II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu https://engineering.purdue.edu/~dbuckmas/ OUTLINE Internal combustion engines Hydraulic power circuits Mechanical power transmission Electrical circuit analysis (briefly)

References Engineering Principles of Agricultural Machinery, 2nd ed. 2006. Srivastava, Goering, Rohrbach, Buckmaster. ASABE. Off-Road Vehicle Engineering Principles. 2003. Goering, Stone, Smith, Turnquist. ASABE.

Other good sources Fluid Power Circuits and Controls: Fundamentals and Applications. 2002. Cundiff. CRC Press. Machine Design for Mobile and Industrial Applications. 1999. Krutz, Schueller, Claar. SAE.

Engines Power and Efficiencies Thermodynamics Performance

Engine Power Flows

Power & Efficiencies Fuel equivalent Pfe,kW = (HgkJ/kg∙ṁf,kg/h)/3600 [Hg = 45,000 kJ/kg for No. 2 diesel] Indicated Pi,kW = pime,kPaDe,lNe,rpm/120000 Brake Pb,kW = 2πTNmNe,rpm/60000 Friction Pf = Pi-Pb

Power & Efficiencies Indicated Thermal Eit = Pi/Pfe Mechanical Em = Pb/Pi Overall (brake thermal) Ebt = Pb/Pfe = Eit*Em Brake Specific Fuel Consumption BSFC= ṁf,kg/h/Pb,kW

Dual Cycle 

Related equations Compression ratio = r r = V1/V2 Displacement De,l = (V1-V2)*(# cylinders) = π(borecm)2(strokecm)*(# cyl)/4000 Ideal gas p1V1/T1 = P2V2/T2 Polytropic compression or expansion p2/p1 = rn [n = 1 (isothermal) to 1.4 (adiabatic), about 1.3 during compression & power strokes]

Related equations Air intake ṁa,kg/h = .03De,lNe,rpmρa,kg/cu mηv,decimal From Stoichiometry (fuel chemistry) A/F = air to fuel mass ratio = 15:1 for cetane

What is the displacement of a 6 cylinder engine having a 116 mm bore and 120 mm stroke?

For this same engine (7.6 l displacement, 2200 rpm rated speed), what is the air consumption if it is naturally aspirated and has a volumetric efficiency of 85%? Assume a typical day with air density of 1.15 kg/m3. With a stoichiometric air to fuel ratio based on cetane, at what rate could fuel theoretically be burned?

Consider the this same (595 Nm, 137 kW @ 2200 rpm) engine which has a high idle speed of 2400 rpm and a torque reserve of 30%; peak torque occurs at 1300 rpm. Sketch the torque and power curves (versus engine speed). Torque (Nm) Power (kW) Speed (rpm)

A quick problem … Diesel engine generating 60 kW at 2300 rpm Q: torque available

Power Hydraulics Principles Pumps, motors Cylinders

About Pressure 14.7 psia STP (approx __ in Hg) Gage is relative to atmospheric Absolute is what it says … absolute & relative to perfect vacuum What causes oil to enter a pump? Typical pressures: Pneumatic system Off-road hydraulic systems

Liquids Have no Shape of their own

Practically Incompressible Liquids are Practically Incompressible

4/1/2017 Pascal’s Law Pressure Exerted on a Confined Fluid is Transmitted Undiminished in All Directions and Acts With Equal Force on Equal Areas and at Right Angles to Them. --Hydraulics is a means of power transmission --Oil is the most commonly used medium because it serves as a lubricant and is practically non-compressible (it will compress approximately 1/2 of a 1 percent per 1000 PSI). --Weight of oil varies with viscosity, but averages between 55 to 55 lbs per cubic foot. (at 100 degrees F). NOTE: A cubic foot of oil is 1728 Cu.In (12x12x12). A gallon is 231 Cu.In., so a Cubic Foot of oil is equivalent to 7.48 Gallons. --A liquid is pushed, NOT DRAWN, into a pump. Atmospheric pressure equals 14.7 PSI at sea level. --Oil takes the course (path) of least resistance. FORMULAS; 1. H.P. = GPM x Pressure x .000583 -or- H.P. = GPM x PSI / 1714 2. One H.P. = 33000 ft./lbs. per minute (33000 lbs raised 1 ft in 1 minute) One H.P. = 746 Watts, One H.P. = 42.4 BTU per minute 3. Required Area of a transmission line; Area = GPM x .3208 / velocity (ft./sec) -or- Velocity (ft./sec) = GPM / 3.117 x Area Pascal’s Law, named after Blaise Pascal (French 1623-1662) 22 Hydraulic Fundamentals

Application Principles 4/1/2017 Application Principles 1 lb (.45kg) Force 1 sq in (.65cm2) Piston Area 10 lbs (4.5kg) 10 sq in (6.5cm2) Piston Area This slide illustrates one of the basic principles of hydraulics; LIQUIDS TRANSMIT APPLIED PRESSURE EQUALLY IN ALL DIRECTIONS. BUILDS: 1. When a one pound force is applied to this handle and the area of the piston is one square inch, with the confined fluid, what PSI pressure will be produced? Note that this pressure is exerted in every direction 2. With a 10 square in piston, how much weight will this system lift? This principle is what allows us to multiple our work efforts. With one lb of down pressure, we are able to lift 10 lbs. --Pressure is caused by a resistance to flow. In this case the 10 lb weight. Point out that resistance to flow is what causes pressure. In this example, if there were a 100 lb weight on the right side (in place of the 10 lb weight), how much pressure would be required to lift it. (10 PSI). 1 psi (6.9kpa) 23 Hydraulic Fundamentals

Hydraulic “lever”

Types of Hydraulic Systems Open Center Closed Center The control valve that regulates the flow from the pump determines if system is open or closed. Do not confuse Hydraulics with the “Closed Loop” of the Power Train. (Hydro)

Closed Center Hydraulics Open Center Flow in Neutral 4/1/2017 Closed Center Hydraulics Open Center Flow in Neutral Trapped Oil Open Center Valve Hydraulic flow continually moves through the system because the hydraulic pump is constantly pumping fluid. The valve is open to return in neutral to allow the fluid to circulate back to the reservoir. Oil is drawn out of the reservoir because atmospheric pressure (14.7 psi) pushes it through the lines into the pump. 26 Hydraulic Fundamentals

4/1/2017 Extend 27 Hydraulic Fundamentals

4/1/2017 Retract 28 Hydraulic Fundamentals

4/1/2017 Neutral Again 29 Hydraulic Fundamentals

Pumps

Pump Inefficiency Leakage: you get less flow from a pump than simple theory suggests. Increases with larger pressure difference Friction: it takes some torque to turn a pump even if there is no pressure rise Is more of a factor at low pressures

Efficiency of pumps & motors Em – mechanical efficiency < 1 due to friction, flow resistance Ev – volumetric efficiency < 1 due to leakage Eo =overall efficiency = Em * Ev Eo = Power out/power in

Qgpm = Dcu in/rev Nrpm /231 Flow Speed

Tinlb = Dcu in/rev ∆Ppsi /(2π) Torque Required Pressure Rise

Theoretical pump Effect of leakage Relief valve or pressure compensator Flow Pressure

Constant power curve Flow Pressure Php = Ppsi Qgpm/1714

1a. If a pump turns at 2000 rpm with a displacement of 3 in3/rev, theoretically, how much flow is created? 1b. If the same pump is 95% volumetrically efficient (5% leakage), how much flow is created? Example pump problems

2a. If 8 gpm is required and the pump is to turn at 1750 rpm, what displacement is theoretically needed? 2b. If the same pump will really be is 90% volumetrically efficient (10% leakage), what is the smallest pump to choose? Example pump problems

Example pump problems 3a. A 7 in3/rev pump is to generate 3000 psi pressure rise; how much torque will it theoretically take to turn the pump? 3b. If the same pump is 91% mechanically efficient (9% friction & drag), how much torque must the prime mover deliver?

Example motor problem If a motor with 2 in3/rev displacement and 90% mechanical and 92% volumetric efficiencies receives 13 gpm at 2000 psi … a. How much fluid power is received? b. What is it’s overall efficiency? c. How fast will it turn? d. How much torque will be generated?

Cylinders Force balance on piston assembly: Fexternal P2 * A2 P1 * A1

Example cylinder problem 3000 psi system 2” bore cylinder Extends 24 inches in 10 seconds Q: max force generated max work done power used flow required

Example cylinder problem Tractor source with 2500 psi and 13 gpm available Return pressure “tax” of 500 psi Cylinder with 3” bore, 1.5” rod diameters Q1: How much force will the cylinder generate? Q2: How long will it take to extend 12 inches? Example cylinder problem

Power Transmission

Transmissions transform power a torque for speed tradeoff

Gears

Planetary Gear Sets

Belt & Chain Drives Speed ratio determined by sprocket teeth or belt sheave diameter ratio

FIRST GEAR

FIRST GEAR First gear speeds … if … Input shaft: 1000 rpm Main countershaft: 1000 (22/61) = 360 rpm Ratio = input speed/output speed = 1000/360 = 2.78 Ratio = output teeth/input teeth = 61/22 = 2.78 Secondary countershaft: 360 rpm (41/42) = 351 rpm Output shaft: 351 rpm (14/45) = 109 rpm RATIO: input speed/output speed = 1000/109 = 9.2 Product of output teeth/input teeth = (61/22)(42/41)(45/14) = 9.2

Example gear problem If 50 kW @ 2400 rpm drives a pinion gear with 30 teeth and the meshing gear has 90 teeth (assume 98% efficiency)… Q1: What is the speed of the output shaft? Q2: How much power leaves the output shaft? Q3: How much torque leaves the output shaft?

Example planetary gear problem If the sun of a planetary gear set turns at 1000 rpm, what speed of the ring would result in a still planet carrier? Teeth on gears are sun: 20 and ring: 100.

Example belt problem Php = Tft-lbNrpm/5252 If a belt drive from a 1750 rpm electric motor is to transmit 5 hp to a driven shaft at 500 rpm and the small sheave has a pitch diameter of 4” … Q1: What should the pitch diameter of the other pulley be? Q2: Which shaft gets the small sheave? Q3: How much torque does the driven shaft receive? Php = Tft-lbNrpm/5252

V I R Electricity Voltage = Current * Resistance Vvolts = Iamps * Rohms Power = voltage times current PWatts = Vvolts*Iamps V I R Ohm’s Law expresses the relationship between the current (I), the voltage (E), and the resistance (R) in a circuit. Ohm’s law can be expressed in three different ways, and can be applied to the entire circuit or to any part of the circuit. WHEN ANY TWO FACTORS ARE KNOWN, THE THIRD UNKNOWN FACTOR CAN BE CALCULATED FROM OHM’S LAW. Virtually no one refers to “Electromotive Force” anymore. Voltage is the MEASURE of the difference in POTENTIAL between two points. The potential is the electromotive force, the unit of measure is the volt. The unit of measure for resistance is the OHM.

Three Types of Circuits Series Same current, voltage divided Parallel Same voltage, current divided Series / Parallel 1. A good example of a series circuit is the “old time” Christmas tree lights where if one bulb burned out, none of the bulbs would light. Of course this was because when there is only one path for current to travel and that path is cut off (bulb filament burns out), there is no longer a complete path, current can’t flow and nothing works. Series circuits are low amps, high resistance. 2. Parallel circuits offer multiple paths for the current to travel each one offering a complete path from positive to negative. To calculate resistance with two 10 ohm resistors in parallel resistance value * resistance value (10 x 10 = 100) ----------------------------------------------- resistance value + resistance value (10 + 10 = 20) 100/20 = 5 ohms Parallel circuits are high amps, low resistance 3. Series / Parallel circuits are a combination of the two resulting in resisters in series as well has parallel. + - 12 v.

Example 12 V DC problem A 12 V DC solenoid a hydraulic valve has a 5 amp fuse in its circuit. Q1: What resistance would you expect to measure as you troubleshoot its condition? Q2: How much electrical power does it consume?

Example 12 V DC problem Q1: Identify specifications for a relay of a 12 V DC lighting circuit on a mobile machine if the circuit has four 60W lamps. Q2: Would the lamps be wired in series or parallel?

Good luck on the PE Exam! My email address: dbuckmas@purdue.edu My web page: https://engineering.purdue.edu/~dbuckmas/ Note … ASABE members can access ASABE texts electronically at: http://asae.frymulti.com/toc.asp