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

Mar 31, 2015

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Page 1: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

II-C Power & Energy SystemsDennis Buckmaster

[email protected]://engineering.purdue.edu/~dbuckmas/

OUTLINE• Internal combustion engines• Hydraulic power circuits• Mechanical power transmission• Electrical circuit analysis (briefly)

Page 2: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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.

Page 3: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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.

Page 4: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Engines

• Power and Efficiencies• Thermodynamics• Performance

Page 5: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Engine Power Flows

Page 6: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 7: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 8: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.
Page 9: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.
Page 10: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.
Page 11: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Dual Cycle

Page 12: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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]

Page 13: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

• 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

Related equations

Page 14: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 15: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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?

Page 16: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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)

Speed (rpm)

Power (kW)

Page 17: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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A quick problem …

• Diesel engine generating 60 kW at 2300 rpm• Q: torque available

Page 18: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Power Hydraulics

• Principles• Pumps, motors• Cylinders

Page 19: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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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

Page 20: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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Liquids Have no Shape of their own

Page 21: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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Liquids are Practically Incompressible

Page 22: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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.

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Page 23: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Application Principles1 lb (.45kg)Force

1 sq in (.65cm2)Piston Area

1 psi

(6.9kpa)

10 sq in (6.5cm2)Piston Area

10 lbs (4.5kg)

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Page 24: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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Hydraulic “lever”

Page 25: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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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)

Page 26: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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Trapped Oil

Closed Center HydraulicsOpen CenterFlow in Neutral

Page 27: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Extend 27

Page 28: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Retract 28

Page 29: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Neutral Again 29

Page 30: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Pumps

Page 31: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 32: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 33: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Speed

Flow

Qgpm = Dcu in/rev Nrpm /231

Page 34: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Pressure Rise

TorqueRequired

Tinlb = Dcu in/rev ∆Ppsi /(2π)

Page 35: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Pressure

Flow

Theoretical pump

Effect of leakage

Relief valve or pressure compensator

Page 36: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Pressure

Flow

Constant power curve

Php = Ppsi Qgpm/1714

Page 37: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 38: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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?

Page 39: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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 pump problems

Page 40: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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?

Page 41: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Cylinders

Force balance on piston assembly: Fexternal

P1 * A1P2 * A2

Page 42: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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• 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

Page 43: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

• 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

Page 44: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Power Transmission

Page 45: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Transmissions transform powera torque for speed tradeoff

Page 46: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Gears

Page 47: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Planetary Gear Sets

Page 48: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Belt & Chain Drives

• Speed ratio determined by sprocket teeth or belt sheave diameter ratio

Page 49: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

FIRST GEAR

Page 50: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

FIRST GEAR

Page 51: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

• 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 gear problem

Page 52: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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 planetary gear problem

Page 53: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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?

Example belt problem

Php = Tft-lbNrpm/5252

Page 54: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

ElectricityElectricity

Voltage = Current * Resistance

Vvolts = Iamps * Rohms

Power = voltage times current

PWatts = Vvolts*Iamps

V

I R

Page 55: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Three Types of CircuitsThree Types of Circuits

Series Same current, voltage divided

+

-12 v.

Parallel Same voltage, current divided

Series / Parallel

Page 56: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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

Page 57: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

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?

Example 12 V DC problem

Page 58: II-C Power & Energy Systems Dennis Buckmaster dbuckmas@purdue.edu dbuckmas/ dbuckmas@purdue.edu OUTLINE Internal combustion.

Good luck on the PE Exam!

• My email address:

[email protected]

• My web page:

https://engineering.purdue.edu/~dbuckmas/

Note … ASABE members can access ASABE texts electronically at:

http://asae.frymulti.com/toc.asp