MAINTAINANCE OF FARM MACHINERY

En Mohd Fauzie JusohLecturer

Agrotechnology ProgrammeFaculty of Agro-Based Industry

Universiti Malaysia Kelantan (Jeli Campus) Locked Bag No.100, 17600 Jeli, Kelantan.

014-2903025/fauzie.j@umk.edu.my/

FARM RECORD

Logbook

• With greater reliance on agricultural machinery in crop production, the ownership cost of tractors and various farm implements are becoming a bigger component of the total cost of production.

• As such, care must be taken not only in the proper selection and purchase of tractors and farm implements but also in their efficiently use.

• Logbook : Daily record/log book of each tractor.

Logbook

Consist of :DateName of operator and attendantWhether a trailer is hitched Job description Starting time and hour meter reading Finishing time and hour meter readingTotal hours workedDistance travelled (km) Load carried (kg) Fuel and oil consumptionParts replacement and repairs

Maintenance schedule

• Breakdowns of machinery at work not only cause direct repair cost but also losses in revenue due to poorer or lesser yield

• As such a farm manager should set up a proper maintenance schedule for all tractors and farm implements under his care

• Different machine manufacturer will propose slightly different maintenance schedule

• A general maintenance schedule will be given in the hard copy

Maintenance schedule

Machines that are well looked after will have:

Longer useful lives

Resale values

Experience less breakdown

Lower fuel and oil consumption

MACHINE CAPACITY

Machine Capacity

The machine capacity is its rate of machine performance

The performance capacity will be measured in terms of hectare per hour (ha/hr), tons per hour, liter per minute, etc

Understanding how to estimate the capacities of machines will help farm managers making a plan in selecting and buying machine for future use

To know machine capacities for selection of power units and an equipment that can complete field operations on time

Important to avoid the added expenses of larger that necessary machines

Factors that influence field efficiency of the machine

Speed – average rate of travel in km/hr

Width – distance in feet or meters across the processing portion of the machine

Efficiency

Ratio of effective capacity of a machine to its theoretical capacity.

an indicator to determine how much working time is spent versus turning, filling hoppers and other jobs

Theoretical Field Capacity (TFC)

• Is the maximum possible capacity obtainable at a given speed, assuming the machine is using its full width

• TFC depend on factor : speed and width

• TFC[Area/Time] = Speed (m/hr) x Width (m)

= m2 /hr or ha/hr

Theoretical field capacity (TFC)

• Ex : Suppose a tractor with a 4.25 meter disk travels 8 km/hr. What is the theoretical field capacity in ha/hr?

The theoretical field capacity (TFC) would be := speed x width= 8000 m/hr x 4.25 m= 34,000 m 2/hr

Convert to hectare --- 1 hectare = 10,000 m 2

TFC = 34,000 m2 /hr10,000 m2

= 3.4 ha/hr

Theoretical Field Capacity (TFC)

Theoretical Field Capacity cannot be sustained for long periods of time because field operations will be interrupted by turns, filling hoppers and breakdowns

The Effective Field Capacity (EFC) is always less than the Theoretical Field Capacity (TFC)

Effective Field Capacity (EFC)

• The number of acres/hectares actually covered over a long period of time

• EFC brings in the factor of efficiency. This capacity determination represent the real life or actual capacity obtainable over a period of time

• EFC [Area/Time] = Total Area (ha)Total Time (hr)

Effective Field Capacity

• Ex : If a 4.25 meter disc actually covers 28 ha while operating for 10 hours with no breakdowns, its EFC would be :-

EFC = Total hectares (ha)

Total hours (hr)

= 28 ha

10 hr

= 2.8 ha/hr

Field Efficiency @ Field Capacity

• EFC for one day’s experience may not give a true picture of the EFC for the season. It will be different depending on the period use.

• The different can be determine by calculating the Field Efficiency

• Field Efficiency (FE) = EFC x 100%

TFC

FACTORS THAT AFFECT FIELD EFFICIENCY

• Not using the full capacity of the tractor

• Refilling procedures

• Transporting procedures

• Repairing of machine in the field / machine breakdown

• Irregular resting time

• Cleanliness of the implements

• Rest of operation

• Adjustment of implement

Example 1

Suppose a tractor with a 4.25 meter disk travels 8 km/hr. If the disc of this tractor is used for 2 weeks period and cover 195 hectares, calculate the Effective field capacity and field capacity. This tractor work 8 hour/day.

Total day = 14

Total working hour/day = 8

Total hours in field = 14 x 8 = 112

Total hectares covered = 195

EFC = Total hectares

Total hours

= 195

112

= 1.74 ha/hr

Field efficiency = 1.74 x 100%

3.4

= 51%

Example 2

• Assume a 6 rows 30 inches row corn head equipment combined is working at a speed of 3.2 km/hr with long row well organizes unloading pattern and no break-down. Field study indicated field efficiency of 70% achieved. Find effective field capacity.

Width = 30 in x 6 rows= 180 in

@ = 4.572 m

TFC = width x speed= 4.572 m x 3200 m/hr

= 14,630 m2/hr

1 hectare = 10,000 m2

TFC = 14,630 m2/hr 10,000 m2

= 1.46 ha/hr

FE = EFC x 100

TFC

EFC = FE x TFC

100

= 0.7 x 1.46 ha/hr

= 1.022 ha/hr

Example 3

• The theoretical field capacity (TFC) of a disc harrow is 2.0 ha/hr. If its field efficiency (FE) is 70%, calculate :

i. The effective field capacity (EFC)

ii. The time taken to plow a 18 ha field, in minutes

TFC = 3.0 ha/hr, FE = 65%

a. EFC = FE x TFC100

= 0.65 x 3.0 ha/hr= 1.95 ha/hr

b. Time taken to plow a 12 ha field, in minutes1.95 ha = 1 hr

12 ha = x1.95x = 12

x = 6.15 hr x 60= 369 minute

Example 4

• One farmer has a working hour of 8 hrs per day using a TS90 Ford tractor to plow 25 hectare of corn farm. If the theoretical field capacity (TFC) of a moldboard plow is 2.5 ha/hr and the overall field efficiency is 70% then can he finish plowing in 2 days time?

• TFC = 2.5 ha/hr, FE = 70%, Area = 25, • Working Hours = 8

EFC = FE x TFC100

= 0.7 x 2.5 ha/hr= 1.75 ha/hr

1.75 ha = 1 hr25 ha = x hr1.75x = 25 ha

x = 14.29 hr (required time to finish plowing)

Working hour for 2 day = 8 x 2

= 16 hr

The farmer can finish plowing in 2 day because he only need 14.29 hr to do the job while working hour in 2 day is 16 hr

MACHINE POWER REQUIREMENT

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 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

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

1. Engine type

The combustion process in the cylinder 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

2. Power rating

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 horsepower1 hp = 550 ft.lb/sec

Work = Force x Distance

Power rating

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 lb x 3 feet

= 60 ft.lb

The amount of work done is 60 ft.lb with no reference to time

• 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

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

1 N = 0.225 lb

1 kN = 224.8 lb force

Metric equivalent

• Drawbar power is the measure of pulling power of the engine by way of tracks, wheels or tyres at a uniform speed

• Formula for Drawbar (kilowatt) is :

Drawbar (kW) = Force (kilonewtons) x Speed (km/hr)

3.6

Draft is the horizontal component of pull, parallel to the line of motion

Metric equivalent

• Ex: If the draft of a trailing implement such as a disc harrow is measured at 11.1 kilonewtonsand is pulled at a speed of 8 km/hr, what is the drawbar in kilowatt?

Drawbar (kW) = Force (kilonewtons) x Speed (km/hr)

3.6= 11.1 kN x 8 km/hr

3.6= 24.7 kW

Metric equivalent

• Ex: 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 = Drawbar x 3.6

Force

= 50 kW x 3.6

22.2 kN

= 8.1 km/hr

Metric equivalent

• Ex : 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 3.6Speed (km/hr)

= 65 kW x 3.68 km/hr

= 29.25 kN

Metric equivalent

Width = Total draft

Draft per meter

= 29.25 kN

4 kN/meter

= 7.3 meters

Metric equivalent

• 1 dbhp = 45000 kgm/min• Drawbar horsepower = Force, lbs x speed, mph

375• 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 is involved• Size of the implement have to be related to the amount of

soil resistance

Metric equivalent

• Ex : If the draft of a trailing implement like a disc harrow is measured at 2500 pounds and is pulled at a speed of 5 mph. What is the drawbar horsepower?

Drawbar horsepower = Force, lbs x speed, mph375

= 2500 x 5375

= 33.3 HP

3. Soil resistance to machines

• With a given tractor, there is a set of amount of power available. This available power is used for :

Moving the tractor over the ground

Pulling the implement over the ground

Powering the implement for useful work

• This reduces the available usable drawbar power

4. 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

E.g:

A tractor is mounted with a plow that is 5 feet wide. The resistance from the soil is 2 500 lb per foot of plow. Calculate the total draft.

Total draft = Soil Resistance x Width

5. Matching tractors and implements

• When matching a tractor and implement, 3 important factors must be considered :

The tractor must not be overloaded or early failure of components will occur

The implement must be pulled at the proper speed or optimum performance cannot be obtained

The soil conditions and their effects on machine performance must be considered

Questions a

• The brake horsepower (BHP) of a tractor is 96 HP. The power available at the drawbar (DHP) is only 75% of the BHP. The tractor is pulling 5000 lbs of harvested crops through a field. What is the maximum speed in miles per hour that the tractor can travel ?

Questions b

• A tractor has a plow that is 3 feet wide. Plowing is carried out at a depth of 8 inches. The resistance from the soil is 20 psi (pounds per square inch). The tractor must be able to plow the field at an average speed of 5 miles per hour (MPH). How much horsepower is required?

Questions c

• The tractor has an indicated power (IHP) of 90 HP where else its brake power (BHP) is 6.6 Hp less than IHP. The average pulling load cause by the moldboard plow is 4500 pound. Calculate the average forward travelling speed if the available power at the drawbar (DHP) is 90% of BHP.

Questions d

• The brake horsepower (BHP) of a tractor is 100 HP. The power available at the drawbar (DHP) is only 60% of the BHP. The tractor is pulling 4000 lbs of harvested crops through a field. What is the maximum speed the tractor can travel, in miles per hour (MPH) ?

Questions e

• Calculate the drawbar horsepower required to pull an 80 inch tiller at a depth of 5 inch in a soil with a resistance of 30 psi and at a forward speed of 3miles per hour. If the approximate PTO horsepower equals to 125% of the drawbar horsepower, what is the approximate PTO horsepower needed?

Q&ATQ