Professional Engineering Exam Review Machinery Management Gary Roberson
Feb 25, 2016
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
MR R D sc • D is implement draft, N (lbf)• Rsc is soil and crop resistance, N
(lbf)• MR is total implement motion
resistance, N (lbf)
Implement Draft
WT2C(S) B(S) A iF D
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
375S x D Pdb
• Pdb = Drawbar Power, HP• D = Draft, lbf• S = Speed, mph
Drawbar Power
3.6S x D Pdb
• Pdb = Drawbar Power, kW• D = Draft, kN• S = Speed, km/hr
Drawbar Power ExampleAn 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
375S x D Pdb
Pdb
Drawbar Power
375S x D Pdb
Hp 113 Pdb
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
c(F)b(w)a Ppto • Ppto = PTO power• W = implement working width, ft• F = material feed rate. t/hr
Table 2, D497.5kW 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
c(F)b(w)a Ppto
Implement PTO Example
1.3(10)5.4 Ppto
c(F)b(w)a Ppto
Implement PTO Example
1.3(10)5.4 Ppto
c(F)b(w)a Ppto
HP 18.4 Ppto
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
1714F x p Phyd
• Phyd = fluid power, HP• P = fluid pressure, psi• F = fluid flow, gpm
Hydraulic Power
1000F x p Phyd
• 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
1714F x p Phyd
Hydraulic Power
171410.5 x 2200 Phyd
1714F x p Phyd
Hydraulic Power
171410.5 x 2200 Phyd
1714F x p Phyd
HP 13.48 Phyd
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
746E x I Pel
• Pel = Electrical Power, HP• I = electrical Current, A• E = Electrical potential (voltage), V
Electrical Power
1000E x I Pel
• 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
746E x I Pel
Electrical Power
74612 x 20 Pel
746E x I Pel
Electrical Power
74612 x 20 Pel
746E x I Pel
HP 0.32 Pel
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
elhydptot m
dbt PPP
Ex EP P
• 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
elhydptot m
dbt PPP
Ex EP P
Implement Power
02.5120.67x 0.96
48 P
t
elhydptot m
dbt PPP
Ex EP P
Implement Power
02.5120.67x 0.96
48 P
t
elhydptot m
dbt PPP
Ex EP P
HP 89.1 Pt
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
PWDG x x BA Ci
• 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
0.51 x 10 x 151000 Ci
Machine Capacity
0.51 x 10 x 151000 Ci
Ac/Hr 3.11 Ci
Machine Capacity
• Ca = available capacity, ac/hr• S = speed, mph• W = width, ft• Ef = Field Efficiency
8.25E x W x S C f
a
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
8.250.80 x 18 x 6 Ca
Machine Capacity Example
8.250.80 x 18 x 6 Ca
Ac/hr 0.471 Ca
Machine Capacity ExerciseYou 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 ExerciseYou 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%.
8.250.80 x FT x36/12)(8 x MPH 6 Ca
AC/HR 13.96 Ca
Machine Capacity ExerciseYou 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.
0.60 x Hrs 10 x Days 10Ac 500C i
Ac/Hr 8.33 C i
Machine Capacity ExerciseIs 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.
Questions?
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 RobersonAssociate Professor and Extension SpecialistBiological and Agricultural EngineeringNorth Carolina State UniversityE-mail: [email protected]: 919-515-6715
Good Luck!