Professional Engineering Exam Review Machinery Management

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 i

F 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


• 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


• D=Fi[A+B(S)+C(S)2]WT



• D=0.85x[52+4.9(5)]12x6



• D=0.85x[52+4.9(5)]12x6

• D= ?



• D=0.85x[52+4.9(5)]12x6

• D= 4682 lbf


• 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


• 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



• D=0.85x[91+5.4(8)]12x15



• D=0.85x[91+5.4(8)]12x15

• D= ?



• 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


• 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.


4959 LB


• 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.


Table 1, D497.5


• 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.


22,291 N

Drawbar Power

375

S x D Pdb

• Pdb = Drawbar Power, HP• D = Draft, lbf• S = Speed, mph

Drawbar Power

3.6

S x D Pdb

• 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

375

S x D Pdb

Pdb

Drawbar Power

375

S 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


c(F)b(w)a Ppto


1.3(10)5.4 Ppto

c(F)b(w)a Ppto


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



PTO Power Exercise



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

1714

F x p Phyd

• Phyd = fluid power, HP• P = fluid pressure, psi• F = fluid flow, gpm

Hydraulic Power

1000

F 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

1714

F x p Phyd

Hydraulic Power

1714

10.5 x 2200 Phyd

1714

F x p Phyd

Hydraulic Power

1714

10.5 x 2200 Phyd

1714

F 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

746

E x I Pel

• Pel = Electrical Power, HP• I = electrical Current, A• E = Electrical potential (voltage), V

Electrical Power

1000

E 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

746

E x I Pel

Electrical Power

746

12 x 20 Pel

746

E x I Pel

Electrical Power

746

12 x 20 Pel

746

E 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 E

P 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 E

P P

Implement Power

02.5120.67x 0.96

48 P

t

elhydptot m

dbt PPP

Ex E

P P

Implement Power

02.5120.67x 0.96

48 P

t

elhydptot m

dbt PPP

Ex E

P 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 B

A 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 15

1000 Ci

Machine Capacity

0.51 x 10 x 15

1000 Ci

Ac/Hr 3.11 Ci

Machine Capacity

• Ca = available capacity, ac/hr• S = speed, mph• W = width, ft• Ef = Field Efficiency

8.25

E x W x S C

fa

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


• 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)


8.25

0.80 x 18 x 6 Ca


8.25

0.80 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 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%.

8.25

0.80 x FT x36/12)(8 x MPH 6 Ca

AC/HR 13.96 Ca


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.

0.60 x Hrs 10 x Days 10

Ac 500C i

Ac/Hr 8.33 C i


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.

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!

mailto:[email protected]

Professional Engineering Exam Review Machinery Management

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