Third year - الجامعة التكنولوجية year... · Third years_ Combustion Engineering _ Chemical engineering 1 Combustion engineering References 1.Gary L.borman

Third year

By

Zainab Yousif Shnean

2009 -2010

Third years______________ Combustion Engineering ____________ Chemical engineering

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Combustion engineering References 1.Gary L.borman , Combustion Engineering , Mc Graw-hill, New-York, 1998 2.Stephen R.Turns,An Introduction to Combustion , Mc Graw-hill, New-York, 1996. 3. Francis W. and Peters M.C., Fuels and Fuels Technology, 2nd edition, 1980,

.4 "تكنولوجيا الوقود " تأليف د.جابر شنشول

Combustion Process The combustion process is the cornerstone to development in our civilization .from burning wood for warmth and cooking ,to modern transportation which burns petroleum products, to generating electricity by burning solid fossil fuels, our modern world would collapse without conversion of fossil fuels to heat .combustion is a complex subject and any substantive changes to the process should only be contemplated after consultation with the regulating bodies having jurisdiction, the manufacturer of the fuel burning equipment ,the control system supplier and other trained specialists. .

Combustion Fundamentals Combustion or burning by definition is a process of conversion of chemical energy to thermal energy by rapid oxidation of the element in fuel s .the three main elements of fuels are :- carbon, hydrogen and sulphur . Oxygen is obtained from combustion air which contains: 21%O2 by volume (23% by weight and 79%N2) by volume .during combustion ,these elements are oxidized into carbon dioxide (CO2),water vapors (H2O) and sulphur dioxide (SO2) accompanied by the release

of heat and light.

Combustion of Carbon Carbon can produce two compounds depending on the availability of the air supply .

If enough air is supplied, carbon dioxide is produced .if the air is exactly right stoichiometeric conditions), the gaseous products equal s the air quantity, i.e. ( 21% CO2 and 79%N2, plus release of heat .With a starved air supply ,the carbon is partially burnt to carbon monoxide and the full calorific value of the fuel is not released. This is known as in complete combustion, a dangerous condition in any fuel burning equipment. Figure (1) provides an estimate of combustion loss due to incomplete combustion which is indicated by the presence of Coin the flue gas. Not the loss indicated in this chart is in addition to normal combustion losses.


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Figure 1

INCOMPLETE COMBUSTION LOSS

Combustion air requirement Stoichiometric air is the theoretical amount of air required for complete combustion. In actual applications, however, it is impossible to get perfect mixing of the fuel and air .thus additional air, termed excess air, is the required to burn the fuel safely and completely

Stoichometry Calculations

1.for hydrogen-air reaction

2.for a fuel containing carbon ,hydrogen and oxygen which is burned to completion:

) (O2+3.76N2)

Where are the numbers of carbon, hydrogen and oxygen atoms in a molecule of fuel

analysis of the fuel.

The mole of stoichiometeric air per mole and fuel are:-

The stoichiometeric fuel/air ratio by weight is


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The percent excess air is the actual air used minus the stoichiometric air all divided by the stoichiometeric air times 100.

Percent theoretical air is the amount of air actually used divided by

the stoichiometeric air:-

Hence

For example 110 % theoretical air is a lean mixture with 10%excess air ;85% % theoretical air is a rich mixture which is 15% deficient in air .Sometimes equivalence ratio as the fuel/air mass (f) divided by the stoichiometeric fuel/air ratio (fs) used instead of excess air to describe a combustible mixture the equivalence ratio

Excess air is directly related to equivalence ratio


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Example 1 For a stoichiom

Solution:- Basis=1 Kgmole H2 From an (H) balance 2=2b From an (O) balance 2a=b a=1/2 From an (N) balance

The partial pressure of water vapour in the products is obtained from the mole fraction.


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Example 2 Bituminous coal is burned to completion with 50% excess air. Find the fuel to air ratio (f) and the volumetric analysis of the products .the as-received ultimate analysis of the coal is 70% with carbon,5% hydrogen 15 %O2, 5% moisture and 5% ash.

Solution:- Basis for 100Kg of coal

N(kmole) M(kg/kgmole) M(kg) species 5.833 12 70 C 5 1 5 H 0.937 16 15 O 0.278 18 5 H20

Solving for a from an oxygen atom balance

Xi dry Xi N species 0.1256 0.1185 5.833 CO2

0.0564 2.778 H2O 0.0712 0.0672 3.307 O2 0.8032 0.7579 37.303 N2


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Home work:-

Q1:- Stationary gas turbine engine operates at full(3590 kW) at an equivalence ratio of 0.236 with an air flow of 15.9 kg/s .the equivalent composition of the fuel

(natural gas)is .determine the full mass flow rate and the operating air-fuel ratio for the engine.

Q2:- A natural gas fired industrial boiler operates with an oxygen concentration of 3 mole percent in the flue gases .determine the operating air-fuel ratio and the equivalence retreat natural gas as methane.

Q3:- Natural gas(97% , 3% by volume is burned with 200% excess air if

85% from methane change to and 10% to CO and 5 %not burned .determined the operating air-fuel ratio .the equivalence ratio and the composition of flue gas.

Chemical Energy

Latent heat of vaporization In many combustion processes, a liquid –vapor phase change is important .for example a liquid fuel droplet must first vaporize before it can burn. And if cooled sufficiently, water vapor can condense from combustion products formally; we define the latent heat of vaporization as the required in a constant –pressure

process to completely vaporize a unit mass of liquid at given temperature.

The latent heat of vaporization is also known as the enthalpy of vaporization.

A.Heat of reaction To understand the heat of reaction, consider the reaction of a fuel and air mixture of mass m. for a constant –volume combustion with heat transfer Qv taken as negative if out of the system the first law gives.

Where subscripts s and c refer to sensible and chemical energy


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Where To is the reference temperature. The quantity of (Cv )r is the specific heat of the reactant mixture .

*If the heat transfer is just large enough to bring the products temperature back to the reactant temperature ,and if this temperature is taken as the reference temperature To for the sensible energy then (u2-u1)=0 by definition ,and Qv is the chemical energy released by the reaction. The quantity is the lower heat value (LHV) of the fuel for constant –volume combustion .if the water in the products is condensed .the value of becomes the higher heating value (HHV)of the fuel for constant volume combustion.

If the reaction takes place at constant pressure and the total heat transfer is Qp, then the energy equation becomes

If T1=T2=To ,then Qp is the chemical energy released for the constant pressure case ,if the moles of gaseous products Np are larger than moles of gaseous reactants Nr then some of the chemical energy is expended to push side the

ambient pressure .thus ,for reactant and products which are ideal gases .

The heat of reaction may be calculated for reactions taking place at temperature other than To, and for cases where the initial and final temperature are not equal, by use of the heat of reactant data taken at To. Consider the reaction at constant pressure with reactant temperature T1 and product temperature T2 assume, for example T2>T1>To. To use the Qp(To) value ,imagine that first the reactants are cooled from T1 to To, then the reaction takes place at To, and finally the products

are heated from To toT2 .


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Where r referee to the reactants if to the products and hs is the sensible enthalpy. Note that Qp(To)is negative for an exothermic reaction.

Example 1

The higher heating value of gaseous methane and air at 25 is 55.5MJ/kg. find the heat of reaction at constant pressure of a stoichiometric mixture of methane

and air if the reactant and products are at 500 . Solution

The reaction is

Analysis of reactants and products using sensible enthalpies appendix yields

Yi MJ His(MJ/KgMole) Xi Reactants 16 0.055 8.2 0.095 CH4 32 0.22 6.07 0.19 O2 28 0.725 5.91 0.715 N2

MJ His(Mj/kgmole) Xi Products 44 8.31 0.095 CO2 18 6.92 0.19 H2O 28 5.91 0.715 N2

Np=10.52 hsp = 6.33 MJ/Kgmole

Mp=27.6 Kg/Kgmole hsp = 229 KJ/kg

Since the water vapor doesn’t condense ,the lower heating value is used

=50.113 kJ/kg fuel

And


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=-2747 kJ/kg reactants

Qp = -2747+ (229-223)=-2741 kJ/kg reactants

The negative sign indicates that the heat flow out of the system.

Example 2

In a flow calorimeter 24 mg/s of graphite particulate reacts completely with oxygen initially at 25 CO to form carbon dioxide at 1atm and 25 0C . The rate at heat absorbed by the calorimeter water is 787.0w.find the heat of formation of CO2

. Solution:-

The energy balance

And since the sensible energies are all zero at the reference temperature at 25 0C and since the C and O2 are in their standard states so that the heats of formation are

zero.

B. Heat of formation and absolute enthalpy

The heat of formation of a particular species is defined as the heat of reaction per mole of product formed isothermally from elements in their standard states.

(1 atm and 25 0C )

The absolute enthalpy of substance is defined as the sensible enthalpy relative to the reference temperature To plus the heat of formation at the reference temperature.


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The absolute (or standard enthalpy) that is the sum of on enthalpy that takes in to account the energy associated with chemical bonds.

The enthalpy of formation that is associated only with the temperature.

The sensible enthalpy change

= +

-

Example A gas stream at 1 atmosphere contains mixture of CO ,CO2 and N2 in which. The CO mole fraction is 0.1 and the CO2 mole fraction is 0.2 the gas stream temperature is 1200k. determine the absolute enthalpy of the mixture on the both mole basis (kJ/kmole) and mass basis (kJ/kg).also determine the mass fraction of the three component gases .

Solution:- Σ

=1-

= Σ

= XCO[ + - ]

+ X [ + - ]

+ X [ + - ]


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Substituting values from Appendix (table for CO ,table for CO2 and table for N2).

] + 0.2[-393.546+44.488] + 0.7[0+28.118]

= -58.339 KJ/Kmole To find ,we need to determine the molecular weight of the mixture,

= 0.1(28.01)+0.2(44.01)+0.7(28.013) =31.212

= =-1869 KJ/kg mix

= 0.1× = 0.0897

= 0.282

= 0.6282

Home work:-

Q1:-The higher heating value of a dry ash free bituminous coal is 12500 Btu /Ibm =29050 kJ/kg .the coal contains 70% wt carbon and 5% wt hydrogen on a dry ,ash free basis .find the enthalpy of formation of this coal.

Enthalpy of Combustion and Heating Values :- Consider the steady –flow reactor shown in fig(1) in which a stoichiometeric mixture of reactants enter and products exit. both at standard state conditions (25 ,1 atm). The combustion process is a assumed to be complete i.e all of the fuel carbon is converted to CO2 and all of the fuel hydrogen is converted to H2O. for the products to exit at the same temperature as the entering reactants heat must be removed from the reactor.

The amount of heat removed can be related to the reactants and product absolute enthalpies by applying

The definition of the enthalpy of reaction or the enthalpy of combustion ΔhR (per mass of mixture)is

reactor Product reactant


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Note:-

The heat of combustion Δhc (known also as the heating value is numerically equal to the enthalpy of reaction but with opposition sign.

Example 2 A. Determine the upper and lower heating values at 298 K of gaseous n-decane C10H22,per kmole of fuel and per kg of fuel. The molecules weight of n-decane is 142.284.

B.if the enthalpy of vaporization of n-decane is 359 kJ/kg. Fuel at 298 k. what are the upper and lower heating values of liquid n-decane.

Solution

A. for 1 mole of C10H22 the combustion equation can be :-

For either the upper or lower heating value

Where the numerical value of H prod depends on whether the H2O in the produced liquid (determining higher heating value )or gases (determining lower heating value).the sensible enthalpies for all species involved are zero since we desire ΔHc at the reference state (298k).furthermore ,the enthalpies of formation the O2 and N2 are also zero at 298k.


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From tables = - =-241.847- 44010=-285857 kJ/kmole

For the lower heating value we use -241847 kJ/kmole

Adiabatic Flame Temperature A. Constant pressure combustion

B. Constant volume combustion


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We define two adiabatic flame temperature : one for constant –pressure combustion and one for constant volume. If a fuel –air mixture burn adiabatically at constant pressure .the absolute enthalpy of the reactants at the initial state T=298K,P=1 atm equal the a absolute enthalpy of the final state T= ,P=1 atm

Or, equivalently on a per mass of mixtures basis

The equation above defines what is called the constant pressure adiabatic flame temperature. The definition is illustrated graphically in fig. Conceptually, the adiabatic flame temperature is simple ;however ,evaluating this quantity required knowledge of the composition of the combustion products .at typical flame temperatures ,the products

dissociate and the mixture comprises many species.

Example Estimate the constant pressure adiabatic flame temperature for the combustion of stoichiometeric -air mixture .the pressure is 1atm and the initial reactant temperature

is 298k.

1. complete combustion the product mixture consists only of , O and

2. the product mixture enthalpy estimated using specific heats evaluated at 1200 =0.5(T1+Tad)where Tad is guessed to be about 2100k.

Solution


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Species Enthalpy formation(298k) Specific heat 1200k (kj/kmole) (kj/kmole.k)

-74831 -

-393546 56.21 O -241845 43.87

0 33.71

0 -

Equating to and solving to yields

For constant –volume adiabatic flame temperature

( , )= ( , )

Where U is the absolute or standardized internal energy of the mixture.

Since most compilations or calculations of thermodynamics properties provide values for H(or h) rather than u we can rearrange eqn above to the following form :-

- -V( - ) =0

We can apply the ideal-gas law to eliminate the PV term is

P,V=

Thus - - ( - )=0


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Example:-

Estimate the constant volume adiabatic flame temperature for a stoichiometeric -air mixture .using the same assumption as in previous example initial conditions are P= 1atm

and the initial reactant temperature is 298k.

solution:-

we note ,however that the CP ,values should be evaluated at a temperature somewhat greater than 1200k since the constant volume Tad will be higher than the constant pressure - - ( - )=0

Or

- ( - )=0

-

-


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Home work

1.determine the enthalpy of formation in Kj/kmole for methane given the lower heating value of 50016 kj/kg at 298k.

2.determine the absolute enthalpy of the mixture for a temperature of 1000k .express your result in kj/kmole of mixture.

(CO=0.095 , =6 , O=7 , =34 ,NO=0.005)BY mole .

3.the higher heating value for liquid octane ( )at 298k is 47893 kj/kg and heat of vaporization is 363 kj/kg .determine the enthalpy of formation at 298k for octane vapour.

4. determine the adiabatic flame temperature for constant pressure combustion of astoichiometeric propane –air mixture assuming reactants at 298k.

Gas- Fired Furnace combustion

Energy Balance and Furnace Efficiency The fuel and air flow rates required foe a given heat output are collected from an energy and mass balance for a control volume placed around the combustion chamber heat exchanger- mixer

Naha + Nfhf = q + qL +

combusion Mixer Reactions q

fuel products

Air combustion chamber heat exchanger

Blower

Where p stands for product in term of the sensible enthalpy fuel heating value and enthalpy of vaporization of water . The energy balance

Foe complete combustion 100% combustion efficiency


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q = useful heat output

Example Propane is burned to completion in a furnace with 5% excess air the fuel and air are at 770F. If 5%

of the heat is lost through the walls of the furnace and the combustion product exit the furnace to

the stack at 3400F . What is the useful heat output of the furnace per pound of propane.

Solution

From table HHV 21670 Btu/Ib or 953480 Btu/Ibmole

Products NJ / Nf K Btu/Ibmole K Btu/Ibmole fuel

CO2 3 2.528 7.584

H2O 4 2.142 8.568

O2 0.25 1.882 0.47

N2 19.74 1.83 36.282

From equation of energy balance over furnace

The useful heat output is 82.4% of the higher heating value and 89.5% of the lower heating value.

Furnace Efficiency The efficiency of the furnace is defined as the ration of the useful heat output to the

energy input. In general for any type of fuel.


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The furnace efficiency can be increased by the following measures

1- Decrease the temperature of the exhaust products

2- Reduce the excess air which will reduce the moles of products per mole of fuel

and also reduces the blower power.

3- Reduce the extraneous heat loss

4- Reduce the blower power requirements

Where

Example Stack gas analysis of a natural gas- fired furnace gave the following volumetric analysis 4% O2 ,

10% CO2 , 17 H2O, 86% N2 all on a dry basis the fuel was 84% CH4 and 16%C2H6 by volume

and the higher heating value was 23.3 Btu/Ibm. The fuel and air entered the furnace at 770F and

the stack gas temperature was 3400F. No blower was used and heat losses were neglected. What is

the operating efficiency of this furnace and what is the excess air.

Solution

The wet product analysis by volume is

4/1.17 = 3.42%O2 , 8.55% CO2 , 14.53% H2O , 73.05% N2

The moles for carbon in the fuel per mole of fuel Nc/N2 = 0.84 + 0.16*2 = 1.16

The molecular weight of fuel is

Mf = 0.84 *16 +0.16*30 = 18.24 Ibmole/Ibmole

The molar higher and lower heating values are

HVV = 233* 18.24 = 425


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LHV = 425 – 0.1453 * 1.03*13.6*18 = 388.4

Homework Calculate the flame temperature producted produced from burnt methane with air at 6000C

Example

A laminar flame propagates through a propane-air mixture with an equivalence ratio of 0.9 pressure of 5atm and temperature of 300K. The flame velocity is 22 cm/s .Find

a)the gas temperature

b) the velocity and c)the pressure behind the flame.

Take: hp = hr = -0.1267*106 kJ/kg , ρr = 5.97 kg/cm3 , Mr = 28.318kg/Kmole

a) b)

this the velocity relative to the flame front

c) Pp – Pr = ρr(v)2 – (ρpvp)2 = 5.97(0.22)2 – 0.784(1.67)2 = -1.9 Pa = -0.0000187 atm

Pp = 4.99981atm

Simplify laminar flame model In order to gain understanding of the laminar flame let us consider a simpler more approximate flame model, following the work at Mallard in 1883 Assumption

1- Heat conduction from the flame to the reactors was the rate limiting step 2- The flame consisted of a preheat zone and reactor zone 3- Neglect any diffusion and chemical reactants in the preheated zone. 4- The boundary between the two zones was assumed to be set by the ignition

temperature. 5- The reaction zone extends over a distance of s with these assumption the

energy equation.


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In the preheat zone only becomes

Solution of equation taking k and cp constant is

Where α is the thermal diffusivity m2/s

Assuming that chemical reaction starts at temperature fig. where x =0 the boundary conditions are

T = Tr at x = -∞

T = Tig at x = 0

Where ------------------------------------------(1)

---------------------------------------------(2

Equation 1 and 2 and using the laminar burning velocity notation for vr gives the result of Mallard

Where

Problem Estimate the laminar burning velocity at a stoichiometric propane-air mixture initially at 298K and 1 atm pressure using the thermal flame theory.


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Flames

There are two basic idealized types of flames

1.permixed flames

2.diffusion flames

Premixed flames:- It arise from the combustion of gaseous reactants which are perfectly mixed. a premixed flame is a rapid essentially constant radiates light and heat ,and propagates as a thin zone with speeds of less than a few –meters per second .

diffusion flames it arise from the combustion of separate gaseous fuel and oxidizer streams which combustion as they mix diffusion flames are dominated by the mixing of the reactants which can be either laminar or turbulent and reaction takes place at the interface between the fuel and oxidizer.

Laminar premixed flames:- Bunsen burner shown in fig provides an example of a stationary laminar premixed flames, provided that the reactants flow through the tube under laminar conditions. Fuel enter under a slight positive pressure at the base of the burner and entrains air.

It is seen from that s figure:-

1.burning velocity for a particular fuel can vary by a factor of 3 depending on the fuel-air ratio.

2. the rich and lean limits of flammability are shown in this figure.

3.laminar flames will not occur above or below these limits.

Fig 1.Bunsen burner flame:(a) schematic of burner ,(b) flow diagram ,and (c) streamlines and temperature ( )for a

laminar slot burner


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4. hydrogen has the highest burning velocity and widest limits of flammability while methane has the lowest burning velocity and narrowest limits.

5. the flame temperature is highest near stoichiometric and lowest near the flammability limits fig 2 .

6. the maximum burning velocities are found just to the rich side of stoichiometeric . 7. higher laminar burning velocity is associated with

higher flame tem

perature.

Flammability limits

They are typically obtained for a given fuel by initiating a flame in a vertical glass tube about 5 cm diameter and I m length and observing if it propagates along tube.

Effect of reactant pressure and temperature on laminar burning velocity

For slow-burning mixture ( the burning velocity decreases with increasing pressure. the observed pressure dependence can be expressed as a power

law:-

Fig 2. Flame temperature as a function of equivalence ratio for

various fuels.


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where :-

From figure for fast burning mixtures (

1. increased pressure increased the flame temperature because there is less dissociation.

Laminar flame theory Consider a laminar premixed flame free from the effects of any walls. The gas flow is uniform and the flame front is planar .let the flame be stationary so that the gas flows into the flame front as noted in fig (4). The burning velocity is always defined as relative to the unburned gas velocity .so in this case the burning velocity is the unburned gas speed. let subscript refer to the mixture of reactants and subscript refer to the mixture of products. Conservation of mass, momentum and energy across the flame front are given by

Fig .3 influence of pressure on laminar burning velocity of propane –air mixtures at ambient temperatures


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Natural and artificial draught The pressure required to supply air to a furnace and to remove the flue gases from the furnace is called "draught".

Measurement of draught Since the pressure required is low ,draught is usually expressed as mm of water gauge (mm H20) ,which is the height of water in a U-tube gauge equivalent to the pressure in the furnace ,flue ,or chimney .manometers used to measure draught are calibrated in in

O.,mm O,mm Hg . Natural draught is produced by a chimney .the resultant flow of gas is controlled by dampers.

Artificial draught is produced by fans and is controlled by the speed of the fans, variation in the pith of the fan blades , or by dampers.

Artificial draught may be employed as:-

1.induced draught .

2. forced draught .

3.balanceed draught .

1.induced draught A fan is installed at the base of the chimney to augment the natural draught of the chimney. the draught from the fan overcomes the resistance of the fuel bed to the passage of primary air and the resistance of the furnace installation to the secondary air drawn over the fire bed. The furnace operates under suction. .

2. forced draught A fan is installed below ,or in front of, the grate to force primary air thought the fire bed and secondary air over the grate .the furnace operates under pressure.

Fig 4. Typical laminar flame profile


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3.balanceed draught One fan is installed at the base of the chimney and anther at the front end of the furnace the pressure and suction effects of the fans balance in the furnace ,which operates at, or near ,atmospheric pressure.

Design of chimney The draught produced by a chimney is proportional to the height of the chimney and to the density of the chimney gases. The latter is determined by their composition and temperature . a chimney and furnace system is a thermo-syphon in which the difference in pressure at the base of the chimney due to the hot gases and that due to a column of air of equal dimensions determines

the draught available.

Chimney height required for a given draught (and temperature )is calculated from the formula:-

Where:-

h:- draught in (mm O)

H:- height of chimney in( m)

h*:- height gases in( m).

T:- the temperature of the fuel gases ( K)

:- the temperature of air (K). Example

The height of chimney is 28m and the temperature of flue gases and air is 320 20 respectively, the amount of air required to burn is 15 kg/kg of fuel.

Calculate:-

a. The drought in mm water .

b. The height of flue gases.


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=353×28[ ]

=9884

h =15.1 mm water

=23.88 m

Fuel fired equipment and applications :- The purpose of a process furnace is to supply heat to the content in controlled manner .the furnace may be used for heating metals to a precisely controlled temperature for heat treatment or for melting. The furnaces are manufactured in many different types and sizes, some of which are described in this section. Furnaces may be batch or continuous type .furnace ,which generate heat by burning fuels, may be of the direct or indirect fired types. Furnace are also heated by electric resistance or induction heaters.

Batch Furnaces The batch furnaces process the product in batches, which means that the furnace doors must be opened and closed at the beginning and end of each batch cycle. Since this is a significant source of energy loss, the loading and unloading times should be minimized. It is also important to load the furnace completely to minimize the energy loss per unit of product.

Continuous Furnaces Continuous furnaces process the product continually by moving it through the heating zones on chains or conveyors. Since the loading and unloading doors are open all or most of the operating time, there is a significant heat loss through these openings. Continuous furnaces also may have a significant heat loss because of the conveying mechanism, which is heated to the operating temperature of the product. If the conveyor cools off outside the furnace before re-entering the loading area, the energy required to heat the conveyor is not used productively. Thus it is better if the conveyor stays within the heated furnace area.

Direct Fired Furnaces The products of combustion are in direct contact with the product being heated in a direct fired furnace. The heat transfer process from the flame to the product is more effective


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than with the indirect heated furnace. The higher rate of heat transfer which can be achieved with direct fired furnaces can lead to a local surface overheating of the product, unless the furnace temperature is properly controlled.

Indirect Heated Furnaces In indirect heated furnaces the heat is transferred through some form of heat exchanger. This type of furnace may be used to provide a controlled environment for oxidizing or reducing, by introducing an artificial atmosphere independent of the combustion process. Since the heat transfer from the flame to the product is not as effective as with the direct fired furnace, it can be expected that the flue gas temperature will be higher, resulting in higher heat losses unless heat recovery is used. There are few special considerations for indirect fired furnaces which affect the heat balance calculations. If the controlled atmosphere is maintained inside the furnace, the heat input and output of the gas entering and leaving the furnace must be included in the heat balance. If heat is required for the preparation of the atmosphere, the energy required in the gas generator must be included as part of the total heat input to the furnace. Electrical energy used for refrigeration or other purposes in the gas generator must also be included.

Dryers Dryers use heat to evaporate water or solvents from materials such as lumber, grain, ceramics, paints and carbon electrodes. The same principles of energy management described for furnaces also apply to dryers and much of the equipment is similar in concept. A major difference is in operating temperature, which is generally much lower than furnaces, as this avoids damage to the product. As a result the direct fired heaters must operate with very high percentage of excess air. This means that excess air cannot be reduced to achieve the energy savings. Indirect fired dryers can operate at normal values of excess air within the combustion chamber. With direct and indirect fired heaters there is a large amount of heat in exhausted air in the form of evaporated water or solvent. Often the solvents must be incinerated before discharge to the atmosphere by burning additional fuel in the dryer discharge and raising the temperature to about 900EC. Recovery of the heat in the dryer exhaust can be achieved by a heat exchanger which is used to preheat the incoming air for drying with indirect fired dryers or the combustion air for firing in the direct dryers.

Kilns There is no fundamental difference between furnaces and kilns from the energy management viewpoint. The ceramic and brick industries use stationary kilns. The rotary kilns are used by the cement and pulp industries. Some rotary kilns burn pulverized coal or refuse-derived fuel. The large heat input to the rotary kilns provides opportunities for the insulation of heat exchangers to recover flue gas heat.

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Third year - الجامعة التكنولوجية year... · Third years_____ Combustion Engineering _____ Chemical engineering 1 Combustion engineering References 1.Gary L.borman

Third year - الجامعة التكنولوجية year... · Third years_ Combustion Engineering _ Chemical engineering 1 Combustion engineering References 1.Gary L.borman