A specially set 40 cm Five-bottom disc plow was used for a ploughing operation. If there was a 5% overlap in cutting width and the entire disc formation left a trapezoidal furrow cross section having a top width of 310 cm and a depth of 10 cm. Unit draft was 0.6 kgcm-2 while operational speed was 5 km/h.
1. Determine the volume of soil worked in 3 hours
2. What horse power would be required to the job?
3. What total draft would have been experienced by the plough if there was no overlap in total cutting width?
4. What volume of soil would be worked in 4 hours if the plough was set normally, that is, giving normal furrow cross section?
1. To determine the volume of soil worked in 3 hours, we need to calculate the area of the furrow cross section and multiply it by the distance traveled by the plow in 3 hours.
First, let's find the area of the trapezoidal furrow cross section. The formula for the area of a trapezoid is (base1 + base2) multiplied by the height, divided by 2.
Base1 = top width = 310 cm
Base2 = base width of the disc plow = 40 cm
Height = depth of the furrow = 10 cm
Area = (310 + 40) * 10 / 2 = 3500 cm^2
Now, let's convert the area to square meters by dividing by 10,000 (since 1 square meter = 10,000 square centimeters).
Area = 3500 cm^2 / 10,000 = 0.35 m^2
Next, let's calculate the distance traveled by the plow in 3 hours. The formula for distance is speed multiplied by time.
Speed = 5 km/h
Time = 3 hours
Distance = 5 km/h * 3 hours = 15 km
Now, we can calculate the volume of soil worked by multiplying the area by the distance.
Volume = Area * Distance = 0.35 m^2 * 15 km = 5.25 m^3
Therefore, the volume of soil worked in 3 hours is 5.25 cubic meters.
2. To calculate the horsepower required for the job, we can use the formula: horsepower = (draft * speed * distance) / (75 * time)
Draft = unit draft = 0.6 kg/cm^2 = 0.6 kg/(10,000 cm^2)
Speed = 5 km/h
Distance = 15 km
Time = 3 hours
Plugging in the values, we get:
Horsepower = (0.6 kg/(10,000 cm^2) * 5 km/h * 15 km) / (75 * 3 hours)
= (0.6 kg/100 m^2 * 5 km/h * 15 km) / (75 * 3 hours)
= (0.003 kg/m^2 * 75 km/h) / 75
= 0.003 kg/m^2 * 1,000 m/3600 s
= 0.000833 kg/m^2/s
So, the horsepower required for the job is approximately 0.000833 horsepower.
3. If there was no overlap in the total cutting width, the total draft would still be the same as the unit draft (0.6 kg/cm^2).
4. To calculate the volume of soil worked in 4 hours if the plow was set normally, we need to calculate the area of the furrow cross section and multiply it by the distance traveled by the plow in 4 hours.
Using the same formulas and values as in question 1 (except for time), we can calculate:
Distance = 5 km/h * 4 hours = 20 km
Volume = Area * Distance = 0.35 m^2 * 20 km = 7 m^3
So, the volume of soil worked in 4 hours with the plow set normally is 7 cubic meters.