Transcript Reaching Equipment Purchasing Decisions

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
2

D  F A  B(S)  C(S) WT
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
Pdb
D xS

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  113Hp
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
Phyd
pxF

1714
• Phyd = fluid power, HP
• P = fluid pressure, psi
• F = fluid flow, gpm
Hydraulic Power
pxF
Phyd 
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
Phyd
p x F

1714
Hydraulic Power
Phyd
p x F

1714
2200 x 10.5
Phyd 
1714
Hydraulic Power
Phyd
p x F

1714
2200 x 10.5
Phyd 
1714
Phyd  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
IxE
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
Pel
IxE

746
Electrical Power
Pel
IxE

746
20 x 12
Pel 
746
Electrical Power
Pel
IxE

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  Phyd  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  Phyd  Pel
Em x Et
Implement Power
Pdb
Pt 
 Ppto  Phyd  Pel
Em x Et
48
Pt 
 12  2.5  0
0.96 x 0.67
Implement Power
Pdb
Pt 
 Ppto  Phyd  Pel
Em x Et
48
Pt 
 12  2.5  0
0.96 x 0.67
Pt  89.1 HP
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  13.1 Ac/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
Ca  13.96AC/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
Ci  8.33Ac/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: [email protected]
Phone: 919-515-6715