1) The document discusses factors to consider when selecting a power unit, such as a tractor, to match the size and type of machines used for field operations. 2) Key factors include the engine type, power ratings, soil resistance, tractor and implement sizes, and matching implements to the tractor's power output to avoid overloading. 3) Power is measured in horsepower or kilowatts, with drawbar, PTO, and brake power representing the tractor's pulling, power take-off, and maximum engine power respectively. The document provides formulas to calculate power needs based on implement size, soil conditions, and operating speed.

1. MACHINE POWER
REQUIREMENT

2. Introduction
• To match power units to the size and type of
machines so all field operations can be carried out
on time with a minimum of cost
• If tractor is oversized for implements, the costs
will be excessive for the work done
• If the implements selected are too large for the
tractor, the quality or quantity of work may be
lessened or the tractor will be overloaded usually
causing expensive breakdowns

3. Factors to consider when selecting
a power unit
1. Engine type
2. Power ratings
3. Soil resistance to machines
4. Tractor size
5. Matching implements
6. Sizing for critical work

4. Engine type
• The combustion process in the cylinders
converts the energy contained in fuel to a
rotating power source
• This rotating power source can be further
converted into 3 forms:
– Drawbar Pull
– PTO output
– Hydraulic System Output

5. Power Ratings
• Power is a measure of the rate at which work is
being done
• The English power unit is defined as 550 foot-
pounds of work per second
• The metric power unit is measured in kilowatts
(kW)
1 kW = 1.34 horsepower

6. If a load require a force of 20 pounds to move it
vertically a distance of 3 feet, the amount of work
done is :
Work = Force x Distance
= 20 lbs x 3 feet
= 60 ft. lb
The amount of work done is 60 ft.lb with no
reference to time.

7. If a 1000 lb force is moved 33 feet in one minute, the rate of doing
work is one horsepower, because one horsepower equals 33,000 ft.
lb. per minute
The equivalent rate of work in one second to equal 1 horsepower is :
1 HP = 33,000 ft.lb = 550 ft. lb per second
60 seconds
When working with field machinery, we usually think of miles per
hour and pounds of draft. For these conditions the formula for
horsepower is :
HP = Force,lb x Speed, mph
375

8. Metric Equivalent
Metric unit for power is kilowatt (kW)
Force is measured in newtons or kilonewtons
1 HP = 0.746 kW
1 kW = 1.34 HP
1N = 0.225 lb
1 kN = 224.8 lb force
Formula for kilowatt is:
KW = Force (kilonewtons) x Speed (km/hr)
3.6

9. Eg: If the draft of a trailing implement, such as a
disk harrow, is measured at 11.1 kilonewtons and
is pulled at a speed of 8 km/hr, what is the
drawbar kilowatt ?
Drawbar, kW = 11.1 kN x 8 km/hr =24.7 kW
3.6

10. The formula can also be used to determine
speed.
Eg: A tractor is pulling a plow with a total draft
of 22.2 kilonewtons. How fast can the plow be
pulled if the tractor has 50 drawbar kilowatts ?
Speed = 50 kW x 3.6 = 8.1 km/hr
22.2 kN

11. To determine draft :
Eg: Given 65 kW tractor, speed 8 km/hr., field cultivator
draft is 4 kN per meter of width when used in a given
field. What width of cultivator could be pulled ?
Draft = Power,kW x 3.6 = 65 kW x 3.6 = 29.25 kN
Speed, km/hr 8 km/hr
Width, meters = Total Draft = 29.25 = 7.3 meters
Draft per meter 4

12. Eg: If the draft of a trailing implement like a
disk harrow is measured at 2,500 pounds and
is pulled at a speed of 5 mph, what is the
drawbar horsepower ?
Drawbar horsepower = Force, lbs x speed,mph
375
= 2500 x 5
375
= 33.3

13. • This formula can be used to determine how fast
an implement could be pulled with a given size of
tractor
Speed = Drawbar horsepower x 375
Draft, lbs
• This formula can also be used to determine how
large an implement can be pulled but an extra step
or two is involved.
• Size of the implement have to be related to the
amount of soil resistance

14. Determining tractor size needed
• There are various kind of power
– Brake
– PTO
– Drawbar
• Tractor power is measured in horsepower
(USA) or in kilowatts (kW)- metric
equivalents

15. Brake Horsepower
is the maximum power the engine can develop without
alterations
the engine can be hooked to a dynamometer to determine
how much brake horsepower can be developed
useful in sizing stationary engines for operating irrigation
pumps, grinders and other large equipments
the same engines used for large tractors are often used as
stationary engines

16. Power Take-Off Horsepower
(PTO hp)
• the power measured at the PTO shaft

17. Drawbar Horsepower *
• is the measure of the pulling power of the engine by way of tracks,
wheels or tires at a uniform speed
• drawbar horsepower varies depending widely on several factors -
soil surface and type of hitch
• drawbar horsepower is the function of drawbar pull and speed of the
various kind of horsepower, maximum PTO hp is the most
commonly used in designating the size of a tractor
• large tractors that do not have PTO shaft, they may be given a brake
hp (flywheel) or maximum drawbar hp rating

18. Matching tractors and implements
When matching a tractor and implement, 3
important factors must be considered :-
1. The tractor must not be overloaded or early
failure of components will occur
2. The implement must be pulled at the proper
speed or optimum performance cannot be obtained
3. The soil conditions and their effects on
machine performance must be considered

19. With a given tractor, there is a set of amount of
power available. This available power is used for:-
I. moving the tractor over the ground
II. pulling the implement over the ground
III. powering the implement for useful work
The softer or looser the soil conditions are, the greater
amount of power that will be consumed because of
greater rolling resistance
This reduces the available usable drawbar power
SOIL CONDITIONS---POWER

20. Condition Usable Drawbar Power Ratio of Maximum
Of Soil As a % of Maximum PTO Power To Usable
PTO Power Drawbar Power
_________ __________________ ___________________
Firm 67 Percent 1.5
Tilled 56 Percent 1.8
Soft or Sandy 48 Percent 2.1
___________________________________________________

21. Eg: A 5-bottom, 40 cm plow operates in medium to
heavy soil with surface conditions considered as
firm. Given 32.5 drawbar kW per meter of width
for gumbo and 27.4 kW per meter for clay. We
will use 30 drawbar kW per meter as the average.
• Plow width = 5 x 40 = 200 = 2 meters
100
• 2 meters x 30 kW per meter = 60 drawbar kW
needed

22. if tractor size is known, determine how large an
implement can be pulled.
Tractor size – 110 PTO kW
Speed - 8 km/h
Soil condition – firm
Draft – 5.83 kN per meter of width
Usable drawbar kW – 67 % of maximum
PTO kW = .67 x 110 = 73.7 kN
Width = 73.7 kW x 3.6 = 5.7 meters
8 x 5.83

23. Position Control System
Hydraulic control of an attached implement eg. spreader
or broadcaster whereby the operator will preselect and
position the implement as determined by the position of
hand control lever
The position of the hand lever and the hydraulic
cylinder are always the same
Pressure is controlled by the relief valve and the
hydraulic cylinder will automatically move the
implement to its predetermined position and maintain it
there

24. Draft
is the horizontal component of pull, parallel to the line of
motion
Power
is the rate of doing work
1hp = 550 ft.lb/sec.
1 kW = 1.34 hp (metric unit)
Drawbar Power (dbhp)
is the measure of pulling power of the engine by way of
tracks, wheels or tires at a uniform speed
1 dbhp = 4500 kg. m / minute

25. THANK YOU