TKP3501 Farm Mechanization

Transcription

1 TKP3501 Farm Mechanization Topic 8: Tractors and Power Units Ahmad Suhaizi, Mat Su

2 ASMS Why we need machineries? Type of machine available Filters, oil, Traditional vs modern lubrication, Introduction parts Maintenance Introduction to a tractor Type of tractor Specification Tractors components & Systems Main components Systems; - Fuel & Intake - Combustion - Cooling - Electric & instruments - Lubrication - Hydraulic Power tiller Other small Compact equipment equipment ** = Calculations Others - Bearing & seal Implemen - Shaft Land Primary ts tillage - Belt & pulley preparatio Secondary tillage - Chain & sprocket Crop type; n - Gear - Oil palm - Lubrication (grease, oil) - Rice - Vegetable Theoretical Field Capacity Effective Field Farm Capacity Efficiency** Field Efficiency How to choose the tractor and implement size** Tractor & power unit Type of power available Crop Production Seedling & Planting Crop type Planter** Livestock Feeding system Milking Aquaculture Forestry Horticultur e Emerging Technologies Sensor Tracking GPS, GNSS GIS, Mapping Cost Optimization analysis** Fertilization & Harvesting Irrigation** Yield, Baller Spreader Transportation Pump Farm Sprinkler maintenanc Grass, road, e drainage

3 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

4 Introduction 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. 4

5 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 Row crop Orchard Vineyard Industrial Garden ( Hp) ( Hp) (< 25Hp) 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) 5

6 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

7 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

8 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

9 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

10 Nebraska Tractor Test Laboratory Report 10

11 Report Example New Holland T80A [Source: 11

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13 2. Power Rating 13 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

14 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

15 Work Work (knm)/(kj) = Force (kn) x Distance (m) 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

16 Power Power (kw) = Force kn x Distance (m) time (s) 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 (km hr )

17 Fuel Consumption Estimates 17

18 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 Drawbar power Measured at the point implements are attached to the tractor, drawbar or 3 point hitch 18

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20 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

21 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. Brake Power (BP) = Indicated Power (IP) - Power loss to friction (Fp) 21

22 PTO Power Power kw = T x N 9549 Power Hp = T x N 5252 Drawbar Power Where Power in kw ( 1 Hp = 0.75 kw) T = Torque ( Nm) * (Ib-ft) N = speed (rpm) * (rpm) Db (Hp) = F Ib x V(mph) 375 Where Db in Hp ( 1 Hp = 0.75 kw) F = Force (Ib) V = speed (mph) Db (kw) = F kn x V(km hr ) 3.6 Where Db in kw F = Force or draft (kn) V = speed (km/hr) 22

23 Best estimate according 86 % rules PTO Hp = E Hp x 0.86 PTO HP = Db Hp 0.86 Power loss Accessories Temperature 10% other than fan 5 % fan and radiator 1% or each 5.6 C above 29 C - Gasoline 1% or each 2.7 C above 29 C Diesel Where E Hp = Engine power Db Hp = Drawbar power Altitude 3% for each 305 m above 152 m Type of service 10% for intermittent loads 20% for continuous loads 23

24 Exercise 1 Estimate the amount of power available at the PTO and drawbar for a tractor rate at 125 engine power. 24

25 Exercise 1 Estimate the amount of power available at the PTO and drawbar for a tractor rate at 125 engine power. PTO Hp = E Hp x 0.86 = 125 x 0.86 = Hp ~ 110 Hp Db Hp = PTO Hp x 0.86 = (E Hp x 0.86) x 0.86 = x 0.86 = Hp 25

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27 Sizing implement and tractor from nomograph Chisel plow-coarse Implement width = 10 Typical speed = 6 mph (see table 3) 27

28 28 Source: Nebraska Tractor Test

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31 31 Indicated Power

32 A Distance A-B B 32

33 33 What is the minimum indicated power requirement?

34 34 What is the minimum indicated power requirement?

35 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

36 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) 36

37 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, 38 Drawbar HP = 0.75 x PTO HP or PTO HP = 1.33 x Drawbar HP

39 39 Chisel plow-coarse

40 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

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

42 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

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. 44

44 45 Drawbar attachment

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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

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

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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

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

51 52 Power Loss Calculations

52 Best estimate according 86 % rules PTO Hp = E Hp x 0.86 PTO HP = Db Hp 0.86 Power loss Accessories Temperature 10% other than fan 5 % fan and radiator 1% or each 5.6 C above 29 C - Gasoline 1% or each 2.7 C above 29 C Diesel Where E Hp = Engine power Db Hp = Drawbar power Altitude 3% for each 305 m above 152 m Type of service 10% for intermittent loads 20% for continuous loads 53

53 Exercise 2 What is the usable power? 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at C air temperature? 54

54 Exercise 2 What is the usable power? 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at C air temperature? Engine Hp = 165 x (1 (0.01 x )) 5.6 Engine Hp = Hp or ~ 160Hp The effect of the C temperature is to reduce the usable power about 1.5% 55

55 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

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? Engine Hp = 165 x (100% (3% x )) 305 Engine Hp = Hp or ~ 160Hp The effect of the 500 m altitude to reduce the usable power about 3.42% 57

57 Exercise 4 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at continuous duty? 58

58 Exercise 4 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at continuous duty? Engine Hp = 165 x (100% 20%) Engine Hp = 132 Hp The effect of the continuous duty is to reduce the usable power about 20% 59

59 Exercise 5 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at ambient tempereature of C, altitude of 500 m for continuous duty? 60

60 Exercise 5 1. How much usable power will a 165Hp spark ignition engine produce if it will be operating at ambient tempereature of C, altitude of 500 m for continuous duty? Adjustment temp = 0.01 x = Adjustment altitude = 3% x = Adjustment continous = 20 % = 0.2 Total adjustment = = Engine Hp = 165 Hp x = Hp ~ 120 Hp 61

61 Assignment 2 On PutraBlast 62

62 Thank you. The winner is men vs 1 tractor