Mass and Energy Balance

1

Lecture-2: Basics of Combustion EngineeringMass and Energy Balance

Contents of this lecture

2.1 General Formulation of Mass and Energy Balance

2.2 The First Law of Thermodynamics

2.3 Example of Mass Balance of a Furnace

2.4 Energy Released in Chemical Reactions2.5 Example of Energy Balance of a Furnace

2.6 Temperature of Adiabatic Combustion2.7 Furnace Exit Temperature

2

2.1 General Formulation of Mass and Energy Balance

First identify a control volume

control volume

inin Em �� ,

storedstored Em �� ,

gg Em �� ,

o utou t Em �� ,

Then, specify a time basisBalance at an instant

Balance over a time interval

3

Mass Balance at an Instant

dtmdmmmm storedoutgin

)(≡=−+ ��

control volume

inm�

storedm�gm�

outm�

dtmdmm outin

)(=− ��

=0

We balance rates in kg/s

4

Energy Balance at an Instant

control volume

inE�

storedE�gE�

o u tE�

dtEdEEEE stoutgin

)(≡=−+ ��

We balance rates in J/s

5

Mass and Energy Balance over a Time Interval 12 ttt −=∆

mdtdtmddtmdtm

t

t

t

tout

t

tin ∆=⋅=⋅−⋅ ∫∫∫

2

1

2

1

2

1

)(��

mmm outin ∆=−

EdtdtEddtEdtEdtE

t

t

t

tout

t

tg

t

tin ∆=⋅=⋅−⋅+⋅ ∫∫∫∫

2

1

2

1

2

1

2

1

)(��

EEEE outgin ∆=−+

We balance amounts in kg

We balance amounts in J

6

Mass and Energy Balance under Steady-State Conditions

outin mm �� =

outgin EEE �� =+

7

2.2 The First Law of Thermodynamics

The first law of thermodynamic is the overall energy balance that is extended into all possible forms of energy.

8

control volume

Systems Considered

control volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volume

control volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volume

9

An Open System with

control volume

Energy Balances

over a time interval

EEE outin ∆+=at an instant

dtEdEE outin

)(=− ��

dEdEdE outin +=

0=gE�

10

An Open System with 0=gE�

Control volume

inEoutEE∆

EEE outin ∆+=

11

The First Law of Thermodynamics

An Isolated System 0=inE 0=outEand

0=∆Eso

“the sum of all energies is constantin an isolated system”

A system operated in a steady-stateoutin EE �� =

“it is not possible to construct a perpetuum mobile of the first kind”

12

2.2.1 System Energy

control volume

•

Center of gravity

2w2ω

control volume

•

Center of gravity

1w1ω

0=zReference level (height)

1z

2z

zgmIwmUE ⋅⋅+⋅+⋅+=22

22 ω

)()(2

)(2 12

21

22

21

2212 zzgmIwwmUUE −⋅⋅+−⋅+−⋅+−=∆ ωω

13

Internal Energy),(),(),( vpumvTumpTumU ⋅=⋅=⋅=

dmudumumddU ⋅+⋅=⋅= )(

Specific internal energy (J/kg)

dvvudT

Tudu

Tv

⋅∂∂+⋅

∂∂=

dvpTpTdTcdu

vv ⋅−

∂∂⋅+⋅= )(

14

It is not possible to calculatean absolute value of internal energy

∫+=vT

vT

duuvTu,

,0

00

),(

Instead specifying uo, the reference enthalpy (h0) is specified

0000 vphu ⋅−=

),( 000 pTvv = Equation of state

15

Specific Internal Energy of an Ideal Gas

pMTRv⋅⋅=

00

000 0 T

MR

pMTR

pu ⋅−=⋅⋅

⋅−=

MvR

Tp

v ⋅=

∂∂

dTTcTMRdTTcuTu

T

Tv

T

Tv ⋅+⋅−=⋅+= ∫∫

00

)()()( 00

16

The Change of Internal Energy

EEE outin ∆+=

)()(2

)(2 12

21

22

21

2212 zzgmIwwmUUE −⋅⋅+−⋅+−⋅+−=∆ ωω

dTTcTMRdTTcuTu

T

Tv

T

Tv ⋅+⋅−=⋅+= ∫∫

00

)()()( 00

E∆ can be calculated

How to calculate outin EE − ?

E∆

E∆

17

Energy Entering and Leaving the System

EEE outin ∆+=

outin EE − ?

(a) Energy of a stream of fluid

(b) Mechanical work(c) Heat(d) Electrical energy

18

Energy of a stream of fluid

dpphdT

Thdh

Tp

⋅∂∂+⋅

∂∂=

Explain why enthalpy is the stream energy

hmH ⋅= vpui ⋅+=

Is enthalpy a state variable ? If so, why?

19

Energy of a stream of fluid

dpphdT

Thdh

Tp

⋅∂∂+⋅

∂∂=

dpTvTvdT

Thdh

pp

⋅

∂∂⋅−+⋅

∂∂=

∫ ⋅+=T

Tp dTTchTh

0

)()( 0

=0 for an ideal gas

h0=0 for T0=198.15 Kp0=1 bar

20

Mechanical Work

Work is done when an object is moved against an opposing force

Work is not energy. Work is means of transmitting energy.

Convention:

control volume control volume

0>L 0<L

21

Mechanical Work

dVpdL ⋅−=

∫ ⋅−=2

1

dVpL

p

V

21−−L

1

2

1V 2V

1p

2p

Is work a state variable?

22

HeatHeat is energy in transit

due to a temperature difference

Heat is not work. Why?Convention:

heat supplied to the system is positiveheat removed from the system is negative

ft QQQ +=

Total heat absorbedby the system Heat supplied from

the surroundings

Heat generated withinthe system due to friction

23

Specific heat capacities

dTdq

c t= c - denotes specific heat in J/kgKC - denotes specific heat in J/kmolK

RCC vp =−

RCv ⋅=2

freedom of degrees ofnumber

24

Specific heat capacities

45,

34,

23,2,1, TCTCTCTCC

RC

pppppp ⋅+⋅+⋅+⋅+=

JANAF polynomials

Cp in kJ/kmolK

25

0 500 1000 1500 2000 2500 30001

2

3

4

5

6

7

8

Vibrational motions

Rotational motions

Translational motions

9/2

7/2

5/2

CO2 H2O

CO N2 O2

H2

N

H O

C p/R

Temperature in K

26

0 500 1000 1500 2000 2500 30000

20000

40000

60000

80000

100000

120000

140000

160000

CH4

CO2H2O

CO

Air

N2

O2

H2

H,O,N

H(T

)-H

(298

) in

kJ/k

mol

Temperature in K

27

Mean Specific Heat Capacities

∫ ⋅⋅=T

T dTTCT

C0

''0

)(1

∫ ⋅⋅−

=2

1

2

1

''

12

)(1 T

T

T

T dTTCTT

C

28

Energy Balance of Thermal Systems (Machines)

control volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volumecontrol volume

dEdEdE outin =−

dUdLdQt =+

dUdVpdQt =⋅−

)()( umdvmdpdqm t ⋅=⋅⋅−⋅

dudvpdqt =⋅−

29


dEdEdE outin =−

control volume

dmhdVpdQdU t ⋅+⋅−=

dmvpuvmdpdqmumd t ⋅⋅++⋅⋅−⋅=⋅ )()()(

dvpdqdu t ⋅−=

30


For any system:

dvpdqdu t ⋅−=

dpvdqdh t ⋅+=

tdqdh = for p=constant

cdqdu = for v=constant

31

J. Warnatz, U. Mass; R. Dibble“Technische Verbrennung”Springer

Lüdecke, LüdeckeThermodynamikSpringer ISBN 3-540-66805-5

32

Mass balance in combustion

Change of mass withtime (kg/s)

= Mass inputkg/s

- Mass outputkg/s

outputinput mmdt

d(mass)�� −=

For a steady state:

outputinput mm �� =

33

The purpose of making a mass balance is :

(a) to calculate the accumulation of mass (if any)(b) to calculate the out-coming streams(c) to calculate composition (make up) of

the outcoming streams

The atoms are conserved (neither created nor destroyed) and the flow rates in kg/s balanceMolecules are not conserved and the flow ratesin kmol/s may or may not balance

A correct mass balance is a prerequisite to a subsequent energy balance

34

BOILER

200 kg/h CH4

4000 kg/h airCombustion products

Example2.1. The task – make a mass balance

Mass balance in kmol/h (Chemical Engineering Approach)

CH4 + 2 O2 = CO2 + 2 H2O

Input:CH4 12.5 kmol/hN2 109.57 kmol/hO2 29.13 kmol/h

165.1=λOutput (wet):CO2 12.5 kmol/hH2O 25 kmol/hN2 109.57 kmol/hO2 4.13 kmol/h

kmol/h 02.151=∑ kmol/h 02.151=∑?

why

35

1.00001.00000.99990.9999∑

0.03520.03150.03270.0273Oxygen

0.81820.73040.86820.7246Nitrogen

-0.1071-0.1653Water vapour

0.14660.1310.0990.0827Carbon dioxide

DryWetDryWet

Mass fractionMolar fractionSpecies

Establising composition of the out-coming streams is anintegral part of any mass balance

Table 2.2 Composition of the combustion products

36

Exercise 2.1. The task – make a mass balance

BOILER

200 kg/h CH4

4000 kg/h air Combustion products

Mass balance in kg/h (Combustion Engineering Approach)CH4 + 2 O2 = CO2 + 2 H2O

16 kg CH4 + 64 kg O2 = 44 kg CO2 + 36 kg H2O

Input:CH4 200 kg/hN2 3,068 kg/hO2 932 kg/h

Output (wet):CO2 550 kg/hH2O 450 kg/hN2 3,068 kg/hO2 132 kg/h

165.1=λ

kg/h 020,4=∑ kg/h 020,4=∑

37

2.3 Reaction Enthalpy (enthalpy of a chemical reaction)

0...........2211 =+++ nn AAA υυυ

01

=∑=

i

n

ni Aυor

The change of the enthalpy in a chemical reaction is called reactionenthalpy and it is calculated as:

∑=∆i

iipressure

etemperaturR HH υ

To calculate Hi a reference state of zero enthalpy is needed.

The pure elements in their most stable state at T=298.15 K and p=1 bar are prescribed zero enthalpy.

The pure elements are: O2, N2, H2, C (graphite)

38

2.4 Standard Enthalpies of FormationThe concept of the standard enthalpy of formation is to „measure“enthalpy of any chemical compound

Chemical reactor operatedat: p=1 bar, T=298.15 K

n pureelements 1 mol of

the compound

0

15.298Hf∆

The standard enthalpy of formation of a compound is the reaction enthalpyof its formation reaction from the pure elements in their most stable stateat the temperature T=298.15 K and the pressure p=1bar (indicated by „0“)

Note: If the enthalpy of formation of a compound is positive heathas to be supplied to form the compound and the formationreaction is endothermic.

39

282.00-235.31C2H5OH(g)Ethanol

269.2082.93C6H6(g)Benzene

219.4552.10C2H4(g)Ethylen

269.91-103.85C3H8(g)Propane

229.49-84.68C2H6(g)Ethane

186.10-74.85CH4(g)Methane

213.7-393.5CO2(g)Carbon dioxide

197.6-110.53CO(g)Carbon monoxide

157.99716.6C(g)Carbon

2.381.9C(s,diamond)Diamond

153.19472.68N(g)Nitrogen atoms

183.639.3OH(g)Hydroxyl radicals

69.95-285.83H2O (l)Water

188.72-241.81H2O (g)Water vapour

114.6218.0H(g)Hydrogen atoms

160.95249.2O(g)Oxygen atoms

5.740C(s,graphite)Carbon

191.500N2(g)Nitrogen

130.570H2(g)Hydrogen

205.040O2(g)Oxygen

(J/mol K)(kJ/mol)Aggregation stateCompound 015.298Hf∆

0298S

Standard enthalpies of formation of some compounds (JANAF Tables)

40

Indirect determination of the enthalpy of formation (Hess law)Since enthalpy is a state function it can be determined using reactionenthalpy of oxidation reactions that are relatively easy to measure.

Task – determine the enthalpy of formation of pure methane knowingthe reaction enthalpies of other oxidation reactions.

C + 2 H2 = CH4 =∆0

15.2984 HCH?

(1) C(s, graphite) + O2 = CO2kJ/mol 5.393

02981 −=∆ HR

(2) H2(g) + 0.5 O2 = H2O kJ/mol 18.24102982 −=∆ HR

(3) CO2 + 2 H2O = CH4 + 2 O2 kJ/mol 52.80202983 =∆ HR__________________________________________________________________

(1)+2(2)+(3) C+O2+2H2+O2+CO2+2H2O = CO2+2H2O+CH4+2O2

C + 2 H2 = CH4

=∆0

15.2984 HCH02981 HR∆

029822 HR∆×+ =∆+

02983 HR

4CH of kJ/mol 78.7425.802)81.241(25.393 −=+−×+−=

41

2.5 Lower Calorific Value (LCV) and Gross Calorific Value (GCV)

H2 + 0.5 O2 = H2O (liquid) + 285,830.00 kJ/kmol of H2GCV (Brennwert)

H2 + 0.5 O2 = H2O (gas) + 241,900.00 kJ/kmol of H2

LCV (Heizwert)

Combustion chamberp=1 bar

The definition of LCV

unit of fuel

oxidizer (oxygen or air)Products of completecombustion; H2O (g)

LCVTemperature T1Temperature T2

T1=T2 (usually 0˚ C, 20˚C or 25˚ C)

42

2.6 What is the relationship between LCV and the reaction enthalpy ?

Example:H2 +0.5 O2 = H2O (g) (p=1 bar, T=298.15 K)

The reaction enthalpy is:

=−−==∆ ∑0

15.298,20

15.298,20

15.298,20 5.0 OHHi

iiR HHHHυ

= -241.81 kJ/mol

but LCV of H2 is 241.81 kJ/mol

∑=∆=i

iiR HHLCV υ015.298

For oxidation reaction writtenfor 1 mol of fuel,water in gas phase

43

LCV Fuel Formula Molar mass g/mol kJ/mol kJ/mn

3 kJ/kg Hydrogen H2 2.016 241.9 10,792 119,990 Carbon oxide CO 28.01 283.0 12,625 10,104 Methane CH4 16.03 802.5 35,805 50,062 Ethane C2H6 30.05 1,428.0 63,707 47,521 Propane C3H8 44.06 2,044.1 91,193 46,394 Butane C4H10 58.08 2,658.5 11,860.4 45,773 Pentane C5H12 72.15 3,272.9 146,014 45,362 Ethylene C2H4 28.03 1,323.1 59,027 47,203 Propylene C3H6 42.05 1,926.0 8,595.1 45,817 Butylene C4H8 56.06 2,542.8 113,442 45,359 Acetylene C2H2 26.06 1,255.9 56,029 48,123

Table 2.5 LCV of some selected fuels (p=1 bar, T=298.15 K)

44

Calculate LCV of methane using values of the formation enthalpies.

CH4 + 2 O2 → CO2 + 2 H2O (g)

-CH4 - 2 O2 + CO2 + 2 H2O (g) = 0

=×+×+

+×−+×−==∆ ∑

OHCO

OCHii

iR

HH

HHHH

22

24

)2()1(

)2()1(υ

kJ27.802)81.241(2)5.393()0()2()85.74()1(

−=−×+−+×−+−×−

40

15.298 CH kJ/mol 802.2727.802 =−=LCV

45

2.7 Dependence of LCV on temperature

0fuel 1 242322 =++−− OHCOO υυυ

=∆ 0

TR H ?--------------------------------------------------------------------------------------------------------

==∆ ∑=

=

04

1

0

T

i

iiiTR HH υ ∫∑ =+

=

=

T

ipTi

ii dTCH

15.298,15.298

4

1

)(υ

∑ ∫=

=

4

1 15.298,

i

i

T

ipi dTCυ

dTCHHi

i

T

ipiRTR ∑ ∫=

=

+∆=∆4

1 15.298,

0

15.298

0 υ

+⋅+∑=

015.298,

4

1ii

iHυ

46

∫∑=

=

−=T

ip

i

iiT dTCVLCVLC

15.298,

4

1

0

15.298

0 υ

∫∑∫∑==

−+=T

ipk

i

T

ipi

iT dTCdTCVLCVLC15.298

,

products

115.298,

reactants

1

0

15.298

0 υυ

ip

i

iiCdT

LCVd,

4

1

)(∑=

=−= υ

kpk

iipi

i CCdTLCVd

,

products

1,

reactants

1

)(∑∑

==

−= υυ

Kirchhoff law

Integral forms

Differentialforms

47

Calculate LCV of methane at T= 1298.15 K. (gases are ideal)

-CH4- 2 O2+CO2+2 H20 = 0

kJ/mol 5.802015.298 =LCV ?0

15.1298 =LCV

∫∑=

=

−=T

ip

i

iiT

dTcVLCVLC15.298

,

4

1

0

15.298

0 υ

with T=1298.15 Kand for CH4, CO2, H2O cp=33.3 kJ/kmol Kand for O2 cp=29.1 kJ/kmol K

)15.29815.1298(103.33)1(5.802 3015.1298 −×××−−= −LCV

1000103.3311000101.29)2( 33 ×××−×××−− −−

molkJ /1.7944.85.8021000103.332 3 =−=×××− −

48

Energy Balance of a Furnace

when p= const (dh=dqc) L = 0(most of furnace and process applications)

QHH outin�� +=

Furnace p=constHeat sink

Hin Hout

Q

49

Energy Balance (cont)

QHH outin�� +=

In order to calculate Hin and Hout both the amount and composition of the incoming and out-coming streams have to be known.

These are obtained from a mass balance

∑ ∫ =+speciesin

i

T

ipifi

in

dTCHn_

15.298,, )(�

∑ ∫ ++speciesout

k

T

kpkfk

in

dTCHn_

15.298,, )(� Q�

where ni is in kmol/s

ifH ,is in kJ/kmol

Cp,i is in kJ/kmol K

Chemical EngineeringApproach

50

Energy Balance (cont)

QHH outin�� +=

=++ ∫∫inin T

oxidiserpoxidiser

T

fuelpfuel dTcmdTcLCVm15.298

,15.298

, )( ��

dTcmproductsall

k

T

kpk

out

∑ ∫_

15.298,� Q�+

wheremfuel is in kg/s

LCV is in kJ/kg

cp,i is in kJ/kg K

CombustionEngineeringApproach

51

Example 2.4 Make the energy balance of the boiler in exercise 2.1 knowing that 2 MW of heat is extracted by the water tube wall (heat sink). Calculate Tout (Tin = 298.15 K).

BOILER

200 kg/h CH4

4000 kg/h air Combustion products

Since the gases are ideal:for O2 and N2 Cp=29.1 kJ/kmol Kfor CH4, CO2, H2O Cp=33.3 kJ/kmol K

To do:Energy balanceCalculate ToutCalculate efficiency of the boiler

52

Energy balance using enthalpies of formation

BOILER

200 kg/h CH4

4000 kg/h air Combustion productsHeat sink

Q=2 MW

CH4 ∫ −=+×−×15.298

15.298

3 kJ/s 259.93600/)3.331085.74(5.12 dT

N2 kJ/s 03600/)1.29(57.10915.298

15.298

=× ∫ dT

O2 ∫ =×15.298

15.298

kJ/s 03600/)1.29(13.29 dT

TOTAL INPUTS -259.9 kJ/s

INPUTS

53

Energy balance using enthalpies of formation (cont.)

BOILER

200 kg/h CH4


Q=2 MW OUTPUTSCO2

12.5(-393.5x103+33.3x(Tout-298.15))/3600 = -1,366.32 + 0.11563 (Tout-298.15)H2O

25(-241.81x103+33.3x(Tout-298.15))/3600 = - 1,679.24 + 0.23125(Tout-298.15)N2

109.57( 29.1x(Tout-298.15))/3600 = 0.8856 (Tout-298.15)O2

4.13( 29.1x(Tout-298.15))/3600 = 0.0334 (Tout-298.15)Heat extracted (Q) = 2,000.00

TOTAL OUTPUTS = -1,045.56 + 1.2659 (Tout-298.15)

54

Energy balance using enthalpies of formation (cont.)

-259.9 = -1,045.56 + 1.2659 (Tout-298.15)

and Tout = 918.81 K

What is the thermal input into the boiler?Physical enthalpy 0 kJ/sChemical enthalpy -(-1,366.32-1,679.24-(259.9)) = 2,785.66 kJ/s

TOTAL ENTHALPY = 2.786 MW

and Hout = 2.786 – 2 = 0.786 MW

BOILERHeat sink2.786 MW

Chemical Enthalpy of CH4

Nil Physical EnthalpyTin=298.15 K

0.786 MWPhysical Enthalpy

of CombustionProducts

Tout=918.81 K

Q=2 MW

72.0786.22 ==η

55

Energy balance using LCV

BOILER

200 kg/h CH4


Q=2 MWINPUTSCH4 kJ/s 22.781,23600/)16/3.33062,50(200

15.298

15.298∫ =+ dT

N2 ∫ =15.298

15.298

kJ/s 03600/)28/1.29(068,39 dT

O2 kJ/s 03600/)32/1.29(93215.298

15.298

=∫ dT

TOTAL INPUTS 2,781.22 kJ/s

56

Energy balance using LCV (cont.)

BOILER

200 kg/h CH4


Q=2 MWOUTPUTS

CO2

550(33.3(Tout-298.15)/44/3600 = 0.11563 (Tout-298.15)H2O450 (33.3 (Tout-298.15)/18/3600 ) = 0.23125 (Tout-298.15)N23,068(29.1(Tout-298.15)/28/3600 = 0.8856 (Tout-2898.15)O2

132(29.1(Tout-298.15)/32/3600 = 0.0334(Tout-298.15)Heat extracted = 2,000.00

TOTAL OUTPUTS = 2,000.00 + 1.2659 (Tout-298.15)

57

Energy balance using LCV (cont.)

BOILER

200 kg/h CH4


Q=2 MW

INPUTS OUTPUTS

22.781,2 = 00.000,2 + )15.298(2659.1 −× outT

K 82.915=outT

72.078.22 ==Efficiency

58

Adiabatic combustion temperature

For adiabatic combustion Q=0

∫+inT

fuelpfuel dTcLCVm15.298

, )(� ∫+inT

oxidiserpoxidiser dTcm15.298

,�

∫×+=outT

productspoxidiserfuel dTcmm15.298

,)( ��

One may calculate Tout for any excess air ratio

Thus, Tout is uniquely defined for any fuel and excess air ratio

59

Adiabatic combustion temperature (cont.)

For stoichiometric combustion 1=λ

and fuelairoxidiser mlm �� ×= max,

∫+inT

fuelp dTcLCV15.298

, ∫×+inT

airpair dTcl15.298

,min,

∫+=adT

productspair dTcl15.298

,min, )1(

adT is a maximum temperature obtainable from combustion of the fuel

adT is uniquely determined for any fuel (it is a fuel property)

60

Adiabatic combustion temperature (cont.)

Adiabatic combustion temperature for stoichiometriccombustion in air (Tin=298.15 K, p= 1bar).

H2 2,473 K CH4 2,285 K

C2H2 2,936 K C2H6 2,357 K

C3H8 2,400 K CO 2,624 K

Assumption:CO2 and H2O are the only combustion products

61

2.11 Furnace exit temperature

Furnace p=const

Heat sink

Hin Hout

QEnergy balance:

++ ∫fuelinT

fuelpfuel dTcLCVm,

298, )(� =∫

airinT

airpfuelair dTcml,

298,min,λ

QdTcmlmoutT

productspfuelairfuel ++ ∫298

,min, )( �� λTout is furnace exit temperature

62

2.11 Furnace exit temperature (cont)

Definition of a new variable TTITTI – TOTAL THERMAL INPUT

+= LCVmTTI fuel�

∫∫ +airinfuelin T

airpfuelair

T

fuelpfuel dTcmldTcm,,

298,min,

298, )( �� λ

PHYSICAL ENTHALPY (PREHEAT)

FUEL THERMAL INPUT

63

2.11 Furnace exit temperature (cont)

∫ ∫

∫

++

+−=

fuelin airin

out

T T

airpfuelp

T

productspair

dTcdTcLCV

dTcl

TTIQ

, ,

298 298,,

298,min, )1(

1λ

�

),( outTfTTIQ λ=�

adout TT →0Q if →�

64

Exercise 2.4. Generate a graph showing the relationship Q/TTI as a function ofexcess air and furnace exit temperature for combustion of methane

Assume:

Tin,fuel=298 K, Tin,air=298 K and LCV = 50,062 kJ/kg of CH4

LCV

dTcl

LCVQ

outT

productsp∫+−= 298

,airmin, )1(1

λ�

65

Exercise 2.4 (cont)

16 kg CH4 + 64 kg O2 = 44 kg CO2 + 36 kg H2O

lmin,air = 64/16/0.233=17.1674 kg of air/kg of CH4

Composition of combustion products:

wCO2 = 2.75/(1+17.1675λ )wH2O = 2.25/(1=17.1675λ )wN2 = 13.1675λ /(1+17.1675λ )wO2 = 4(λ –1)/(1+17.1675λ )

322818442

2

2

2

2

2

2

2

,,,,,

OpO

NpN

OHpOH

COpCOproductsp

cw

cw

cw

cwc +++=

66

Exercise 2.4 (cont)

Assumption: cps are constant (independent of temperature)

for CO2, H2O Cp=33.3 kJ/kmol Kfor O2,N2 Cp=29.1 kJ/kmol K

∫ =outT

productsp dTc298

,

)298()32

1.2928

1.2918

3.3344

3.33( 2222 −×+++ outONOHCO Twwww

67

500 1000 1500 2000 2500 30000.0

0.2

0.4

0.6

0.8

1.0

constant cp valuesCombustion of pure CH4

Tin=298 K

λλλλ=3.0

λλλλ=2.0

λλλλ=1.5

λλλλ=1.3

λλλλ=1.2

λλλλ=1.15

λλλλ=1.1

λλλλ=1.05

λλλλ=1.0

Q/T

TI

Furnace exit temperature in K

68

Exercise 2.4 (cont)

Assumption: cps are calculated using JANAF polynomials

∫ =outT

productsp dTc298

,

+− +−∑ )298(2

21)298(( 22

,1,_

outpioutpispeciesproduct i

i TCTCMwR

+−+−+ )298(41)298(

31 44

4,33

3, outpioutpi TCTC

})298(51 55

5, −+ outpi TC

69

500 1000 1500 2000 2500 30000.0

0.2

0.4

0.6

0.8

1.0

Tin=298 K

cp values from JANAF polynomials

Combustion of pure CH4

λ=λ=λ=λ=3.0

λλλλ=2.0 λλλλ=1.3

λλλλ=1.5

λλλλ=1.2

λλλλ=1.15

λλλλ=1.1

λλλλ=1.05

λλλλ=1.0

Q/T

TI


70

Exercise 2.4 (cont)

Adiabatic temperature for stoichiometric oxidationof CH4 in air (Tin = 298) have been calculated as:

constant cp values cp from JANAF polynomials

Tad 2,810 K 2,285 K

71

500 1000 1500 2000 2500 30000.0

0.2

0.4

0.6

0.8

1.0

Tin= 298 K λλλλ=1.1

λλλλ=1.0

λλλλ=1.1

λλλλ=1.0

Groningen Natural Gas

Coke Oven Gas

λλλλ=1.1

λλλλ=1.0

Blast Furnace Gas

Combustion of various gaseous fuels in airQ

/LC

V


72

A BLAST FURNACE

75

SMELTING of IRON ORE

Iron oreFe2O3 (Hematite)Fe3O4 (Magnetite)

Pig IronFe

Reducing agent

CO

SMELTING

A chemical reactor

BLAST FURNACE

76

REMOVAL OF OXYGEN FROM IRON OXIDES

2 Fe2O3 + CO = CO2 + 2 Fe3O4 Begins at 450 ºC

Fe3O4 + CO = CO2 + 3 FeO Begins at 600 ºC

FeO + CO = CO2 + Fe Begins at 700 ºC

Production of CO and heat

C + O2 = CO2

CO2 + C =2 CO

77

REMOVAL OF SULPHUR FROM IRON OXIDES

PyritesFeS, FeS2

Pig IronFe

Limestone CaCO3is added to remove S

Reactions:

CaCO3 = CaO + CO2

FeS + CaO + C = CaS + FeO + CO

78

BLAST FURNACEincoming and out-coming streams

INCOMING

Iron ore (0.5 – 1.5 “)Pellets (lower iron content ore)Sinter (fine ore, small coke, fine limestone)

Coke

Limestone (flux)

Hot blast (hot air)Oxygen

OUT-COMING

Pig Iron (molten)SlagBlast Furnace Gas

79

A TYPICAL COMPOSITION OF A PIG IRON

Iron (Fe) 93.5 - 95 %

Carbon (C) 4.1- 4.4 %

Silicon (Si) 0.30 – 0.90 %

Sulphur (S) 0.025 – 0.050 %

Manganese (Mn) 0.55 – 0.75 %

Titanium (Ti) 0.02 – 0.06 %

80

A COKE-OVEN

82

COAL to COKE TRANSFORMATION

Steps:

Heat is transferred into the coal charge

A plastic layer near walls is formed (375-475 C)

Evolution of tars and aromatic hydrocarbons

Coke stabilization phase(600 – 1000 C)

83

COKE OVENincoming and out-coming streams

INCOMING

Blended coals

Air (reducing atmosphere)

Heat

OUT-COMING

Coke

Coke oven gas

84

A TYPICAL COMPOSITION OF A BLAST FURNCACE COKE

Carbon (C) 96-97 %

Hydrogen (H) 0.4-0.6 %

Oxygen (O) 0.5 %

Nitrogen (N) 1 %

Sulphur (S) 0.9 %

85

500 1000 1500 2000 2500 30000.0

0.5

1.0

λλλλ=1.1

λλλλ=1.0Vietnamese Antracite λλλλ=1.1

λλλλ=1.0Coal Fettnuss mvb

λλλλ=1.1

λλλλ=1.0

Light Oil

Oil and Coals

Q/L

CV


Mass and Energy Balance

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