Professional Engineering Exam Review Machinery Management

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Professional Engineering Exam Review Machinery Management

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Professional Engineering Exam Review Machinery Management

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Professional Engineering Exam ReviewMachinery Management

Gary Roberson

- Implement performance
- Draft and power estimation
- Fuel consumption
- Machine capacity

- 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.

- 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.

- ASAE EP496.3 (FEB2006)Agricultural Machinery Management
- Management decisions related to machine power requirements, capacities, cost, selection and replacement

- ASAE D497.6 (JUN2009) Agricultural Machinery Management Data
- Data for use with decision tools from ASAE EP496.3

- 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

- 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

- D is implement draft, N (lbf)
- Rsc is soil and crop resistance, N (lbf)
- MR is total implement motion resistance, N (lbf)

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)

- 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.

- Chisel plow with straight points
- Table 1 in D497.5
- A = 52, B = 4.9, and C = 0

- Table 1 in D497.5
- Medium soil texture
- Table 1 in D497.5
- F2 = .85

- Table 1 in D497.5
- S = 5 mph
- W = 12 ft or 12 tools
- T = 6 in

- D=Fi[A+B(S)+C(S)2]WT

- D=Fi[A+B(S)+C(S)2]WT
- D=0.85x[52+4.9(5)]12x6

- D=Fi[A+B(S)+C(S)2]WT
- D=0.85x[52+4.9(5)]12x6
- D= ?

- D=Fi[A+B(S)+C(S)2]WT
- 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.

- Chisel plow with straight points
- Table 1 in D497.5
- A = 91, B = 5.4, and C = 0

- Table 1 in D497.5
- Medium soil texture
- Table 1 in D497.5
- F2 = .85

- Table 1 in D497.5
- S = 8 km/hr
- W = 12 shanks
- T = 0.15 meters = 15 cm

- D=Fi[A+B(S)+C(S)2]WT

- D=Fi[A+B(S)+C(S)2]WT
- D=0.85x[91+5.4(8)]12x15

- D=Fi[A+B(S)+C(S)2]WT
- D=0.85x[91+5.4(8)]12x15
- D= ?

- D=Fi[A+B(S)+C(S)2]WT
- D=0.85x[91+5.4(8)]12x15
- D= 20,533 N

- 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?

- 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

- 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?

- 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

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

- Pdb = Drawbar Power, kW
- D = Draft, kN
- S = Speed, km/hr

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?

- 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.

- Ppto = PTO power
- W = implement working width, ft
- F = material feed rate. t/hr

kWh/t

kW

kW/m

- A large round baler has a capacity of 10 tons per hour. The baler has a variable bale chamber

- Variable Chamber Round Baler
- Table 2 in D497.5
- A = 5.4, B = 0, and C = 1.3

- 10 t/hr capacity

- Table 2 in D497.5

- 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?

- 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?

- 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

- Fluid power requirement from the tractor for the implement
- Hydraulic motors and cylinders used to drive implement functions

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

- Phyd = fluid power, kW
- P = fluid pressure, kPa
- F = fluid flow, L/s

- A harvester uses hydraulic power to drive a conveyor. The requirements were measured at 10.5 gpm at a pressure of 2200 PSI.

- 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.

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

- Pel = Electrical Power, kW
- I = electrical Current, A
- E = Electrical potential (voltage), V

- A sprayer uses electrical power to drive a pump. The requirements were measured at 20 amps at 12 volts.

- 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

- 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

- 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

- 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

- 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.

- Determine the tractive efficiency anticipated.
- From Figure 1, D497.5
- 2WD on tilled soil surface, TE = 0.67

- From Figure 1, D497.5
- Assume a mechanical efficiency of 0.96

- Determine the implement power requirement
- Apply the 80 % rule
- Example:
- Implement power = 89.1 HP
- Tractor power = 89.1/.8 = 111.4 HP

- 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?

- 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

- 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

- 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

- 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

- Required Capacity
- Size of machine necessary to get the job done in the time available.
- Acres/Hour

- Size of machine necessary to get the job done in the time available.
- Effective Capacity
- Available capacity of equipment in operation
- Acres/Hour

- Available capacity of equipment in operation

- 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

- 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

- 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

- Ca = available capacity, ac/hr
- S = speed, mph
- W = width, ft
- Ef = 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 width
- Measured width of the working portion of a machine
- For row crops, it is row spacing times number of rows

- Measured width of the working portion of a machine
- Effective width
- Actual machine working width, may be more or less than the theoretical width

- 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)

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?

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?

Available Capacity > Required Capacity

13.96 ac/hr > 8.33 ac/hr

Yes, the implement is large enough.

Questions?

- 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

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

Good Luck!