I-1c FarmMech 20-60

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TILLAGE EQUIPMENT 

Tillage – any physical soil manipulation which changes the structure of the soil, kills weeds, and rearranges dead plant materials.



Objectives of tillage: 1.

To develop a desirable soil structure for a seedbed: a. b. c. d.

2. 3.

higher water infiltration rate. decreased water surface runoff. greater water holding capacity. promote root penetration.

To control weeds, cut roots and bury green materials. To incorporate manure and chemical fertilizer.

For wetland conditions: 4. 5. 

To turn the soil into a soft “puddle”. To form a hard layer which reduces water leaching.

Classification of tillage: 1. Primary tillage – initial cutting, breaking and usually inversion of the soil. Implements used are moldboard, disc and chisel plows and subsoilers cutting the soil to a depth of 6” to 36”. Often referred to as plowing. 2. Secondary tillage – subsequent breaking, pulverization and leveling of the soil making it ready for planting. Implements used are disc, spiketooth and spring-tooth harrows preparing the soil to a depth of 3” to 6”. Often referred to as harrowing. 3. General-purpose tillage – Combined primary and secondary tillage in one operation. Implements used are the rotavators and floating tillers cutting the soil to a depth of up to 6”. Often referred to as rotavating.



Terms used in plowing (Figure 1): 1. 2. 3. 4. 5. 6.

Land – unplowed soil. Furrowslice – soil cut, lifted, inverted and thrown to one side of the plow bottom. Furrow – trench or canal left by the furrowslice. Backfurrow – raised ridged when two furrowslices overlap each other. Deadfurrow – trench left equal to two furrows when furrowslices are thrown on opposite sides. Furrow wall – unbroken side of the furrow.

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Figure 1. Furrows made by different types of plows 

Primary Tillage Equipment: 1. Moldboard plow – one of the oldest and most important agricultural equipment for land preparation. It cuts the soil to a depth of 6” to 18 “. 

Basic parts of moldboard plow bottom (Figure 2): a. b. c.



Share – provides the cutting edge of the plow bottom. Moldboard – receives the furrowslice from the share; lifts, inverts and throws it to one side of the plow bottom. Landside – counteracts the side pressure exerted by the furrowslice on the plow bottom.

Figure 2. Moldboard plow bottom Clearances of the moldboard plow bottom: a.

Vertical clearance – bend downward of the point of the share to make the plow penetrate the soil to the proper depth.

b.

Horizontal clearance – bend outward or sideward of the point of the share towards the unplowed land to make the plow cut the proper width.

Agricultural Machinery and Mechanization

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Size of the moldboard plow bottom – perpendicular distance between the wing and the point of the share. This also expresses the width of cut of the plow bottom.

2. Disc plow – it was developed in an effort to reduce friction in the sliding moldboard plow bottom by using a rotating disc plow bottom. 

Basic parts of disc plow bottom (Figure 3): a. b.

c.

Disc – round and concave disc of heat-hardened steel with sharpened edges to cut the soil. It is rotated by the soil during soil cutting. Scraper – curved plate placed on the concave side of the disc to scrape off the soil and throw it on one side of the plow bottom. Rear furrow wheel – rear wheel that presses against the furrow wall to resist the side pressure exerted by the soil against the plow.

Figure 3. Three-bottom Disc Plow (Deere & Co.) 



Angles of the disc plow: a.

Tilt angle – angle the disc makes with the vertical to make the disc penetrate the soil to the proper depth (15 – 25 degrees).

b.

Disc angle – angle the disc makes with the direction of travel to make the disc cut the proper width and allow the disc to rotate (42 – 45 degrees).

Size of the disc plow bottom – expressed in terms of diameter of the disc. The width of cut, however, is influenced by disc diameter, depth of cut and disc angle.

3. Draft of Plows – determined by the resistance of the soil to tillage operation, the speed of plowing, sharpness of the cutting edge and over-all adjustment of the plow. Agricultural Machinery and Mechanization

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Table 1.

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

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.



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.

Specific draft of different soils SOIL TYPE

SPECIFIC DRAFT, SD Lbs/in2 Kg/cm2 3 0.21 3-6 0.21-0.42 5-7 0.35-0.49 6-8 0.42-0.56 10-11 0.70-0.77 12-15 0.85-1.06 16-18 1.13-1.27 18-20 1.27-1.41

Sandy soil Sandy loam Silty loam Clay loam Heavy clay Virgin soil, clay Gumbo, moist Dry adobe Table 2.

Increase in draft due to speed SPEED MPH 1 2 3 4 5 6

DRAFT, % KPH 1.6 3.2 4.8 6.4 8.0 9.6

Sample Problem 1:

100 114 128 142 156 170

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:

C

SWEff 10

where:

C = Field Capacity, Ha/hr S = Speed, Kph W = Width of cut, m Ef = Field Efficiency, decimal

C

(6.4 kph )(4 X 0.36 m)(0.78) (10 h)  7.2 ha 10

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b. Draft requirement based on soil type:

Ds  SpecificDraftxWxd

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:

Hp (draft ) 

Da xS 274

where:

Da = Adjusted Draft, Kg S = Speed, Kph

Hp (draft ) 

( 2,004 Kg)(6.4 Kph) 274

=

46.8 hp

e. Tractor Horsepower:

Hp (tractor )  Sample Problem 2:

46.8 hp  58.5 hp 0.8

A four-wheel tractor with 3 – 36 cm moldboard plow is to operate on silty loam soil at a depth of 25 cm. The maximum drafts of the tractor at different working speeds are given below. Determine the tractor horsepower.

GEAR SETTING 1L 2L 3L 4L 1H

SPEED (KPH) 2.4 4.0 5.6 7.2 9.6

MAX. DRAFT (KG) 4,000 3,100 2,200 1,300 1,000

Solution: a.

Draft =- Sp.D xW x d = 0.42 kg/cm2 x (3 x 36 cm) x 25 cm = 1,134 Kg

b.

From the table above: From Table 2: Therefore:  Not okay

at 1,300 Kg Max Draft  7.2 Kph at 7.2 Kph  149 % Increase in Draft Da = 1,134 Kg x 1.49 = 1,690 kg

From the table above: Agricultural Machinery and Mechanization

at 2,200 Kg Max Draft  5.6 Kph

PSAE Region IV - Agricultural Engineering Board Review Materials From Table 2: Therefore:  Okay



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at 5.6 Kph  135 % Increase in Draft Da = 1,134 Kg x 1.35 = 1,531 Kg

Da xS 1,531 Kg x 5.6 Kph   31.3 hp 274 274

c.

Hp (draft ) 

d.

Hp (tractor ) 

31.3 hp  39 hp 0.8

Secondary Tillage Equipment: 1.

Disc harrow – consists of two or more sets of disc gangs cutting the soil to a depth of 3” to 6”. A disc gang is made up of a number of discs mounted on a common axle. 

Basic parts of disc harrow: a. b. c. d.



Discs – round and concave discs Disc gang – an assembly of discs all rotating together on a common shaft (3 -13 discs/gang at 6” to 12 “ spacing per disc). Bumpers – a washer on one of the common shaft to absorb the end thrust of the gang. Bearings – two to three bearings per gang.

Types of disc harrows (Figure 4): a. Single-action disc harrow – consists of two gangs of discs placed end-to-end which throws the soil in opposite directions. b. Double-action disc harrow – consists of four gangs of discs. Two gangs in front as in single –action and two gangs at the rear. c. Offset disc harrow – consists of two gangs of disc. One gang located behind the other and operated in an offset position in relation to the tractor.

Figure 4. Types of disc harrows

Agricultural Machinery and Mechanization

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Spike-tooth harrow – consists of long rigid spikes clamped or welded to cross bars on a staggered arrangement to attain maximum stirring and raking of the soil. The cross bars can be rotated to change the angle of the teeth (Figure 5).

Figure 5. Spike-tooth harrow with folding sections 3.

Spring-tooth harrow – consists of long, flat and curved teeth made from spring steel. The curved teeth are welded to cross bars on a staggered arrangement. The spring characteristic allows the teeth to flex and slide over obstructions (Figure 6).

Agricultural Machinery and Mechanization

PSAE Region IV - Agricultural Engineering Board Review Materials Figure 6.



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Tractor-mounted spring-tooth harrow

General-purpose tillage equipment: 1. Rotavators – constructed with a set of cutting knives or tines mounted on a horizontal power-driven shaft rotating at high speed. The knives slice thin sections of the soil and pulverize them by centrifugal force (Figure 7).  

Rotavators are mounted on 2-wheel or 4-wheel tractors and driven by the tractor PTO. Rotavators are suited for both dry and wet land operations.

Figure 7.

4-Wheel tractor-mounted rotavator

2. Floating power tillers – specially designed power tillers for wet land operation fitted with front-mounted cutting blades on a cagewheel (Figure 8).

Figure 8. Floating tiller

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PLANTING EQUIPMENT 

A mechanical device used to place seeds or plants in the soil for crop production.



Classification of planting equipment: 1. Row-crop planters – planters designed to plant seeds or plants in rows far enough apart to permit the subsequent entry of machinery. a. Row-crop drill planter – designed to plant seeds continuously in rows with row spacing greater than 36 cm. b. Hill-drop planter – designed to plant one or more seeds or plants in rows and in hills. 2. Solid planters – planters designed to plant seeds or plants in patterns which do not allow subsequent entry of machinery. a. Solid drill planter – designed to plant seeds continuously in rows with row spacing less than 36 cm. b. Broadcaster – designed to scatter the seeds uniformly over the surface of the field without definite rows and hills. Seeds are covered by harrowing.



Mechanical functions of a planter: 1. 2. 3. 4.

Make a furrow Meter the seeds Deposit the seeds Cover the seeds

PART (Figure 1) -

Figure 1. Typical Planter Agricultural Machinery and Mechanization

Furrow opener Metering device Seed tube Furrow closer

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

Different types of furrow openers (Figure 2)

Figure 2. Furrow openers 

Different types of metering devices (Figures 3 to 7)

Figure 3. Adjustable orifice with agitator

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Figure 4

Rotating orifice

Figure 5. Vertical roller

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Figure 6. Fluted roller

Figure 7. Horizontal seed plate 

Different types of seed tubes (Figure 8)

Figure 8. Seed tubes 

Different types of furrow closers and press wheels (Figures 9 and 10)

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Figure 9. Furrow closers

Figure 10.Press wheels



Calibration of planters – a procedure of determining the performance of a planter under the different seeding rates and row and hill spacings it is designed to operate. 

Planters are calibrated when newly acquired and at regular intervals.

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Sample Problem 1: The fertility of a field is such that maximum corn yields are obtained with a population of 54,000 plants per hectare. The rows are 0.75 m apart and an average emergence of 85 percent is expected. How many seeds per hill should be planted if the hills are 0.5 m apart? Given:

Plant population = 54,000 plants/ha Row spacing = 0.75 m Ave. emergence = 85 % Hill spacing = 0.5 m

Required: No. of seeds per hill Solution: A = area per hill = RS X HS = 0.75 m X 0.5 m = 0.375 m 2/hill 10,000 m 2 /ha = 26,667 hills/ha 0.375 m 2 /hill

NH = No. of hills/ha =

NS = No. of seeds/ha =

54,000 plants/ha = 63,529 seeds/ha 0.85

NS/H = No. of seeds/hill =

63,529 seeds/ha = 2.38 seeds/hill 26,667 hills/ha = 2 to 3 seeds/hill



Sample Problem 2: Using the results of the calibration test of the 9 X 7 grain drill, determine the seeding rate adjustment to use if it is desired to plant at the rate of 100 kg per hectare. Given:

Required:

Ground wheel (GW) diameter = 1.22 m Width = 9 rows X 7 inches per row No. of GW revolutions = 10 Adjustment Discharge/10 rev (gm) 0 0 ¼ 140 ½ 460 ¾ 740 Full 1100 a. b.

Calibration curve Adjustment at 100 kg/ha

Solution: Agricultural Machinery and Mechanization

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W = Width = 9 rows X 7 inches/row X 0.0254 m/inch = 1.6 m S10 = Distance = ∏ X D X 10 = 3.14 X 1.22 m X 10 = 38.31 m A10 = Area = W X S10 = 1.6 m X 38.31 m = 61.32 m2 Discharge at (using ratio and proportion): ¼ setting:

X1/4 = 0.14 kg X

10,000 m 2 /ha = 23 kg/ha 61.32 m 2

½ setting:

X1/2 = 0.46 kg X

10,000 m 2 /ha = 75 kg/ha 61.32 m 2

¾ setting:

X3/4 = 0.74 kg X

10,000 m 2 /ha = 120 kg/ha 61.32 m 2

Full setting:

XF = 1.1 kg X

10,000 m 2 /ha = 179 kg/ha 61.32 m 2

Calibration Curve 

Sample Problem 3: Using the specifications of the transmission system of the metering device of the two-row corn planter, determine the hill spacings. If the row spacing is 0.75 m, average seeds per hill is two and emergence is 90 percent, what are the expected plant populations per hectare? Given:

Row spacing = 0.75 m Seeds per hill = 2

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Emergence = 90 percent No. of rows = 2 No. of cells of seed plate (SP) = 20 Ground wheel diameter (GW) = 0.60 m No. of teeth Speed ratio (GW/SP) T=6 6/1 T=8 5/1 T = 10 4/1 T = 12 3/1 Required: a. b.

Hill spacings Expected plant populations

Solution: C = Circumference of GW = ∏ X D = 3.14 X 0.6 m = 1.88 m Hill spacing (HS) and expected plant population (EPP) for: T1:

HS1 =

C X SR 1.88 m X 6 = = 0.56 m SP 20

A1 = Area/hill = RS X HS = 0.75 m X 0.56 m = 0.42 m2/hill 2 EPP1 = 10,000 m / ha X Seeds/hill X Emergence Area/hill =

T2:

HS2 =

10,000 m 2 /ha X 2 seeds/hill X 0.90 = 42,857 plts/ha 0.42 m 2 /hill

1.88 m X 5 = 0.47 m 20

A2 = 0.75 m X 0.47 m = 0.35 m2/hill EPP2 =

T3:

HS3 =

10,000 m 2 /ha X 2 seeds/hill X 0.90 = 51,428 plts/ha 0.35 m 2 /hill

1.88 m X 4 = 0.38 m 20

A3 = 0.75 m X 0.38 m = 0.28 m2/hill 10,000 m 2 /ha X 2 seeds/hill X 0.90 EPP3 = = 64,286 plts/ha 0.28 m 2 /hill T4:

HS4 =

1.88 m X 3 = 0.28 m 20

A4 = 0.75 m X 0.28 m = 0.21 m2/hill

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EPP4 =

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10,000 m 2 /ha X 2 seeds/hill X 0.90 = 85,714 plts/ha 0.21 m 2 /hill

CROP PROTECTION EQUIPMENT 

Crop protection – process of providing plants the conditions that will make them free of weeds, pests and diseases.  the growing stage between planting and flowering constitutes the longest stay in the field and the most vulnerable period of the crop.  weeds compete with uncontrolled.

the

available

moisture

and

nutrients

if

left

 pests and diseases multiply above the economic threshold levels if left uncontrolled. 

Methods of pest control: 1. Cultural control – modification of cultural practices such as time of planting and crop rotation. 2. Ecological control – change in the environment of the crop and the pest which favors the survival of the crop such as flooding to kill insects and weeds. 3. Biological control – introduction of certain insects which feed on pests, application of chemosterilants to render the male sterile, or planting of certain plants whose odor drive pests away. 4. Physiological control – breeding and planting of pest resistant varieties. 5. Chemical control – application of chemicals to control weeds, pests and diseases. 6. Flame control – use of flame for the selective burning of weeds in crops whose stem is not injured by a short exposure to intense heat.

7. Mechanical control – use of tools, implements and machines to reduce or eliminate weeds and insects such as in land preparation, cultivation and weeding. MECHANICAL CONTROL: 

Yield losses due to uncontrolled weeds alone are significantly high (Table 1).

Table 1.

Yield losses due to uncontrolled weeds Crop

Lowland rice Upland rice Corn

Percent Yield Losses Average 69 79 50

Agricultural Machinery and Mechanization

Range 37 – 97 41 – 100 18 – 80

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Soybean 69 55 – 100 Peanut 65 35 – 94  Labor for weeding is highest among the different operations for lowland rice production (Table 2) Table 2.

Manual labor use by operation for lowland rice

Operation Seedling preparation Irrigation Land preparation (carabao) Transplanting (hand transplanting) Fertilizer application (broadcasting) Weeding (push-type weeder) Rodent control Chemical application (spraying) Harvesting Threshing Bagging and hauling Drying Milling TOTAL: 

Man-days/ha 7.33 6.24 27.98 15.52 2.76 75.0 – (38%) 1.84 4.5 27.89 11.40 5.07 8.95 3.94 198.42

Classification of weeders: A. By design of soil working part: 1. Blade type – rectangular, triangular and trapezoidal shapes with cutting edges sharpened and hardened for soil cutting and weed uprooting. 2. Tine type – straight or curved, round or square cross-section steel rods sharply pointed and hardened at the soil working end. 3. Rotary type – curved spikes or paddles attached radially to a common axle which rotate when pushed forward to uproot and bury weeds. B.

By power source: 1. Manual weeders – hand-held or push-type weeders for upland or lowland. a) Hand-held weeders – utilize the blade and tine type of soil working parts with short (< 0.5 m), medium (< 1 m) and long (> 1 m) handles. b) Push-type weeders – utilize any of the three types of soil working parts for upland (wheel hoe) and lowland (rotary weeder) weeding. 2. Animal-drawn weeders – soil working parts are mounted on a frame or tool bar pulled by an animal for upland weeding. 3. Tractor-drawn weeders – soil working parts are mounted on a frame or tool bar pulled by a two-wheel or four-wheel tractor for upland weeding.

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4. Power weeders – rotary type weeder driven by its own engine for lowland weeding. 

Capacity and performance of weeders (Table 3)

Table 3.

Capacity and performance of some weeders tested in the Philippines Name of weeder

Lowland weeders (rice): Spin tiller Single-row rotary Single-row rotary and paddle Double-row rotary Power weeder Upland weeders (corn): Light blade hoe Blade hoe V-blade Wheel hoe

Hours per Ha

Weeding Index (%)

Plant Damage (%)

63 70 80 45 99

93 91 88 88 78

3 5.4 5.5 5.6 4.3

120 114 180 110

86 86 81 94

6.2 6.0 12.3 11.5



Types of hand held weeders (Figure 1)



Figure 1. Hand held weeders Types of push-type weeders (Figures 2 and 3)

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Figure 2. Wheel hoe (upland)

Figure 3. Rotary weeder (lowland) CHEMICAL CONTROL: 

Most popular way of applying chemicals in the Philippines is with the use of knapsack sprayers.



Sprayers are used primarily for the application of herbicides, insecticides, fungicides and plant nutrients.



Functions of a sprayer: 1. Break the liquid into droplets of effective size 2. Distribute the spray solution uniformly 3. Regulate the spraying rate



Basic components: 1. Chemical supply 2. Energy source 3. Atomizer 4. Control device 5. Conductors lance

: : : :

Agricultural Machinery and Mechanization

Tank Pump and pressure chamber Nozzles Cut-off valve : Flexible hose and rigid

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Principles of creating hydraulic energy: 1. Pressure is created by adding liquid to a chamber with fixed volume of air (Ex. Knapsack sprayer) (Figure 4) 2. Pressure is created by adding air to a container with a fixed volume of liquid (Ex. Compression sprayer)(Figure 5)

Figure 4. Knapsack sprayer Compression sprayer



Figure 5.

Operation of knapsack and compression sprayers (Figures 6 and 7)

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Figure 6. Operation of knapsack sprayer

Figure 7. Operation of compression sprayer 

Nozzles:  Performance is dependent on hydraulic energy as follows: 1. 2. 3.

The higher the pressure, the smaller the droplets The higher the pressure, the higher the flow rate The higher the pressure, the wider the spray angle

 Types of nozzles (Figures 8 and 9): 1. Cone type – produces a cone shaped pattern of spray which could be hollow or solid cone. Agricultural Machinery and Mechanization

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



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Best suited for spraying crops because it produces a spray in which droplets approach the leaves from several angles. The pressure is high and then orifice is small.

2. Fan type – produces a flat pattern.  

Best suited for spraying flat surfaces as in the application of herbicides. The pressure is low and the orifice is large.

Figure 8. Cone nozzle pattern

Figure 9. Fan nozzle pattern 

Sprayer calibration:  For a given sprayer, the following factors should be measured: 1. 2. 3.

Nozzle discharge at different pumping rates (L/min) Effective working width or swath (m) Walking speed (m/min)

 Compute for the application rate (L/ha) 

Sample problem 1:

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Using the results of the calibration test of a knapsack sprayer, determine the application rate in l/ha. Effective width or swath is 1.5 m, average walking speed is 20 m/min, and discharge rate is 1.5 l/min. Given:

Effective width or swath = W = 1.5 m Walking speed = S = 20 m/min Discharge rate = q = 1.5 l/min

Required: Solution:

Application rate (l/ha)

A = area per minute = W X S = 1.5 m X 20 m/min = 30 m2/min T = time per hectare =

10,000 m 2 / ha = 333.33 min/ha 30 m 2 /min

Q = application rate = q X T = 1.5 l/min X 333.33 min/ha = 500 l/min 

Sample problem 2: The application rate of a certain insecticide is recommended at 4 l/ha. Using the results of the calibration test of a 16-liter knapsack sprayer, determine the amount of insecticide to mix with water per loading of the sprayer. Discharge rate of the sprayer is 1.2 l/min, effective width or swath is 1.5 m, and average walking speed is 20 m/min. Given:

Insecticide application rate = 4 l/ha Sprayer capacity = 16 liters Discharge rate = 1.2 l/min Swath = W = 1.5 m Walking speed = S = 20 m/min

Required: Solution:

Amount of insecticide to mix with water per sprayer loading T1 = Time per load =

a.

16 l/load 1.2 l/min

= 13.33

min/load b.

c.

proportion): =

d.

A1 = Area per load = W X S X T1 = 1.5 m X 20 m/min X 13.33 min/load = 400 m2/load X1 = Amount of insecticide per load (ratio and

4 l/ha X 400 m 2 /load = 0.16 l/load 2 10,000 m /ha EXTRA:

NL = Number of loadings =

10,000 m 2 /ha = 25 loads/ha 400 m 2 /load

VT = Total volume = 25 loads/ha X 16 l/load = 400 l/ha Agricultural Machinery and Mechanization

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WATER PUMPING EQUIPMENT 

Mechanical device used for transferring fluids from one place to another



The source of fluid is usually of lower elevation than the point of delivery



Source of water are usually wells, rivers, lakes, reservoirs and canals



Pumps are essential for good health and sustained agricultural production



Pumps are either hand-operated for domestic water supply or power-operated for both domestic and irrigation purposes.



Pump classification (Figure 1): A. Positive Displacement Pump – discharges the same volume of water regardless of the head against which they operate. 1.

Reciprocating Pump – having to-and-fro motion 

Basic parts:

Piston or plunger Inlet or admission valve Outlet or discharge valve



Types of reciprocating pumps: a. Lift Pump – single-acting pump consisting of an open cylinder, piston with built-in bucket valve. It lifts the water to flow out from a spout (Figure 2). b. Force Pump – single-acting or double-acting pump consisting of a plunger, inlet valve, and outlet valve. It forces the water above atmospheric pressure as distinguished from the lift pump (Figure 3).

B. Variable Displacement Pump – inverse relationship between discharge rate and pressure head. High pressure head will result in low discharge rate and vice versa. 1. Centrifugal pump – most commonly used type for domestic and irrigation purposes. Depends on centrifugal force for their operation. Consists of an impeller inside an involute casing (Figure 4). 2.

Propeller Pump – used for low pressure head and high discharge. Extensively used for drainage pumping and the transfer of water from canals and rivers to adjacent fields (Figure 5).

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Figure 2. Lift pump

Figure 4. 

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Figure 3. Force pump

Centrifugal pump

Figure 5. Propeller pump

Sizing of Pumps and Prime Movers:  A criterion that is well accepted in computing the pump capacity is to base it on the highest daily water requirement of the crop.  According to the National Irrigation Administration (NIA), the water requirement for rice production is 10 mm/day.  Table 1 shows the combination of pump capacity, area to be irrigated and time of irrigation that can supply the daily water requirement for rice production.

Table 1. Operating Hrs/day 8 12 16 20 24

Area (ha) that can be irrigated at different capacities (gpm) and pumping time (hrs/day) Hectare for different gpm 20 30 40 50 60 70 80 90 100 0.33 0.50 0.67 0.83 1.00

0.50 0.75 1.00 1.25 1.50

0.67 1.00 1.33 1.67 2.00

0.85 1.25 1.67 2.08 2.50

Agricultural Machinery and Mechanization

1.00 1.50 2.00 2.50 3.00

1.17 1.75 2.33 2.91 3.50

1.33 2.00 2.67 3.33 4.00

1.50 2.25 3.00 3.75 4.50

1.67 2.50 3.33 4.17 5.00

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 Table 2.

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The size of the pump can be determined by the pump output using Table 2. Sizes and outputs of centrifugal pumps

Pump Size, inches (inlet and outlet diameter) 1.25 1.50 2 2.50 3 4

Pump output (gpm) 20 – 30 30 – 50 50 – 70 70 – 100 100 – 200 200 - 300

 The horsepower needed to drive the pump is computed as follows: Hp =

QXH 3960 X Eff s

Where: Q = pump output, gpm H = total head, ft. Effs = system efficiency = Effp X Efft X Effpm Effp = pump efficiency Efft = transmission efficiency Effpm = prime mover efficiency  The total head H may be estimated by adding the friction and velocity head losses to the total static lift.  Pump efficiency varies with the size of the pump with bigger ones being more efficient (Table 3). Table 3. Normal efficiencies of centrifugal pumps Pump Output (gpm) Efficiency (%) 20 32 30 37 40 40 60 45 100 50 150 55  Table 4 gives the transmission efficiencies of four types of transmission systems Table 4.

Normal efficiencies of transmission systems Type of Transmission Direct Gear V-belt Flatbelt

Efficiency (%) 100 98 95 90

 For continuous operation, electric motors and engines are expected to deliver only a certain percentage of their rated power as shown in Table 5. Agricultural Machinery and Mechanization

PSAE Region IV - Agricultural Engineering Board Review Materials Table 5.

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Continuous delivery rating of power units Type of Power Unit Electric Motor < 1 hp > 1 hp Internal Combustion Engine Air-cooled gasoline Air-cooled diesel Water-cooled diesel

Delivery Rating 65 – 72 72 – 90 60 70 80

 The power unit for driving pumps may be electric motors or engines.  Electric motors provide an economical installation when an adequate and reliable source of electric power is available at reasonable cost.  Engine should be used when the source of electric power is not reliable or too costly.  When an engine is considered, the decision is to use either a gasoline engine or a diesel engine.



 Where the annual use is more then 800 hours, the high cost of the diesel engine may be overcomed. Specific Yield of well – the discharge rate of a well per foot drawdown.  Example: Specific yield = 3 gpm/ft drawdown) Discharge rate of pump = 30 gpm Total drawdown = 10 ft  Total drawdown is added to the total head for the computation of power to drive the pump.



Sample Problem: Determine the size of the centrifugal pump and air-cooled diesel engine to use for a 1.5-ha field planted to lowland rice. The field is to be irrigated for 12 hours every 2 days. Source of water is a well with the following data: a. b. c. d. e.

Static water level = 8 ft Specific yield = 12 gpm/ft DD Discharge head = 0 ft Friction and velocity head = 2 ft Power transmission uses v-belt

Given: Area = 1.5 ha Operating time = 12 hr/2 days Static suction lift = 8 ft Specific yield = 12 gpm/ft DD Friction/velocity head = 2 ft Discharge head = 0 ft Type of transmission = v-belt Agricultural Machinery and Mechanization

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Required: a. b.

Size of centrifugal pump Size of diesel engine

Solution: a. Size of centrifugal pump: From Table 1, at 12 hr/day and 1.5 ha, the required pump capacity is 60 gpm. Since pumping operation is to be carried out every two days, then the pump capacity should be doubled to 120 gpm. From Table 2, at 120 gpm, the required pump size is 3X3 centrifugal pump. b. Size of diesel engine:

QXH Hp = 3960 X Eff X Eff X Eff p pm t Q = pump discharge = 120 gpm H = Total head = static head + drawdown + friction/velocity head + discharge head = 8 ft +

120 gpm + 2 ft + 0 ft = 20 ft 12 gpm/ft

From Table 3 and interpolating, Effp = 0.52 From Table 4, Efft = 0.95 From Table 5, Effpm = 0.70 Therefore: Hp = =

120 gpm X 20 ft 1.75 hp 3690 X 0.52 X 0.95 X 0.70

Since power units are available in standard sizes only, select the next higher standard size of air cooled diesel engine.

HARVESTING AND THRESHING EQUIPMENT  

Harvesting – process of gathering the useful portion of the crop from the field Threshing – process of separating the grains from the straw. Additionally, it may include cleaning the grains. Systems of harvesting and threshing of rice:

 1.

Manual harvesting

Agricultural Machinery and Mechanization

Manual threshing

PSAE Region IV - Agricultural Engineering Board Review Materials  Use of hand tools: scythe, yatab  80 – 160 man-hrs/ha  Figure 1

2.

3.

4.

5.



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 Use of flails or sticks  Hampasan  Foot threshing  Animal treading  140 – 220 man-hrs/ha  Figure 1  Field losses = 5 – 16% Manual harvesting Mechanized threshing  Same as above  Use of pedal thresher at 100 man-hrs/ha  Power thresher at 12 manhrs/ha  Field losses = 3 – 10 % Mechanical harvesting Mechanical threshing  Use of manually-operated  Same as above harvester at 50 manhrs/ha  Power harvester at 5 manhrs/ha  Field losses = 3 – 6 % Combine harvesting – both operations done by a single machine – combine  3 – 21 man-hrs/ha  Field losses – 1.5 – 6 % Stripping harvesting – removal of grains from panicle without cutting the straw - stripper  3 – 21 man-hrs/ha  Field losses = 2 – 6 % sickle,

Operations done by a combine and stripper: COMBINE 1. Separating the rows 2. Cutting the standing crop 3. Conveying the cut materials into the threshing section 4. Threshing 5. Shaking the straw to separate loose grains from the straw 6. Cleaning the grains of chaff 7. Conveying the grains to the tank or sack

1. 2. 3. 4.

Agricultural Machinery and Mechanization

STRIPPER Separating the rows Stripping Cleaning the grains Conveying the grains to the tank or sacks

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Figure 1. Traditional harvesting, threshing and cleaning 

Factors affecting choice of system: 1. 2. 3. 4. 5. 6. 7.



Kind of crop Timeliness of operation Topography Farm size Type of culture (row or broadcasted, upland or lowland) Availability and cost of labor Availability of capital

A major factor affecting choice of system is timeliness of operation because it affects field losses, time available for the next crop and grain quality

Time of harvesting One week before maturity At maturity One week after maturity Two weeks after maturity Three weeks after maturity Four weeks after maturity

Field losses (%) 0.77 3.35 5.63 8.64 40.70 60.46



If the crop is harvested too early, it will have a large percentage of imperfectly formed kernels.  The field should be drained 1 – 1.5 weeks before harvesting to harden the soil HARVESTINBG EQUIPMENT: 

Groupings of harvesters: 1. Hand tools – include the sickle, scythe, yatab, lingkao and cradle (Figure 2) 2. Reapers-windrower – a machine that cuts the standing crop, conveys the cut crop to one side, and lays them down in an orderly manner.

Agricultural Machinery and Mechanization

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3. Reaper-binder – a machine that cuts the standing crop, binds the cut crop, and lays down the bound crop to one side (Figure 3). 

Cutting mechanisms: 1.

Slicing/tearing action – employed in hand tools (Figure 4)

2.

High velocity, single-element, impact action – use of sharp or dulledged blades moving at high velocity of 2,000 fpm to 9,000 fpm (Figure 5)

3.

Two-element, scissor type action – shearing action between the moving and stationary blades (Figure 6).

Figure 2. Harvesting hand tools

Agricultural Machinery and Mechanization

PSAE Region IV - Agricultural Engineering Board Review Materials

Figure 3. Reaper-binder

Slicing action (Sharp smooth edge)

Tearing action (Serrated edge)

Figure 4. Slicing and tearing action

Figure 5. High velocity, single-element, impact action Agricultural Machinery and Mechanization

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Figure 6. Double-element Scissor type action THRESHING EQUIPMENT: 

Methods of threshing: 1. Rubbing action– grains are detached from their panicles because of a rubbing action as in treading by man, animal and vehicle. 

Output of man treading is 14 kg/hr

2. Impact action – grains are accelerated faster than their panicles and are detached as in hampasan and mechanical threshers. 

Output of hampasan is 34 kg/hr



Output of mechanical threshers vary with size of machine and power source.

3. Stripping action – grains are detached from their panicles when the straw is pulled through a “V” configuration or a comb-like device is passed through the panicles. 

Mechanical threshers:  Mechanical threshers employ the impact method  Variability of threshers come from: 1. Power source – manual as in pedal thresher or power thresher as in engine-driven thresher (Figures 7 and 8). 2. Type of feeding: a.

Hold-on feeding – Straws do not pass through the threshing section (Figure 9).   

Low power requirement Lightweight construction Examples: Pedal thresher combine

Agricultural Machinery and Mechanization

and

Japanese

PSAE Region IV - Agricultural Engineering Board Review Materials b.

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Throw-in feeding – Straws pass through the threshing section (Figure 10).   

High power requirement Heavyweight construction Examples: Axial-flow thresher and US combine

3. Direction of threshing materials (Figure 11): a.

Tangential-flow – Materials are feed between the revolving cylinder and stationary concave and go straight out of the thresher tangentially. 

b.

About 60% of the grains pass through the concave and the rest are separated in subsequent operations.

Axial-flow – Materials are fed between the revolving cylinder and stationary concave on one end, go around the cylinder several times axially and discharge at the other end. 

About 90% of the grains are separated from the straw at the cylinder.

4. Type of cylinder teeth (Figure 12): a. Wire-loop b. Peg-tooth c. Rasp-bar 

Cleaning systems:  Separation of the bulky straw, chaff, empty kernels and very light impurities from the grains.  Light materials can be separated from the grains by winnowing using natural wind or blower.  For hold-on threshers, straws do not pass through the thresher and only the removal of the chaff and light materials are needed using blowers and screens  For throw-in threshers, straws pass through the thresher and cleaning is done using a straw walker, blower and screens.

Agricultural Machinery and Mechanization

PSAE Region IV - Agricultural Engineering Board Review Materials

Figure 7. Pedal thresher

Figure 8. Axial-flow thresher

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PSAE Region IV - Agricultural Engineering Board Review Materials

Figure 9. Hold-on feeding type of thresher

Figure 10.Throw-in feeding type of thresher

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Figure 11.Flow of materials

Figure 12.Types of threshing teeth

Sample problem 1: A 5-m self-propelled combine makes an average stop of 4 minutes everytime its 2-ton grain tank is to be unloaded. The yield of the 20-ha field is 40 tons. The operating speed is 4.8 kph. The time for turning on the headland at the ends of the 500-m field is 15 seconds. Find:

a. b. c.

theoretical field capacity actual field capacity Field efficiency

Solution: a.

Theoretical field capacity = CT =

Agricultural Machinery and Mechanization

SWEff 10

where Eff = 1.0

PSAE Region IV - Agricultural Engineering Board Review Materials

= b) Actual field capacity = CA =

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(4.8 kph)(5 m)(1.0) = 2.4 Ha/hr 10

A T

where A = 20 ha

T = t1 + t2 + t3 where t1 = actual working time in the rows at 100% efficiency t2 = time for turning at headland t3 = time for unloading of tank No. of rows = NR =

400 m = 80 rows 5m/row

t1 = 80 rows X 500 m/row X 1km/1,000m X 1/4.8 kph = 8.333 hr t2 = 80 rows X 1 turn/row X 15 sec/turn X 1 hr/3600 sec = 0.333 hr t3 = 40 tons X

1 X 4 min/tank X 1 hr/60 min = 1.333 hr 2 tons/tank

T = t1 + t2 + t3 = 8.333 hr + 0.333 hr + 1.333 hr = 10 hr CA =

b.

A 20 ha = = 2 ha/hr T 10 hr

Field efficiency =

CA CT

X 100 =

2 ha/hr = 83.33 % 2.4 ha/hr

Sample problem 2: A 4-m combine travelling at 5 kph can empty its 1.64-ton grain tank in 60 seconds. When unloading on-the-go, it has an 82 % field efficiency. Field yield is 3 tons/ha. Find: What would be the field efficiency if the combine stopped to load? Solution: 1.

Unloading on-the-go: Actual field capacity (on-the-go) =

(5 kph)(4 m)(0.82) SWEff = 10 10 = 1.64 ha/hr

2.

No. of unloadings/hr = 3 tons/hr X 1.64 ha/hr X

1 1.64 tons/load

= 3 unloadings/hr 3.

Time for unloading = t2 = 3 unloadings/hr X 60 sec/unloading = 180 sec

4.

Eff =

t1 T

→

t1 = operating time = Eff X T

Agricultural Machinery and Mechanization

PSAE Region IV - Agricultural Engineering Board Review Materials = 0.82 X 3600 sec = 2952 sec 5.

New Eff =

t1 2952 sec X 100 = X 100 = 78.10 % T  t2 3600 sec  180 sec

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