tkp3501 farm mechanization - universiti putra malaysia 8_tractors and power... · a specific task...

TKP3501

Farm Mechanization

Topic 8:

Tractors and Power Units

Ahmad Suhaizi, Mat Su

Email: asuhaizi@upm.edu.my

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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27

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

29

30

http://www.howardmy.com/

Indicated Power

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A BDistance A-B

32

33

What is the minimum indicated power requirement?

34

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.

37

38

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.

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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.

47

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…..