11

Heat Release in Heat Release in CombustionCombustion

朱　信朱　信Hsin ChuHsin ChuProfessorProfessor

Dept. of Environmental EngineeringDept. of Environmental EngineeringNational Cheng Kung UniversityNational Cheng Kung University

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1. Introducton1. Introducton

It is important to be able to quantify the thermal changes It is important to be able to quantify the thermal changes which take place on the combustion of a fuel-hence we which take place on the combustion of a fuel-hence we must look at the thermodynamics of the combustion must look at the thermodynamics of the combustion process.process.

Next slide (Fig. 3.1)Next slide (Fig. 3.1)An elementary open system within a An elementary open system within a boundary containing a working fluid, which is initially a boundary containing a working fluid, which is initially a mixture of fuel und air. mixture of fuel und air. The mixture could be ignited and burned at constant The mixture could be ignited and burned at constant pressure (as occurs in boilers), or combustion could take pressure (as occurs in boilers), or combustion could take place at constant volume.place at constant volume.

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2. Constant-pressure Combustion2. Constant-pressure Combustion

A combustion system must obey the First Law of A combustion system must obey the First Law of Thermodynamics, but the application of this law may at Thermodynamics, but the application of this law may at first sight appear difficult owing to the change in chemical first sight appear difficult owing to the change in chemical composition of the “working fluid” from a mixture of fuel composition of the “working fluid” from a mixture of fuel and air to the resulting combustion products.and air to the resulting combustion products.

The application of the First Law to the open system of The application of the First Law to the open system of Fig. 3.1 gives, for each kg of working fluid, the following Fig. 3.1 gives, for each kg of working fluid, the following equation:equation:(U(U22 – U – U11) + (P) + (P22VV22 – P – P11VV11) + g(Z) + g(Z22 – Z – Z11) + ½(C) + ½(C22

22-C-C1122) = Q-W) = Q-W (1) (1)

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The terms on the left-hand side are respectively the chanThe terms on the left-hand side are respectively the change in internal energy of the fluid (u), the “flow work” involge in internal energy of the fluid (u), the “flow work” involved in propelling the fluid through the system (pv), the chved in propelling the fluid through the system (pv), the change in potential energy of the fluid (z) and finally its chaange in potential energy of the fluid (z) and finally its change in kinetic energy (c).nge in kinetic energy (c).

The work done by the system (W), in the case of a boiler, The work done by the system (W), in the case of a boiler, is zero, so the right-hand side of the equation represents is zero, so the right-hand side of the equation represents the amount of heat transferred into the system (Q).the amount of heat transferred into the system (Q).

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The changes in potential and kinetic energy of the fluid The changes in potential and kinetic energy of the fluid can be considered negligible, so equation (1) simplifies can be considered negligible, so equation (1) simplifies to:to:(U(U22 - U - U11) + (P) + (P22VV22 – P – P11VV11) = Q) = Q

The enthalpy of the fluid, H, is given byThe enthalpy of the fluid, H, is given by H = U+PV H = U+PVleading to a final form of the steady flow equation:leading to a final form of the steady flow equation: H H22 – H – H11 =Q =Q (2) (2)

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This simple expression relates the heat transferred This simple expression relates the heat transferred across the boundary of the system to the change in across the boundary of the system to the change in enthalpy of the fluid as it enters and leaves the system.enthalpy of the fluid as it enters and leaves the system.

We will see later how it leads to a very convenient We will see later how it leads to a very convenient method of measuring the efficiency of a boiler.method of measuring the efficiency of a boiler.We can approach this problem by considering separately We can approach this problem by considering separately the enthalpies of the reactants and the products.the enthalpies of the reactants and the products.

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3. Enthalpy of a Mixture of Gases3. Enthalpy of a Mixture of Gases

A change in the enthalpy of the reactants can be A change in the enthalpy of the reactants can be calculated by summing the enthalpy changes of each of calculated by summing the enthalpy changes of each of the constituent gases.the constituent gases.

The change in enthalpy of a gas as a function of The change in enthalpy of a gas as a function of temperature is given by temperature is given by

H = c△H = c△ pp ( t)△ ( t)△Which would at first sight appear to give a straight line Which would at first sight appear to give a straight line relationship between enthalpy change and temperature relationship between enthalpy change and temperature difference.difference.

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The temperature changes which occur in combustion are The temperature changes which occur in combustion are considerable; nitrogen, for example, could enter the considerable; nitrogen, for example, could enter the combustion system at ambient temperature (25 ) and ℃combustion system at ambient temperature (25 ) and ℃be heated in the flame to around 2,000 .℃be heated in the flame to around 2,000 .℃

At the lower temperature nitrogen has a specific heat at At the lower temperature nitrogen has a specific heat at constant pressure of 1.04 kJ/kg/K, rising to 1.30 kJ/kg/K constant pressure of 1.04 kJ/kg/K, rising to 1.30 kJ/kg/K at flame temperature. at flame temperature.

All gases have values of cAll gases have values of cpp which increase with which increase with temperature, hence a property diagram relating the temperature, hence a property diagram relating the enthalpy of the reactants Henthalpy of the reactants HRR to their temperature will be to their temperature will be curve similar to that in Fig. 3.2 (next slide).curve similar to that in Fig. 3.2 (next slide).

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The enthalpy change of the mixture can be obtained by summing The enthalpy change of the mixture can be obtained by summing the changes of each of its constituents, an equation describing the the changes of each of its constituents, an equation describing the curve is curve is

For most purposes, an averaged value of CFor most purposes, an averaged value of CPP taken over the taken over the

temperature interval is adequate, but the variation of specific heat temperature interval is adequate, but the variation of specific heat with temperature is not linear and for accurate work tabulated with temperature is not linear and for accurate work tabulated enthalpy values should be used.enthalpy values should be used.

( ) ( )R P RH m c t

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4. Enthalpy of Combustion 4. Enthalpy of Combustion

Similarly, the enthalpy-temperature diagram for the products Similarly, the enthalpy-temperature diagram for the products will be:will be:

If a quantity of heat has left the system, the enthalpy of the If a quantity of heat has left the system, the enthalpy of the products is less than the enthalpy of the reactants at the products is less than the enthalpy of the reactants at the same temperature.same temperature.

We can thus sketch two curves on an enthalpy-temperature We can thus sketch two curves on an enthalpy-temperature diagram, one for the reactants and one for the products, with diagram, one for the reactants and one for the products, with the curve for the products lying below that for the reactants. the curve for the products lying below that for the reactants. (Fig. 3.3, next slide).(Fig. 3.3, next slide).

( ) ( )P P PH m c t

En

thal

py

△H25

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At any given temperature, the vertical distance between At any given temperature, the vertical distance between the two curves in Fig. 3.3 represents the enthalpy the two curves in Fig. 3.3 represents the enthalpy released by the combustion process.released by the combustion process.In general the magnitude of this quantity will depend on In general the magnitude of this quantity will depend on the temperature chosen.the temperature chosen.

The standard temperature adopted is 25 and the ℃The standard temperature adopted is 25 and the ℃quantity H△quantity H△ 2525 is the enthalpy of combustion for the fuel in is the enthalpy of combustion for the fuel in

question.question.

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Some examples of enthalpy of combustion for gaseous fuels Some examples of enthalpy of combustion for gaseous fuels are given in Table 3.1.are given in Table 3.1.

Table 3.1 Standard enthalpies of combustionTable 3.1 Standard enthalpies of combustion

FuelFuel △△HH2525 (kJ/kmol) (kJ/kmol)

Hydrogen (HHydrogen (H22))

Methane (CHMethane (CH44))

Propane (CPropane (C33HH88))

Acetylene (CAcetylene (C22HH22))

-241,800-241,800

-802,300-802,300

-2,045,400-2,045,400

-1,256,400-1,256,400

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The combustion of all the above fuels will produce water in thThe combustion of all the above fuels will produce water in the flue gases, which can be considered as existing in either the e flue gases, which can be considered as existing in either the liquid or vapor phases.liquid or vapor phases.

All the figures quoted for the enthalpy of combustion given abAll the figures quoted for the enthalpy of combustion given above are for water in the vapor phase.ove are for water in the vapor phase.

Should the enthalpy of combustion be required with water in tShould the enthalpy of combustion be required with water in the liquid phase, then the latent heat of evaporation for the wathe liquid phase, then the latent heat of evaporation for the water vapor produced in the combustion products (Her vapor produced in the combustion products (H fgfg) must be a) must be a

ccounted for:ccounted for:( H△( H△ 2525))ff = ( H△ = ( H△ 2525))gg – n × H – n × Hfgfg

where n represents the number of kmoles of water produced pwhere n represents the number of kmoles of water produced per kmole of fuel.er kmole of fuel.

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The latent heat at 25 has a value of 44,000 kJ/kmol of water ℃The latent heat at 25 has a value of 44,000 kJ/kmol of water ℃produced. produced.

Example 1:Example 1:the value of H△the value of H△ 2525 for methane is -802,300 kJ/kmol with water i for methane is -802,300 kJ/kmol with water in the vapor phase. Calculate H△n the vapor phase. Calculate H△ 25 25 for the case when all the wafor the case when all the water vapor is condensed.ter vapor is condensed.

Solution:Solution:the relevant stoichiometric combustion equation for methane is:the relevant stoichiometric combustion equation for methane is:CHCH44 + 2 O + 2 O22 → CO → CO22 + 2 H + 2 H22OOi.e. 2 kmoles of water vapor are produced for each kmole of fuei.e. 2 kmoles of water vapor are produced for each kmole of fuel burnt.l burnt.( H△( H△ 2525 ) )ff = ( H△ = ( H△ 2525 ) )gg – 2 × H – 2 × Hfgfg

= -802,300 – 2 × 44,000 = -802,300 – 2 × 44,000 = -890,300 kJ/kmol = -890,300 kJ/kmol

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5. Constant-volume Combustion5. Constant-volume Combustion

The heat released when a solid or liquid fuel is burnt is The heat released when a solid or liquid fuel is burnt is measured by burning a sample of the fuel at constant volume, measured by burning a sample of the fuel at constant volume, hence it is important to look at the thermodynamics of this hence it is important to look at the thermodynamics of this process as well as that of combustion at constant pressure.process as well as that of combustion at constant pressure.

Internal combustion engine (vehicles) is a constant-volume Internal combustion engine (vehicles) is a constant-volume combustion system.combustion system.

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6. Internal Energy of Combustion6. Internal Energy of Combustion

The combustion of a mixture of fuel and air at constant The combustion of a mixture of fuel and air at constant volume is most easily envisaged as taking place inside a rigid volume is most easily envisaged as taking place inside a rigid closed container.closed container.

The First Law energy equation for a closed system is:The First Law energy equation for a closed system is: (U (U22 – U – U11) = Q –W ) = Q –W (3) (3)

As the system boundary is fixed, no work can cross it, hence As the system boundary is fixed, no work can cross it, hence equation (3) becomes:equation (3) becomes: (U (U22 – U – U11) = Q ) = Q

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The change in internal energy can be related to temperature riThe change in internal energy can be related to temperature rise by:se by:

U = c△U = c△ VV ( t)△ ( t)△ Thus we can plot an internal energy-temperature diagram for tThus we can plot an internal energy-temperature diagram for t

he reactants and products of combustion in the same way as he reactants and products of combustion in the same way as an enthalpy-temperature plot was made in section 3. (Fig. 3.4, an enthalpy-temperature plot was made in section 3. (Fig. 3.4, next slide)next slide)

inte

rna

l

en

erg

y

△U25

2222

The curves of Fig. 3.4 are described by:The curves of Fig. 3.4 are described by:

Once again, at any given temperature the internal energy of Once again, at any given temperature the internal energy of the products must be less than that of the reactants (since the products must be less than that of the reactants (since heat has left the system).heat has left the system).

The difference between the two curves at the reference The difference between the two curves at the reference temperature of 25 is referred to as the internal energy of ℃temperature of 25 is referred to as the internal energy of ℃combustion for that fuel.combustion for that fuel.

( ) ( ) and

( ) ( ) R v R

P v P

U m c t

U m c t

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Each fuel will have two values for internal energy of comEach fuel will have two values for internal energy of combustion: one for water in the vapor phase and one for wabustion: one for water in the vapor phase and one for water in the liquid phase.ter in the liquid phase.

The two values will differ by the internal energy of evaporThe two values will differ by the internal energy of evaporation (Uation (Ufgfg) of water at 25 , which is 37,602 kJ/kmol.℃) of water at 25 , which is 37,602 kJ/kmol.℃

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7. Relationship between H△7. Relationship between H△ 2525 and and

U△U△ 2525 These two quantities are associated with fuel types, These two quantities are associated with fuel types,

because calorific measurements are usually performed because calorific measurements are usually performed at constant pressure for a gaseous fuel, and at constant at constant pressure for a gaseous fuel, and at constant volume for liquid and solid fuels.volume for liquid and solid fuels.

As all types of fuels are burned at constant pressure in As all types of fuels are burned at constant pressure in heating plants, the question naturally arises as to the heating plants, the question naturally arises as to the nature and magnitude of the difference between the two nature and magnitude of the difference between the two values.values.

2525

We getWe get

The (PV) terms for solids and liquids are very small, so for anThe (PV) terms for solids and liquids are very small, so for any reactant or product in the gas phase: y reactant or product in the gas phase: PV = nRT PV = nRThence:hence: H△ H△ 2525 = U△ = U△ 2525 + RT (n + RT (nPP - n - nRR))where nwhere nPP and n and nRR represent the total number of kmoles of com represent the total number of kmoles of combustion products and reactants in the gas phase, respectively.bustion products and reactants in the gas phase, respectively.

25 25( ) andP RH H H

H U PV

25

25

( ) ( ) ( )

( ) ( )

P R P R

P R

H U U PV PV

U PV PV

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If nIf nPP = n = nRR, then H△, then H△ 2525 and U△ and U△ 2525 will have the same value. will have the same value. Example 2:Example 2:

H△ H△ 2525 for propane (C for propane (C33HH88) is -2,045,400 kJ/kmol with water in t) is -2,045,400 kJ/kmol with water in t

he vapor phase. Calculate the corresponding internal energy he vapor phase. Calculate the corresponding internal energy of combustion.of combustion.

Solution:Solution:We can write the stoichiometric equation for propane ignoring We can write the stoichiometric equation for propane ignoring the presence of nitrogen and any excess oxygen as they will athe presence of nitrogen and any excess oxygen as they will appear in both the reactants and the products:ppear in both the reactants and the products: C C33HH88 + 5 O + 5 O22 → 3 CO → 3 CO22 + 4 H + 4 H22OO

So nSo nPP – n – nRR = 7 – 6 = 1 = 7 – 6 = 1

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△ △HH2525 = U△ = U△ 2525 + RT (n + RT (nPP - n - nRR))

-2,045,400 = U△ -2,045,400 = U△ 2525 + (8.314 × 298 × 1) + (8.314 × 298 × 1)

U△ U△ 2525 = -2,047,878 kJ/kmol = -2,047,878 kJ/kmol In conclusion, there is normally little numerical difference betwIn conclusion, there is normally little numerical difference betw

een the enthalpy and internal energy of combustion of a fuel.een the enthalpy and internal energy of combustion of a fuel.However, the phase of any water produced by combustion is However, the phase of any water produced by combustion is highly significant. highly significant.

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8. Calorific Values8. Calorific Values

The terms “enthalpy of combustion” and “internal energy of The terms “enthalpy of combustion” and “internal energy of combustion” have precise thermodynamic definitions. combustion” have precise thermodynamic definitions.

However, in more practical situations engineers generally use However, in more practical situations engineers generally use calorific values as a measure of the heat released when unit calorific values as a measure of the heat released when unit quantity of a fuel is burned.quantity of a fuel is burned.

The calorific value for a liquid fuel measured at constant The calorific value for a liquid fuel measured at constant volume differs from its internal energy of combustion.volume differs from its internal energy of combustion.Such discrepancies are small.Such discrepancies are small.

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The calorific value of a fuel is normally expressed as kJ (or The calorific value of a fuel is normally expressed as kJ (or MJ) per mMJ) per m33 (gaseous fuels) or kJ (MJ) per kg which is (gaseous fuels) or kJ (MJ) per kg which is applicable to all types of fuel.applicable to all types of fuel.

If the calorific value includes the latent heat of condensation If the calorific value includes the latent heat of condensation of the water produced it is usually referred to as the “gross of the water produced it is usually referred to as the “gross calorific value”, where as if the water is in the vapor phase, calorific value”, where as if the water is in the vapor phase, the term “net calorific value” is used.the term “net calorific value” is used.

The expressions “higher calorific value” and “lower calorific The expressions “higher calorific value” and “lower calorific value” are also in use.value” are also in use.

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A fuel can have four calorific values:A fuel can have four calorific values:(1)(1) gross calorific value at constant volume;gross calorific value at constant volume;(2)(2) gross calorific value at constant pressure;gross calorific value at constant pressure;(3)(3) net calorific value at constant volume;net calorific value at constant volume;(4)(4) net calorific value at constant pressure.net calorific value at constant pressure.

The numerical values of (1) and (3) correspond closely to the The numerical values of (1) and (3) correspond closely to the two values of U△two values of U△ 2525 while the values of (2) and (4) are close to while the values of (2) and (4) are close to

the two values for H△the two values for H△ 2525..

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The value for hThe value for hfgfg for water at 25 is 2,442 kJ/kg.℃ for water at 25 is 2,442 kJ/kg.℃ Example 3:Example 3:

A liquid fuel consists of 86% carbon and 14% hydrogen (by mA liquid fuel consists of 86% carbon and 14% hydrogen (by mass). Its gross calorific value is 43.5 MJ/kg. calculate the net cass). Its gross calorific value is 43.5 MJ/kg. calculate the net calorific value.alorific value.

Solution:Solution:HH22 + ½ O + ½ O22 → H → H22OO

hence 2 kg hydrogen produce 18 kg water vapor. The mass of hence 2 kg hydrogen produce 18 kg water vapor. The mass of water vapor produced by the combustion of 1 kg fuel is thuswater vapor produced by the combustion of 1 kg fuel is thus0.14 × 9 = 1.26 kg 0.14 × 9 = 1.26 kg the net calorific value of the fuel is then the net calorific value of the fuel is then

43.5 – 1.26 × 2.442 = 40.42 MJ/kg43.5 – 1.26 × 2.442 = 40.42 MJ/kg

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On a per kilogram basis, hydrogen has a much higher calorific On a per kilogram basis, hydrogen has a much higher calorific value than carbon. Carbon has a calorific value of 32.8 MJ/kg value than carbon. Carbon has a calorific value of 32.8 MJ/kg whereas molecular hydrogen has a net calorific value of 120.9 whereas molecular hydrogen has a net calorific value of 120.9 MJ/kg.MJ/kg.

The calorific value of a hydrocarbon fuel cannot be predicted The calorific value of a hydrocarbon fuel cannot be predicted with any degree of accuracy from these figures, but there is with any degree of accuracy from these figures, but there is an implicit suggestion that the calorific value of a fuel may be an implicit suggestion that the calorific value of a fuel may be roughly related to the relative quantities of carbon and roughly related to the relative quantities of carbon and hydrogen from which it is made up.hydrogen from which it is made up.

The broad trend is illustrated in Fig. 3.5 (next slide).The broad trend is illustrated in Fig. 3.5 (next slide).The C:H ratio range runs from anthracite and bituminous coal The C:H ratio range runs from anthracite and bituminous coal (the highest C:H ratio) to natural gas (the lowest).(the highest C:H ratio) to natural gas (the lowest).

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A much more consistent trend is exhibited by the liquid and A much more consistent trend is exhibited by the liquid and gaseous fuels, which have a much higher proportion of gaseous fuels, which have a much higher proportion of hydrogen than do coals. hydrogen than do coals.

Fig. 3.6 (next slide) show the gross calorific values of these Fig. 3.6 (next slide) show the gross calorific values of these fuels as a function of their C:H ratio.fuels as a function of their C:H ratio.Residual fuel oils (H, M, and Z)Residual fuel oils (H, M, and Z)Distillate oil (G and K)Distillate oil (G and K)Propane (P)Propane (P)Natural gas (N)Natural gas (N)

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It is possible to obtain a rough estimate of the calorific value It is possible to obtain a rough estimate of the calorific value of a solid or liquid fuel from its ultimate analysis and the of a solid or liquid fuel from its ultimate analysis and the calorific values of its combustible constituents.calorific values of its combustible constituents.

For this purpose, the relevant gross calorific values are:For this purpose, the relevant gross calorific values are:carboncarbon 33.9 MJ/kg33.9 MJ/kghydrogenhydrogen 144.4 MJ/kg144.4 MJ/kgsulfursulfur 9.4 MJ/kg9.4 MJ/kg

The estimated calorific value is given byThe estimated calorific value is given by

where C, H, O and S are the masses of these elements where C, H, O and S are the masses of these elements present in 1 kg of fuel. Ipresent in 1 kg of fuel. It assumes that the oxygen in the fuel t assumes that the oxygen in the fuel is already bound to the hydrogen present.is already bound to the hydrogen present.

33.9 144.4 - 9.48

OCV C H S

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The calorific value of a gaseous fuel is measured in a deThe calorific value of a gaseous fuel is measured in a device known as the Boys’ calorimeter. The operation of thivice known as the Boys’ calorimeter. The operation of this device is outlined diagrammatically in Fig. 3.7 (next slids device is outlined diagrammatically in Fig. 3.7 (next slide). e). It is basically a counterflow heat exchanger.It is basically a counterflow heat exchanger.

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When Steady-State conditions are obtained, the heat When Steady-State conditions are obtained, the heat released by burning the fuel is calculated from the released by burning the fuel is calculated from the temperature rise of the cooling water and its mass flow temperature rise of the cooling water and its mass flow rate.rate.

The apparatus is designed so that the average of the The apparatus is designed so that the average of the starting and finishing gas temperatures is close to the starting and finishing gas temperatures is close to the reference value of 25 .℃reference value of 25 .℃

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This temperature is well below the dew point of the This temperature is well below the dew point of the combustion products, so the value measured is the combustion products, so the value measured is the gross calorific value of the fuel.gross calorific value of the fuel.

Not all the water vapor generated in the combustion of Not all the water vapor generated in the combustion of the fuel may condense out in the calorimeter, but this the fuel may condense out in the calorimeter, but this can be corrected for by supplying the calorimeter with can be corrected for by supplying the calorimeter with near-saturated air.near-saturated air.

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The calorific value of a liquid or solid fuel is measured in The calorific value of a liquid or solid fuel is measured in the bomb calorimeter (Fig. 3.8, next slide).the bomb calorimeter (Fig. 3.8, next slide).

This device consists of a spherical vessel containing a This device consists of a spherical vessel containing a small sample of the fuel in an atmosphere of pressurized small sample of the fuel in an atmosphere of pressurized oxygen.oxygen.The bomb is immersed in an isolated water bath which The bomb is immersed in an isolated water bath which acts as the calorimeter for the heat released when acts as the calorimeter for the heat released when combustion is initiated.combustion is initiated.

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The temperature change in the water is measured by a The temperature change in the water is measured by a sensitive thermometer, and knowing the thermal capacity sensitive thermometer, and knowing the thermal capacity of the apparatus the heat released can be determined.of the apparatus the heat released can be determined.

Once again, the temperature rise is arranged to take Once again, the temperature rise is arranged to take place from just below, to just above the reference place from just below, to just above the reference temperature of 25 with a maximum rise of about 4 .℃ ℃temperature of 25 with a maximum rise of about 4 .℃ ℃

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In practice, the thermal capacity of the apparatus is In practice, the thermal capacity of the apparatus is evaluated experimentally by burning a known quantity of evaluated experimentally by burning a known quantity of a reference fuel, typically benzoic acid, in the device.a reference fuel, typically benzoic acid, in the device.

A quantity of water is introduced into the bomb to ensure A quantity of water is introduced into the bomb to ensure saturation, hence the condensation of all the water saturation, hence the condensation of all the water produced by combustion of the fuel.produced by combustion of the fuel.