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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
DF 
   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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