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TKP3501
Farm Mechanization
Topic 8:
Tractors and Power Units
Ahmad Suhaizi, Mat Su
Email: [email protected]
Introduction
Why we need machineries?Type of machine available Traditional vs modern
Introductionto a tractor
Type of tractorSpecification
Tractors’ components & Systems
Main componentsSystems;- Fuel & Intake- Combustion- Cooling
- Electric & instruments- Lubrication- Hydraulic
Others- Bearing & seal- Shaft
- Belt & pulley- Chain & sprocket- Gear- Lubrication (grease, oil)
Maintenance
Filters, oil,
lubrication, parts
ImplementsPrimary tillage Secondary tillage
Compact equipment
Power tillerOther small equipment
** = CalculationsHow to choose the tractor and implement size**Tractor & power unitType of power available
Crop Production
Land preparatio
nCrop type;- Oil palm- Rice - Vegetable
Crop typePlanter**
Farm Efficiency**
Theoretical Field CapacityEffective Field CapacityField Efficiency
Fertilization &
Irrigation**
Farm
maintenance
SpreaderPumpSprinkler
Grass, road, drainage
Harvesting
Yield, BallerTransportationSeedling
& Planting
Cost analysis**
Emerging Technologies
SensorTracking GPS, GNSSGIS, Mapping
Livestock
Optimization
Feeding systemMilkingAquaculture
ForestryHorticulture
©A
SM
S
Learning outcome
Be able to describe the common design of tractors
Be able to estimate drawbar and PTO power using the
86% rule
Be able to calculate and estimate a proper tractor
and implement size at safe operating speed
3
Introduction
4
A major task : To match power units to the size and
type of machines so all field operations can be
carried out on time with a minimum cost.
If the tractor is oversized for implements, the costs
will be excessive for work done.
If the implements selected are too large for the
tractor, the quality and quantity of the work may be
lessened or the tractor will be overloaded, usually
causing expensive breakdowns.
Categories of Tractors
Many design of the tractor according
to the different manufacturer and for
a specific task e.g. for vegetable
farming, forestry and construction.
Categories:
General purpose (25 – 400 Hp)
Row crop (50 – 100 Hp)
Orchard
Vineyard
Industrial
Garden (< 25Hp)
5
Propulsion system
Rear wheel drive (RWD)-general
purpose
Four wheel drive articulating
steering (4WDAS)
Four wheel drive four wheel steer
(4WD)
Tracks (T)
Rear wheel drive front wheel assist
(FWA)
Some of the factors to consider when
selecting a power unit include:
1. Engine type
2. Power ratings
3. Soil resistance to
machines
4. Tractor size
5. Matching implement
6. Sizing for critical work
6
1. Engine type
The three general types of engine currently use:
i. Diesel
ii. LP-Gas
iii. Gasoline/ Petrol
All three types are classified as
Internal combustion engine
7
Diesel engine is a compression-ignition engine
with only air being compressed in the cylinder.
Diesel fuel is then injected into the cylinder and
the fuel-air mixture is ignited by the head of
compression.
Gasoline and LP-Gas (Liquid Petroleum Gas)
engines are spark-ignition engines. In both cases,
the fuel-air mixture is drawn into the cylinder and
ignited by a spark.
8
2. Tractor Power Rating
Used to evaluate the size of the tractors
and engines
Engine power claimed by manufacturer not
always true – marketing strategy
Nebraska Tractor Test – test for engine,
PTO, drawbar power & three point hitch
tractortestlab.unl.edu/
Useable power usually less than advertised
power
9
Nebraska Tractor Test Laboratory
Report
10
Report Example
New Holland T80A [Source:
http://tractortestlab.unl.edu/documents/TL80A.pdf]
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2. Power Rating
Power is a measure of the
rate at which work being done.
The English power unit is defined as 550
foot pounds of work per seconds or
called “HORSEPOWER ” (Hp).
The Metric power unit is measured in
KILOWATTS (kW).
1 Kilowatts ( kW) = 1.34 Horse power
(hp). Or 1Hp = 0.75kW
13
When working with machinery, we usually thinks of
miles per hour and pounds of draft. For this conditions
the formula for power is:
𝑯𝒐𝒓𝒔𝒆𝒑𝒐𝒘𝒆𝒓 =𝒇𝒐𝒓𝒄𝒆,𝒑𝒐𝒖𝒏𝒅𝒔 𝑿 𝒔𝒑𝒆𝒆𝒅,𝒎𝒑𝒉
𝟑𝟕𝟓, (English/US unit)
𝑲𝒊𝒍𝒐𝒘𝒂𝒕𝒕𝒔 =𝒇𝒐𝒓𝒄𝒆, 𝒌𝑵 𝑿 𝒔𝒑𝒆𝒆𝒅,𝒌𝒎/𝒉𝒓
𝟑.𝟔(Metric unit)
14
Work
The amount of work done is not
referring to time.
It would not matter whether the
time move the load was one
minute or one hour; the amount of
work stays the same.
15
Work (kNm)/(kJ) = Force (kN) x Distance (m)
Power
When the rate of work is considered, then the
power can be determined
𝑷𝒐𝒘𝒆𝒓 (𝒌𝑾) =𝑭𝒐𝒓𝒄𝒆 𝒌𝑵 𝒙 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 (𝒎)
𝒕𝒊𝒎𝒆 (𝒔)
𝑷𝒐𝒘𝒆𝒓 (𝒌𝑾) =𝑭𝒐𝒓𝒄𝒆 𝒌𝑵 𝒙 𝑺𝒑𝒆𝒆𝒅 (
𝒌𝒎𝒉𝒓
)
𝟑. 𝟔
16
𝑷𝒐𝒘𝒆𝒓 (𝒌𝑾) =𝑭𝒐𝒓𝒄𝒆 𝒌𝑵 𝒙 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 (𝒎)
𝒕𝒊𝒎𝒆 (𝒔)
Fuel Consumption Estimates
17
18
Drawbar power
Measured at the
point implements
are attached to
the tractor,
drawbar or 3
point hitch
PTO power
Rating of the
power available
at the PTO of an
agriculture
tractor
Brake power
The power available at the flywheel
Of the engine. Rate the irrigation pump
and other similar external uses
Engine power
Calculate power based on
the bore, cylinder pressure,
and speed of the engine
19
Engine power
Engine power is an Indicated Power (IP)
The power that engine would develop from fuel injected
(neglecting all the losses)
The power that developed inside the combustion
chambers during combustion that pushes the piston
down.
20
Brake Power
Brake Power (BP) is the power that actually used
to do useful work (rotating the crankshaft)
The BP always lower than IP because some of the
power developed in cylinders (combustion
chamber) is lost in overcoming the internal
friction in the engine.
21
Brake Power (BP) = Indicated Power (IP) - Power loss to friction (Fp)
PTO Power
22
𝑃𝑜𝑤𝑒𝑟 𝑘𝑊 =𝑇 𝑥 𝑁
9549Where
Power in kW ( 1 Hp = 0.75 kW)
T = Torque ( Nm) * (Ib-ft)
N = speed (rpm) * (rpm)Drawbar Power
𝐷𝑏 (𝐻𝑝) =𝐹 𝐼𝑏 𝑥 𝑉(𝑚𝑝ℎ)
375
Where
Db in Hp ( 1 Hp = 0.75 kW)
F = Force (Ib)
V = speed (mph)
∗ 𝑃𝑜𝑤𝑒𝑟 𝐻𝑝 =𝑇 𝑥 𝑁
5252
𝐷𝑏 (𝑘𝑊) =𝐹 𝑘𝑁 𝑥 𝑉(
𝑘𝑚ℎ𝑟
)
3.6
Where
Db in kW
F = Force or draft (kN)
V = speed (km/hr)
Best estimate according 86 % rules
23
Where
𝐸𝐻𝑝= Engine power
𝐷𝑏𝐻𝑝 = Drawbar power
𝑃𝑇𝑂𝐻𝑝 = 𝐸𝐻𝑝 𝑥 0.86
𝑃𝑇𝑂𝐻𝑃 =𝐷𝑏𝐻𝑝0.86
Power loss
Accessories 10% other than fan
5 % fan and radiator
Temperature 1% or each 5.6˚C above 29 ˚C - Gasoline
1% or each 2.7˚C above 29 ˚C – Diesel
Altitude 3% for each 305 m above 152 m
Type of service 10% for intermittent loads
20% for continuous loads
Exercise 1
Estimate the amount of power available at the PTO and
drawbar for a tractor rate at 125 engine power.
24
Exercise 1
Estimate the amount of power available at the PTO and
drawbar for a tractor rate at 125 engine power.
25
𝑃𝑇𝑂𝐻𝑝 = 𝐸𝐻𝑝 𝑥 0.86= 125 x 0.86
= 107.5 Hp
~ 110 Hp
𝐷𝑏𝐻𝑝 = 𝑃𝑇𝑂𝐻𝑝 𝑥 0.86= (𝐸𝐻𝑝 𝑥 0.86) 𝑥 0.86= 107.5 x 0.86
= 92.45 Hp
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Chisel plow-coarse
Implement width = 10’
Typical speed = 6 mph
(see table 3)
Sizing implement and tractor
from nomograph
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Source: Nebraska Tractor Test
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http://www.howardmy.com/
Indicated Power
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A BDistance A-B
32
33
What is the minimum indicated power requirement?
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What is the minimum indicated power requirement?
The formula for determining how fast an implement
could be pulled with a given size of tractor.
Speed mph =Drawbar,Hp X 375
Draft,Ib−pounds(English/US unit)
Speed (km/hr) =Drawbar,kW X 3.6
Draft, kN(metric unit)
35
The formula can also be used to determine how large an
implement can be pulled.
The resistance is usually given as kN per meter of width.
Step 1: Determine the draft
Draft (kN) = Drawbar power (kW) x 3.6
Speed (km/hr)
Step 2: Determine width
Width = Draft (kN)
Draft (kN/m)
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3. Soil resistance to machine
The SOIL RESISTANCE TABLE, adapted from
the Agricultural Engineering Year Book lists
some units draft range for those implements
having the highest power requirements per
foot of width, and horsepower required per
foot of width and typical field speed.
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E.g.
For 4-wheel drive tractor, operates on firm soil,
DrawbarHP = 0.75 x PTOHP
or
PTOHP= 1.33 x DrawbarHP
39
Chisel plow-coarse
The softer or looser the soil conditions are greater the
amount of power that will be consumed because of
greater rolling resistance.
This reduce usable drawbar power.
Table soil conditions—power can be used to estimate
the usable drawbar horsepower for various soil
conditions.
41
Soil Conditions and Power
Condition of soil Useable Drawbar
Power as a
percentage of
maximum PTO
Power
Ratio of Maximum
PTO Power to
Usable Drawbar
Power
Firm 67 percent 1.5
Tilled 56 percent 1.8
Soft or sandy 48 percent 2.1
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4 & 5. Matching Tractors and
implement
Even when it is known how much power is needed for a given
field operation, knowing the size of tractor to use still
presents problem.
There are several different kinds of power measurements, all
applying to the same tractor.
1. Brake Horsepower
2. Power Take Off (PTO) Horsepower
3. Drawbar Horsepower.
43
4 & 5. Matching Tractors and
implement
When matching a tractor with implement, three
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 proper speed or
optimum performance cannot be obtained.
3. The soil conditions and their effect on machine
performance must be considered.
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Drawbar attachment
45
46
With a given a tractor, there is a set amount of power
available. The available power is use for:
1. Moving tractor over the ground
2. Pulling the implement over the ground.
3. Powering the implement for useful work.
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6. Sizing Power Units for Critical Work
Periods
Matching machine to fit time available
Let’s take example we want to select a plow and tractor to
plant corn. The time available 225 hours.
Experts says, the best total time will be used for the
operation is 85% from total time available.
48
49
Example
If you plow 1000 acres every year in light clay loam with firm
tractive conditions, how large a tractor and plow would be
needed?
Assume the total working time available is 225 hours every
year. Plowing speed of 4.5 miles per hour. Field efficiency is
80 %.
Actual total time used by the farmer is 85%. Power required
per feet according to table soil resistance is 12 hp.
50
Questions/Review
Why do we need to estimate the power requirement for a
specific attachment to the tractor?
Name two main types of power available from the farm
tractor.
What is the engine power? What is the brake power?
Why is the Brake power < PTO power?
51
Power Loss Calculations
52
Best estimate according 86 % rules
53
Where
𝐸𝐻𝑝= Engine power
𝐷𝑏𝐻𝑝 = Drawbar power
𝑃𝑇𝑂𝐻𝑝 = 𝐸𝐻𝑝 𝑥 0.86
𝑃𝑇𝑂𝐻𝑃 =𝐷𝑏𝐻𝑝0.86
Power loss
Accessories 10% other than fan
5 % fan and radiator
Temperature 1% or each 5.6˚C above 29 ˚C - Gasoline
1% or each 2.7˚C above 29 ˚C – Diesel
Altitude 3% for each 305 m above 152 m
Type of service 10% for intermittent loads
20% for continuous loads
Exercise 2
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at 37.78˚C air temperature?
54
What is the usable power?
Exercise 2
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at 37.78˚C air temperature?
55
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 165 𝑥 (1 − (0.01 𝑥37.78 − 29
5.6))
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 162.413 𝐻𝑝 𝑜𝑟 ~ 160𝐻𝑝
The effect of the 37.78˚C temperature is to reduce the usable power about 1.5%
What is the usable power?
Exercise 3
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at 500 m above sea level?
56
Exercise 3
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at 500 m above sea level?
57
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 165 𝑥 (100% − (3% 𝑥500 − 152
305))
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 159.35 𝐻𝑝 𝑜𝑟 ~ 160𝐻𝑝
The effect of the 500 m altitude to reduce the usable power about 3.42%
Exercise 4
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at continuous duty?
58
Exercise 4
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at continuous duty?
59
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 165 𝑥 (100% − 20%)
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 132 𝐻𝑝
The effect of the continuous duty is to reduce the usable power about 20%
Exercise 5
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at ambient tempereature of
37.78˚C, altitude of 500 m for continuous duty?
60
Exercise 5
1. How much usable power will a 165Hp spark ignition engine
produce if it will be operating at ambient tempereature of
37.78˚C, altitude of 500 m for continuous duty?
61
𝐴𝑑𝑗𝑢𝑠𝑡𝑚𝑒𝑛𝑡𝑡𝑒𝑚𝑝 = 0.01 𝑥37.78−29
5.6= 0.0156
𝐴𝑑𝑗𝑢𝑠𝑡𝑚𝑒𝑛𝑡𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 = 3% 𝑥500−152
305= 0.0342
𝐴𝑑𝑗𝑢𝑠𝑡𝑚𝑒𝑛𝑡𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 = 20% = 0.2
𝑇𝑜𝑡𝑎𝑙 𝑎𝑑𝑗𝑢𝑠𝑡𝑚𝑒𝑛𝑡 = 0.0156 + 0.0342 + 0.2 = 0.2498
𝐸𝑛𝑔𝑖𝑛𝑒𝐻𝑝 = 165 𝐻𝑝 𝑥 1 − 0.2498 = 123.78 𝐻𝑝 ~ 120 𝐻𝑝
Assignment 2
On PutraBlast
62
Thank you.
63
30 men vs 1 tractor
The winner is…..