ABE TUTORIALS

AGRICULTURAL AND BIOSYSTEMS ENGINEERING

PROBLEM:
THE TOTAL DRAFT OF A 4-BOTTOM 41-CM MOLDBOARD PLOW WHEN PLOWING 18 CM DEEP AT 6 KPH WAS 15 KN.

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Calculate the specific draft (N/cm²).

a. 20.3
b. 1.27
c. 720
d. 5.08

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Step 1: Calculate the total cross-sectional area.
Area = Number of bottoms * Width * Depth
Area = 4 * 41 cm * 18 cm = 2,952 cm²

Step 2: Calculate Specific Draft.
Specific Draft = Total Draft (converted to N) / Area
Specific Draft = 15,000 N / 2,952 cm² ≈ 5.08 N/cm²

Result: d. 5.08

What is the actual power requirement for the above problem in kW?

a. 33.5
b. 25
c. 67
d. 50

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Step 1: Convert speed to meters per second (m/s).
Speed = 6 kph / 3.6 = 1.667 m/s

Step 2: Calculate Power.
Power = Draft (kN) * Speed (m/s)
Power = 15 kN * 1.667 m/s = 25 kW

Result: b. 25

If the field efficiency for the above problem is 57%, what is the rate of work in ha/hr?

a. 1.3
b. 0.33
c. 0.19
d. 0.56

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Step 1: Calculate Theoretical Field Capacity (TFC).
TFC = (Total Width [m] * Speed [kph]) / 10
Total Width = (4 bottoms * 41 cm) = 1.64 m
TFC = (1.64 * 6) / 10 = 0.984 ha/hr

Step 2: Calculate Actual Field Capacity (AFC).
AFC = TFC * Field Efficiency
AFC = 0.984 * 0.57 ≈ 0.56 ha/hr

Note: 0.56 ha/hr is the mathematically correct answer for the given parameters.