gas power cycles

Presented by:Presented by:Damon OgdenDamon Ogden

22--1313--0101

ME 372 Thermodynamics II

Chapter 8Gas Power Cycles

ME 372 Thermodynamics II

Chapter 8Gas Power Cycles

Reciprocating EnginesReciprocating Engines

CH 8

Sections

8.4 - 8.6

Short VideoShort Video

• 3.3 Liter• V 12• 1965 Ferrari 275• Water Brake Dynamometer • Over 300 hp

Discussion TopicsDiscussion Topics

1. - History2. - General Knowledge3. - 4 Stroke Cycle4. - Otto Cycle5. - Diesel Cycle

HistoryHistory• Nikolaus Otto patented the “Otto Cycle”

Engine 1876Spark ignition engines use the Otto Cycle

• Rudolf Diesel received patent for “Diesel Cycle” Engine in 1892

Compression Ignition engines use theDiesel Cycle and hence the name “diesel” engines

How much do you know?How much do you know?

• What are are some differences between gasoline and diesel engines?

Ways to describe an engineWays to describe an engine

• Application• Design Geometry• Working Cycle• Fuel• Method of load control• Method of ignition

Piston -CylinderNomenclature

Otto / Diesel Cycle DifferencesOtto / Diesel Cycle DifferencesFour Stroke Cycle

Otto Cycle Diesel Cycle

Throttle Air Load Control Meter Fuel

Fuel/Air mix Working Fluid Air During Compression

Spark Start Combustion Inject Fuel

Four-Stroke CycleFour-Stroke Cycle

• Intake

http://www.howstuffworks.com/engine.htm

• Compression• Power• Exhaust

SuckSqueezeBangBlow

Thermodynamic Properties

Thermodynamic Properties

• The working fluid is air, and behaves as an ideal gas.

• Combustion processes are modeled as heat addition from an external source.

• Heat reject process to the surroundings is used to restore working fluid to its initial state.

• All processes are internally reversible.

Four Stroke Cycle P-V diagram

Intake Valve Opens

Combustion

Volume

Pressure Exhaust Valve Opens

Intake Valve Closes

Exhaust Valve Closes

TDC BDC

Intake

Exhaust

Power

Spark Ignition, Otto Cycle Engines

Spark Ignition, Otto Cycle Engines

• From < 1 hp to > 3000 hp• Cars• Trucks• Motor Cycles• Small Engines, i.e.;

Lawn Mower, • Mems

Spark Ignition Engine Characteristics

Spark Ignition Engine Characteristics

• Compression Ratio of 6 - 12=• RPM 900 - 8000• Power per unit Volume

3 - 60 kW / LiterFerrari in video @ 68 kW / liter

• Bore = .05 - .45 m

@TDC

@BDC

VV

BBDC

TDC

p

v

1

2

3

4

Otto Cycle

s = constant

A Quick Review of TermsA Quick Review of Terms• s = entropy, h = enthalpy, • u = internal energy• Isometric = Constant Volume• Isobaric = Constant Pressure• Isothermal = Constant Temperature• Isentropic = No irreversibilities,

Ideal process foradiabatic processes(adiabatic = no heat transfer)

Otto Cycle ModelOtto Cycle Model

• Intake and exhaust strokes not modeled

• Compression modeled as isentropic• Heat input modeled as isometric,

occurring at TDC• Expansion modeled as isentropic• Heat rejection modeled as isometric,

occurring at BDC

• Process 1-2: Isentropic Compression

• Apply First Lawq w u− = ∆

rvv

vv

r

r 11

2

1

2 ==

Since ∆s = 0

p

v

1

2

Otto Cycle AnalysisOtto Cycle Analysis

• Process 2-3 Isometric Heating• Apply First Law

q w u− = ∆

q u uin = −3 2

but v = constant, therefore w = 0

p

v

1

2

3

Otto Cycle AnalysisOtto Cycle Analysis

• Process 3-4 Isentropic Expansion• Apply First Law

q w u− = ∆

vv

vv

rr

r

4

3

4

3= =

p

v

1

2

3

4But since ∆s = 0

Otto Cycle AnalysisOtto Cycle Analysis

• Process 4-1 Isometric Cooling• Apply First Law

q w u− = ∆

q u uout = −1 4

p

v

1

2

3

4but v = constant, therefore:

Otto Cycle AnalysisOtto Cycle Analysis

Compression Ignition Diesel Cycle EnginesCompression Ignition Diesel Cycle Engines

• From < 50 hp to > 6000 hp• Cars• Trucks• Trains• Boats• Power Plants• Construction Equipment 1.47 L, 50 hp

Volkswagen

Combustion Ignition Engine Characteristics

Combustion Ignition Engine Characteristics

• Compression Ratios, rc of 12 - 23• Power per unit volume

2 - 26 kW / Liter• Bore .075 - 1 m• RPM 110 - 5000

Caterpillar

BBDC

TDC

@TDC

@BDC

VV

=cr 8 Liter

400 hp

Diesel Cycle ModelDiesel Cycle Model

• Intake and exhaust strokes not modeled

• Compression modeled as isentropic• Heat input modeled as isobaric,

occurring from TDC to appropriate volume

• Expansion modeled as isentropic• Heat rejection modeled as isometric,

occurring at BDC

T

s

1

2

3

4

p=const

v=const

Diesel Cycle

p

v1

23

4s=c

s=c

q w u− = ∆

rvv

vv

r

r 11

2

1

2 ==

p

v1

2

Since ∆s = 0

Diesel Cycle AnalysisDiesel Cycle Analysis• Process 1-2: Isentropic

Compression• Apply First Law

p

v1

2

Diesel Cycle AnalysisDiesel Cycle Analysis

• Process 2-3 Isobaric Heating• Apply First Law• q - w = ∆u• but p = constant, • therefore w = p ∆∆∆∆v• qin = h3 - h2

• Cutoff Ratio = rc

• rc = v3 - v2• Only process where Otto and Diesel Cycle differ

• Process 3-4 Isentropic Expansion

• Apply First Law• q - w = ∆u

Diesel Cycle AnalysisDiesel Cycle Analysis

p

v1

23

4•But since ∆∆∆∆s = 0

cr

r

rr

vv

vv ==

3

4

3

4

q w u− = ∆

Diesel Cycle AnalysisDiesel Cycle Analysis

• Process 4-1 Isometric Cooling• Apply First Law

• but v = constant, therefore:

p

v1

23

4

q u uout = −1 4

Comparison Of the Two Cycles

Comparison Of the Two Cycles

• Otto cycle utilizes an external energy source to initiate Combustion (spark plug)

• Diesel cycle relies on temp and pressure to start combustion

• Diesels only compress air, during compression stroke

• Otto cycle, spark engines compress both air and fuel during compression stroke

Comparison ContinuedComparison ContinuedFuel plays major role in cycle differences

• Spark ignition engines need fuels that are resistant to (knock)

• Diesel engines require fuels that will auto ignite under proper pressure and temperature

• Diesel fuel allows for higher Comp Ratios, = higher efficiencies.

Summary of Reciprocating Engines

Summary of Reciprocating Engines

• Been around for 125 years• Haven’t changed much:

Computers, Lower Emissions• Good for man, bad for the earth:• Not going anywhere soon• Future is Hydrogen, exhaust = H2O

Future = Mems Engines?Future = Mems Engines?

.001 kW

The EndThe End

Any Questions ?

Blank SlideBlank Slide

w u uon = −1 2

w u u p v vby = − + −3 4 2 3 2( )

q u uout = −1 4

q h hin = −3 2

w w w u u p v v u unet by on= + = − + − + −( ) ( ) ( )3 4 2 3 2 1 2

η thnet

in

wq

u u p v v u uh h

= = − + − + −−

3 4 2 3 2 1 2

3 2

( )

Diesel Cycle AnalysisDiesel Cycle Analysis• Combining above equations:

RTpv =uwq ∆=−

For isentropic processes:

rvv

vv

r

r 1

1

2

1

2 ==

in

netth q

w=η

vwMEP net

∆=

Equations:Equations:

Example: 9-2 Diesel CycleAn air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of compression, p1= 14.2 psi, V1 = 0.5 ft3, and T1 = 520oR. Determine:(a) the heat added, [Btu](b) the heat rejected, [Btu](c) the thermal efficiency(d) the Carnot efficiency(e) the mean effective pressure, [psi]

point 1 2 3 4p - psia 14.2

T oR 520v ft3/lb

u - Btu/lbh - Btu/lb

vr

V - ft3 0.5

Summary of Conditions - Diesel Cycle

r = 16, rc = 2

( )lbft

ftin

inlb

RRlb

lbftp

RTv

f

oo

f3

2

2

2

1

11

56.13

11442.14

52097.28

1545

=

==

Solution: (point 1)Solution: (point 1)

(isentropic compression from point 1:

911.916

58.15812

1

2

1

2 ==⇒== rr

r vrv

vvv

from air table, T2 is between 1480 & 1520 o

563.034.10578.934.10911.9

)1480()1520(

)1480(2 =−−=

−−

=rr

rr

vvvv

fr

( )( ) RT o150214801520563.014802 =−+=

Solution cont'd (point 2)Solution cont'd (point 2)

( )( )

lbBtu

u

84.266

44.26226.270563.044.2622

=

−+=

( )( )

lbBtu

h

4.369

89.36347.374563.089.3632

=

−+=

Point 2 cont'dPoint 2 cont'd

( ) 222

1

21

2

1

1

212

1

11

2

22

65616520

15022.14inlb

inlb

p

rTTp

vv

TTpp

Tvp

Tvp

ff =

=

==⇒=

Point 3

( )( ) RRT

rTvv

ppTT

Tvp

Tvp

oo

c

3005215023

22

3

2

323

2

22

3

33

==

==⇒=

Point 2 cont'dPoint 2 cont'd

from Air Table, T3 is between 3000 & 3050oRusing appropriate linear interpolation, determine

( )( )lb

Btuu 16.58604.58528.5961.004.5853 =−+=

( )( )lb

Btuh 15.79268.79034.8051.068.7903 =−+=

( )( ) 174.1180.1118.11.0180.13 =−+=rv

Point 3 cont'dPoint 3 cont'd

Point 4Point 4Isentropic expansion from point 3

( ) 390.90625.0

5.0174.13

434 =

==

VVvv rr

from table, 1520 & 1560oR, interpolation

273.0578.9890.8578.9390.9

)1520()1560(

)1520(4 =−−=

−−

=rr

rr

vvvv

fr

( )( )( )( )

lbBtuu

RT o

4.27226.27013.178273.026.270

153115201560273.01520

4

4

=−+=

=−+=

Point 1 2 3 4Start of Start of Start of Start of

Compressio Heating Expansion Cooling

p - psia 14.2 656T oR 520 1502 3005 1531

v ft3/lb 13.56u - Btu/lb 88.62 266.84 586.16 272.4h - Btu/lb 124.7 369.84 792.15

vr 158.58 9.911 9.39V - ft3 0.5 0.0313 0.0625 0.5

Summary of Conditions - Diesel Cycle

r = 16, rc = 2

Calculations:Calculations:(a)

lb

lbft

ftvVm 2

3

3

1

1 1069.356.13

5.0 −×===

lbBtuhhqin 31.42284.36915.79223 =−=−=

( ) Btulb

BtulbqmQ inin 57.153.4220369.0 =

==

BtuQin 57.15=

Calculations cont'dCalculations cont'd(b)

( ) ( )( )lb

BtulbuumQout 4.27262.880369.041 −=−=

BtuQout 78.6−=(c)

57.1578.611 −=−=

in

outth Q

Qη

564.0=thη

w u uon = −1 2

w u uby = −3 4

q u uout = −1 4

q u uin = −3 2

w w w u u u unet by on= + = − + −( ) ( )3 4 1 2

η thnet

in

wq

u u u uu u

= = − + −−

3 4 1 2

3 2

Otto Cycle AnalysisOtto Cycle Analysis• Combining above equations:

Example: 9-1At the beginning of the compression process in an air-standard Otto Cycle, p1 = 14.7 psi, T1 = 530oR. The compression ratio is 8. Determine for a maximum cycle temperature of 2000oR: thermal efficiency, Carnotefficiency, and mean effective pressure.

Point 1 2 3 4start start start start

of of of ofcompression heating expansion cooling

p-psia 14.7T oR 530 2000

v-ft3/lbu-Btu/lb

vr

Summary of Conditions

r = 8

Point 1 2 3 4start start start start

of of of ofcompression heating expansion cooling

p-psia 14.7T oR 530 2000

v-ft3/lbu-Btu/lb 90.3 367.61

vr 151.38 4.258

Summary of Conditions - data from air table

r = 8

RTpv =uwq ∆=−

For isentropic processes:

rvv

vv

r

r 1

1

2

1

2 ==

in

netth q

w=η

vwMEP net

∆=

Equations:Equations:

( )lbft

inlb

inftR

Rlblbft

pRTv

f

oo

f3

2

2

2

1

11 35.13

7.14

1441530

97.281545

=

==

lbftlbft

rvv

331

2 67.18

/35.13 ===

92.188

38.15112 ===

rvv r

r

Solution:Solution:

From air tables, note that T2 is between1160 and 1200 oR.doing the interpolation:T2 = 1191 oRu2 = 207.33 Btu/lb

Solution cont'dSolution cont'd

Point 1 2 3 4start start start start

of of of ofcompression heating expansion cooling

p-psia 14.7T oR 530 1191 2000 954

v-ft3/lb 13.35 1.67u-Btu/lb 90.3 207.3 367.61 164.2

vr 151.38 18.92 4.258 34.064

Summary of Conditions - calculations

r = 8

( ) ( )

net

net

wlb

Btu

wwwww

=

=+−++−=+++= −−−−

4.86

02.1647.36703.2073.9014433221

( ) in

in

qlb

Btuuuqq

==−

=−== −

3.1603.2076.367

2332

Solution - cont'dSolution - cont'd

thin

netth q

w ηη ==== 54.03.1604.86

Carnoth

cCarnot T

T ηη ==−=−= 74.0200053011

( )

2

2

2321

9.39

14467.135.13

7784.86

inlb

MEP

ftin

lbft

Btulbft

lbBtu

vvwMEP

f

f

net

=

−

=−

=

Solution - cont'dSolution - cont'd

Calculations cont'dCalculations cont'd(d)

300552011 −=−=

h

cCarnot T

Tη

827.0=Carnotη(e)

−

−

=−+=

−=

2

233

2121

144

778

0313.05.078.657.15

ftinBtulbft

ftftBtuBtu

VVQQ

VVWMEP

f

outinnet

2101inlb

MEP f=

Internal Combustion Engine Models

Internal Combustion Engine Models

• Spark Ignition Engine -Otto Cycle

• Compression Ignition Engine -Diesel Cycle

The piston moves the length The piston moves the length of the cylinder four times for one of the cylinder four times for one complete cycle.complete cycle.

Hence the name “four stroke” orHence the name “four stroke” or“four cycle engine”.“four cycle engine”.

Otto Cycle Model

Diesel Cycle Model

Piston-Cylinder NomenclaturePiston-Cylinder Nomenclature• Bore = B, Stroke = S• Bottom dead center = BDC• Top dead center = TDC• Mean Piston Speed = Sp = 2SN ,

N = engine speed in rev/s• Clearance volume = VTDC

• Displacement volume = VBDC - VTDC

• Compression ratio r = VBDC/VTDC

Method of Load ControlMethod of Load Control

• Throttle Air • (Otto cycle, spark ignition)

• Fuel Metering• (Diesel Cycle, compression

ignition)

T

s

1

2

3

4

p=const

v=const

Diesel Cycle

p

v1

23

4s=c

s=c

q w u− = ∆

Diesel Cycle AnalysisDiesel Cycle Analysis• Process 3-4 Isentropic

Expansion• Apply First Law

• But since ∆∆∆∆s = 0

p

v1

23

4

q w u− = ∆

cr

r

rr

vv

vv ==

3

4

3

4

p

v1

23

4

Diesel Cycle AnalysisDiesel Cycle Analysis• Process 3-4 Isentropic

Expansion• Apply First Law

• But since ∆∆∆∆s = 0

q w u− = ∆

q u uout = −1 4

p

v1

23

4

Diesel Cycle AnalysisDiesel Cycle Analysis• Process 4-1 Isometric Cooling• Apply First Law

• but v = constant, therefore:

Intake Stroke(Suck)

Intake Stroke(Suck)

• Intake valve open

• Exhaust valve closed

• Piston moves from TDC to BDC drawing in mixture of fuel and air (Only air for diesel cycle)

Compression Stroke(Squeeze)

Compression Stroke(Squeeze)

• Intake valve closes

• Exhaust valve remains closed

• Piston moves from BDC to TDC compressing fuel/air mixture, or air only

Otto Cycle

• Near top of stroke, spark ignites fuel/air mixture, causing heat input

Diesel Cycle

• Fuel injected into compressed air, in the cylinder

Power Stroke(Bang)

Power Stroke(Bang)

• Both valves closed

• Hot combustion products expand causing piston to move from TDC to BDC, producing work output

Exhaust Stroke(Blow)

Exhaust Stroke(Blow)

• Exhaust valve opens

• Intake valve remains closed

• Piston moves from BDC to TDC, forcing spent combustion products out of cylinder

ApplicationsApplications

• Auto• Semi-Truck• Locomotive• Marine• Aircraft• Stationary Power

Design Geometry Operating CycleDesign Geometry Operating Cycle

• V• Inline• Opposed• Rotary• Radial

• 4 stroke(There are others, but

those will be for a discussion)

FuelsFuels

• Gasoline• Diesel• Natural Gas• Liquid Propane• Methanol• Mixed gases