Chapter 4 PROPERTIES OF WORKING FLUIDS - t...

1

CHAPTER 4 PROPERTIES OF WORKING

FLUIDS

4.1 Introduction

4.2 Burned and Unburned Mixture Composition

4.3 Gas Property Relationships

4.4 A Simple Analytic Ideal Gas Model

4.5 Relation Between Unburned and Burned Mixture Charts

4.6 Exhaust Gas Composition

4.1 Introduction

The study of engine operation through an

analysis of processes inside engine requires models

for compositions and working fluids inside the

engine, as well as, models for individual processes-

induction, compression, combustion, expansion,

and exhaust. This chapter deals with models for

working fluid composition, and thermodynamics

and transport properties.

2

Process Spark-ignition engine Compression-ignition engine

Intake AirFuel1

Recycle exhaust2

Residual gas

Air Recycle exhaust2

Residual gas

Compression AirFuel vaporRecycle exhaustResidual gas

AirRecycle exhaustResidual gas

Expansion Combustion products (mixture of N2, H2O, CO2, CO, H2, O2, NO, OH, O, H, …)

Combustion products (mixture of N2, H2O, CO2, CO, H2, O2, NO, OH, O, H)

Exhaust Combustion products (mainly N2, CO2, H2O and either O2 if < 1 or CO, and H2, if > 1)

Combustion products (mainly N2, CO2, H2O and O2)

Table 4.1 Working fluid constituents

1 Liquid and vapor in the intake; mainly vapor within cylinder2 Sometimes used to control NOx emissions

3

EXHAUST GAS RECIRCULATION (EGR)

SYSTEM

4

Unburned mixture Burned mixture

1.Single ideal gas throughout operating cycle with cv and cp constant

2.Ideal gas; cv,u constant Ideal gas; cv,b constant3.Frozen mixture of ideal gases cv,u(T)

4.Frozen mixture of ideal gases cv,u(T)

Approximations fitted to equilibrium thermodynamic properties

5.Frozen mixture of ideal gases cv,u(T)

Mixture of reacting ideal gases in thermodynamic equilibrium

Note: Subscript i,u, and b denote species i in the gas mixture, the unburned mixture,

and burned mixture properties respectively.

Table 4.2 Categories of models for thermodynamics properties

5

4.2 Burned and unburned mixture composition

The mass of charge trapped in the cylinder (mc) is the

inducted mass per cycle (mi) plus the residual mass (mr) left

over from the previous cycle.

6

mc = mi + mr

Residual fraction xr

c

rr

m

mx

In some engines, a fraction of engine exhaust gases is

recycled to the intake to dilute the fresh mixture for control

of NOx

Percent of exhaust gas recycled (% EGR)

100(%)

i

EGR

m

mEGR

EGRm

im

: the mass of Exhaust Gas Recycled

: inducted mass per cylinder

Burned gas fraction, xb

c

rEGRb

m

mmx

rr xxEGR )1)(100/( 7

mc = mi + mr

► Burned mixture composition

Combustion of CHy with air

y = H/C ratio

Stoichiometric Combustion Equation:

22222 )4

1(773.32

)773.3)(4

1( Ny

OHy

CONOy

CH y

8

Combustion equation (with equivalence ratio, )

written per mole O2:

222

2

222

222

222

)1(2NnOnHn

COnOHnCOnNOHc

NOH

COOHCO

(1)

where : N/O molar ratio (= 3.773 for air)

: 4/(4+y)ni : moles of species i per mole O2 reactant

9

The ni are determined using these assumptions

1) CO and H2 can be neglected

2) O2 can be neglected3) water gas reaction can be assumed to be in equilibrium

OHCOHCO 222

with the equilibrium constant

22

2)(HCO

COOH

Pnn

nnTK

111

10

or Kp can be determined from

3

9

2

63 102803.010611.110761.1743.2ln

TTTKP

Kp of 3.5 is often assumed which corresponds to the value at temperature T = 1740 K

The ni is obtained from element balance of equation (1), and

ni is shown in table 4.3

11

Table 4.3 burned gas composition under 1700 K

species ni, moles/moles O2 reactant

CO2

H2OCOH2

O2

N2

Sum: (nb)

00

c

0

1 1

)1(2

1

1)1(

c

c )1(2

c )1(2

)2(

12

absent !

absent !

absent !

Summation of moles above

The value of c is obtained by solving the quadratic

0)1(2)1(2)1(2)1( 2 KKccK

i

ib nn is given in the bottom line of Table 4.3

Hence the mole fractions are

ix~

b

i

n

n

13

For alcohols, alcohol-hydrocarbon blends, the

))(24

1(1

22 NOzy

OCH zy

per mole O2:

222 )2

1()1(2 Nz

OHC

(2)

where )1(2

2

z

14

*

21*

zIf we write and

The reactant expression (2) becomes:

222 *)1(2* NOHC

which is same form as the reactant expression for hydrocarbon fuel (1)

15

Table 4.3 can still be used to give the burned gas composition, except

*Replacing with

*,, ,,

Example 4.1 Calculate the low-temperature burned gas

composition resulting from the combustion of 7 g/s air

with 0.48 g/s ethane (C2H6)

Kp is assumed 3.5.

Stoichiometric combustion:

2222262 773.35.332)773.3(5.3 NOHCONOHC

16

sFA )/( 1.16

sAF )/(1.16

1

actualAF )/(g/sg/s

7

48.0

You’re supposed to

know how to

find (A/F)s!!given

(again )

s

actual

AF

AF

)/(

)/( 11.1)1.16(

7

48.0

1We need c since based on Table 4.3

018.046.12 ccThus

134.0c (use smaller positive value)

mole fraction:

b

CO

COn

nx 2

2

~ 17

y = 6/3 from C3H6

Remember ??

0)1(2)1(2)1(2)1( 2 PPP KKccK

Where Kp = 3.5 and y

4

4

We can find c by solving the following equation

bn )2(From Table 4.3,

345.5773.31.1)57.02(

2COn 495.0134.0)1.1(57.0)( c

2

~COx

0.4950.093

5.345 Ans

OHn2

c )1(2

88.0134.0)1.157.01(2

OHx2

~164.0

345.5

88.0 Ans

18

b

CO

COn

nx 2

2

~

► Unburned mixture composition

Number of moles of fuel per mole O2 in mixture

depends on the molecular weight of fuel, Mf

Molecular formula of fuel )( yCH

)12( yM f

The fresh mixture per mole O2

22)( NOCH y

Substituting )12/( yM f from (3), then

the reactant expression becomes

22)()21(4

NOCHM

y

f

19

The unburned mixture (fuel, air, and a burned gas fraction),

per mole O2 can be written:

)(22222 NOHCOOHCOb nnnnnnx

unburned reactants unburned residuals

The number of moles of each species in the unburned

mixture, per mole O2, is summarized in Table 4.4.

20

22)()21(

4)1( NOCH

Mx y

f

b

Table 4.4 Unburned mixture composition

Species ni, moles/mole O2 reactant

fuelO2

N2

CO2

H2OCOH2

Sum1

00

1 1

fb Mx /)21)(1(4

bx1

bx

)1(2 bx

fb Mx /)21)(1(4

bx1

)( cxb

cxb )1(2

cxb

cxb )1(2

un un21

The total number of moles of unburned mixture nu,

bb

f

bu nxM

xn

1

)21(4)1( , and nb is given in Table 4.3

Molecular weight of (burned and unburned) mixture

The mass of mixture (burned or unburned) per mole O2

16.28)21(432 RPm

Molecular weight

- burned mixture, Mb bRPb nmM /

- unburned mixture, Mu uRPu nmM / 22

Previous table for burned mixture !!!

Figure 4.1 Molecular weight of unburned and burned

isooctane-air mixtures as a function of fuel/air

equivalence ratio and burned gas fraction

23

Mo

lec

ula

r w

eig

ht

Equivalence ratio

Table 4.5 Factors for relating properties on molar and mass basis

Quantity, per mole O2 in the

mixtureGeneral equation1

Equation for C8H18-air mixtures

Moles of burned mixture (nb)

Moles of unburned mixture (nu)Mass of mixture2

Mass of air2 138.2

1,1)1( bn

1,)2( bn bn

773.436.0 bn

773.336.1

1

11

)21(4)1(

bb

f

b nxM

x

un

773.408.0

bx28.0

un

773.408.0

)128.1( bx

16.28)21(432 RPm 12.92.138

16.2832

* *

1For hydrocarbon fuel, for air = 3.773; for fuels containing oxygen, and

are replaced for and respectively

2Unit: kg/kmol

24

25

4.3 Gas Property Relationships

Individual species in Unburned and Burned gas mixtures can

be modeled as ideal gases.

The most important relationship for property determination

for engine calculations are summarized below

Internal Energy u(T,v) and Enthalpy h(T,p)

Specific heat

TdT

duc

T

uc v

v

v

TdT

dhc

T

hc p

p

p

Constant volume

Constant pressure

26

Specific heat ratioc

cv

p

0

0 v

Tc dT

Tu u

0

0 p

TdTc

Th h

The change from its value at the reference condition

T0, v0, p0 to the value at T, v, p

pv

o

dv dpccs s ds dT R dT R

T v v p

entropy

27

Integrate

00 0

0

00ln ln

v

v

s vTdvc

ds dT RT v

s T v

Tc

dT R v vT

Ts s

Similarly,

0

0

0

lnp

Tp

dT RT p

T

cs s

The integral terms in the above equation are functions of

temperature only, we define

v

dvRdT

T

cds v

0

lnv

v

0

0

( ) ( )

( ) ( )

v

p

TdT

T TT

T

TdT

T TT

T

c

c

then0

0

0

0

ln

ln

R

pR

p

vs s

v

s s

Thus the entropy change between state 1 and 2

22 12 1

1

lnp

Rp

s s

28

o

v

vRdT

T

TT

vcss ln

0

0

0

0

0

lnp

Tp

dT RT p

T

cs s

define Ψ(T) Ф(T)

29

ss 12

Isentropic process:

2 12

1

lnp

Rp

Mixture properties are determined either on mass or molar basis

sx

hx

ux

ii

ii

ii

s

h

u

sx

hx

ux

ii

ii

ii

s

h

u

and

cxc

cxc

ipip

iviv

,

,

cxc

cxc

ipip

iviv

,

,

22 12 1

1

lnp

Rp

s s

Mass

basis

Molar

basis

30

4.4 A SIMPLE ANALYTIC IDEAL GAS MODEL

- heat capacity cc vp, are constant but different for unburned

& burned gas mixture

hTcu

hTcu

bfbbvb

ufuuvu

,,

,,

hTch

hTch

bfbbpb

ufuupu

,,

,,

h uf , h bf ,& : enthalpies of formation of unburned

and burned gas mixture at 0 K

unburned

burned

31

uu bu

hTchTc bfbbvufuuv .,,, (1)

Combustion process

• Constant volume adiabatic

We solve for Tb by using relations

MM

RR

b

u

u

b

/( 1)vC R

(2)

(3)

32

From (1) hTchTc bfbbvufuuv .,,,

, , ,

,

1b uv u f u f b

v b

c h hT Tc

uv,u f

v,b

fv,u

u

v,b v,b

1+

= +

c hTc

hcT

c c

u fb

u

v,bb u

-1= +

-1

γR hT

cγR

vC R /( 1)

33

u fb

u

v,bb u

-1= +

-1

γR hT

cγR

fb u

b bu uu

M= -1 +

-1M

hTγTγ R

Tb

b fb b

u

buu

b fb bb

uu uu

u

-1 -1M= +

-1M

-1 -1M M= +

M-1M

γ γhT

Rγ

γ γhT

Rγ

MM

RR

b

u

u

b

vC R /( 1)

MM

RR

b

u

u

b

34

- Constant pressure adiabatic

hh bu

R

hT

MM

Tu

f

u

u

u

u

b

b

b

b

1

1

35Figure 4.2 Internal energy versus temperature plot for

stoichiometric unburned and burned gas mixtures: iso-

octane fuel; unburned residual fraction 0.1.

unburned

burned

TdT

duc

T

uc v

v

v

36

4.5 THERMODYNAMIC CHARTS (แผนภมิู)

1. Chart that relates the mixture temperature, pressure and

volume in “compression process”2. Chart that presents internal energy and enthalpy as

functions of temperature

ASSUMPTIONS :

•Compression process is reversible and adiabatic

•Fuel is in vapor phase

•Mixture composition is homogeneous and frozen

•Each species can be modeled as an ideal gas

•The burned gas fraction is zero

4.5.1 unburned mixture charts

37

22 1

1

22 1

1

( ) ( ) ln

( ) ln

u

u

vRnT T

v

pRnT T

p

un : Moles of unburned gas per kilogram of fuel

3

uair air

Jmp Pa v n R T Kkg kg .K

38Figure 4.3 Sensible enthalpy and internal energy of unburned

isooctane-air mixtures as function of temperature.

Units:kJ/kg air in mixture.

hs

us

39Figure 4.4 Isentropic compression functions, and , as

function of temperature for unburned isooctane-air mixtures. Units: J/kg airK.

40

Table 4.6 Unburned mixture composition for charts

22 1

1

22 1

1

( ) ( ) ln

( ) ln

u

u

vRnT T

v

pRnT T

p

Example 4.2 The compression process in an internal combustion

engine can be modeled approximately as adiabatic and reversible

(i.e., as an isentropic process). A spark-ignition engine with a

compression ratio of 8 operates with a stochiometric fuel vapor-air

mixture which is at 350 K and 1 atm at the start if the compression

stroke. Find the temperature, pressure, and volume per unit mass

of air at the end of the compression stroke. Calculate the

compression stroke work.

Give = 350 K at the start of compression, find at the end

= 350 K,

= 150 J/kg air K. From Eq. (4.25a),

of compression using the isentropic chart, Fig. 4-4 For

42

Figure 4-4 then gives T2 = 682 K

The ideal gas law gives

and 2 12 1

1 2

6828 15.5 atm

350

T vp p

T v

1150 292ln 757 J / kg air K

8

From table 4.6

Figure 4-4

This is wrong in

your handout. Please correct it !!



350 K

150



757

682 K350 K

150

45

Note that p2 can also be obtained from Fig. 4-4:

The compression stroke work, assuming the process

is adiabatic and using the data in Fig. 4-3, is

2 2 1

1

2

980 180ln 2.74

292

15.5 atm

u

p

p n R

p

46Figure 4.3 Sensible enthalpy and internal energy of unburned

isooctane-air mixtures as function of temperature.

Units:kJ/kg air in mixture.

T1=350 K

u1=40

T2=682 K

u2=350

47

Problem 2 (20 points)

A 3-liter, six cylinder SI engine is running at 3600 rpm. The engine has an equal size

of bore and stroke. Its compression ratio is 9.5. The connecting rod length is 16.6 cm.

If combustion ends at 20o after top center (ATC), calculate

•Bore

•Mean piston speed

•Clearance volume

•Piston velocity as soon as the combustion ends• Distant between the crank axis and the piston pin axis as soon as the combustion

ends

48


In an analysis of combustion of propane (C3H8)with air, it is found that the exhaust

gas composes of CO2, O2, H2O and N2 in which CO2 is found to be 9 % by mole.

Compute the fuel/air equivalence ratio and the corresponding A/F.

(Molecular weight of air: 28.96, atomic weight H: 1, O: 16, C: 12, N: 14)

49


Gaseous mixture in a closed tank composes of CO2, CO and O2. These gases

are in chemical equilibrium based on

2

Kp

2 O2CO2CO

Pressure of the mixturein the tank is 1.4 atm. If there are 6 moles of CO,

calculate the equilibrium constant Kp of this reaction.

50

4.5.2 Burned mixture charts

1. For products of combustion at high temperature2. During expansion process

ASSUMPTIONS :

•Expansion process is reversible and adiabatic.

•Each species in mixture can be modeled as an ideal gas.

•Mixture is in thermodynamic equilibrium at temperature

above 1700 K. Mixture composition is frozen below 1700 K.

•At the datum state of 298.15 K and 1 atm chemical

elements in their naturally occurring form (N2, O2, H2 as

diatomic gases and C as solid graphite) are assigned zero

enthalpy and entropy.

51Figure 4.5 Internal energy versus entropy chart for

equilibrium burned gas mixture, isooctane fuel, equivalence

ratio 1.0

52

Example 4.3 the expansion process in an internal combustion

engine, following completion of combustion, can be modeledapproximately as an adiabatic and reversible process (i.e.,

isentropic). Under full-center immediately following

combustion is 7100 kPa. Find the gas state at the end of the

expansion stroke and the expansion stroke work. The

compression ratio is 8, the mixture is stoichimetric, and the

volume per unit mass of air at the start of expansion is 0.125

m3/kg air

Locate 7100 kPa and 0.125 m3/kg air on the

= 1.0 burned gass chart (Fig. 4-8). This gives 2825 K,

and s1 = 9.33 kJ/kg airK. The gas expands at

m3/kg air. Following a constantconstant entropy to

entropy process from state 1 on Fig. 4-8 gives



ratio 1.0

1

2

54

= 1840 K, = 570 kPa, and = -1540 kJ/kg air

The expansion stroke work,

assuming the process is adiabatic, is

kJ/kg air

Looked up from chart at point 2,

55

4.5 Relation between unburned and burned mixture charts (ความสัมพนัธ์ระหว่างแผนภูมสิารผสมเผาไหม้และไม่เผาไหม้)

1T 1p1vGiven unburned mixture at , , , the combustion could

be considered as

(1) constant-volume adiabatic combustion

(2) constant-pressure adiabatic combustion

Charts

Unburned Burned

- Expansion process

- Enthalpy and entropy of the

chemical elements in their naturally occurring form form ( , and and C (solid)) are

assigned zero2O 2N

2H

- Compression process

- Enthalpy and internal

energy of the mixture = 0 at T = 298 K

GOING FROM UNBURNED TO BURNED

56

Combustion

End of compression

(unburned)

Start of expansion

(burned)

Adiabatic

Constant volume Constant pressure

ub = uu hb = hu

Relations between unburned and burned

57

ub = uu hb = hu

Constant volume Constant pressure

ufusu uuu ,,

ufusu hhh ,,

Fig. 4-3 Fig. 4-3

4.0buf xh 11819.51,

buf xu 11833.47, , ,

6.0buf xh 17718.77,

buf xu 1779.70, , ,

8.0buf xh 23618.103,

buf xu 23656.94, , ,

0.1buf xh 29517.129,

buf xu 29562.118,

, ,

2.1buf xh 27596.155,

buf xu 27699.141, , ,

58

The datum for internal energy and enthalpy for the

unburned mixture in unburned chart is different from

that used in the burned chart. In order to locate the

state of the unburned mixture in the burned chart, the

reference datum has to be changed. These data can be

related through the enthalpies of formation.

The unburned mixture enthalpy hu with the same datum as

the burned mixture is

ufusu hhh ,,

where hs,u is the sensible enthalpy

59

is the enthalpy of formation of unburned mixture per

kilogram of air, defined as

ifh ,

~

ufh ,

~

if

i

i hn ,

~ ,

where ni = number of kilomoles of species i per kilogram of air

Similarly, the internal energy uu is given by

ufusu uuu ,,

where if

i

iuf unu ,,~

Alternatively,

ufu , can be obtained from

TRnnhu RPufuf

~)(,,

60

ufh ,

ufu , Expressions of and (kJ/kgair) at several :

4.0buf xh 11819.51,

buf xu 11833.47, , ,

6.0buf xh 17718.77,

buf xu 1779.70, , ,

8.0buf xh 23618.103,

buf xu 23656.94, , ,

0.1buf xh 29517.129,

buf xu 29562.118,

, ,

2.1buf xh 27596.155,

buf xu 27699.141, , ,

61

bx

uT

usu , upuv

Example 4.4 Calculate the temperature and pressure after constant pressure adiabatic combustion of unburned mixture (with = 1.0 and = 0.08) at the state corresponding

= 682 K,

= 350 kJ/kgair, = 1.57 MPa and = 0.125 m3/kgair

to the end of compression process:

Analysis

b uh h

, ,b s u f uh h h (1)

For = 1.0

ufh , bx29517.129 (2)

2367.129 airkJ/kg366

From figure 4.3 at T = 682 K, = 1.0,

ush , airkJ/kg465we obtain (3)

Replacing (2) and (3) into (1), resulting in

bh = 465 - 366 = 99 kJ/kgair

ANS

bvbu

kPa1570p

airkJ/kg665bu 2440KbT 3

air0.485 m /kgbv

Using a trial and error procedure to find and at

, from the burned chart (figure 4.5) for = 1.0

, we finally get

,

,

62

For constant pressure combustion, Pb = Pu = 1.57 MPa

b b b bu h p v airkJ/kgbv31057.199

, ,b s u f uh h h



ratio 1.0

1570

TRIAL AND ERROR PROCEDURE

64

1. Make a reasonable guess of vb

2. Compute ub

3. For ub and pb = 1570 kPa locate point on the

burned chart

4. Find vb from the chart and check if

this vb is equal to vb in (1.)

5. Use vb in (4.) for computing ub as steps (2.) to

(4.) repeat.

65

4.6 Intro. to Exhaust Gas Composition

While the formulas for the products of combustion used

in chapter 3 are useful for determining unburned mixture

stoichiometry, they do not correspond closely to the actual

burned gas composition. In the actual case, not all the fuel

which enters the engine is fully burned inside the chamber; the

combustion inefficiency even when excess air is present in a

few percent. Also, the contents of each cylinder are not

necessarily uniform in composition, and the amounts of fuel

and air fed to each cylinder of a multi-cylinder engine are not

exactly the same. For all these reasons, the composition of the

engine exhaust gases cannot be easily calculated.

66

It is now routine to measure the composition of engine

exhaust gases (e.g., CO2, NOx and unburned hydrocarbonsand particulates). This is done to determine engine emissions.

It is also done to determine the relative proportions of fuel and

air which enter the engine so that its operating equivalence

ratio can be computed.

67Figure 4.6 Spark-ignition engine exhaust gas composition data in

mole fraction as a function of fuel/air equivalence ratio.

Fuel-air equivalence ratio

Mole

fra

ctions o

f

68Figure 4.7 Hydrogen concentration in spark-ignition engine

exhaust as a function of carbon monoxide concentration. Units:

percent by volume.

69Figure 4.8 Exhaust gas composition from several diesel engines

in mole fractions on a dry basis as a function of fuel/airequivalence ratio.

Chapter 4 PROPERTIES OF WORKING FLUIDS - t...

Documents

Transcript of Chapter 4 PROPERTIES OF WORKING FLUIDS - t...