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Professional Engineering Exam Review Machinery Management Gary Roberson Topics for Discussion • Implement performance • Draft and power estimation • Fuel consumption • Machine capacity Documents to Review • ASAE S296.5 (DEC2003) General Terminology for Traction of Agricultural Traction and Transport Devices and Vehicles – terminology to assist in the standardized reporting of information on traction and transport devices and vehicles. Documents to Review • ASAE S495.1 (NOV2005) Uniform Terminology for Agricultural Machinery Management – Uniform use of machinery management terms. – Definitions used in system analysis, economic analysis, and mechanical concepts. Documents to Review • ASAE EP496.3 (FEB2006)Agricultural Machinery Management – Management decisions related to machine power requirements, capacities, cost, selection and replacement Documents to Review • ASAE D497.6 (JUN2009) Agricultural Machinery Management Data – Data for use with decision tools from ASAE EP496.3 Books of Interest • Machinery Management, W. Bowers, Deere and Co. • Farm Power and Machinery Management, D. Hunt, Iowa State University Press. • Engineering Principles of Agricultural Machines, A. Srivastava, et al , ASABE • Engineering Models for Agricultural Production, D. Hunt, AVI Publishing Co. • Agricultural Systems Management, R. Peart and W. Shoup, Marcel Dekker Implement Power Requirement • Drawbar power – Power developed by the drive wheels or tracks and transmitted through the hitch or drawbar to move the implement. – Power is the result of draft (force) and speed Implement Draft D Rsc MR • D is implement draft, N (lbf) • Rsc is soil and crop resistance, N (lbf) • MR is total implement motion resistance, N (lbf) Implement Draft A B(S) C(S)2 WT DF i Where: – D=draft, N (lbf) – F=soil texture parameter – i=texture indicator: • 1=fine, 2=medium, 3=coarse – A, B, And C = machine parameters (Table 1, D497) – S=speed, km/h (mph) – W=width, m (ft) or number of tools – T=tillage depth, cm (in), • (1 for tools that are not depth specific) Implement Draft Example • A 12 foot wide chisel plow with straight points and shanks spaced 1 foot apart is used at a depth of 6 inches in medium textured soil at a speed of 5 mph. Table 1, D497.5 Implement Draft Example • Chisel plow with straight points – Table 1 in D497.5 • A = 52, B = 4.9, and C = 0 • Medium soil texture – Table 1 in D497.5 • F2 = .85 • S = 5 mph • W = 12 ft or 12 tools • T = 6 in Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[52+4.9(5)]12x6 Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[52+4.9(5)]12x6 • D= ? Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[52+4.9(5)]12x6 • D= 4682 lbf Implement Draft Example • A 12 shank chisel plow with straight points and shanks spaced 0.3 meters apart is used at a depth of 0.15 meters in medium textured soil at a speed of 8 km/hr. Table 1, D497.5 Implement Draft Example • Chisel plow with straight points – Table 1 in D497.5 • A = 91, B = 5.4, and C = 0 • Medium soil texture – Table 1 in D497.5 • F2 = .85 • S = 8 km/hr • W = 12 shanks • T = 0.15 meters = 15 cm Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[91+5.4(8)]12x15 Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[91+5.4(8)]12x15 • D= ? Implement Draft Example • D=Fi[A+B(S)+C(S)2]WT • D=0.85x[91+5.4(8)]12x15 • D= 20,533 N Implement Draft Exercise • A 4 shank subsoiler with straight points is used at a depth of 16 inches in coarse textured soil at a speed of 4 mph. • What’s the Draft? Table 1, D497.5 Implement Draft Exercise • A 4 shank subsoiler with straight points is used at a depth of 16 inches in coarse textured soil at a speed of 4 mph. • What’s the Draft? 4959 LB Implement Draft Exercise • A 4 shank subsoiler with straight points is used at a depth of 0.41 meters in coarse textured soil at a speed of 6.5 km/hr. • What’s the Draft? Table 1, D497.5 Implement Draft Exercise • A 4 shank subsoiler with straight points is used at a depth of 0.41 meters in coarse textured soil at a speed of 6.5 km/hr. • What’s the Draft? 22,291 N Drawbar Power D xS Pdb 375 • Pdb = Drawbar Power, HP • D = Draft, lbf • S = Speed, mph Drawbar Power D xS Pdb 3.6 • Pdb = Drawbar Power, kW • D = Draft, kN • S = Speed, km/hr Drawbar Power Example An Implement with a draft of 8,500 lbf is operated at a net or true ground speed of 5.0 MPH with 10 percent wheel slippage. What is the implement drawbar power? Drawbar Power D xS Pdb 375 Pdb Drawbar Power D xS Pdb 375 Pdb 113 Hp PTO Power • PTO power is required from some implements and is delivered through the tractor PTO via a driveline to the implement. • The rotary power requirement is a function of the size and feed rate of the implement. PTO Power Ppto a b(w) c(F) • Ppto = PTO power • W = implement working width, ft • F = material feed rate. t/hr Table 2, D497.5 kW kW/m kWh/t PTO Power Example • A large round baler has a capacity of 10 tons per hour. The baler has a variable bale chamber Table 2, D497.5 Implement PTO Example • Variable Chamber Round Baler – Table 2 in D497.5 • A = 5.4, B = 0, and C = 1.3 – 10 t/hr capacity Implement PTO Example Ppto a b(w) c(F) Implement PTO Example Ppto a b(w) c(F) Ppto 5.4 1.3(10) Implement PTO Example Ppto a b(w) c(F) Ppto 5.4 1.3(10) Ppto 18.4 HP PTO Power Exercise • A rectangular baler has a capacity of 3 tons per hour. Bale dimensions (cross section) are 16” x 18”. • What’s the PTO power requirement? Table 2, D497.5 PTO Power Exercise • A rectangular baler has a capacity of 3 tons per hour. Bale dimensions (cross section) are 16” x 18”. • What’s the PTO power requirement? PTO Power Exercise • A rectangular baler has a capacity of 3 tons per hour. Bale dimensions (cross section) are 16” x 18”. • What’s the PTO power requirement? 6.3 Hp Hydraulic Power • Fluid power requirement from the tractor for the implement • Hydraulic motors and cylinders used to drive implement functions Hydraulic Power p xF Phy d 1714 • Phyd = fluid power, HP • P = fluid pressure, psi • F = fluid flow, gpm Hydraulic Power pxF Phy d 1000 • Phyd = fluid power, kW • P = fluid pressure, kPa • F = fluid flow, L/s Hydraulic Power Example • A harvester uses hydraulic power to drive a conveyor. The requirements were measured at 10.5 gpm at a pressure of 2200 PSI. Hydraulic Power p xF Phy d 1714 Hydraulic Power p xF Phy d 1714 2200 x 10.5 Phy d 1714 Hydraulic Power p xF Phy d 1714 2200 x 10.5 Phy d 1714 Phy d 13.48 HP Electrical Power • Some implements require electrical power supplied by the tractor for certain functions. – Typically electrical power for control functions is small and can be neglected. – Electrical power for pumps and motors should be accounted for. Electrical Power I xE Pel 746 • Pel = Electrical Power, HP • I = electrical Current, A • E = Electrical potential (voltage), V Electrical Power IxE Pel 1000 • Pel = Electrical Power, kW • I = electrical Current, A • E = Electrical potential (voltage), V Electrical Power Example • A sprayer uses electrical power to drive a pump. The requirements were measured at 20 amps at 12 volts. Electrical Power I xE Pel 746 Electrical Power I xE Pel 746 20 x 12 Pel 746 Electrical Power I xE Pel 746 20 x 12 Pel 746 Pel 0.32 HP Implement Power • Combined total of drawbar, PTO, Hydraulic and Electrical power – Drawbar power adjusted by tractive and mechanical efficiencies • 80% rule – Implement power should not exceed 80% of rated tractor power Tractive Efficiency • Ratio of drawbar power to axle power • Takes into account the added resistance the tractor will encounter in moving through the soil. – Firmer soil, higher TE – Softer soil, lower TE Mechanical Efficiency • Accounts for power losses in the tractor drive train. – Accounts for friction loss, slippage in a clutch, torque converters, etc. • Usually constant for a given tractor – Typically 0.96 for tractors with mechanical transmissions – 0.80 to 0.90 for hydrostatic transmissions Power Efficiency Chart Implement Power Pdb Pt Ppto Phy d Pel Em x Et • Pt = total power, HP • Pdb = drawbar power, HP • Em = mechanical efficiency • Et = tractive efficiency • Ppto = PTO power, HP • Phyd = Hydraulic power, HP • Pel = electrical power, HP Implement Power Problem • Determine the recommended tractor size for an implement that requires 48 drawbar horsepower, 12 PTO horsepower and 2.5 hydraulic horsepower. The tractor should be 2 wheel drive with a mechanical transmission and you will operate on a tilled soil surface. Power Efficiency Chart Drawbar Power Conditions • Determine the tractive efficiency anticipated. – From Figure 1, D497.5 • 2WD on tilled soil surface, TE = 0.67 • Assume a mechanical efficiency of 0.96 Implement Power Pdb Pt Ppto Phy d Pel Em x Et Implement Power Pdb Pt Ppto Phy d Pel Em x Et 48 Pt 12 2.5 0 0.96x 0.67 Implement Power Pdb Pt Ppto Phy d Pel Em x Et 48 Pt 12 2.5 0 0.96x 0.67 Pt 89.1HP Tractor Size • Determine the implement power requirement • Apply the 80 % rule • Example: – Implement power = 89.1 HP – Tractor power = 89.1/.8 = 111.4 HP Tractor Size Exercise • An implement uses 25 PTO horsepower, 3.6 horsepower through the hydraulic system and 1.9 horsepower in the electrical system. What is the minimum recommended tractor size? Tractor Size Exercise • An implement uses 25 PTO horsepower, 3.6 horsepower through the hydraulic system and 1.9 horsepower in the electrical system. What is the minimum recommended tractor size? 38.1 Hp Tractor Fuel Consumption • Fuel consumption can be estimated for tractors used in various operations. – Specific fuel consumption is quoted in units of gal/hp-hr • Average fuel Consumption (Diesel) – Qs = 0.52X + 0.77 - 0.04(738X + 173)1/2 – where X = ratio of equivalent PTO power to rated tractor power Tractor Fuel Consumption Example • A 95 PTO horsepower tractor is used with a 55 horsepower load. How much fuel will be consumed in one day (10 hours)? – X = 55/95 = 0.58 • Qs = 0.52x0.58 + 0.77 - 0.04((738 x 0.58) + 173)1/2 Qs = 0.092 gal/hp-hr Tractor Fuel Consumption Example • Estimated Fuel Consumption – Qi = Qs x Pt – Qi = 0.092 x 55 – Qi = 5.06 gal/hr • Total Fuel Consumption – 5.06 gal/hr x 10 hrs = 50.6 gal Equipment Economics • Required Capacity – Size of machine necessary to get the job done in the time available. • Acres/Hour • Effective Capacity – Available capacity of equipment in operation • Acres/Hour Machine Capacity • Required capacity will tell you how large the machine should be • Effective capacity will tell you what a given piece of equipment can deliver • Effective capacity should equal or exceed required capacity for most applications Machine Capacity A Ci B x G x PWD • Ci = required capacity, ac/hr • A = Area to be covered, ac • B = days available • G = working hours per day • PWD = probability of a day suitable for field work in the given time frame Machine Capacity Example • What size machine is needed to cover 1000 acres in a three week (5 days per week) window in August in North Carolina. You can work up to 10 hours per day. • From Table 5. D497.5 – PWD = 0.51 Machine Capacity 1000 Ci 15 x 10 x 0.51 Machine Capacity 1000 Ci 15 x 10 x 0.51 Ci 13.1Ac/Hr Machine Capacity S x W x Ef Ca 8.25 • Ca = available capacity, ac/hr • S = speed, mph • W = width, ft • Ef = Field Efficiency Field Efficiency • Ratio of effective field capacity to theoretical field capacity • Effective field capacity is the actual rate at which an operation is performed • Theoretical field capacity is the rate which could be achieved if a machine operated 100% of the time available at the required speed and used 100% of its theoretical width Theoretical vs. Effective Width • Theoretical width – Measured width of the working portion of a machine • For row crops, it is row spacing times number of rows • Effective width – Actual machine working width, may be more or less than the theoretical width Field Efficiency and Speed Machine Capacity Example • What is the capacity of disc harrow that operates at 6 mph with a working width of 18 ft? • From Table 3. D497.5 – Typical field efficiency is 80% (0.80) Machine Capacity Example 6 x 18 x 0.80 Ca 8.25 Machine Capacity Example 6 x 18 x 0.80 Ca 8.25 Ca 10.47 Ac/hr Machine Capacity Exercise You are given an implement that covers 8 rows on a 36 inch row spacing. This implement is effective at 6 miles per hour with a field efficiency of 80%. You have a 2 week window working 5 days a week, 10 hours per day. Probability of a working day is 60%. You have 500 acres to cover. Is this implement large enough to get the job done? Machine Capacity Exercise You are given an implement that covers 8 rows on a 36 inch row spacing. This implement is effective at 6 miles per hour with a field efficiency of 80%. 6 MPH x (8 x36/12) FT x 0.80 Ca 8.25 C a 13.96 AC/HR Machine Capacity Exercise You have a 2 week window working 5 days a week, 10 hours per day. Probability of a working day is 60%. You have 500 acres to cover. 500 Ac Ci 10 Days x 10 Hrs x 0.60 C i 8.33 Ac/Hr Machine Capacity Exercise Is this implement large enough to get the job done? Available Capacity > Required Capacity 13.96 ac/hr > 8.33 ac/hr Yes, the implement is large enough. General Problem Solving Guides • Study the problem • Determine the critical information • Decide on a solution method or equation • State all assumptions, cite data sources • Solve the problem • Indicate solution clearly Contact Information Gary Roberson Associate Professor and Extension Specialist Biological and Agricultural Engineering North Carolina State University E-mail: gary_roberson@ncsu.edu Phone: 919-515-6715

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