Ø Resistance of the soil is expressed as specific draft. This is the force required to cut a unit cross-sectional area of soil as shown in Table 1. The specific draft is multiplied by the width and depth of cut to get the draft.
Ø The increase in draft due to speed as shown in Table 2 is applied to that part of the total required for turning and pulverizing the furrow slice.
Table 1. Specific draft of different soils
|
SOIL TYPE
|
SPECIFIC
DRAFT, SD
|
|
|
Lbs/in2
|
Kg/cm2
|
|
|
Silty loam
Clay loam
Heavy clay
Virgin soil, clay
Gumbo, moist
Dry adobe
|
3
3-6
5-7
6-8
10-11
12-15
16-18
18-20
|
0.21
0.21-0.42
0.35-0.49
0.42-0.56
0.70-0.77
0.85-1.06
1.13-1.27
1.27-1.41
|
Table 2. Increase in draft due to speed
|
SPEED
|
DRAFT, %
|
|
|
MPH
|
KPH
|
|
|
1
2
3
4
5
6
|
1.6
3.2
4.8
6.4
8.0
9.6
|
100
114
128
142
156
170
|
Sample
Problem 1: Determine the hectares plowed per
hour when a tractor is operating at 6.4 kph and is pulling four 36-cm moldboard
bottoms at a depth of 20 cm. How many
hectares can be plowed in 10 hours if field efficiency is 78 percent? If the soil is clay loam, what is the draft
required to work the soil? Draft hp requirement? Tractor hp requirement?
Solution:
a.
Hectares
plowed in 10 hours:
where: C = Field Capacity, Ha/hr
S = Speed, Kph
W = Width of cut, m
Eff = Field Efficiency, decimal
b. Draft
requirement based on soil type:
where:
Ds = Draft, kg
Specific
Draft = Kg/cm2
from Table 1
W = width of cut, cm
D
= depth of cut, cm
Ds = (0.49
Kg/cm2)(4x36 cm)(20 cm) = 1,411 Kg
c. Adjusted
draft requirement due to speed of plowing (from Table 2):
Da
= 1,411 kgx1.42 =
2,004 Kg
d. Draft
horsepower requirement:
where: Da
= Adjusted Draft, Kg
S = Speed, Kph
=
46.8 hp
e. Tractor
Horsepower:
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