- - AGRICULTURE CORE CURRICULUM - - (CLF2000) Advanced Core Cluster: AGRICULTURE MECHANICS (CLF2900) Unit Title: WORK AND POWER ____________________________________________________________________________ (CLF2913) Topic: PROBLEMS IN WORK Time Year(s) AND POWER 3 Hours 1 / 2 / 3 / 4 ____________________________________________________________________________ Topic Objectives: Upon completion of this lesson the student will be able to: Learning Outcome #: (T-3) - Demonstrate knowledge of the transmission of power using chains, belts, gears, and shafts. (T-1) - Define linear motion and rotational motion. Special Materials and Equipment: Examples of the six simple machines, i.e. levers, inclined plane, wedge, screw, pulley, wheel and axle. References: Jacobs, C. O., & Harrell, W. R. (1983). AGRICULTURAL POWER AND MACHINERY. New York: McGraw-Hill. Smith, H. P. and Lambert, H. W. (1976). FARM MACHINERY AND EQUIPMENT . McGraw-Hill Book Company. Evaluation: Quiz by instructor and evaluation of power problems in homework and lab. TOPIC PRESENTATION: PROBLEMS IN WORK AND POWER A. Leverage Problems 1. While using a pry bar 7 ft. long, a man weighing 220 lbs. pushes with all his weight at the end of the bar. The fulcrum is placed 2 ft. from the rock which he is just able to move. How much does the rock weigh? Formula: Force X Force Arm = Weight X Weight Arm 220 X 5 ft. = W X 2 ft. 220 X 5 ft. --------------- = W = 550 pounds 2 ft. 2. A man lifts a wheelbarrow which he has just filled with concrete. He wishes to know how much he is lifting so he estimates the weight of the wheelbarrow and concrete at 300 lbs. The handles are 3.5 ft. long and the wheel is located at the opposite end from where he is lifting. If he determines that the center of the load is 1.5 ft. from the wheel, how much is he lifting? Formula: Force X Force Arm = Weight X Weight Arm F X 3.5 ft. = 300 X 1.5 ft. 300 X 1.5 ft. F = --------------- = 128.6 lbs. 3.5 ft. 3. A tractor with a front-end loader has a lifting capacity of 2100 pounds. How much force must the hydraulic cylinders develop in order to reach the 2100 lb. lift capacity? The loader arms are 9 ft. long to the center of the bucket load and the hydraulic rams attach to the arms at a point 3 ft. from the pivot point (fulcrum). Formula: Force X Force Arm = Weight X Weight Arm F X 3 ft. = 2100 lbs. X 9 ft. 2100 X 9 ft. F = ---------------- = 6300 lbs. 3 ft. Note: A third class lever is always a leverage disadvantage situation, i.e. it has more input than output. B. Gear Problems 1. Two gears are meshed together and rotate in a ratio of 4:1. The drive gear has 15 teeth and the driven gear has 60 teeth. If the speed (RPM) of the drive gear is 2200, how many RPM is the driven gear going? Formula: Drive Size X RPM = Driven Size X RPM 15 teeth X 2200 = 60 teeth X RPM. 15 teeth X 2200 RPM = -------------------- = 550 RPM 60 teeth 2. A tractor has a PTO (power take off) shaft speed rated at 540 RPM. If the mower it is pulling is driven by a reduction belt type drive and the pulley sizes are as follows, how fast will the mower spin when the rated PTO speed is reached? a. Drive Pulley = 6 inches diameter b. Driven pulley = 14 inches diameter Formula: Drive Size X RPM = Driven Size X RPM 6 in. X 540 RPM = ------------------ = 231.4 RPM 14 in. 3. On a ground driven planter unit, the drive wheel is 50 inches in circumference and turns a 13-tooth sprocket driving a chain up to a 20-tooth sprocket. The 20-tooth sprocket is connected to a shaft which turns a 17-tooth sprocket that is connected to a 28-tooth sprocket which in turn spins the 45-cell planter plate. This is a compound drive system and will set the spacing of the seeds down the row. (A compound drive problem is handled as two individual problems.) How many seeds will drop per revolution of the 50 inch drive wheel? Formula: Drive Size X RPM = Driven Size X RPM Part #1 13-tooth X 1 = 20-tooth X RPM 13 X 1 RPM = ---------------- = .65 RPM 20 Part #2 17-tooth X .65 = 28-tooth X RPM 17 X .65 RPM = --------------- = .395 RPM 28 Cell Plate (45 cell) X .395 RPM = 17.76 seeds drop in one revolution of 50 inch drive wheel 50 in. Seed spacing down the row then is: ---------- = 2.82 inches 17.76 seeds C. Work Problems 1. A horse is able to lift a pallet of bricks (330 lbs. per pallet) to a second story window (10 feet up) once each hour and works for ten hours. How much work has been accomplished at the end of 10 hours? Formula: Work = Force X Distance (feet) Work = 330 lbs. X 10 ft. X 10 times Work = 33,000 lbs.ft. 2. The horse is replaced with an engine which accomplishes the same amount of work in 48 seconds. How many HP is the engine? Force X Distance (ft.) Formula: HP = -------------------------- 550 X Seconds 33,000 lbs. ft. HP = ---------------------------- = 1.25 HP 550 X 48 seconds D. Energy Problems 1. Heat energy The amount of energy needed to heat a given amount of water is calculated by: Q = (M)(C)(TF - TI) Where, Q = Amount of heat required in BTU'S M = Weight of water to be heated in lbs. C = Constant (equals 1 for water) TF = Desired final water temperature TI = Initial water temperature From the above formula, determine the energy required to heat 100 gallons of water to 140 degrees F. (by setting the thermostat on a hot water heater) compared to the energy required to heat 100 gallons of water to 180 degrees F. Assume that the initial water temperature is 60 degrees F. in both cases. Determine the cost differential if electricity costs $0.10 per KWH. Given: Water weighs 8.34 pounds per gallon TF = 140 degrees F. and 180 degrees F. TI = 60 degrees F. To convert BTU'S to KWH, divide by 3413 Solution: Q = (M)(C)(TF - TI) Where, M = 100 gallons X 8.34 lbs/gal = 834 pounds C = 1 TF = 140 degrees F. and 180 degrees F. TI = 60 degrees F. Therefore, 140 degrees F. Q = (834)(1)(140-60) = 66,720 BTU's Q = 66,720 BTU X 1/3413 KWH/BTU = 19.55 KWH Cost = 19.55 KWH X $0.10/KWH = $1.96 80 degrees F. Q = (834)(1)(180-60) = 100,080 BTU's Q = 100,080 BTU X 1/3413 KWH/BTU = 29.32 KWH Cost = 29.32 KWH X $0.10/KWH = $2.93 Cost differential = $2.93 - $1.96 = $0.97 2. Irrigation Pumping Energy Costs. The energy to pump irrigation water is calculated by the formula: KWH/ACFT = (1.024)(H)/E Where, KWH/ACFT = pump energy in kilowatt hours per acre foot of water pumped H = total lift of the pump in feet E = pump efficiency as a decimal Pump efficiencies may vary between 30 and 70 percent. Assume that electricity costs $0.13 per KWH. Also assume that the pump lift is 100 feet. Finish the chart below to calculate the pump energy per acre foot and the pumping cost per acre foot for the various pump efficiencies. Given: H = 100 feet E = 0.30, 0.40, 0.50, 0.60 and 0.70 Cost of electricity is $0.13/KWH PUMP ENERGY REQ'D COST EFFICIENCY (KWH/ACFT) $/ACFT 30% 341.3 $44.37 40% 256.0 $33.28 50% 204.8 $26.62 60% 170.7 $22.19 70% 146.3 $19.02 E. Power Problems 1. Hydraulic horsepower. The rated flow of hydraulic oil from your 90 PTO horsepower wheel tractor is 26 GPM at 2320 psi. What percentage of the 90 PTO horsepower is used to pump hydraulic oil at maximum flow and rated pressure? Solution: HHP = (F)(P) divided by 1714 Where, F = 26 gpm P = 2320 psi Therefore, HHP = (26)(2320) divided by 1714 HHP = 35.2 Percentage of rated horsepower = 35.2 X 100 divided by 90 = 39.1% 2. Drawbar and PTO Horsepower. Calculate the drawbar and PTO horsepower (for a 2-WD tractor) requirements for a 15 foot wide landplane operated at 3.0 miles per hour in a medium soil. Solution: DBHP = (UD)(W)(S) divided by 375 Where, UD = 500 lbs/ft of width W = 15 feet S = 3.0 mph Therefore, D = 500 lbs/ft. X 15 ft = 7,500 lbs. Thus, DBHP = (7,500)(3.0)/375 = 60.0 HP PTOHP = (K)(DBHP) Where, K = 1.80 DBHP = 60.0 HP Therefore, PTOHP = (1.80)(60) = 108.0 HP F. Pressure Problems 1. A person wants to build a log splitter to split wood for the fireplace. If the cylinder is 5 inches in diameter and a force of 25,000 pounds must be developed, how much pressure must the pump develop in order to split the logs? Formula: Force = Area (sq. in.) X pressure (psi) 2 Force = (3.14 X R ) X psi 2 25,000 lbs. = (3.14 X 2.5 ) X psi 25,000 lbs. = 3.14 X 6.25 X psi 25,000 lbs. psi = --------------- = 1274 psi 3.14 X 6.25 2. How much lift can be created using a hydraulic jack when the force at the end of the jack handle is 35 pounds, the handle length is 15 inches, the jack pump attaches to the handle 1 inch from the fulcrum, the hand grasps the end of the handle 12 inches farther out, the pump diameter is .5 inches, and the lift ram outer diameter is 2.5 inches? Calculate the lift in pounds. Part #1 - Handle Leverage Force Formula: Force X Force Arm = Weight X Weight Arm 35 lbs. X 13 in. = W X 1 in. 455 W = --------- = 455 lbs. 1 Part #2 - Pump Generated Pressure Formula: Force = Area X psi 455 lbs. = (3.14 X .25 X .25) X psi 455 lbs. psi = ----------------- = 2318.5 psi 3.14 X .0625 Part #3 - Lift Ram Force Formula: Force = Area X psi Force = (3.14 X 1.25 X 1.25) X 2318.5 psi Force = 11,375 lbs. (rated - 5 ton jack) G. Torque 1. The farmer's daughter buys a new high torque V-8 fuel injected engine to drop into her old 1969 Chevy Short-bed 4X4 she is restoring. The book claims it should put out approximately 275 foot/lbs. of torque. Prove this claim given the following facts: Fact #1 - When she straps the engine into the school engine dynamometer, it records a maximum twisting force of 118 pounds at 2200 RPM Fact #2 - She remembers that the torque arm in the dynamometer is actually 30" long. Torque Calculation: Formula: Torque = Force (lbs) X Lever Arm Length (ft) Torque = 118 lbs. X 2.5 ft. Torque = 295 pounds foot H. Hydraulic System Problems 1. Calculate the pressure required to create a force of 15,000 pounds in a log splitter with a 4.5" hydraulic cylinder if the push is exerted on the extension stroke. Force 15,000 Formula: psi = -------- = ------------------- = 943.62 psi Area 3.14 X 2.25 X 2.25 2. How much force is created on the retraction stroke in the previous problem if the cylinder has a 2" rod and the pressure is 944 psi? Area used must be the net area available. Formula: Force = Area X psi Force = 15.89 - (3.14 X 1 X 1) X 944 psi Force = 12.75 X 944 = 12,036 lbs. 3. How big must a cylinder be if the available pump produces 2500 psi and a lifting force of at least 17,000 pounds is needed. Force 17,000 Formula: Area = --------- = --------- = 6.8 square inches psi 2500 Area = 3.14 X R X R 6.8 sq.in. R X R = ------------- = 2.17 3.14 R = Square root of 2.17 = 1.47 inch radius cylinder Diameter = 1.47 X 2 Diameter = 2.94 inches A 3-inch diameter cylinder will be sufficient. 7/24/91 YNJ/JR/tf #%&C