × 60 min/hr = 4218 min

Result: e. nota (Calculated value is 4218 min. If choices were in hours, 70.3 would be 'b')

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IF THE DEPTH OF CUT OF PROBLEM ABOVE IS 20.32 CM AND THE UNIT DRAFT IS 0.493 KG/CM², WHAT IS THE REQUIRED DRAW-BAR HORSEPOWER?

a. 7.8 | b. 0.71 | c. 784 | d. 3920 | e. nota

View Solution

Step 1: Cross-sectional Area of Cut (A)
A = Total Width (cm) × Depth (cm)
A = (5 × 35.56 cm) × 20.32 cm = 177.8 cm × 20.32 cm ≈ 3612.9 cm²

Step 2: Total Draft (D)
D = Area × Unit Draft = 3612.9 cm² × 0.493 kg/cm² ≈ 1781.16 kg

Step 3: Drawbar Horsepower (DBHP)
DBHP = (Draft × Speed) / 270
DBHP = (1781.16 kg × 6 kph) / 270 ≈ 39.58 hp

Result: e. nota (Closest choice c/d are off by factors of 10 or 100)