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# Agriculture Engineering-chptr15

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### Agriculture Engineering-chptr15

1. 1. MACHINE POWER REQUIREMENT
2. 2. 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 of 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
3. 3. Factors to consider when selecting a power unit1. Engine type2. Power ratings3. Soil resistance to machines4. Tractor size5. Matching implements6. Sizing for critical work
4. 4. Engine type• The combustion process in the cylinders 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
5. 5. Power Ratings• 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 horsepower
6. 6. If a load require a force of 20 pounds to move itvertically a distance of 3 feet, the amount of workdone is :Work = Force x Distance = 20 lbs x 3 feet = 60 ft. lbThe amount of work done is 60 ft.lb with noreference to time.
7. 7. If a 1000 lb force is moved 33 feet in one minute, the rate of doingwork is one horsepower, because one horsepower equals 33,000 ft.lb. per minuteThe equivalent rate of work in one second to equal 1 horsepower is :1 HP = 33,000 ft.lb = 550 ft. lb per second 60 secondsWhen working with field machinery, we usually think of miles perhour and pounds of draft. For these conditions the formula forhorsepower is : HP = Force,lb x Speed, mph 375
8. 8. Metric EquivalentMetric unit for power is kilowatt (kW)Force is measured in newtons or kilonewtons1 HP = 0.746 kW1 kW = 1.34 HP1N = 0.225 lb1 kN = 224.8 lb forceFormula for kilowatt is: KW = Force (kilonewtons) x Speed (km/hr) 3.6
9. 9. Eg: If the draft of a trailing implement, such as adisk harrow, is measured at 11.1 kilonewtons andis pulled at a speed of 8 km/hr, what is thedrawbar kilowatt ?Drawbar, kW = 11.1 kN x 8 km/hr =24.7 kW 3.6
10. 10. The formula can also be used to determine speed.Eg: A tractor is pulling a plow with a total draftof 22.2 kilonewtons. How fast can the plow bepulled if the tractor has 50 drawbar kilowatts ? Speed = 50 kW x 3.6 = 8.1 km/hr 22.2 kN
11. 11. To determine draft :Eg: Given 65 kW tractor, speed 8 km/hr., field cultivatordraft is 4 kN per meter of width when used in a givenfield. What width of cultivator could be pulled ?Draft = Power,kW x 3.6 = 65 kW x 3.6 = 29.25 kN Speed, km/hr 8 km/hrWidth, meters = Total Draft = 29.25 = 7.3 meters Draft per meter 4
12. 12. Eg: If the draft of a trailing implement like a disk harrow is measured at 2,500 pounds and is pulled at a speed of 5 mph, what is the drawbar horsepower ?Drawbar horsepower = Force, lbs x speed,mph 375 = 2500 x 5 375 = 33.3
13. 13. • 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 or two is involved.• Size of the implement have to be related to the amount of soil resistance
14. 14. 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
15. 15. Brake Horsepoweris the maximum power the engine can develop withoutalterationsthe engine can be hooked to a dynamometer to determinehow much brake horsepower can be developeduseful in sizing stationary engines for operating irrigationpumps, grinders and other large equipmentsthe same engines used for large tractors are often used asstationary engines
16. 16. Power Take-Off Horsepower (PTO hp)• the power measured at the PTO shaft
17. 17. Drawbar Horsepower *• is the measure of the pulling power of the engine by way of tracks, wheels or tires at a uniform speed• drawbar horsepower varies depending widely on several factors - soil surface and type of hitch• drawbar horsepower is the function of drawbar pull and speed of the various kind of horsepower, maximum PTO hp is the most commonly used in designating the size of a tractor• large tractors that do not have PTO shaft, they may be given a brake hp (flywheel) or maximum drawbar hp rating
18. 18. Matching tractors and implements When matching a tractor and implement, 3 important factors must be considered :-1. The tractor must not be overloaded or early failure of components will occur2. The implement must be pulled at the proper speed or optimum performance cannot be obtained3. The soil conditions and their effects on machine performance must be considered
19. 19. With a given tractor, there is a set of amount ofpower available. This available power is used for:-I. moving the tractor over the groundII. pulling the implement over the groundIII. powering the implement for useful work The softer or looser the soil conditions are, the greater amount of power that will be consumed because of greater rolling resistance This reduces the available usable drawbar power SOIL CONDITIONS---POWER
20. 20. Condition Usable Drawbar Power Ratio of MaximumOf Soil As a % of Maximum PTO Power To Usable PTO Power Drawbar Power_________ __________________ ___________________Firm 67 Percent 1.5Tilled 56 Percent 1.8Soft or Sandy 48 Percent 2.1___________________________________________________
21. 21. Eg: A 5-bottom, 40 cm plow operates in medium to heavy soil with surface conditions considered as firm. Given 32.5 drawbar kW per meter of width for gumbo and 27.4 kW per meter for clay. We will use 30 drawbar kW per meter as the average.• Plow width = 5 x 40 = 200 = 2 meters 100• 2 meters x 30 kW per meter = 60 drawbar kW needed
22. 22. if tractor size is known, determine how large animplement can be pulled.Tractor size – 110 PTO kWSpeed - 8 km/hSoil condition – firmDraft – 5.83 kN per meter of widthUsable drawbar kW – 67 % of maximum PTO kW = .67 x 110 = 73.7 kN Width = 73.7 kW x 3.6 = 5.7 meters 8 x 5.83
23. 23. Position Control SystemHydraulic control of an attached implement eg. spreaderor broadcaster whereby the operator will preselect andposition the implement as determined by the position ofhand control leverThe position of the hand lever and the hydrauliccylinder are always the samePressure is controlled by the relief valve and thehydraulic cylinder will automatically move theimplement to its predetermined position and maintain itthere
24. 24. Draft is the horizontal component of pull, parallel to the line of motionPoweris the rate of doing work 1hp = 550 ft.lb/sec. 1 kW = 1.34 hp (metric unit)Drawbar Power (dbhp) is the measure of pulling power of the engine by way of tracks, wheels or tires at a uniform speed 1 dbhp = 4500 kg. m / minute
25. 25. THANK YOU