notes1

11

COMBUSTION THEORY

ME 576

33

ME 576

• THERMODYNAMICS PRIMER• THE 3 T’s OF COMBUSTION• GAS POWER CYCLES• INTERNAL COMBUSTION ENGINES• CHEMICAL KINETICS• WASTE INCINERATION• DETONATION WAVES• LAMINAR PRE-MIXED FLAMES• DIFFUSION (NON PRE-MIXED) FLAMES• TURBULENT CHEMICALLY REACTING FLOW

MODELING• NUMERICAL SIMULATION OF COMBUSTION• ABLATION

HISTORICAL JOURNEY OF THERMO

Sir Willard Gibbs (circa 1900)

Prof. Laszlo Tisza (MIT)

Prof. Herb Callen (Univ. Penn.)

Prof. Mel Branch (CU Boulder, 1993)

Prof. K.R. Anderson, Ph.D.,P.E.4

Prof. Gary Koonce,Cal Poly Pomona, 1989

THERMODYNAMICS REVIEW

extensive,,

intensive,,

system of mass

;;

Properties Extensive vs.Intensive

Postulate)Nernst the(aka reached benot can K 0 T Zero,Absolute :Law 3

leirreversib , reversible;:Law 2

dependentpath ) (~

;~

:Law 1

re, temperatusame at the

are then theym,equilibriuin are and bodies if :Law 0

micsThermodyna of Laws

rd

nd

st

th

==

=

===

=

>=≥

=+=

==

SHU

shu

mm

Ss

m

Hh

m

Uu

T

dQdS

dWddQdU

TTT

BA

equilBA

5

IDEAL GAS RELATIONS

vp

vp

CCk

CCR

RTp

v

RTpv

mVv

mRTpV

/

/1

/

==

−===

==

=

γ

ρρ

RTRTupvuh

RTu

MWRRk

RC

Rk

kC

u

v

p

2

52

3

:Gases Ideal Monatomic

/1

1

=+=+=

=

=−

=

−=

6

SPECIFIC HEAT DEFINITIONS

• CONSTANT VOLUME CASE

• CONSTANT PRESSURE CASE

V

VV

CT

U

T

Q

UWQ

≡∂∆∂=

∂∂

∆=−0, NO MOVING BOUNDARIES

P

pp

CT

H

T

Q

HQ

pVUH

UUVpVpQ

UWQ

≡∂∆∂=

∂∂

∆=+=

−+−=∆+=

121122

7

IDEAL GAS ENTROPY CHANGE

v

Rdv

T

dTCds

T

dvp

T

duds

v

RTp

dTCdu

v

v

+=

∴

+=

=

=

Relation1st sGibb'

Law Gas Ideal

C of Definitionv

1

2

2

1

12

2

1

2

1

2

1

ln)(

)(

:affordsn Integratio

v

vR

T

dTTCsss

v

Rdv

T

dTTCds

v

v

+=−=∆

+=

∫

∫∫∫

8


∫ −=−=∆

−=

−=

=

=

2

1 1

2

12ln

)(

Relation 2nd sGibb'

/

Law Gas Ideal

of Definition

ely,Alternativ

p

pR

T

dTTCsss

p

dpR

T

dTCds

T

vdp

T

dhds

pdTv

dTCdh

C

p

p

p

p

9


• In order to proceed– Assume constant specific heats– Assume variable specific heats

• Use constant specific heats for monatomic (i.e. He) ideal gases where Cp ≠ Cp (T) or when Cp & Cv vary linearly over ∆T range of interest and use average values

• Use variable specific heats when ∆T during the process is large and thus, the Cp & Cv vary non-linearly within the ∆T range of interest

10


• Constant Specific Heats

• Average Specific Heats

• Variable Specific Heats

∫

∫

=−

=

+=−=+=∆

−=+=∆

2

1

12

0

21

1

2

1

2,

1

2

1

2,

1

2

1

2

1

2

1

2

)()()(

)()( Define

2;lnlnlnln

lnlnlnln

dTTCTsTs

dTTCTs

TTT

p

pR

T

TC

v

vR

T

TCs

p

pR

T

TC

v

vR

T

TCs

poo

T

p

avgavgpavgv

pv

o

11


• Table A-17 “Ideal Gas Properties of Air” tabulates T (K) vs. so(T) [kJ/kg-K]

• Thus, for variable specific heats

1

2

12ln

p

pRsss oo −−=∆

12

IDEAL GAS ISENTROPIC PROCESSES

avgvC

R

avgv

v

avgv

v

v

T

T

v

v

C

R

T

T

v

v

C

R

T

T

v

vR

T

TC

s

,

1

2

1

2

1

2

,1

2

1

2

1

2

1

2

1

2,

lnexplnexp

lnln

lnln0

const.) adiabatic & e(reversibl isentropic heats specific Const.

−

=

−=

−=

+=

=⇒

13


1

2

1

1

2

,

,

,

,

,

,,

lyconsequent

11

) k use books (some ratioheat specific

Recall,

−

=

−=−=

∴

==≡

−≡

k

avgv

avgp

avgv

avgv

avgp

avgvavgp

v

v

T

T

kC

C

C

R

C

Ck

CCR

γ

“1st Isentropic ideal gas relationship”

14


k

k

avgp

avgvavgpC

R

avgp

avgp

avgp

p

p

T

T

k

k

kC

CCR

p

p

T

T

p

p

C

R

T

T

p

p

C

R

T

T

p

pR

T

TC∆s

avgp

1

1

2

1

2

,

,,

1

2

1

2

1

2

,1

2

1

2

,1

2

1

2

1

2,

111Using,

lnexplnexp

lnln

lnln0

process isentropic andfor Next,

,

−

=

−=−=−

=

=

=

=

−==

“2nd Isentropic ideal gas relationship”15


k

k

kk

k

kk

v

v

p

p

v

v

p

p

p

p

v

v

T

T

=

=

=

=

−−

−−

2

1

1

2

11

2

1

1

2

1

1

2

1

2

1

1

2

Finally,

“3rd Isentropic ideal gas relationship”

16


• Summary

.

.

.1

1

constpv

constTp

constTv

k

k

k

k

==

=−

−

“Ideal Gas Isentropic Process Relationships”

k

vpvpW

v

v

p

p

T

T

v

v

p

pconstpv

kkk

k

kk

−−=

=

=

=

=

=

−−−

1

.;

1122

)1(

1

2

)1(

2

1

)1(

2

1

2

1

1

2

1

2

1

ρρ

17

VARIABLE SPECIFIC HEATS

• Start with an isentropic ideal gas process

:follows as volume"specific relative"

and pressure" relative" introduce wes,remedy thi To

ratio pressure oflieu in given

is ratio volumeerror when & trialrequires

ln

ln0

1

2

12

1

2

12

p

pRss

p

pRss

oo

oo

+=

−−=

18


2

1

1

2

2

1

1

2

211

1

2

.1

2

1

212

1

2

read4 Step

compute 3 Step

read 2 Step

at enter 1 Step

via findcan then we

process isentropican for

,given 17-A TablesAir

only offunction )/exp(

)/exp(

Pressure Relative

)/exp(

)/exp(exp

Ratio Pressure

T

pp

pp

p

T

pT

, & p, Tp

TRs

Rs

p

pp

Rs

Rs

R

ss

p

p

rr

r

r

o

o

consts

r

o

ooo

⇒

=⇒

⇒

⇒

==≡

=−=

=

cf. Cengel & Boles

19


11

22

1

2

.1

2

11

22

2

1

1

2

21

12

1

2

1212

/

/

Now,

only offunction

a is exp sinceonly offunction /

/

/ Ratio Volume

Thus,

analysis engine automotivein i.e.

of insteadknown is Sometimes

r

r

r

r

consts

o

rr

r

r

r

r

pT

pT

v

v

v

v

T

/R)(sT/T/pTpT

pT

pT

p

p

T

T

pT

pT

v

v

/pp/vv

==

==

====

=

20


air tables ofcolumn in

given is / :NOTE

read 4 Step

compute 3 Step

read 2 Step

with 17-A Tableenter 1 Step

find process, isentropican and ,Given

2

1

1

2

2

1

1

2211

r

rr

rr

r

v

pTv

T

vv

vv

v

T

T, v,Tv

=

⇒

=⇒

⇒

⇒

cf. Cengel & Boles

21

3 T’s of Combustion• Combustion efficiency can be further explained in terms of the three

T’s;– Time– Temperature– Turbulence

• Time (aka Residence Time, or time the gases reside in the combustion chamber) For example, in oil heat systems, the amount of time oil vapor has to combust (or reside in the flame front or burning zone) has been improved dramatically with the advent of flame retention burners. The primary difference between this type burner and the older conventional style burner is that the flame retention burners violently spin the air/fuel mixture resulting in better mixing

• This reduces the amount of excess combustion air necessary to insure each fuel droplet is completely surrounded by oxygen and burns completely

• As the amount of combustion air is reduced, efficiency increases

3 T’s of Combustion

• O2 readings within the manufacturer’s specifications document that the burner’s flame retention properties are operating as designed and engineered

• With gas fired systems, air and fuel mixing occurs inside the burner. Again, increasing the amount of time the fuel and air have the opportunity to thoroughly mix will help insure complete combustion

• Adjustable air intake shutters on atmospheric burners primarily control the velocity at which the fuel/air mixture moves down the burner throat or ‘mixing area’


• Increasing the amount of time flue gases are in the heat exchanger also increases the amount of time for heat transfer to occur

• O2 and stack temperature readings within the manufacturer’s specifications document that the flue gases are moving through the heat exchanger at a velocity which allows time for maximum heat transfer while still permitting the introduction of additional combustion air

• As the temperature difference (∆T) between the source of heat and the material being heated increases, so does the rate of heat transfer. This heat transfer rate is measurable in forced air systems and boilers. By reducing the amount of combustion air introduced into the combustion process to the absolute minimum necessary, we increase the DT between the flame/flue gases and the distribution air or boiler water


• Radiant heat energy is much more effective in transferring heat than convective/conductive heat transfer. It is primarily put off by the flame itself and is directly related to flame temperature. As excess air is reduced, the higher flame temperature generates more radiant heat energy

• The burn rate in combustion process is very sensitive to temperature. If flame temperature is increased by 10%, the rate of combustion more than doubles. Unfortunately, the same increase in flame temperature also increases production of NOx gases by more than 10 times when sufficient O2 is available


• To estimate the actual flame temperature, continue the horizontal line from the top of fuel to the point where it intersects the line corresponding to the percent of excess air. From the point of intersection, draw a vertical line up to the point where it intersects the combustion air temperature line (80 DEG F); Then draw a horizontal line to the vertical axis – temperature rise


• For example:– A natural gas burner operating with 25% excess air (4.5%

O2) has an estimated flame temperature of 2930 °°°°F plus 80 °°°°F (combustion air temperature) equals 3010 °°°°F

• Note: Deduct approximately 11% for gas and 4% to 6% for oil due to water vapor in the flue gases

• Note that in this example burning natural gas, a 25% excess air translates to a 4.5% O2 reading and a 20% excess air translates to a 4.0% O2 reading on the analyzer

• O2 readings within the manufacturer’s specifications document that the flame temperature has been maximized for the most efficient radiant heat production and transfer

Efficiency Range

Effect of off-on operations on firetube efficiency

3 T’s of Combustion• ‘Turbulation’ of the fuel, air and heat

source provides for more complete combustion by keeping these components in contact with each other for a longer period of time

• Agitation of flue gases in a heat exchanger serves to provide a continual circulation of hotter flue gasses in contact with the heat exchanger surfaces. Typically, heat exchanger surfaces have a wide variety of irregular surfaces incorporating bumps, ridges etc. to provide this effect. Boilers and domestic hot water heaters commonly have turbulators that provide for this mixing process. These surfaces also produce eddy currents that recirculate flue gases and increase


• These ‘turbulators’ and irregular heat exchanger surfaces are designed to be most effective at the appliance’s full firing rate. Under firing a burner results in the smaller volume of flue gases taking the path of least resistance through the boiler thus reducing the scrubbing and mixing of the flue gases against the fire side of the heat exchanger

• Again, O2 readings within the manufacturer’s specifications verify that the burner is firing at full capacity, maximizing efficient heat transfer

32

GAS POWER CYCLES

ME 576

33

THE CARNOT CYCLE

• The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature TH and a sink at temperature TL

• We can never get better than Carnot’s efficiency

H

L

Carnotth T

T−=1,

η

cf. Cengel & Boles

34

Example

Consider a Carnot cycle executed in a closed system with 0.003 kg of air. The temperature limits of the cycle are 300 K and 900 K, while the minimum and maximum pressures are 20 kPa and 2000 kPa, respectively. Determine the net work per cycle.

p

v

2

3

41

2000 kPa

20 kPa

35kJ 393.01960.003(2/3)

kJ/kg 1960

kJ/kg 196935

2000ln)900(287.0

kPa 935

900300

20kPa, 2000

lnlnln

3

2

3

2

900

30011

3232

32

4.0

4.1

1

3

4

4

32

3

2

2

3

2

3

2232

==∴==⇒=−

==

=

=

==

===

==

=−=−=

−−

−

−

−

W

wqwq

w

T

T

ppp

p

pRT

v

vRT

v

vvpw

qqw

T

T

HH

k

k

HH

HHCarnotnet

H

L

Carnot

η

η

36

AIR-STANDARD CYCLE ASSUMPTIONS

• The working fluid is air• Processes are reversible• Combustion process is replaced by a heat

addition process• Exhaust process is replaced by a heat

rejection process

GEOMETRIC PROPERTIES OF RECIPROCATING ENGINES

37speed crankshaft speed;piston mean 2

ratio radiuscrank length / rod/

ratio ebore/strok/

ration compressioolumecylinder v min.

olumecylinder v max.

Center Dead Bottom

Center Dead Top

===

====

==+

=

====

NLNS

alR

LBR

V

VVr

BDCBC

TDCTC

p

bs

c

cdc

44444 344444 21end lowat diesels marine end,high at Automobile

m/s) 15~8( fpm 300 ~1500

engines C.I. speed low large 9~5

engines, medium small, 4~3

engines C.I. speed low 0.5

engines, medium small, 1.2~ 0.8

Engines C.I. 24~12

Engines S.I. 12~8

Typ.

2

=

=

===

=

p

c

c

S

R

B/L

r

r

aL

p

p

S

S

4-STROKE ENGINE OPERATING CYCLES

38

1) INTAKE STROKE – STARTS WITH PISTON AT TDC AND ENDS WITH THE PISTON AT BDC DRAWING MIXTURE INTO THE CYLINDER, TO INDUCE MASS THE INLET VALVE OPENS SHORTLY BEFORE THE STROKE BEGINS, AND CLOSES AFTER IT ENDS

2) COMPRESSION STROKE – WHEN BOTH VALVES ARE CLOSED AND THE MIXTURE INSIDE THE CYLINDER IS COMPRESSED, AT THE END OF THE COMPRESSION STROKE COMBUSTION IS INITIATED AND THE CYLINDER PRESSURE RISES RAPIDLY

3) POWER STROKE (AKA EXPANSION STROKE) – STARTS WITH PISTON AT TDC AND ENDS AT BDC AS THE HIGH TEMP. , HIGH PRESS. GASES PUSH THE PISTON DOWN AND FORCE THE CRANK TO ROTATE, APPROX. 5 TIMES AS MUCH WORK IS DONE ON THE PISTON DURING THE POWER STROKE AS WHAT HAD TO BE DONE DURING COMPRESSION, AS THE PISTON NEARS BDC THE EXHAUST VALVE OPENS TO BEGIN THE EXHAUST PROCESS AND DROP THE CYLINDER PRESSURE CLOSE TO THE EXHAUST PRESSURE

4) EXAUST STROKE – THE REMAINING UNBURNED GASES EXIT THE CYLINDER, JUST AFTER TD THE EXHAUST VALVE CLOSES AND THE CYCLE REPEATS ITSELF

CF. HEYWOOD

2-STROKE ENGINE OPERATING CYCLES

39

1) COMPRESSION STROKE – BOTH INLET AND EXHAUST PORTS ARE CLOSED, THEN THE CYLINDER CONTENTS ARE COMPRESSED AND FRESH CHARGE IS DRAWN INTO THE CRANKCASE, AS THE PISTON NEARS TDC COMBUSTION IS INITIATED

2) SCAVENGING (AKA POWER AKA EXPANSION) STROKE – MOST OF THE BURNT GASES EXIT THE CYLINDER IN AN EXHAUST BLOWDOWN PROCESS, WHEN THE INLET PORTS ARE UNCOVERED THE FRESH CHARGE WHICH HAS BEEN COMPRESSED IN THE CRANKCASE FLOWS INTO THE CYLINDER

*EACH ENGINE CYCLE WITH ONE POWER STROKE IS COMPLETED IN ONE CRANKSHAFT REVOLUTION, BUT IT IS DIFFICULT TO COMPLETELY FILL THE DISPLACE VOLUME WITH FRESH CHARGE, AND SOME OF THE FRESH MIXTURE FLOWS DIRECTLY OUT OF THE CYLINDER DURING THE SCAVENGING PROCESS (THE FIGURE ABOVE SHOWS A CROSS-SCAVENGED DESIGN)

Reed springInlet valve

CF. HEYWOOD

40

COMPRESSION RATIO

FLpALSTROKEAREAMEPW

VV

WMEP

V

V

V

Vr

net

net

TDC

BDC

===−

=

==

**

Pressure EffectiveMean

Ration Compressio

minmax

min

max

cf. Cengel & Boles

cf. Cengel & Boles

MEP

• MEP = PnR/VdN– nR = number of crank revs for each power

stroke (nR = 2 for 4 stroke, 1 for 2 stroke engines)

– Vd=displaced volume– N = speed– P = cylinder pressure

41

MEP

• Formulas for MEP

42

][in

ft]-[lbf 4.75[psi]

][dm

m]-[N 28.6[kPa]

torque,of In terms

]rpm[ ][in

396,000*[hp] [psi]

]rev/s[ ][dm

10*[kW] [kPa]

3

3

3

3

3

d

R

d

R

d

R

d

R

V

TnMEP

V

TnMEP

NV

nPMEP

NV

nPMEP

=

=

=

=

MEP• Typ. Values for MEP

– Naturally aspirated spark-ignition engines 125 ~ 150 psi (850 ~1050 kPa) @ 3000 rpm where engine max. torque is reached

• At max. rated power, MEP values are 10~15% lower than above

– Turbocharged spark-ignition engines180 ~ 250 psi (1250 ~ 1700 kPa)

• At max. rated power 130 ~ 200 psi (900 ~ 1400kPa)– Naturally aspirated 4-stroke diesel engines 100 ~ 130 psi (700 ~

900 kPa)• At max. rated power ~100 psi (~ 700kPa)

– Turbocharged 4-stroke diesel engines 145 ~ 175 psi (1000 ~ 1200 kPa), for aftercooled MEP ~200 psi (~1400 kPa)

• At max. rated power 125~140 psi (850~ 950 kPa)

– Two-stroke diesel engines 225 psi (~1600 kPa)43

Example• A four cylinder automotive spark-ignition engine provides a max.

brake torque of 150 N-m (110 lbf-ft) in the mid-speed range of about 3000 rpm. Estimate the required engine displacement, bore, stroke and the max. brake power the engine delivers

44

hp 94kW 701000*2

87*2*800

1000*

**

kPa 800 power max. @ MEP Assume

rpm 5200rev/sec 87m/s 15 assume 2

mm 86,2 assume

4*4

engine,cylinder -four aFor

dm 2925

150*2*28.628.6

kPa 925 MEP Assume

maxmax

maxmax

3

2

3max

====

=

==⇒==

==⇒=

=

===

=

r

pp

d

r

n

NVMEPP

NSLNS

BBLB

LBV

MEP

TnV

π

π

45

PLAN FORMULA

• Four-stroke cycle– Intake– Combustion– Power– Exhaust

• A four-stroke cycle produces one power stroke for each piston for 2 revolutions of the main shaft

• i.e. a 4000 rpm 4-stroke engine gives N=4000/2 = 2000 power strokes/min

• Four stroke cycle produces one power stroke for each piston for two revolutions of the crankshaft

46

PLAN FORMULA

• Two-stroke cycle• In one revolution of the crankshaft

– Air intake– Compression– Expansion– Exhaust

• Every outward stroke of the piston is a power producing stroke

47

PLAN FORMULA

• PLAN Formula (ENG)

[ ] [ ] [ ] [ ]

[ ][ ][ ][ ] /minstrokespower ofnumber nstrokes/mipower

area sectoinal-crosspiston in

strokeft

psi

cylinder a ofoutput horsepower

minhpblfft

000,33

nstrokes/mipower inftpsi

2

2

==

==

=

−−

=

N

A

L

MEPP

HP

NALPHP

48

PLAN FORMULA

• PLAN Formula (SI)

• Example: compute the HP of a 2.0 L engine that has a MEP = 1 MPa operating @ 4000 rpm. The engine is a four-stroke engine

Solution: Four-stroke →N=4000/2=2000 power strokes/min

HP=1000 kPa(2 L) (2000 power strokes/min)/ (44760 N-m/min-hp)

HP =89.4 hp

[ ] [ ] [ ]

−−

=

hpminmN

760,44

nstrokes/mipower LiterskPa NLAPHP

49

PLAN FORMULA

( )

psia 117

minstrokespower

2000)in ft(42.412 3/1

psiahpin

minstrokespower

hp) 100(000,33

2)by divide engine, stroke-four a isit (becausemin

strokespower 2/4000

in 412.42cyl) 6(cyl 1

in 34ft, 3/1

in/ft 12

inches 4

000,33

:Solution

inches. 4 is stroke theand inches 3 is borecylinder

The rpm. 4000 @output hp 100 a has that engine

cylindersix stroke-four a of MEP theDetermine

Example

2

3

2

2

=

==

=

====

=

MEPP

N

AL

PLANHP

π

50

OTTO CYCLE• The Ideal cycle for spark-ignition engines

cf. Cengel & Boles

51

OTTO CYCLE

1-2: isentropic compression

2-3: isochoric heat addition

3-4: isentropic expansion

4-1: isochoric heat rejection

cf. Cengel & Boles

52

OTTO CYCLE

( )( )1/

1/1

11

)(

)(

0) PEKE system, (closed Energy ofon Conservati

232

141

,

23

14

,

1414

2323

−−−=

−−−=−==

−=−=−=−=

∆=−+−

=∆=∆

TTT

TTT

TT

TT

q

q

q

w

TTCuuq

TTCuuq

uwwqq

Ottoth

in

out

in

net

Ottoth

vout

vin

outinoutin

η

η

53

OTTO CYCLE

1,

3

4

1

4

3

1

1

2

2

1

1423

11

processes isochoric

are 1-4 and 3-2 since & also

process, isentropic are 4-3 & 2-1

−

−−

−=

=

=

=

∴

==

kOttoth

kk

r

T

T

v

v

v

v

T

T

v v vv

η

cf. Cengel & Boles

54

DIESEL CYCLE

• Ideal cycle for compression ignition engines

1-2: isentropic compression

2-3: isobaric heat addition

3-4: isentropic expansion

4-1: isochoric heat rejection

cf. Cengel & Boles

cf. Cengel & Boles

55

DIESEL CYCLE

)1/(

)1/(1

)(

)(1

)(

)(11

)(

)(

)()(

232

141

23

14

,

23

14

,

1414

41

2323

23232

23

−−−=

−−−=

−−−=−==

−=−=−=−

−=−=−+−=

−=−

TTkT

TTT

TTk

TT

TTC

TTC

q

q

q

w

TTCuuq

uuq

TTChhq

uuvvpq

uuwq

Dieselth

p

v

in

out

in

net

Dieselth

vout

out

pin

in

outin

η

η

56

DIESEL CYCLE

−−−=

−−

==≡

− )1(

111

thusprocesses, gas ideal

isentropic are 43&21

combustion before volume

combustionafter volumeratio off-cut

Define

1,

2

3

c

kc

kDieselth

c

rk

r

r

v

vr

η

cf. Cengel & Boles

57

DUAL CYCLE

• Models the combustion process as a combination of a constant volume and a constant pressure process

cf. Cengel & Boles

58

Example

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17 C, and 800 kJ/kg of heat is transferred to the air during the constant-volume heat addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum T and p that occur during the cycle, (b) the net work output, (c) the thermal efficiency, (d) MEP

cf. Cengel & Boles

59

kPa7.17998290

4.652100

kJ/kg 475.12C, 4.652 :ionInterpolat

481.01 660 1.598

51.84

473.25 650 34.85

:17-A Table

51.848

1.6761

gas idealan ofn compressio isentropic :2-1

676.1kJ/kg, 206.9117-A TableK 290 (a)

21

1212

1

11

2

22

22

22

12

1

2

1

2

111

===⇒=

==

===⇒==

==⇒=

vT

vTpp

T

vp

T

vp

u T

uT

u T v

r

vv

rv

v

v

v

vuT

r

rr

r

r

r

o

60

MPa 345.414.652

14.15757.1799

6.103K, 1575.4 :affordsion Interpolat

6.046 1580 65.1279

12.1275

6.301 1560 99.1260

17-A Table

kJ/kg 12.1275

12.475800

additionheat olumeconstant v :3-2

32

23

23

2

22

3

33

33

33

3

3

23

===⇒=

==

=−=

−=

vT

vTpp

T

vp

T

vp

v T

v T

v T u

u

u

uuq

r

r

r

in

61kJ/kg 38291.20692.588

rejectionheat olumeconstant v :1-4

kJ/kg 588.92 K, 795.16:yieldsion Interpolat

592.30 800 48.08

8.8244

576.12 780 64.51

17-A Table

824.48)103.6(8

gas ideal of

expansion isentropic 4-3 cycle, during done Work (b)

14

44

44

343

4

3

4

=−=−=

==

===⇒==

uuq

uT

u T

u T v

rvvrv

v

v

v

out

r

rrr

r

62

( )

kPa 2.574kJ 1

m-kPa 1)8/11(/kgm 832./kJ/kg 418

/kgm 832.100/290*287./

11

(d)

cycle Otto idealan smaller th is efficiency

i.e. realistic, more is assumptionheat specific variable

%52%5.56811

1

sassumptionheat specificconstant under versus

%52800

418

kJ/kg 418382800

3

3

3

111

1

1

121

4.11

1,

=−=

===

−=

−=

−=

>=−=−=

===

=−=−==

−

−

MEP

pRTv

rv

w

r

vv

w

vv

wMEP

r

q

w

qqqw

netnetnet

kOttoth

in

net

th

outinnetnet

η

η

63

Example

An air standard Diesel cycle has a compression ratio of

18.2. The air is at 80 °F and 14.7 psia at the beginning of

the compression process and at 3000 R at the end of the

heat addition process. Accounting for variable specific

heats with temperature, determine:

(a) cut-off ratio

(b) Heat rejection/unit mass

(c) Thermal efficiency of the cycle

64BTU/lbm 05.402R, 1623.5

395.74 1600 8.263

7.93

409.13 1650 7.556

7.932144.32/18.

144.32 BTU/lbm, 92.04 R, 54017E,-A Table

ncompressio isentropic:21

22

22

11122

111

==

=======

−

h T

h T

h T v

/rv/vvvv

vuT

r

rrr

r

cf. Cengel & Boles

65%1.59

63.388

87/15811

BTU/lbm 87.15804.9291.250

processrejection heat olumeconstant v :4-1

BTU/lbm 19.2506212.11 17E-A Table

6212.11848.1/)18.1(2.18848.1/848.1//

expansion isentropic:43

BTU/lbm 53.38805.40258.790

18.1 BTU/lbm, 58.907 R, 000317E,-A Table

1848.05.1623

3000

additionheat isobaric:32

th

14

44

32343344

23

333

2

3

2

3

2

22

3

33

=−=−=

=−=−=

=⇒======

−

=−=−====

===⇒=

−

in

out

out

r

rrrr

in

r

q

q

uuq

uv

rvvvvvvvv

hhq

vhT

T

T

v

v

T

vp

T

vp

η

66

% 59.1 efficiency thermal(c)

0.4089 ratiorejection heat (b)

0.1848 ratio off-cut (a)

:Summary

21

===

===

inout

c

/qq

r/vv

TURBOCHARGING

ME 576

67

TURBOCHARGING / SUPERCHARGING

68CF. HEYWOOD

COMBUSTION IN COMPRESSION-IGNITION (C.I.) ENGINES

CF. HEYWOOD

CF. HEYWOOD

COMBUSTION IN COMPRESSION-IGNITION (C.I.) ENGINES

CF. HEYWOOD

CF. HEYWOOD

CF. HEYWOOD

TURBOCHARGING

• BASIC CONCEPT OF TURBOCHARGING

72CF. Stone

TURBOCHARGING

73

COMPRESSOR

INTERCOOLERCARBURETOR

INTAKE MANIFOLD

6 =INLET VALVE7 = EXHAUST VALVE8 = EXHAUST MANIFOLD

TURBINE

AIR FLOW PASSES THROUGH THE COMPRESSOR, INTERCOOLER, CARBURETOR, INTAKE MANIFOLD, AND INLET VALVEEXHAUST FLOWS THROUGH EXHAUST VALVE EXHAUST MANIFOLD, WHICH DRIVES THE TURBINE WHICH POWERS THE COMPRESSOR

11 = WASTEGATE LINKAGE CONTROLLED BY A BOOST PRESSURE REGULATOR

ENGINE INLET PRESSURES (AKA BOOST) OF ~ 100 kPa ABOVE ATM. PRESSURES ARE TYPICAL

CF. HEYWOOD

TURBOCHARGING

• TURBOCHARGING IS A FORM OF SUPERCHARGING IN WHICH A COMPRESSOR IS DRIVEN BY AN EXHAUST GAS TURBINE, THUS SUPPLYING PRESSURIZED AIR TO THE ENGINE

• BY PRESSURIZING THE AIR AT INLET TO THE ENGINE, THE MASS FLOW RATE OF AIR INCREASES AND THUS AN INCREASE IN THE FUEL FLOW RATE OCCURS, LEADING TO AN INCREASE IN THE POWER OUTPUT, AND AN IMPROVEMENT IN THE EFFICIENCY

74

TURBOCHARGING

• AXIAL VS. RADIAL FLOW COMPRESSORS

75

CF. Stone

TURBOCHARGING

• THE OVERALL IMPROVEMENT IN EFFICIENCY ULTIMATELY DEPENDS ON THE MATCHING OF THE SUPERCHARGER TO THE ENGINE

• SUPERCHARGING DOES NOT HAVE A SIGNIFICANT EFFECT ON EXHAUST EMISSIONS

• AXIAL AND RADIAL COMPRESSORS CAN BE DRIVEN BY A TURBINE, THUS FORMING A TURBOCHARGER

76

TURBOCHARGING• THE THERMODYNAMIC ADVANTAGE OF

TURBOCHARGERS OVER SUPERCHARGERS STEMS FROM THEIR USE OF THE EXHAUST GAS ENERGY DURING BLOW-DOWN AS SHOWN BELOW

77

CF. Stone

TURBOCHARGING

• THE CHARACTERISTICS OF TURBOCHARGERS ARE FUNDAMENTALLY DIFFERENT FROM THOSE OF RECIPROCATING I.C.E.’S, THUS LEADING TO MATCHING PROBLEMS WHEN THE TWO ARE COMBINED

• SUPERCHARGERS HAVE A MECHANICAL DRIVE, AND THE COMPRESSOR EFFICIENCIES CAUSE THE OVERALL ECONOMY TO DECREASE, BUT THE FLOW CHARACTERISTICS ARE BETTER MATCHED

78

TURBOCHARGING

• TYPICAL AUTOMOTIVE TURBOCHARGER IS SHOWN BELOW

79CF. Stone

TURBOCHARGING

• CHARACTERISTICS– RADIAL FLOW COMPRESSOR AND TURBINE– NO STATOR BLADES FOR SMALL

TURBOCHARGERS• OPERATION SATISFACTORY OVER A WIDER FLOW

RANGE, BUT WITH A LOWER PEAK EFFICIENCY

• WITH STATOR BLADES PRESENT, THE FLOW ANGLE FROM THE ROTOR TO THE STATOR HAS TO MATCH THE BLADE ANGLE, WHICH MEANS FOR EVERY VOLUMETRIC FLOW RATE, THERE IS ONLY ONE ROTOR SPEED AT WHICH THERE IS PROPER MATCHING (SEE NEXT CHART)

80

TURBOCHARGING

• MATCHING:

81CF. Stone

TURBOCHARGING

• RADIAL FLOW COMPRESSOR WORK WELL WITH OR WITHOUT STATOR BLADES, DUE TO THECONSERVATION OF THE MOMENT OF ANGULAR MOMENTUM IN THE DIFFUSER AND THE INCREASE OF THE FLOW AREA AS THE RADIUS INCREASES

82

TURBOCHARGING

• VELOCITY TRIANGLES FOR RADIAL FLOW COMPRESSOR

83CF. Stone

THERMODYNAMICS OF TURBOCHARGING

• T-S DIAGRAM OF A TURBOCHARGER:

84CF. Stone


• GOVERNING THERMODYNAMIC RELATIONSHIPS

( )( )

12

12

42

12

1212

4343

:ies EfficiencIsentropicnery Turbomachi

heat specific mean

:gasesperfect -semiFor

TT

TT

hh

hh

c

TTchhw

TTchhw

ssc

p

pc

pt

−−=

−−=η

=

−=−=

−=−=

85

%9080

rbinescharger tuflow turbo axialfor

&

%8070

eger turbin turbocharradiala for Typically,

%7665

compressorger turbocharradiala for Typically,

43

43

43

43

<η<

<η<

−−=

−−=η

<η<

t

t

sst

c

TT

TT

hh

hh


• IN A TURBOCHARGER, THE COMPRESSOR IS DRIVEN SOLELY BY THE TURBINE, CONSEQUENTLY, A MECHANICAL EFFICIENCY CAN BE DEFINED AS FOLLOWS

( )( )433434

121212

TTcm

TTcm

w

w

,p

,p

t

cmech −

−==η

&

&

86


• AS IN GAS TURBINES, THE PRESSURE RATIOS ACROSS THE COMPRESSOR AND TURBINE ARE PROFOUND AND CAN BE MODELED VIA THE ISENTROPIC RELATIONSHIPS

87

vp

s

s

c/c

p

p

T

T

p

p

T

T

=γ

=

=

γ−γ

γ−γ

1

4

3

4

3

1

1

2

1

2


• IN CONSTANT PRESSURE TURBOCHARGING, IT IS BEST TO HAVE THE INLET PRESSURE > THE EXHAUST PRESSURE (P2/P3 > 1) SO THAT GOOD SCAVENGING WILL OCCUR– SCAVENING = THE OPERATION OF CLEARING

THE CYLINDER OR BURNED GASES AND FILLING IT WITH FRESH MIXTURE (OR AIR) THIS COMIBINED INTAKE AND EXHAUST PROCESS IS CALLED SCAVENGING

• THIS PLACES A LIMIT ON THE OVERALL TURBOCHARGER EFFICIENCY η= η mechηt ηcFOR DIFFERENT ENGINE EXHAUST TEMPERATURES, T3 AS SHOWN ON THE NEXT CHART

88

TURBOCHARGING

89

CF. Stone

TURBOCHARGING

• THE FLOW CHARACTERISTICS OF AN AXIAL VS. RADIAL FLOW COMPRESSOR ARE GIVEN ON THE NEXT CHART– IT CAN BE SEEN THAT THE RADIAL FLOW

COMPRESSOR HAS A WIDER RANGE OF OPERATION DUE AS STATED BEFORE WHEN CONSIDERING MATCHING ISSUES

• THE SURGE LINE MARKS THE REGION OF UNSTABLE OPERATION, WHERE FLOW REVERSAL, ETC. CAN OCCUR

• THE POSITION OF THE SURGE LINE IS INFLUENCED BY THE INSTALLATION OF THE ENGINE

90

TURBOCHARGING

91CF. Stone

TURBOCHARGING THE COMPRESSION IGNITION ENGINE

• COMPRESSION IGNITION ENGINE –NEAR THE END OF THE COMPRESSION STROKE, LIQUID FUEL IS INJECTED AS ONE OR MORE JETS– THE INJECTOR RECEIVES FUEL AT VERY HIGH

PRESSURES IN ORDER TO PRODUCE RAPID INJECTION WITH HIGH VELOCITY JETS OR SMALL CROSS-SECTIONAL AREA

– THE FUEL JETS ENTRAIN AND BREAK UP INTO DROPLETS, LEADING TO RAPID MIXING

92


– RAPID MIXING IS ESSENTIAL FOR THE COMBUSTION TO OCCUR SUFFICIENTLY FAST

– SOMETIMES THE FUEL JET IMPINGES ON THE WALL TO VAPORIZE AND BREAK UP THE JET

– THERE IS A LARGE VARIATION IN FUEL/AIR MIXTURES ON BOTH THE LARGE AND SMALL SCALES WITHIN THE COMBUSTION CHAMBER

93


• PLATE 2 OF STONE PP. 78-79 SHOWS COMBUSTION IN A HIGH-SPEED DIRECT INJECTION DIESEL ENGINE

94

95CF. Stone


• PRESSURE DIAGRAM FO COMPRESSION IGNITION ENGINE

96CF. Stone


• RECALL, TURBOCHARGING IS USED TO INCREASE THE ENGINE OUTPUT BY INCREASING THE DENSITY OF THE AIR DRAWN INTO THE ENGINE

• THE PRESSURE RISE ACROSS THE COMPRESSOR INCREASES THE DENSITY, BUT THE TEMPERATURE RISE REDUCES THE DENSITY

97


• THE LOWER THE ISENTROPIC EFFICIENCY OF THE COMPRESSOR, THE GREATER THE TEMPERATURE RISE FOR A GIVEN PRESSURE RATIO

• NOW, WE DEVELOP A THERMODYNAMIC MODEL…

98


99

η

−

+=

η=−−

=

γ−γ

γ−γ

c

cs

s

p

p

TT

TT

TT

p

pTT

1

1

Tfor solve

into

Plug

1

1

2

12

2

12

12

1

1

212

11

1

2

1

2

1

2

1

1

gas ideal anfor

−

γ−γ

η

−

+=ρρ

ρ=

c

p

p

p

p

RTp


• THE EFFECT OF COMPRESSOR EFFICIENCY ON CHARGE DENSITY IF SHOWN BELOW

100

CF. Stone


• THE EFFECT OF FULL COOLING (ISOTHERMAL COMPRESSION) IS SHOWN ON THE PREVIOUS CHART

• IT IS SEEN THAT THE TEMPERATURE RISE IN FTHE COMPRESSOR SUBSTANTIALLY DECREASES THE DENSITY RATIO, PARTICULARLY AT HIGH PRESSURE RATIOS

101


• ENSURING THAT THE COMPRESSOR OPERATES IN THE EFFICIENT PART OF THE REGIME, MINIMIZES THE WORK INPUT AND THE TEMPERATURE RISE

• HIGHER ENGINE INLET TEMPERATURES RAISE THE TEMPERATURE THROUGHOUT THE CYCLE, AND WHILE THIS REDUCES IGNITION DELAY, IT INCREASES THE THERMAL LOADING ON THE ENGINE

102


• THE ADVANTAGES OF CHARGE COOLING LEAD TO THE USE OF INTERCOOLERS, WHICH HAVE THE EFFECTIVENESS METRIC

103

( )( ) ( )

( ) ( )

η−ε−+=

ε+ε−

η+=

ε+ε−=

=−−==ε

γ−γ

γ−γ

c

/

c

/

p/pTT

p/pTT

TTT

T

TT

TT

111

11

1

gasesperfect for cooler,-inter thefromExit

mperatureambient teat or water)(air medium cooling

ferheat trans possible max.

ferheat trans actual

112

13

112

13

123

1

12

32


• NEGLECTING THE PRESSURE DROP IN THE INTER-COOLER

• EFFECT OF CHARGE COOLING ON THE DENSITY RATIO IS SHOWN ON THE NEXT CHART FOR EFF. 70%, AND AMB. TEMP. = 20 C…

104

( ) ( ) 1112

1

3

1

3 111

−γ−γ

η−ε−+=

ρρ

c

/p/p

p

p


105

cf. Stone


• INTERCOOLING INCREASES THE AIR FLOW RATE AND WEAKENS THE A/F RATIO FOR A FIXED FUELING RATE– THE TEMPERATURES WILL BE REDUCED

THROUGHOUT THE CYCLE, INCLUDING THE EXHAUST STAGE

– THE TURBINE OUTPUT WILL THEN BE REDUCED, UNLESS IT IS REMATCHED

– THE REDUCED HEAT TRANSFER AND CHANGES IN COMBUSTION LEAD TO AN INCREASE IN BMEP AND A REDUCTION IN SPECIFIC FUEL CONSUMPTION ON THE ORDER OF 6% FOR P2/P1= 2.25 & ε =0.7

106


• LOW HEAT LOSS ENGINES IMPROVE THE PERFORMANCE OF DIESEL ENGINES– FIRST, THERE IS IMPROVEMENT IN EXPANSION WORK AND

HIGHER EXHAUST TEMPERATURE, WHICH LEADS TO GAINS WHEN THE ENGINE IS TURBOCHARGED

– THE REDUCED COOLING REQUIREMENTS ALLOW FOR A SMALLER CAPACITY COOLING SYSTEM

– THE ASSOCIATED REDUCTION IN POWER CONSUMED BY THE COOLING SYSTEM IS A SIGNIFICANT PART OF THE LOAD

– REDUCING THE HEAT TRANSFER FROM THE COMBUSTION CHAMBER LEADS TO REDUCED IGNITION DELAY, AND HENCE REDUCED DIESEL KNOCK (SPONTANEOUS IGNITION OF UNBURNT GASES LEADS TO RAPID PRESSURE RISE WHICH IS AUDIBLE, CAUSED BY RESONANCES OF THE COMBUSTION CHAMBER WALLS), BUT THE HIGHER COMBUSTION TEMPEATURES WILL LEAD TO AN INCREASE IN NOX EMISSIONS

107


• STUDY OF AUGMENTED HEAT TRANSFER ON DIESEL ENGINE

108

cf. Stone


• ENGINE PERFORMANCE CHARACTERISTICS

109

cf. Stone


110naturally aspi-

cf. Heywood


111cf. Heywood


112cf. Heywood


113cf. Heywood

114

JET PROPULSION CYCLES

ME 576

115115

SPECIFIC IMPULSE

• A metric used to compare propulsion systems is the specific impulse, defined as

• Example, say a Nitrogen rocket nozzle has an exit velocity of 550 m/s (Ma = 550/340 = 1.62), then:

slbm

lbf

skg

N

g

V

m

FI

c

exitsp /

~/

~rate flow mass

thrust ===&

(kg/s)

N550

s-Nm-kg

1

m/s 550

2

==spI

116116

SPECIFIC IMPULSE

second ingper thrust N of kg 0.18 consume We

kg/s 18.0

(kg/s)N

550

N 100

:ust) their throf

in terms rated are (nozzles nozzle N 100 afor So

(kg/s)

N550

2

=

=spI

117117

SPECIFIC IMPULSE

consumed) is fuel of (lots lbm/s 161

(lbm/s)lbf

124.2

lbf 20,000

thenused, is thruster lbf 20,000 aSay

(lbm/sec)

lbf2.124

s-lbfft-lbm

32.2

sft

4000

)7.3~1090/4000~(Ma ft/s 4000 of

ityexit velocan h rocket witHydrogen aConsider

:Example

2

=

===c

exitSP g

VI

118118

THE GAS TURBINE FOR JET PROPULSION

• The turbojet engine used extensively for aircraft propulsion is a simple modification of the open gas-turbine cycle:

cf. Reynolds & Perkins

119119

THE GAS TURBINE FOR JET PROPULSION

• Hot combustion gases are expanded in the turbine only far enough to generate work to drive the compressor• The remaining energy is then converted into high velocity K.E. by expanding it in a nozzle downstream of the turbine• The jet thrust results from the difference in momentum of the entering air flow to the compressor and the high-velocity exhaust gases

)( 05 VVg

mF

c

−=&

120120

TURBOJET WITH AFTERBURNER

• Afterburners are used to provide greater nozzle exhaust velocities and substantially increase the engine thrust• The afterburner is essentially a reheat device:


121121

TURBOJET WITH AFTERBURNER

• Typically, specific impulses ~ 600 N/(kg/s) hold for modern jet engines• Afterburners increase this number somewhat, however, fuel consumption is greatly increased• Thus, afterburners are “cut in” only for short periods of time for high thrust

122122

THE RAMJET

• The ramjet is a device employed for aircraft propulsion at very high speeds• Uses the momentum of the on-rushing air for compression, thus eliminating any moving parts• Operates on the same principle as the turbojet, except for the use of a flow passage which converts K.E. of incoming flow into pressure


123123

THE RAMJET

• The incoming flow velocity is reduced in the inlet nozzle in process 0-1• Fuel is burned at constant pressure in process 1-2, thus increasing the temperature•Finally, the flow is expanded to a high velocity, V3 in the outlet nozzle•Due to the energy which has been added to the flow, the velocity at exit is greater than that at inlet

124124

THE RAMJET

• The net thrust is given as the change in momentum flux

•The ramjet is only feasible for high inlet-flow velocities, otherwise it would not be possible to achieve a high enough compression ratio in the inlet to counteract the losses incurred due to friction and engine irreversibilities

)( 03 VVg

mF

c

−=&

125125

THE CHEMICAL ROCKET ENGINE

• A rocket is a simple device whereby a fuel and an oxidizer are supplied to a combustion chamber (a.k.a. burner) at high pressure

126126



127127


• The high-pressure products of combustion are then expanded through a nozzle to high velocities, thus producing thrust

• Specific Impulse ranges from 1500 ~ 3000 N/(kg/s)

4Vg

mF

c

&=

128128

THE NUCLEAR HYDROGEN ROCKET ENGINE

• Highest specific impulse is obtained for the lightest possible particles• Specific impulse ~ 9000 N/(kg/s)


129129

THE ION ENGINE

• Ion engines used on long-term space voyages


130130

THE ION ENGINE

• Easily ionized element (e.g. cesium or xenon) is evaporated and then brought into contact with a high-temperature surface (grid)• Electrons are removed at the surface, and the cesium ions and electrons are accelerated by appropriate electric fields and then mixed to form a neutral plasma beam which is shot from the end of the engine• Specific impulses ~ 100,000 N/(kg/s) are possible• XIPS (Xenon Ion Propulsion System) Deep Space 1 NASA

131131

SPECIFIC IMPULSE

Propulsion Device Specific Impulse N/(kg/s)

[lbf/(lbm/s)]

Pros/Cons

Turbojet 600 [5886] Inoperable in space flight

Chemical Rocket 1500~3000[14,715~29,340]

Consume large amounts of expensive fuel

Hydrogen Nuclear 9000 [88,290] Important in space flights, risks of nuclear power source

Ion Propulsion 100,000[981,000]

Small flow rate, small thrust, used for accelerating a vehicleonce into deep space

Note: 9.81 N/(kg/s) = 1 lbf/(lbm/s)

PROPERTIES OF LIQUID PROPELLANT SYSTEMS

132

CF. BARNARD & BRADLEY “FLAME & COMBUSTION”

133

Example

134

R 20684208122460

Then,

toreduces heats specificconstant for which,

losses) no (assuming equal are works turbineand compressor theSince

R 8125

50420

ncompressio Isentropic:2-1

psia 50 psia, 5

fps, 300 R, 2460 R, 420

4

4312

4312

4.1

4.01

1

2

12

3251

131

=+−=

−=−

−=−

=

=

=

=======

−

T

TTTT

hhhh

p

pTT

pppp

VTT

k

k

135

( )

mph 2109hr 1

sec 3600

ft 5280

mile 1ft/sec 3093

R12722068BTU

lbfft778

R-lbm

BTU24.0

s-lbf

ft-lbm 2.322

)(2)(2

: theAssuming

22

:nozzle theacrossEnergy ofon Conservati

R 127250

52460

processexpansion nozzleexhaust - turbineIsentropic :5-3

5

25

54545

54

2

5

5

2

4

4

4.1

4.01

3

5

35

==

−−

=

−=−=

<<<

+=+

=

=

=

−

V

V

TTJCghhgV

V V

g

Vh

g

Vh

p

pTT

pcc

cc

k

k

136

( )

rates flowhigher for larger ally proportion

become wouldand rate, flow lbm/sec 1.0for are nscalculatio above The

kW 35 hp 47lbf/sec-ft 26,000ft/sec) lbf(300 87

is force thrust by the developedpower thecity,inlet velo the

of that toequal velocity a with moves turbojet that theassume weIf

kW 417BTU/hr 1,422,000BTU/sec 395

R 812)-R)(2460-BTU/lbm 24lbm/sec(0. 1)(

:is mass-poundper addedheat theNow,

turbojet theof velocity theof that todirection opposite

in the is thrust that theindicatessign minus thewhere

lbf 87s-ft/lbf-lbm 2.32

ft/sec 3093)-0lbm/sec(30 1

:follows as computed isThrust

1

23

251

=====

====−=

−=−=

FVP

Q

TTCmQ

VVg

mF

H

pH

c

&

&

notes1

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Transcript of notes1