Farm Machinery Operation and Performance

Fertilizer drill metering rate

A fertilizer or seed drill meters a fixed mass of material per revolution of its metering wheel, so the application rate depends on how many metering revolutions occur over a given area. The basic relation is rate = (mass discharged per metering revolution) × (metering revolutions per unit area). The metering wheel is normally driven by the ground wheel through a chain or gear train, so the speed ratio of ground wheel to metering wheel fixes the number of metering revolutions per ground wheel revolution. The area covered per ground wheel revolution is the working width (number of rows × row spacing) multiplied by the distance travelled, which is \pi D for a ground wheel of diameter D. For units, 1 kg/ha = 0.1 g/m^2, because 1 kg = 1000 g and 1 ha = 10000 m^2. Ground wheel skid reduces the distance travelled per revolution, so the seed actually dropped in the field is lower than a laboratory calibration predicts. The percentage variation is computed relative to the predicted drop. Soft soil reduces the effective rolling circumference, so the area covered per revolution falls while the mass metered per revolution stays the same, and the rate per unit area rises. Volume flow of seed, needed for sizing the seed box, equals mass flow divided by bulk density, and mass flow equals seed rate × field capacity. For example, a drill with an effective width of 1.2 m moving at 3 km/h covers 0.36 ha/h, so at a seed rate of 10 kg/ha it meters 3.6 kg/h.

Machinery fixed and variable cost

Machinery costs are split into fixed costs, which accrue every year whether or not the machine is used (depreciation, interest, shelter and insurance), and variable costs, which depend on hours or area worked (fuel, lubricants, repairs and labour). The annual usage cost per hour is modelled as C = F/H + V, where F is the annual fixed cost, H is the annual working hours and V is the variable cost per hour. Because the fixed part is spread over more hours, the cost per hour falls as use rises. Two estimates made at different annual hours can be solved simultaneously for F and V. For custom hiring, the break-even point is where revenue equals total cost. With annual fixed cost F, operating cost c per hectare and custom rate R per hectare, the break-even area is A = F/(R-c). For example, with a fixed cost of Rs. 1,50,000 per year, an operating cost of Rs. 400 per hour and a custom rate of Rs. 900 per hour, the break-even use is 300 hours. If the annual cost is treated as wholly fixed, the cost per hour varies inversely with annual hours. A 15 % reduction in cost per hour therefore requires the annual hours to rise by the factor 1/0.85, and the area covered rises in the same proportion at constant field efficiency.

Sprayer nozzle spray width

A nozzle with spray angle \theta held at height h above the target gives a swath of S = 2h\tan(\theta/2). The swath is directly proportional to height, so lowering the boom narrows the footprint of each nozzle. The number of nozzles and the nozzle spacing do not change the swath of an individual nozzle, but the spacing determines the overlap between adjacent footprints. For flat fan nozzles the same geometry gives the footprint width. If each adjacent pair overlaps by a fraction f of the footprint width W, the total width covered by n nozzles is W + (n-1)(1-f)W. Nozzle discharge varies with the square root of nozzle pressure, Q \propto \sqrt{P}. The total boom flow follows from the area covered per hour and the application rate. When part of the pump flow is bypassed to the tank, the pump discharge is the nozzle discharge divided by the fraction that reaches the nozzles, and the pump pressure exceeds the nozzle pressure by the fraction lost in the lines. Pump power is P = pQ/\eta. As a quick check, a nozzle 1.0 m above the target with a 90° spray angle gives a swath of 2(1.0)\tan 45^\circ = 2.0 m.

Reaper cutting and propulsion power

A reaper draws power for three separate tasks: cutting the crop, conveying the cut material through the header, and propelling the machine. The header power is the sum of the cutting and conveying powers, and the propulsion power is separate and is not counted as header power. When the conveying power is a fixed fraction of the cutting power, the header power is the cutting power multiplied by one plus that fraction. For a mower or rotary cutter, the power is computed per unit of swath. The cutting energy per unit area in kJ/m^2 multiplied by the area rate (swath width × speed) gives the cutting power, and the air, stubble and gear train losses are taken per metre of swath width. The propulsion power is the rolling resistance multiplied by the speed. The engine power is found by dividing the PTO branch by \eta_{PTO} and the drawbar branch by \eta_{drawbar}, and adding the two results. For example, a cutter with a 1.0 m swath moving at 3.6 km/h (1 m/s) with cutting energy 2.5 kJ/m^2 requires 2.5 kW for cutting.

Mouldboard plough soil reaction forces

The soil reaction on a mouldboard plough is resolved into three components: longitudinal, along the direction of travel, which is the draft; transverse, the lateral thrust that pushes the plough sideways; and vertical, which lifts or presses the plough. The transverse component acts normal to the mouldboard surface. Friction between soil and metal, characterised by the soil-metal friction angle \phi, adds a component R_T\tan\phi along the direction of travel, so the draft is D = R_L + R_T\tan\phi. When the resultant soil force and the ratios of the lateral and vertical components to the longitudinal component are known, the longitudinal component is the resultant divided by \sqrt{1 + a^2 + b^2}, where a and b are those ratios. Tractor drawbar power follows from the theoretical forward speed of the driving wheel, \pi D N/60, reduced by the wheel slip fraction. The drawbar power is the pull multiplied by this actual speed.

Rotavator blade power requirement

The blades of a rotavator are fixed to brackets on the rotor, and each blade experiences a tangential soil force. The torque on a blade is the tangential force multiplied by its lever arm about the point of attachment. When the resultant force acts at the middle of the blade width, the lever arm used in this model is half the blade width, so T = F \times w/2. The length of soil slice, which is the distance travelled between successive cutting engagements of the blades, is c = v/(nk), where v is the forward speed, n is the rotor speed and k is the number of blade engagements per rotor revolution. Only the ratio of forward speed to rotor speed determines the slice length, so raising both by the same percentage leaves it unchanged. For example, a forward speed of 2 km/h (33.3 m/min) with a rotor speed of 200 rpm and 4 engagements per revolution gives a slice length of about 41.7 mm.

Disc harrow soil reaction force

Each disc of a harrow is a cutting element whose soil reaction acts parallel to the face of the disc. Because the disc is set at the gang angle \alpha to the direction of travel, the forward component of the reaction is F\cos\alpha. The total pull is the sum of these forward components over all discs in the harrow. The working depth does not enter the pull calculation directly, although it affects the magnitude of the force on each disc. Drawbar power is the pull multiplied by the forward speed, with the speed converted to m/s (1 km/h = 0.2778 m/s). In an offset disc harrow, the lateral soil forces of the front and rear gangs produce moments about the hitch point, and these moments are combined algebraically. The offset of the centre of cut equals the net moment of the lateral forces about the hitch divided by the total longitudinal soil force, since the longitudinal forces are balanced at the hitch. For example, a lateral force of 2 kN acting 3 m from the hitch opposed by 1 kN acting 2 m from the hitch gives a net moment of 4 kN·m; with a total longitudinal force of 5 kN, the offset is 0.8 m.

Decibel level of vibration and noise

Vibration acceleration level and sound pressure level are expressed in decibels on a logarithmic scale. For an RMS amplitude x relative to a reference x_0, the level is L = 20\log_{10}(x/x_0). A change of \Delta L decibels corresponds to an amplitude ratio of 10^{\Delta L/20}. A decrease of 5 dB in acceleration level therefore reduces the RMS acceleration to 10^{-0.25} \approx 0.562 of its original value. Doubling the RMS pressure adds 6.02 dB, and tripling it adds 20\log_{10}3 \approx 9.54 dB. Power or intensity ratios use 10\log_{10} instead, so doubling power adds only 3.01 dB. A pressure-based level must never be combined with the factor of 10.

Impact cutting of crop stems

A stem can be severed by a striking blade through bending failure. The impact force acts at a height above the soil, so the stem behaves as a cantilever with the force applied at a lever arm L from the base. The bending moment at the base is M = FL. For a circular stem of diameter d, I = \pi d^4/64 and c = d/2, so the bending stress is \sigma = 32FL/(\pi d^3). Failure occurs when \sigma reaches the ultimate tensile strength, and solving gives F = \sigma\pi d^3/(32L). The force needed to cut the stem falls rapidly as diameter decreases, because it scales with d^3. For example, a stem of 4 mm diameter with a 40 mm lever arm and a strength of 30 N/mm^2 fails at about 4.7 N.