boiler combustion theory and efficiency

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Engineering Encyclopedia Saudi Aramco DeskTop Standards

BOILER COMBUSTION THEORY AND EFFICIENCY

Engineering Encyclopedia Boiler Control

Boiler Combustion Theory and Efficiency

Saudi Aramco DeskTop Standards i

Section Page

INTRODUCTION............................................................................................................. 5

CALCULATING THE TOTAL AIR FLOW TO MEET STOICHIOMETRIC CONDITIONS AND EXCESS AIR REQUIREMENTS FOR A FUEL OF KNOWN COMPOSITION............................................................................................... 6

Combustion Chemistry and Products of Combustion................................................. 6 Boiler Fuels........................................................................................................... 6 Three T’s of Combustion ...................................................................................... 7 Carbon Combustion Formula................................................................................ 9 Hydrogen Combustion Formula.......................................................................... 10 Higher Heating Value & Lower Heating Value .................................................... 12 Fuel Combustion Equation ................................................................................. 13

By-Products of Incomplete Combustion and Resulting Inefficiency ......................... 14 Partial Combustion of Carbon............................................................................. 14 Carbon Monoxide to Carbon Dioxide.................................................................. 15

Excess Combustion Air ............................................................................................ 16 Flue Gas Analysis to Determine Excess Air Requirements ................................ 17 %O2 vs. %CO2 to Determine Excess Air Requirements.................................... 18 Total Air Flow Requirements .............................................................................. 19 Example Problem. Calculating Excess Air......................................................... 19 Excess Air Shortcut Equations ........................................................................... 20

EXCESS AIR TRIM CONTROL..................................................................................... 25

Oxygen Trim Control .......................................................................................... 25 Development of % Oxygen Requirement ........................................................... 27 Oxygen & CO Trim Control................................................................................. 30

CALCULATING BOILER EFFICIENCY BY THE INPUT/OUTPUT METHOD ............... 33

Boiler Efficiency vs. Excess Air ................................................................................ 33

Boiler Inputs and Outputs ........................................................................................ 35 Heat Added by the Fuel...................................................................................... 36 Heat Added to the Incoming Feedwater ............................................................. 36

First Law of Thermodynamics: Applications to Boiler Efficiency ............................. 39

Example Boiler Efficiency Calculation–Input/Output Method ................................... 40

CALCULATE BOILER EFFICIENCY BY THE HEAT LOSS METHOD ......................... 52

Advantages and Disadvantages of the Heat Loss Method ...................................... 52



Saudi Aramco DeskTop Standards ii

Example Boiler Efficiency Calculation–Heat Loss Method....................................... 53

OPTIMUM DISPATCHING OF MULTIPLE BOILERS................................................... 62

Economic Load Allocation........................................................................................ 62

Methods ................................................................................................................... 64 Iterative Solutions ............................................................................................... 65 Model-Based Linear Program (LP) Solutions ..................................................... 66 Model-Based Nonlinear Program Solutions........................................................ 68 Incremental Cost Determination ......................................................................... 69

WORK AID 1: RESOURCES REQUIRED TO CALCULATE BOILER EFFICIENCY BY THE INPUT/OUTPUT METHOD FOR A GIVEN SET OF BOILER OPERATING CONDITIONS................................................................................................................ 72

Work Aid 1A: Example of Boiler Operating Parameters.......................................... 72

Work Aid 1B: ASME PTC 4.1.................................................................................. 73

Work Aid 1C: Combustion Engineering Fuel Burning and Steam Generation Handbook ................................................................................................................ 73

WORK AID 2: RESOURCES REQUIRED TO CALCULATE BOILER EFFICIENCY BY THE HEAT LOSS METHOD FOR A GIVEN SET OF BOILER OPERATING CONDITIONS................................................................................................................ 74

Work Aid 2A: Example of Boiler Operating Parameters.......................................... 74

Work Aid 2B: ASME PTC 4.1.................................................................................. 75

Work Aid 2C: Combustion Engineering Fuel Burning and Steam Generation Handbook ................................................................................................................ 75

WORK AID 3. RESOURCES REQUIRED TO CALCULATE TOTAL REQUIRED AIR FLOW TO MEET STOICHIOMETRIC CONDITIONS AND EXCESS AIR REQUIREMENTS FOR A GIVEN FUEL OF KNOWN COMPOSITION ........................ 76

Work Aid 3A: Combustion Engineering Fuel Burning and Steam Generation Handbook ................................................................................................................ 76

Work Aid 3B: Procedure to Calculate Excess Air.................................................... 76

GLOSSARY .................................................................................................................. 77

ADDENDUM ................................................................................................................. 79



Saudi Aramco DeskTop Standards iii

LIST OF FIGURES

Number Page

Figure 1. Liquid Fuel Viscosity vs. Temperature ............................................................ 8

Figure 2. Increased Requirement for Excess Air at Low Boiler Loads.......................... 17

Figure 3. Excess Air vs. O2 (plus other flue gases) for Gas Fuel ................................. 22

Figure 4. Excess Air vs. O2 (plus other flue gases) for Fuel Oil.................................... 23

Figure 5. % Excess Air vs. Furnace and Flue Gas Temperatures................................ 24

Figure 6. % Oxygen Trim Controller ............................................................................. 26

Figure 7. ShGP SAMA Diagram of ShGP % O2 Trim Controller.................................. 27

Figure 8. BGP Boiler F2D O2 In Flue Gas versus Load Curve .................................... 29

Figure 9. O2% vs. CO at Increasing Load.................................................................... 31

Figure 10. %CO to O2 Cascade Trim Control .............................................................. 32

Figure 11. Boiler Efficiency vs. Excess Air ................................................................... 34

Figure 12. Boiler Unit Energy Input & Output Diagram................................................. 37

Figure 13. Boiler Unit Heat Balance ............................................................................. 38

Figure 14. Economic Load Allocation ........................................................................... 63

Figure 15. Typical Boiler Operating Line ...................................................................... 64

Figure 16. Linear Boiler Operating Line........................................................................ 66

Figure 17. Segmented Linear Boiler Operating Line .................................................... 67

Figure 18. Local versus Overall Optimum .................................................................... 68

Figure 19. Economic Load Allocation ........................................................................... 69

Figure 20. Boiler Load vs Fuel Cost ............................................................................. 70

LIST OF TABLES

Number Page

Table 1. Common Chemical Reactions in Combustion ................................................ 11

Table 2. Combustion Constants ................................................................................... 14

Table 3. Excess Air Required at Full Capacity ............................................................. 16



Saudi Aramco DeskTop Standards iv

LIST OF EQUATIONS

Number Page

Equation 1. Complete Combustion of Carbon (C) .......................................................... 9

Equation 2. Formula for Complete Combustion of Hydrogen (H) ................................. 10

Equation 3. Calculation of Theoretical Air Requirements ............................................. 13

Equation 4. Incomplete Combustion (Insufficient Air)................................................... 15

Equation 5. Carbon Monoxide to Carbon Dioxide ........................................................ 15

Equation 6. Method to Calculate Excess Air ................................................................ 19

Equation 7. Boiler Efficiency - Input/Output Method..................................................... 35

Equation 8. Total Heat Input Calculation ...................................................................... 40

Equation 9. Total Heat Output Calculation ................................................................... 41

Equation 10. Absolute Heat per Pound of Steam Calculation ...................................... 42

Equation 11. Heat Output in Boiler Blowdown Water Calculation................................. 42

Equation 12. Total Heat Output Calculation ................................................................. 43

Equation 13. Dry Refuse per Pound As Fired Fuel Calculation.................................... 48

Equation 14. Combustibles in Refuse Calculation........................................................ 48

Equation 15. Boiler Efficiency Calculation - Heat Loss Method.................................... 52

Equation 16. Carbon Burned per lb As-Fired Fuel Calculation..................................... 53

Equation 17. Dry Gas per lb As-Fired Fuel Burned Calculation ................................... 54

Equation 18. Equation for Heat Loss Due to Dry Flue Gas .......................................... 54

Equation 19. Equation for Heat Loss Due to Moisture in Fuel...................................... 55

Equation 20. Equation for Heat Loss Due to H2O from Combustion of H2 .................. 56

Equation 21. Equation for Heat Loss Due to Combustible in Refuse ........................... 57



Saudi Aramco DeskTop Standards 5

INTRODUCTION

This module discusses combustion chemistry, products of combustion, byproducts of combustion, and excess combustion air. It also discusses the relationship between boiler efficiency and excess air, boiler inputs and outputs, fuel higher heat value, and the applications of thermodynamic laws and principles to boiler efficiency.

This module discusses and gives examples of boiler efficiency calculations by the input/output and heat loss methods. This module discusses the advantages and disadvantages of using the heat loss method to calculate boiler efficiency. Additionally, this module discusses the purpose and methods for the optimum dispatching of multiple boilers.

The sections of this module include:

• Calculating the total air flow to meet stoichiometric conditions and excess air requirements for a given fuel of known composition.

• Calculate boiler efficiency by the input/output method.

• Calculate boiler efficiency by the heat loss method.

• Purpose and methods for optimum dispatching of multiple boilers.




CALCULATING THE TOTAL AIR FLOW TO MEET STOICHIOMETRIC CONDITIONS AND EXCESS AIR REQUIREMENTS FOR A FUEL OF KNOWN COMPOSITION

Combustion Chemistry and Products of Combustion

Combustion is the process in which hydrogen, carbon, and sulfur in a fuel become oxidized by being combined with oxygen from the air. Of these elements, carbon and hydrogen are the major sources of heat when oxidized. Sulfur oxidation is more significant as a source of corrosion and pollution.

The products of complete combustion include water, carbon dioxide, and nitrous oxides and sulfur oxides that pollute the air. The products of incomplete combustion include water, carbon dioxide, carbon monoxide, hydrogen, carbon, aldehydes, nitrous oxides, and sulfur oxides.

Complete combustion can occur when the exact amount of air necessary to furnish the oxygen for complete combustion of a fuel's carbon and hydrogen is present. Incomplete combustion occurs because of insufficient combustion air and/or incomplete turbulence for complete mixing of fuel and air. Table 5-7 on page 63 of The Control of Boilers, 2nd Edition, by Sam G. Dukelow (Course Handout 8) provides a list of common chemical reactions in combustion with molecular weights.

Boiler Fuels Typical fuels for use in boilers around the world are natural and fuel gas, fuel oils, and coal. In Saudi Aramco, we only use gases and fuel oils, not coal. The typical energy contents of boiler fuels are:

Natural Gas: 1000 Btu/scf or 22000 Btu/lb

Refinery or Fuel Gas: 1000 - 3000 Btu/scf

No. 2 Fuel Oil: 19,500 Btu/lb

No. 6 Fuel Oil: 18,000 Btu/lb

Coal. 13,000 Btu/lb




Gas is the only fuel that can be delivered as it is being used. It is delivered via pipeline and pressure controlled to the point of use.

Fuel Oils must be stored, heated, and pumped to the point of use. There, oils must be atomized with either steam or air before combustion can take place in the burner.

Coal must be crushed, then either used in a suspension of air or liquid, burner on a grate, or burned in a fluidized bed.

Burners for liquid fuel oils typically require a viscosity of 135 - 150 Saybolt universal seconds (SSU). Therefore, the more viscous liquid fuels must be heated before they are pumped.

Three T’s of Combustion

Combustion of any fuel requires three T’s, e.g. time temperature, and turbulence. A short time, a high temperature, and a very turbulent mixture indicate a rapid and complete combustion. However, when turbulence is low, the flame is cooler and it may take longer to get complete combustion. Longer burning time and less turbulence has been found to create less nitrogen oxides, a pollutant. Sometimes, complete combustion does not take place because enough time is not given for the flame to burn completely before hot gases contact cooler heat transfer surfaces in the boiler.




Figure 1. Liquid Fuel Viscosity vs. Temperature




Carbon Combustion Formula

Equation 1 illustrates the formula for complete combustion of carbon. One mole of carbon (weighing 12 pounds) combines with one mole of oxygen (containing 2 atoms and weighing 32 pounds) to produce one mole of carbon dioxide weighing 44 pounds and containing 14,093 BTU per pound C of heat energy.

(Carbon) + (Oxygen) = (Carbon Dioxide)

C + O2 = C O2 + 14,093 BTU/lb C

12 lbs + 32 lbs = 44 lbs

Equation 1. Complete Combustion of Carbon (C)




Hydrogen Combustion Formula

Equation 2 illustrates the formula for the complete combustion of hydrogen. Two moles of hydrogen weighing 4 pounds combine with one mole of oxygen weighing 32 pounds to produce 2 moles of water weighing 36 pounds and containing 61,100 BTU per pound H2 of heat energy.

(Hydrogen) + (Oxygen) = (Water)

2H2 + O2 = 2H2O + 61,100 BTU/lb H2

4 lbs + 32 lbs = 36 lbs

Equation 2. Formula for Complete Combustion of Hydrogen (H)




Table 1. Common Chemical Reactions in Combustion

Combustible

Reaction

Moles

Pounds

Heat of combustion

(high) BTU/lb of

fuel

Carbon (to CO) 2C – O2 = 2CO 2 + 1 = 2 24 + 32 = 56 4000

Carbon (to CO2) C + O2 = CO2 1 + 1 = 2 12 + 32 = 44 14,100

Carbon monoxide 2CO – O2 = 2CO2 2 + 1 = 2 56 + 32 = 88 4,345

Hydrogen 2H2 + O2 = 2H2O 2 + 1 = 2 4 + 32 = 36 61,100

Sulfur (to SO2) S + O2 = SO2 1 + 1 = 1 32 + 32 = 64 3,980

Methane CH4 + 2 O2 = CO2 = 2H2O

1 + 2 = 1 + 2 16 + 64 = 80 23,875

Acetylene 2C2H2 + 5 O2 = 4CO2 + 2H2O

2 + 5 = 4 + 2 52 + 160 = 212 21,500

Ethylene CH4 – 3O2 = 2CO2 + 2H2O

1 +3 = 2 + 2 28 + 96 = 124 21, 635

Ethane 2C2H6 + 7O2 = 4CO2 + 6H2O

2 + 7 = 4 + 6 60 + 224 = 284 22,325

Hydrogen Sulfide 2H2S + 3O2 = 2SO2 + 2H2O

2 + 3 = 2 + 2 68 + 96 = 164 7,100

From Steam, It’s Generation and Use, © Babcock & Wilcox




Higher Heating Value & Lower Heating Value

Combustion of Hydrogen always creates WATER! Since this water is in the hot combustion zone, it is immediately vaporized after it is formed. It absorbs heat to vaporize, e.g. equal to the latent heat of vaporization. The ASTM procedures typically specify the latent heat of vaporization for this combustion water as 1040 Btu/lb. Because of the vaporization of combustion water, there is less heat available to make steam in the boiler.

The higher heating value (HHV) is the heat of combustion without considering the vaporization of combustion water. When one subtracts the latent heat of combustion water vaporization from the HHV, it is called the Lower Heating Value (LHV). Therefore, for any fuel that contains hydrogen, there are 2 fuel energies identified, e.g.:

• HHV: The total heat liberated from combustion

• LHV: The total heat liberated from combustion MINUS the latent heat of vaporization of combustion water.

The more hydrogen in the fuel, the lower the actual energy available from combustion. Since gas has more hydrogen than fuel oil than coal, gas is inherently less efficient to burn than oil than coal.

The actual efficiency losses are described below. Note that boiler efficiency calculations generally use HHV and process heater calculations use LHV.

• Coal: 5 to 7%; higher efficiency than gas 1 to 3%; higher efficiency than oil

• Oil: 3 to 5%; higher efficiency than gas 1 to 3%; lower efficiency than coal

• Gas: 5 to 7%; lower efficiency than coal 3 to 5%; lower efficiency than oil




Fuel Combustion Equation

The exact amount of air required to furnish the oxygen for complete combustion of a fuel's hydrogen, carbon, and sulfur content is called the theoretical air. Given the combustion chemistry formulas, given the known content of oxygen in air, and given a known fuel analysis, the theoretical air requirement can be calculated. An example of this calculation is illustrated in Equation 3. This example uses a formula developed from the combustion equations along with the known content of oxygen in air. Additionally, the calculation shows the amount of air theoretically required to produce 10,000 BTUs.

lbsair/lbfuel = (11.53 x 0.902) + 34.34 0.12 - 08

+ 4.29 x 0.035

lbsair/lbfuel = 11.53C + 34.34

Fuel analysis for #6 fuel oil:

3.5% by weight of Sulfer (S)12.0% by weight of Hydrogen (H)90.2% by weight of Carbon (C)

10.40 + 4.12 + 0.15 = 14.67

lbs air/10,000 BTU = 10,000 x 14.6718,500* (Oil) = 7.9

*This value is for pure C.

H2 -O28 + 4.29S

Equation 3. Calculation of Theoretical Air Requirements

The combustion constants for typical combustion reactions are defined in Table 2.




Table 2. Combustion Constants

By-Products of Incomplete Combustion and Resulting Inefficiency

Carbon Monoxide, Hydrogen, and Carbon are by-products of incomplete combustion.

Partial Combustion of Carbon

Equation 4 illustrates a formula for the incomplete combustion of carbon. Two moles of carbon combine with one mole of oxygen to produce two moles of carbon monoxide and 4,000 BTU per pound C of heat energy. Some of the potential heat energy is in the carbon monoxide.




(Carbon) + (Oxygen) = (Carbon Monoxide)

2C + O2 = 2CO + 4,000 BTU/lb C

24 lbs + 32 lbs = 56 lbs

Equation 4. Incomplete Combustion (Insufficient Air)

Carbon Monoxide to Carbon Dioxide

Given the right conditions and more oxygen, carbon monoxide can be converted to carbon dioxide to release the remaining heat energy. The formula for the conversion of carbon monoxide to carbon dioxide is shown in Equation 5.

(Carbon Monoxide) + (Oxygen) = (Carbon Dioxide)

2CO + O2 = 2CO2 + 4,435 BTU/lb CO

56 lbs + 32 lbs = 88 lbs

Equation 5. Carbon Monoxide to Carbon Dioxide




Excess Combustion Air

In reality, if just the theoretical amount of air needed to burn a fuel were supplied, it would not be enough to complete the combustion. A certain amount of additional or excess air is required to ensure complete mixing and optimum heat-release characteristics. Excess air is also necessary from a safety perspective. If the amount of oxygen at the burner drops below the theoretical amount, an explosion may occur because of the buildup of unburned hydrocarbons. Table 3 indicates the minimum typical excess air required for combustion at full load.

Table 3. Excess Air Required at Full Capacity

Fuel %Oxygen in Flue Gas % Excess Air, minimum

Natural Gas 1.5 to 3 7 – 15

Fuel Oil 0.6 to 3 3 – 15

Coal 4.0 to 6.5 25 - 40

As load on a boiler decreases, the turbulence in a burner is reduced, thus increasing the need for excess air. In all boilers, the excess air required for complete combustion is increased at reduced loads. This fact is indicated in Figure 2 that shows an excess air curve for a specific boiler.




Figure 2. Increased Requirement for Excess Air at Low Boiler Loads

Flue Gas Analysis to Determine Excess Air Requirements

Flue gas percentages of O2, of opacity, and of CO2, CO, SO2, and NOx in the flue gas can be all used to determine excess air requirements. Most commonly used to determine the percentage of excess air are the percentage of O2 and the percentage of CO2.




Percent O2. The Procedure to Calculate Oxygen and Carbon Dioxide in Flue Gas (Work Aid 1D) describes the calculations necessary to determine the percent of excess oxygen and carbon dioxide in boiler flue gas, on a dry basis. The inputs required for these calculations include the measured oxygen in flue gas and the measurement basis (i.e., wet vs. dry), an ultimate (elemental) analysis of the fuel, and the absolute humidity in the air.

Percent opacity. This is a function of incomplete combustion and/or the amount of particulate matter in the plume. A dark plume coming out of the stack usually indicates incomplete combustion. A white plume typically is the result of sulfuric acid in the gas.

Percent CO2 and PPM CO . Percent carbon dioxide, CO2, and percent carbon monoxide, CO, are a function of how much carbon in the fuel is converted to CO2.

PPM SO2 and Nox. Sulfur that is present in fuel oils is converted to either sulfur dioxide (SO2) or sulfur trioxide (SO3). Nitrogen in combustion air and/or nitrogen in the fuel is converted to nitrogen oxides (NO or NO2).

%O2 vs. %CO2 to Determine Excess Air Requirements

The measurement of O2 is preferred over the measurement of CO2 to determine excess air requirements. One reason for the preference is that greater precision of measurement is required for CO2 than for O2 to determine excess air requirements with the same degree of precision. Another reason is that the presence of oxygen always indicates the presence of excess air because oxygen is a component of air. With a given CO2, there may be one of two different percentages of total air. Typically most controllers are 'reverse', because the process itself is 'direct'.




Total Air Flow Requirements

The Procedure to Calculate Oxygen and Carbon Dioxide in Flue Gas (Work Aid 1D) describes the necessary calculations to determine the percent excess oxygen and carbon dioxide in boiler flue gas, on a dry basis. The inputs required for these calculations include the measured oxygen in the flue gas and the measurement basis (i.e., wet versus dry), an ultimate analysis of the fuel, and the absolute humidity in the air.

Excess air equals excess oxygen. Excess air equals the %O2 by volume in the flue gas divided by the total air required for stoichiometric combustion. Equation 6 is a procedure to calculate excess air:

1. Obtain flue gas analyses CO2, CO, O2, N2.

2. From the percent N2, calculate the total O2 into the furnace.

3. Reduce the free O2 by the amount required to burn the CO to CO2. The remaining free O2 is excess. (CO is usually negligible)

4. O2 required = (total in) less (excess)

( )( )

( ) x100excess-total

excess=x100O requiredO excess

=O excess Percent 5.2

22

Equation 6. Method to Calculate Excess Air

Example Problem. Calculating Excess Air

Given the following lab flue gas analysis and the composition of air, determine the excess air in combustion:




Lab Flue gas analysis: CO2 9.5

CO 1.8

O2 2.0

N2 86.7

100.0

Air composition. 21% O2, 79% N2

sx gafluemolesmoles/1000.2379.021.07.86furnace ointO2 ==

CO Correction:

Net O2 = 2.0 - 0.9 = 1.1 moles/100 moles flue gas

%02.5100)1.123(

1.1O excess Percent 2 =−

= x

If no CO Correction:

5.9)0.223()100(0.2O excess Percent 2 =

−=

Excess Air Shortcut Equations

The following are shortcut methods of determining excess air based on nominal fuel hydrogen to carbon ratio. Note that the calculation is different depending if the O2 is measure on a wet or dry basis.

Wet Basis Stack Oxygen

)%O-0.952(111.4x%O AirExcess

2

2=




Dry Basis Stack Oxygen

)%O-0.952(91.2x%O AirExcess

2

2=

Another method to determine excess air from oxygen measure in the flue gas is by chart. Excess air versus Oxygen in flue gas Charts have been created for nominal fuel compositions of natural gas, fuel oils, and coal. Figure 3 and Figure 4 show curves for excess air versus oxygen in the flue gas (both wet and dry basis) for both gas and liquid fuels.

Note that the amount of excess air also affects the furnace and flue gas temperatures. With increased excess air, there is more air to heat up, so firebox temperature is lower, and the flue gas velocity is increased, so heat transfer efficiency is decreased and stack temperature goes up. The are both additional heat losses incurred when running with more excess air than essential. Figure 5 graphically shows the relationship between % excess air and furnace/flue gas temperatures.




Figure 3. Excess Air vs. O2 (plus other flue gases) for Gas Fuel




Figure 4. Excess Air vs. O2 (plus other flue gases) for Fuel Oil




Figure 5. % Excess Air vs. Furnace and Flue Gas Temperatures




EXCESS AIR TRIM CONTROL

Since we’ve been discussing the need for excess air and how to measure it, it is a good time to review boiler flue gas excess air trim control.

Oxygen Trim Control

Excess air, which is related to oxygen in the flue gas, is controlled by a standard PID trim controller. This trim controller adjusts the raw combustion air flow rate input from the air flow measuring device. If the trim controller finds that the oxygen in the flue gas is too low (based on it’s set point), it adjusts the total airflow rate down such that the combustion air flow controller will increase total air flow rate. The actual combustion control system is discussed in a different module.

In the typical Saudi Aramco O2 trim controller, the oxygen in the flue gas is measured on a wet basis with a in situ zirconium oxide sensor. Note that this sensor also measures NET oxygen, in that all combustibles are first burned with the available oxygen before the O2 measurement is taken.

As we said earlier, the amount of excess air, and thus O2 in the flue gas, must increase with decreasing load. Thus, to keep the boiler operating at its optimum combustion air rate, the set point to the oxygen trim controller must be characterized by load.

Figure 6 shows a % oxygen trim controller.




Figure 6. % Oxygen Trim Controller




Figure 7 is the SAMA drawing for the Shedgum Gas Plant % oxygen trim controller. Notice that the controller set point is characterized by the steam flow, and in this case the controller uses integral only action.

Figure 7. ShGP SAMA Diagram of ShGP % O2 Trim Controller

Development of % Oxygen Requirement

The next question on the O2 trim controller is. How do you determine the % oxygen in flue gas that is needed at each particular load? That can be answered multiple ways.




The most common way to get this data is to do a load test on the boiler while measuring both %CO and %O2 in the flue gas. The boiler is brought up in load in manual in (typically) 20% increments, and is allowed to stabilize at each load point. The combustion air is controlled such that minimum %CO and %O2 is measured in the flue gas. This is optimum combustion for that load. The %O2 is recorded for this particular load (steam flow rate). A curve of %O2 versus load is then developed for the boiler. This is the data that is used in the O2 trim controller f(x) set point characterizer .

Remember that boiler stack smoking (opacity) MAY commence prior to reaching minimum CO operation. If this is the case, the %O2 at the smoke point should be recorded rather than at optimum CO. Also, an additional 0.5 – 1.0 %O2 is added to the curve as a safety cushion.

Figure 8 shows this curve for Berri Gas Plant boiler F2D. Notice the large increase in % O2 at loads below 100 Mlb/hr. The reason for this is probably that the boiler has reached the minimum air flow limit (typically 25% of MCR airflow), where the airflow cannot be reduced further.

The combustion air data is usually only taken after the boiler is first commissioned, and used for characterization data throughout the life of the boiler. However, a new test should be conducted after any major modifications to the boiler’s burners or combustion air system.




Figure 8. BGP Boiler F2D O2 In Flue Gas versus Load Curve




Oxygen & CO Trim Control

As we have seen, the %O2 signal itself does not determine optimum combustion. Optimum combustion must be determined from boiler tests measuring BOTH %CO and O2. Note that It is also possible to continually monitor both %CO and O2 in the flue gas. By knowing what %CO best indicates optimal combustion, CO can be cascaded to the %O2 controller to give more precise combustion control. Changes in boiler operation, (e.g. when the boiler combustion system degrades over time and requires more excess air), can be overcome with this system since it can continually adjust %O2 in the flue gas to provide best combustion.

Figure 9 shows that the optimum %CO is not constant with boiler load, and should be characterized against load as is %O2. Figure 10 shows a typical %CO to O2 cascade control scheme.

The CO measurement usually comes from a infrared absorption analyzer directing its IR beam across the flue gas stack. CO absorbs IR radiation at specific frequencies, and the attenuation of those IR frequencies determines the % CO in the flue gas.




ActualO2

versusCO atLoads

Figure 9. O2% vs. CO at Increasing Load




Figure 10. %CO to O2 Cascade Trim Control




CALCULATING BOILER EFFICIENCY BY THE INPUT/OUTPUT METHOD

Boiler Efficiency vs. Excess Air

Figure 11 is a diagram of boiler efficiency versus excess air. Flue gas analysis is used to indicate the air/fuel ratio and the degree of completeness of combustion. The gas components usually measured are CO2, O2, and CO. Figure 11 illustrates that as the amount of excess air increases, the amount of O2 in the flue gas increase.

Figure 11 also illustrates that both the amounts of CO and CO2 in the flue gas decrease as excess air increases. Excess oxygen is the residual quantity remaining in the gases exiting the boiler. As more and more combustion (excess) air is mixed with a fixed rate of fuel, a greater fraction of that air (thus excess oxygen) will remain after combustion is complete.

Carbon monoxide increases sharply at low excess air levels, where incomplete combustion of the fuel occurs, i.e. where carbon does not completely oxidize to carbon dioxide. At higher excess air levels, little CO is generated, and is further diluted by the large volume of air. Carbon dioxide approaches its ultimate or maximum value at stoichiometric conditions, where carbon is fully oxidized with little air remaining. Like carbon monoxide, CO2 also becomes diluted at higher excess air levels.

Figure 11 illustrates the relationship between boiler efficiency and excess air. As excess air increases from 0 to about 7 percent, boiler efficiency increases. As excess air increases beyond 7% excess air, boiler efficiency decreases. This trade-off between unburned fuel and stack heat losses, as a function of excess air, is a relationship that exists at all boiler loads. The control engineer who wishes to maximize boiler performance, must first understand the methods of calculating efficiency and the measurement data necessary for these calculations.




CO

82.5

%83

.0%

83.5

%84

.0%

84.5

%85

.0%

Effi

cien

cy

100

1% 13%

200

300

400

500

600

CO

ppm

2%3%

4%5%

6%O

xyge

n

14%

15%

16%

17%

18%

CO

2

33670

50%40%30%20%10%10% 0%

CO2Eff.

O2

% Excess Air

Figure 11. Boiler Efficiency vs. Excess Air




Boiler Inputs and Outputs

Two acceptable methods are used to calculate boiler efficiency. One method is called the input/output method, and the other method is called the heat loss method. The input/output method is less accurate than the heat loss method but is often preferred, because it is the more simple method of the two.

A simplified form of the basic equation used by the input/output method is illustrated in Equation 7, Boiler Efficiency - Input/Output Method. The equation for the input/output method in Equation 7 differs from the equation for the input/output method show on page 13 of ASME PTC 4.1. The equation in Equation 7 of this module does not include a heat credits term.

Efficiency =

Total Heat OutputTotal Heat Input

((Steam Flow) x (Steam Enthalpy - Feedwater Enthalpy) +Heat Output in Blowdown Water)

(Fuel Flow) x (Fuel Higher Heating Value)

Heat Added to Incoming FeedwaterHeat Added by the Fuel

x 100

Equation 7. Boiler Efficiency - Input/Output Method




Heat Added by the Fuel

Boiler inputs are defined as the heat added by the fuel. To calculate the heat added by the fuel, the flow of the fuel is measured over a period of time and is multiplied by the heat content of the fuel.

The heat content of the fuel, also known as the higher-heat value (HHV), is equal to the amount of heat liberated by the fuel per unit quantity of the fuel. Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7) contains higher-heat values for gas and oil expressed in BTU per ft3 for gas and BTU per pound for oil.

The HHV for fuel oil number 6, which is 150,000 BTU per gallon, is on pages 14 and 15 of Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7).

The HHV for natural gas, which is equal to 1061 BTU per ft3, is on page 36 of Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7).

Heat Added to the Incoming Feedwater

Boiler outputs are defined as the heat added to the steam-water circuit. Equation 7, Boiler Efficiency - Input/Output Method, illustrates the calculation for boiler outputs. Steam flow is multiplied by the difference between steam enthalpy and feedwater enthalpy. Heat added to the incoming feedwater also includes the non-useful heat added to the blowdown flow.

Figure 12 is a drawing from the ASME PTC 4.1 document showing a steam generation unit and all energy inputs and outputs. Figure 13, also from ASME PTC 4.1, is heat balance around a steam generation unit, showing all inputs and outputs.




Figure 12. Boiler Unit Energy Input & Output Diagram




Figure 13. Boiler Unit Heat Balance




First Law of Thermodynamics: Applications to Boiler Efficiency

The First Law of Thermodynamics is based on the conservation of energy. The First Law of Thermodynamics states, "Energy can neither be created nor destroyed." Figure 3-1 on page 37 of The Control of Boilers, 2nd Edition, by Sam. G. Dukelow illustrates boiler steam-water mass balance. The mass of the feedwater is balanced with the mass of the steam plus blowdown. Figure 3-2 of The Control of Boilers, 2nd Edition, by Sam. G. Dukelow illustrates boiler fuel, air-flue gas mass balance. The mass of the fuel gas and the mass of the air are balanced with the mass of the flue gas and the mass of the ash or particulate.

Figure 3-5 on page 39 of The Control of Boilers, 2nd Edition, by Sam G. Dukelow illustrates boiler energy-heat balance. Boiler performance is related to the boiler's ability to transfer heat from the fuel to the water while still meeting operational specifications. Boiler efficiency measures how effectively the boiler converts the energy available in the fuel to energy in the steam and blowdown.




Example Boiler Efficiency Calculation–Input/Output Method

Figures 1 and 2 of ASME PTC 4.1 represent a Steam Generating Unit and Heat Balance of Steam Generator, respectively. Equation 8 is the boiler input-output efficiency calculation. In order to calculate boiler efficiency using the input/output method, steam flow, feedwater flow, blowdown flow, and fuel flow must be known. Steam, feedwater, and blowdown pressure, temperature, or percentage moisture of the steam are necessary to determine heat content. Also the fuel higher heat value must be determined.

The ASME Test Form for Abbreviated Efficiency Test is on pages 16 and 17 of ASME PTC 4.1. Item numbers 1, 2 ,5, 8, 11, 13, 15-17, 20, 21, 26, 28, 37, 40, 41, 43-47, and 64 are necessary for an efficiency calculation using the input/output method. Equation 8 of this module, Total Heat Input Calculation, illustrates that in order to calculate Total Heat Input (item 29 of ASME Test Form), the Rate of Fuel Firing (item 28) must be multiplied by the BTU per Pound as Fired (item 41.)

Total Heat Input (Item 29), KBTU/hr =

Rate of Fuel Firing (Item 28), lb/hr BTU x per lb as Fired (Item 41)1000 BTU/KBTU

Equation 8. Total Heat Input Calculation




Equation 9, Total Heat Output Calculation, sums the heat contributions from Actual Water Evaporated or Main Steam (item 26), Reheat Steam (item 27), and Blowdown Flow (item 30). A reheat cycle generally exists on electric utility boilers, where main steam is partially expanded through a steam turbine then returned to the boiler, where it picks up additional superheat. When this cycle does not exist, as on many industrial boilers, the reheat portion of the calculation can be ignored.

Please refer to the note on Blowdown on page 9 of this module, as it relates to Total Heat Output.

Total Heat Output (Item 31), KBTU/hr =

(Actual Water Evaporated (Item 26), lb/hr x

Absolute Heat per Pound of Steam (Item 20), ) ÷10000 BTUlb

KBTUhr

+ Heat Output in Blowdown Water (Item 30),

Equation 9. Total Heat Output Calculation




Equation 10 shows the calculation for heat absorbed by each pound of steam. This represents, the heat added to each pound of fluid from the inlet conditions of the (compressed liquid) feedwater, to the outlet conditions of the (saturated or superheated vapor) final steam.

=Absolute Heat per Pound of Steam (Item 20),lb

BTU

Enthalpy of Saturated or Superheated Steam (Item 16),lb

BTU

- Enthalpy of Saturated Feed to Boiler (Item 17), lbBTU

Equation 10. Absolute Heat per Pound of Steam Calculation

Equation 11, Heat Output in Boiler Blowdown Water Calculation, measures the heat absorbed by the fluid - from (compressed liquid) inlet feedwater conditions to (saturated liquid) blowdown water conditions. Note the ASME procedure treats blowdown flow, used to minimize water solids, as a valid boiler heat output rather than a loss.

Heat Output in Boiler Blowdown Water (Item 30) KB/hr =

lb of Water Blowdown/hr x

(Enthalpy of Saturated Liquid (Item 15), - BTUlb

BTUlbEnthalpy of Saturated Feed to Boiler (Item 17), )

1000

Equation 11. Heat Output in Boiler Blowdown Water Calculation




Be advised the ASME Short Form erroneously shows the Blowdown Heat in the numerator of Total Heat Output Calculation. Blowdown Heat Flow is already in units of KBTU/Hr, as calculated in item 30, at the top of page 2 on the ASME form. This quantity should not be divided by 1000; rather it is directly added to the Total Heat Output Calculation, as shown in Equation 12.

Total Evap. Ht./lb Reheat Ht. in Ht. inWater Stm. Stm. Flow Reheat Stm. Blowdown

( Item 26 x Item 20) + (Item 27 x Item 21 )1000

Heat = + Item 30

Output

Equation 12. Total Heat Output Calculation




The following example is of a boiler efficiency calculation using the input/output method. The ASME heat loss efficiency calculation requires that percent oxygen and carbon dioxide in the flue gas be measured on a dry basis. The flue gas concentrations appearing in this module's efficiency calculations were determined in a separate procedure as documented in Work Aid 1D (Procedure to Calculate Oxygen and Carbon Dioxide in Flue Gas) of module PCI201.03. Boiler efficiency test data are as follows:

Water and Steam Data Fuel Data Gas

1. Feedwater - 350 degF 1. Flow - 11508 LB/Hr

2. Steam Drum - 1273.9 psig 2. Analysis:

3. Superheater Out - 1250.0 psig a. Carbon 72.03

- 900 degF b. Hydrogen 22.88

- 180.4 MLB/Hr c. Oxygen 1.70

d. Nitrogen 3.39

e. Sulfur -

f. Ash -

g. Moisture -

Total 100.00

h. Heat Value 22,322 BTU/lb

Air and Flue Gas Data

1. Combustion Air - 80 degF

2. Flue Gas - 400 degF

3. Oxygen (dry) - 5.12%




4. Carbon Dioxide - 7.26%

5. Carbon Monoxide - 0%

6. Nigrogen - 85.9%

7. Excess Air - 30.6%

Boiler Data

1. Rated Output - 553.90 MMBTU/Hr

2. No. of Waterwalls - 4

The ASME Test form (Addendum A) is filled out as follows:

Pressure and Temperatures

Item # Description Value

1 Steam Pressure in Boiler Drum 1288.6 psia

2 Steam Pressure at Superheater Outlet 1264.7 psia

5 Steam Temperature at Superheater Outlet 900 degF

8 Water Temperature Entering the Boiler 350 degF

11 Temperature Air for Combustion 80 degF

13 Stack Gas Temperature 400 degF




Items 15 to 17 are computed from the Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7):

Unit Quantities


15 Enthalpy of Saturated Liquid (Total Heat)

(from the steam tables, page 46, for steam

at pressure 1288.6 psia) 584.8 BTU/lb

16 Enthalpy of Saturated, Superheated Steam


at pressure 1264 psia and steam temperature

900 degF) 1438.7 BTU/lb

17 Enthalpy of Saturated Feed to Boiler

(from the steam tables, page 44, for feedwater

at 350 degF) 321.6 BTU/lb




The remaining items are filled out in the ASME Test Form:

Hourly Quantities


26 Actual Water Evaporated 180,400.000 lb/hr 28 Rate of Fuel Firing 11508 lb/hr 41 BTU per Pound as Fired 22322 BTU/lb

Flue Gas Analysis

32 CO2 7.26 % 33 O2 5.12 % 34 CO 0 % 35 N2 85.9 % 36 Excess Air 30.6 %

Oil as Fired Ultimate Analysis

43 Carbon 72.03 % 44 Hydrogen 22.88 % 45 Oxygen 1.70 % 46 Nitrogen 3.39 %




Notes and special considerations in performing the calculations:

Carbon Burned per LB of As Fired Fuel (item 24) - This straightforward calculation relies upon an Ultimate analysis of the fuel to determine carbon content. An analysis of ash samples, for solid fuels, is also required to determine the unburned carbon remaining in the refuse, BTU/LB (item 23). The quantity of dry refuse per LB As Fired Fuel (item 22) is usually determined from the calculation shown near the top of page 2 of the ASME short form (Equation 13).

Item 22 = Ash in as Fired Fuel, %100 - Combustibles in Refuse Sample, %

Equation 13. Dry Refuse per Pound As Fired Fuel Calculation

The value in the numerator, Ash in As Fired Fuel, %, is taken directly from the fuel Ultimate Analysis. The unknown in the denominator, Combustibles in Refuse Sample, %, is calculated using the Unburned Carbon in Refuse quantity, as obtained from the laboratory ash analysis (Equation 14).

Combustibles in Refuse, % = 100% x Unburned Carbon in Refuse, BTU/LBHigh Heating Value of Fuel, BTU/LB

Equation 14. Combustibles in Refuse Calculation




Dry Gas per LB As Fired Fuel Burned (item 25) - This determines the quantity of flue gases generated by each pound of fuel burned. It requires both fuel and flue gas analysis data. For the latter analysis, the percent Oxygen, Carbon Dioxide, Carbon Monoxide, and Nitrogen in the flue gas must be determined volumetrically - on a Dry basis.

A Dry Basis analysis has traditionally been determined via manual sampling of the flue gases - using an Orsat kit. This approach permits the determination O2, CO2, and C0; N2 then being determined by difference, i.e. 100% minus all other constituents.

Be advised, when analyzing flue gas, that many zirconium oxide-type oxygen analyzers measure the flue gas in-situ, on a Wet Basis - yielding an apparently lower oxygen reading. Refer to conversion procedures in the ASME companion document, Power Test Code PTC 19.10 - “Flue and Exhaust Gas Analyses”.

Heat Loss Due to Dry Gas (item 65) - This loss tends to be the largest single loss for most boilers. The calculation uses measured and previously-calculated values, along with a constant for the Specific Heat of Flue Gas, 0.24. If a high degree of accuracy is sought in the overall efficiency calculation, this Specific Heat value may be determined based on fuel carbon/hydrogen ratio, flue gas CO2 content, and flue gas temperature. Refer to the curves on Figure 7, page 66, in ASME PTC 4.1. In general, this Specific Heat value will range from 0.23 to 0.25 BTU/LB-DegF.




Heat Loss Due to Moisture in Fuel (item 66) - This loss is significant for fuels with entrained moisture, such as biomass (wood) and some coals. The calculation requires that additional enthalpy values be determined from the steam tables, namely for the superheated vapor in the flue gas and for the saturated liquid in the incoming combustion air (humidity).

Heat Loss Due to Moisture from the Combustion of Hydrogen (item 67) - This loss is large for natural gas and other fuel gases, since about 9 pounds of moisture are generated for each pound of hydrogen burned. The calculation uses the same enthalpy values as for item 67.

Heat Loss Due to Combustibles in Refuse (item 68) - This loss is significant only for solid fuels with measurable ash content. The calculation is based mainly on the lab analysis of the ash.

Heat Loss Due to Radiation (item 69) - This loss must be determined from the ABMA Radiation Loss Chart (Course Handout 11), included in ASME PTC 4.1 (Figure 8, page 67). The Maximum Continuous Rating (MCR) and Actual Boiler Output, both in MBTU/Hr, must first be calculated.

Next, examine the ABMA chart to determine if an MCR line already exists for this boiler, or if it must be added to the chart. If a new line must be added, locate this boiler’s MCR value on the X-Axis, (e.g. if MCR is 600 MBTU/HR, find this value on the X-Axis). Start the new MCR line at the point where this MCR/Actual Boiler Output value intersects with the curved “Radiation Loss at Max. Cont.. Output” line. Draw the new line up and to the left, keeping it parallel with the existing MCR lines. Extend the line until it reaches the 12% Radiation Loss value, as measured on the Y-Axis.




To determine the radiation loss for this steam load, locate the boiler’s Actual Output on the X-Axis, and draw a vertical line so that it intersects with this boiler’s MCR line. Next, draw a horizontal line from the above intersection point to the Y-Axis. Read the boiler’s Gross Radiation Loss, % from the Y-Axis. This gross loss is multiplied by a water wall or air-cooled wall factor as shown at the origin (lower left of the chart). The appropriate factor is determined by the number of water or air-cooled walls as shown at the upper left portion of the chart, i.e. 0, 1, 2, 3, or 4 water or air-cooled walls.

Enter the net radiation loss value directly in the “% Loss” column of the ASME short form. If desired, the equivalent BTU loss may be determined, by reverse calculation.

Unmeasured Losses (item 70) - Since the ASME Short Form method is not a rigorous treatment of the complete efficiency procedure, minor losses are not calculated. Instead, they are estimated using this “Unmeasured Loss” category. This value is as generally agreed upon by boiler owner and manufacturer, when the boiler is commissioned. It can vary from 0.5% to 2%, but is most often assumed at 1%.

Total (item 72) and Efficiency (item 72) - Sum all losses, items 65 through 70, and list the total as item 71. Heat Loss Efficiency, item 72, is defined as 100% minus the total losses (item 71).




CALCULATE BOILER EFFICIENCY BY THE HEAT LOSS METHOD

A simplified version of the heat loss efficiency calculation equation is shown in Equation 15. Each major loss of heat energy is calculated and totaled. The total heat loss is converted to per cent by dividing the heat losses by the sum of the heat in the fuel and multiplying this quotient by 100. The total heat loss in per cent is subtracted from 100% to equal boiler efficiency. The equation for the heat loss method in Equation 15 differs from the equation for the heat loss method show on page 13 of ASME PTC 4.1. Note: Equation 15, does not include a heat credits term.

Efficiency (Percent) = 100 -Heat Losses, BTU

lb x 100Heat in Fuel, BTU

lb

Equation 15. Boiler Efficiency Calculation - Heat Loss Method

Advantages and Disadvantages of the Heat Loss Method

The heat loss method for boiler efficiency calculation is more accurate than the input/output method because the variation of measurements needed have less effect on the percentage efficiency. A major advantage of the heat loss method is the identification of the losses. As conditions change, and boiler efficiency deteriorates, increased losses can be identified and taken care of. Limitations of the input/output method include the accuracy of flow meters and the measurement accuracy of the heat content of the fuel. Another advantage to using the heat loss method over the input/output method is that, for the input/output method, data on load changes is inaccurate. A disadvantage to the heat loss method is that it is more complicated than the input/output method.




Example Boiler Efficiency Calculation–Heat Loss Method

The following is a list of the major heat losses calculated when determining boiler efficiency using the heat loss method:

• Heat loss due to dry flue gas

• Heat loss due to moisture in the fuel

• Heat loss to water from the combustion of hydrogen

• Heat loss due to the combustibles in refuse

• Heat loss due to radiation

The ASME Test Form for Abbreviated Efficiency Test on pages 16 and 17 of ASME PTC 4.1 (Course Handout 10) contain the formulas necessary for the calculation of boiler efficiency using the heat loss method. All the heat losses are calculated in terms of per cent of the sum of the higher-heat value of the as-fired fuel (items 65-70 of the ASME Test Form.) The heat losses are totaled (item 71) and the total is subtracted from 100% to calculate boiler efficiency (item 72.)

Equation 16 is used to calculate carbon burned per pound of as-fired fuel (item 24.) Carbon burned per pound of as-fired fuel is calculated as follows:

• The remainder of dry refuse per lb A.F. fuel (item 22) times BTU per lb in refuse (item 23) is divided by 14,500.

• The above quotient is subtracted from the quotient of percent carbon (item 43) divided by 100.

= Carbon Burned per LBAs-Fired Fuel (Item 24)

(Dry Refuse perLB As-Fired Fuel,

lb/lb (Item 22))% Carbon(Item 43)

(BTU per LBin Refuse,

lb/lb (Item 23))x

14,500 100

Equation 16. Carbon Burned per lb As-Fired Fuel Calculation




Equation 17 illustrates the equation used to calculate dry gas per pound of as-fired fuel burned (item 25.)

(Item 34)

= Dry Gas per LB As-Fired

Fuel Burned (Item 25)

x LB Carbon Burned per LB As-Fired Fuel +

11 x CO2 + 8 x O2 + 7 + (N2 + CO)

Moisture267

3 x ( CO2 + CO )

(Item 32) (Item 33) (Item 35)

(Item 32) (Item 34)

(Item 37)(Item 24)

Equation 17. Dry Gas per lb As-Fired Fuel Burned Calculation

Equation 18 is used to calculate the heat loss due to dry flue gas (item 65 of ASME Test Form.)

=

Where CP = Specific Heat at Constant Pressure in BTU per LB ºF

x

Temp Air forCombustion

(Item 11) ºF

Heat Loss due to DryFlue Gas (Item 65),

BTULB of As-Fired Fuel

Dry Gas perLB As-FiredFuel Burned

(Item 25)

Flue Gas LeavingBoiler (Item 13) ºF

CPFlueGas

x

Equation 18. Equation for Heat Loss Due to Dry Flue Gas




Heat loss due to dry flue gas (item 65) is calculated as follows:

• The dry gas per lb as fired fuel (item 25) is calculated by dividing the amount of pound of dry flue gas by the amount of pound of fired fuel.

• The value for dry gas per lb as fired fuel is then multiplied by the specific heat at constant pressure (CP expressed in BTU per lb.)

• The above product is then multiplied by the remainder of the temperature (in degF) of the flue gas leaving the boiler (item 13) minus the temperature (in degF) of air combustion (item 11).

Equation 19 is used to calculate the heat loss due to the moisture in the fuel (item 66 of ASME Test Form.)

=

Enthalpy of Vapor at 1 PSIA &Temperature of Gas Leaving (Item 13*),

x

Heat Loss due to H2

(Item 67) from Combustion of H2

LB of As-Fired Fuel

LB H2O

LB of As-Fired Fuel(Item 37)

100

* Thes items come from Steam Tables from Combustion Engineering Fuel Burning and Steam Generation Handbook

BTULB

Enthalpy of Liquid Water at TemperatureAir Inlet Temperature (Item 11*),

BTULB

BTU

Equation 19. Equation for Heat Loss Due to Moisture in Fuel




Heat loss due to the moisture in the fuel is calculated by multiplying amount of pound of water per pound of as-fired fuel (item 37/100) by the difference between the enthalpy of vapor at the temperature of the air leaving the boiler (item 13) and the enthalpy of liquid at the temperature of air for combustion (item 11).

Equation 20 is used to calculate the heat loss due to water from the combustion of hydrogen (item 67 of ASME Test Form.)

= 9 x H2 (Item 44) x

Enthalpy of Vapor at 1 PSIA & Temperature (Item 13),

Heat Loss due to H2

BTU

from Combustion of H2

LB of As-Fired Fuel

BTULB

Enthalpy of Liquid Water at Temperature (Item 11),BTULB

(Item 67)

Equation 20. Equation for Heat Loss Due to H2O from Combustion of H2

Heat loss due to water from the combustion of hydrogen is calculated by multiplying 9 times the per cent hydrogen in the fuel (item 44) by the difference between the enthalpy of vapor at the temperature of the air leaving the boiler (item 13) and the enthalpy of liquid at the temperature of air for combustion (item 11).




Equation 21 is used to calculate the heat loss due to combustible in the refuse (item 68 of ASME Test Form.)

= Dry Refuse per LB As-Fired Fuel (Item 22),

x BTU per LB in refuse (Item 23),

Heat Loss due to

(Item 68)Combustible in Refuse

LB of As-Fired FuelLBLB

BTULB

BTU

Equation 21. Equation for Heat Loss Due to Combustible in Refuse

Heat loss due to combustible in the refuse is calculated by multiplying the amount of dry refuse per lb of A.F. fuel (item 22) by the BTU per lb in refuse (item 23).

All heat losses (items 65-69) are converted to per cent of as-fired fuel by dividing the heat loss (BTU/LB of A.F. fuel) by the fuel higher heat value. Heat losses are totaled (item 71) and subtracted from 100 to calculate boiler efficiency (item 72).

Radiation heat losses cannot be measured directly and are computed using the ABMA (American Boiler Manufacturer's Association) Standard Radiation Loss Chart (attached.) This chart uses the inputs of the boiler MCR (i.e. Maximum Continuous Rating), actual boiler heat output, and number of water-cooled furnace walls in determining the radiation loss.

The lines are drawn on the chart representing MCR and actual boiler heat output to determine % radiation heat loss. The percentage radiation loss is multiplied by the waterwall factor also found on the ABMA Standard Radiation Loss Chart.




The following example is of a boiler efficiency calculation using the heat loss method. The ASME Test Form for Abbreviated Efficiency Test is on pages 16 and 17 of ASME PTC 4.1 (Addendum B). Boiler efficiency test data are the same as the data were used in the previous example, Example Boiler Efficiency Calculation–Input/Output Method.



2. Steam Drum - 1273.9 psig 2. Analysis:

3. Superheater Out - 1250.0 psig a. Carbon 72.03

- 900 degF b. Hydrogen 22.88

- 180.4 MLB/Hr c. Oxygen 1.70

d. Nitrogen 3.39

e. Sulfur -

f. Ash -

g. Moisture -

Total 100.00



1. Combustion Air - 80 degF

2. Flue Gas - 400 degF

3. Oxygen (dry) - 5.12%





5. Carbon Monoxide - 0%

6. Nigrogen - 85.9%

7. Excess Air - 30.6%

Boiler Data

1. Rated Output - 553.90 MMBTU/Hr


3. MCR - 385 Mlb/Hr

4. Assume unmeasured losses = 366.1 BTU/lb (As agreed upon by parties to the test)

The ASME Test form (Addendum A) is filled out as follows:

Pressure and Temperatures


1 Steam Pressure in Boiler Drum 1288.6 psia

2 Steam Pressure at Superheater Outlet 1264.7 psia

5 Steam Temperature at Superheater Outlet 900 degF

8 Water Temperature Entering the Boiler 350 degF

11 Temperature Air for Combustion 80 degF

13 Stack Gas Temperature 400 degF




Items 15 to 17 are computed from the Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7):

Unit Quantities


15 Enthalpy of Saturated Liquid (Total Heat)


at pressure 1288.6 psia) 584.8 BTU/lb

16 Enthalpy of Saturated, Superheated Steam


at pressure 1264 psia and steam temperature

900 degF) 1438.7 BTU/lb

17 Enthalpy of Saturated Feed to Boiler

(from the steam tables, page 44, for feedwater

at 350 degF) 321.6 BTU/lb

The remaining items are filled out in the ASME Test Form:




Hourly Quantities


26 Actual Water Evaporated 180,400.000 lb/hr

28 Rate of Fuel Firing 11508 lb/hr

41 BTU per Pound as Fired 22322 BTU/lb

Flue Gas Analysis

32 CO2 7.26 %

33 O2 5.12 %

34 CO 0 %

35 N2 85.9 %

36 Excess Air 30.6 %

Oil as Fired Ultimate Analysis

43 Carbon 72.03 %

44 Hydrogen 22.88 %

45 Oxygen 1.70 %

46 Nitrogen 3.39 %




OPTIMUM DISPATCHING OF MULTIPLE BOILERS

The purpose of the optimum dispatching of multiple boilers is to deliver steam at the lowest operating cost.

Economic Load Allocation

New methods are often searched for to reduce overall steam production costs. One way to dramatically reduce costs is to intelligently allocate steam loads to boilers in multiple boiler powerhouses. Economic load allocation is the minimization of cost by proper allocation of steam demand to a set of boilers.

Boiler operators have no trouble allocating steam production the following situation.

• Boiler No. 1 always operates at 87% efficiency.

• Boiler No. 2 always operates at 86% efficiency.

The solution is, it would seem, obvious. Maximize the steam production from Boiler No. 1 and let Boiler No. 2 make up the difference.

Unfortunately, in most applications the solution is not this simple. The relationship between boiler efficiency and steam load is decidedly nonlinear. The solution of "loading up on the most efficient boiler" may not always be the correct solution. This approach may force the other boiler into an inefficient operating range, increasing overall steam costs.

The varied effects and costs of different fuels complicate matters further, making it very difficult to find a complete and timely optimum dispatching solution (before another load change occurs).




The solution to steam production problems must address more than the performance of each boiler as it operates over its load range. It must also include how the loading of each boiler affects the operation of the other boilers. (See Figure 14.)

BoilerEfficiencies ImplementSolutionLimits and

Constraints

MinimizeOverall

OperatingCost

FuelCost

BoilerStatus

Advise

Boiler 1 Boiler 2 Boiler 3

$/lb $/lb $/lbSteam toProcess -Avg $/lb

FuelAir

FuelAir

FuelAir

Figure 14. Economic Load Allocation




Methods

A number of methods for optimally distributing boiler loads have been developed over the years. These methods include iterative searches, model-based solutions using both linear and non-linear optimization algorithms, and incremental cost determination. Each method has distinct advantages and disadvantages regarding their ease of implementation, ability to handle large applications, execution time, and solution accuracy.

One common characteristic of all optimization methods is the need to mathematically represent or model the boiler’s performance, over its load range. This is necessary so that each boiler can be evaluated - not only at its present load, but at all other potential loads. Figure 15 shows a typical boiler operating line.

100

200

300

400

500

600

700

0

Cos

t, $/

Hr

20 40

Steam Flow %

60 80 100

Figure 15. Typical Boiler Operating Line




This would be represented mathematically with a polynomial function such as:

f(X) = A * X + B

Where: X = Steam Flow, %

A, B = Constants

f(X) = Operating Cost, $/Hr

The polynomial function might be of a higher order, i.e. with second and third order terms, depending on the required accuracy of the model.

Iterative Solutions As its name implies, this method solves the optimization problem (finds the best solution) by repetitively computing all possible solutions. The solution set includes all boiler load combinations that will satisfy the plant steam demand, while maintaining boilers within their respective valid load range.

This approach is probably the simplest to implement. A logic algorithm determines the various combinations of steam loads that will be considered, given the present demand. The appropriate steam load combinations are then evaluated, using a suitable load increment - typically 1000 Lb/Hr of steam flow. The operating line (operating cost versus steam load) curves for each boiler are repeatedly accessed during the search.

Iteration also can produce accurate solutions, since the operating line curve can be any math function necessary to model or represent the boiler.

This optimization approach quickly becomes impractical, though, as the number of boilers increases. With each new boiler, the number of iterations increases exponentially. When four or more boilers are being dispatched, the time required to find an optimum solution may become prohibitive, depending upon the system being used to perform the computations.




Model-Based Linear Program (LP) Solutions

Model-based solutions make use of a linear program (LP) algorithm, typically based on the Simplex technique. A mathematical model of the steam process is developed, considering individual boiler constraints, interactions, and balances.

The steam plant model, current data, and objective function (to minimize operating cost) are developed as a series of equations. These equations are arranged in a prescribed format and the coefficients of all the equation terms are placed in a 2 dimensional matrix. This matrix is then presented to the optimizer algorithm for solutions.

It should be noted that “Linear” in the name Linear Programming indicates that only linear equations are involved. This is unfortunate, since boiler operating costs cannot be accurately represented by a linear or straight-line function. Figure 16 demonstrates the inadequacy of a straight line fit of a boiler’s performance versus load data.

100806040200

700

600

500

400

300

200

100

Steam Flow %

Cos

t, $/

Hr.

Figure 16. Linear Boiler Operating Line




One alternative is to model the boilers with a series of straight line segments, usually three or more, instead of just one. This method of segmentation, shown graphically in Figure 17, will allow for a more accurate representation of the boiler’s performance. Segmentation complicates the modeling effort, though, as many more equations must be introduced and considered by the optimization algorithm - requiring additional computation time.

100806040200

700

600

500

400

300

200

100

Steam Flow %

Cos

t, $/

Hr.

Figure 17. Segmented Linear Boiler Operating Line

Consideration must be given to how the boiler’s performance vs. load data points are used to form multiple line segments. Experience has shown that an optimizer will often arrive at a solution which places a boiler precisely at the junction of two adjacent line segments.




When operating line segmentation is employed and/or additional boilers are to be optimized, the initial matrix size increases accordingly. The solution of large matrices requires significantly more algebraic manipulation - increasing LP execution time and raising concerns over rounding errors.

Model-Based Nonlinear Program Solutions

When approaching nonlinear optimization problems, numerous solution methods have been employed. These search techniques are subdivided in many ways: discrete search versus continuous search; nonsequential search versus sequential search; local search versus global search; search with quadratic convergence versus search without quadratic convergence; etc.

This class of optimization allows for nonlinear modeling of the boilers but takes longer to execute and may, in some instances, identify false optimums. For instance, one search technique might be faster than another, but might erroneously converge on a local maxima (peak) or minima (valley) as an optimum point, as illustrated in Figure 18.

Optimum

LocalMaxima

Figure 18. Local versus Overall Optimum




Employing the appropriate nonlinear solution technique is important for another reason. The concern here is not just to solve the problem, but to solve it efficiently. In the interest of finding solutions in something close to real-time, it is often necessary to direct the optimizer algorithm to use a smaller number of trials and a larger step size. This type of optimizer, when used on-line, typically requires a sophisticated computing platform and a highly-trained system administrator.

Incremental Cost Determination

It has been proven, using Calculus, that two or more boilers are optimally loaded when their incremental operating costs are made equal. For a boiler, incremental cost is represented by the slope of the operating line - such as we have already examined.

Figure 19 illustrates the relationship between boiler load versus fuel cost for Boiler 1 and Boiler 2. The question to ask is, "How do we split the load between the two boilers at low cost of fuel?"

100806040200

600

500

400

300

200

100

Boiler Load (KLB/HR)

Fuel

Cos

t ($

/HR

)

BLR 1Cost = 5 x Load + 100

BLR 2Cost = 5 x Load + 0

Figure 19. Economic Load Allocation




The data from Figure 19 reveals that no matter how the load is split between the boilers, the total cost to operate the boilers is the same:

Boiler 1 Boiler 2 Total Cost

20 K LB/HR 80 K LB/HR $600/HR



Because the slopes of the lines that represent boiler load vs. fuel cost for boiler 1 and boiler 2 are equal, the total cost to operate the boilers is the same for a given load. If the boilers can be set so that the slopes of the lines that represent boiler load vs. fuel cost are equal, the math theory says that the total fuel cost for a given load will be the same no matter how the load is allocated between the boilers.

Figure 20 illustrates the relationship between boiler load versus fuel cost for a second set of boilers.

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500

400

300

200

100

Steam Flow (KLB/HR)

Boi

ler

Ope

ratin

g C

ost

($/H

R)

Boiler 1Cost = 5 x Steam Flow

Boiler 2Cost = 3 x Steam Flow + 100

Figure 20. Boiler Load vs Fuel Cost




The data from Figure 20 reveal that the total fuel cost increases as we shift the larger share of the load from boiler 2 to boiler 1.

Boiler 1 Boiler 2 Total Cost




The slope of the line that represents boiler load vs. fuel cost for boiler 1 is steeper than that of boiler 2. The solution is to make the maximum use of boiler 2 to decrease operating cost.




WORK AID 1: RESOURCES REQUIRED TO CALCULATE BOILER EFFICIENCY BY THE INPUT/OUTPUT METHOD FOR A GIVEN SET OF BOILER OPERATING CONDITIONS

Work Aid 1A: Example of Boiler Operating Parameters



2. Water Pressure - 160.3 psig 2. Analysis:

3A. Water Temp at S.H. Out - 356 degF a. Carbon 69.26

3B. Output Flow - 180.4 MLB/Hr b. Hydrogen 22.68

4. Blowdown - 0.0 MLB/Hr c. Oxygen -

d. Nitrogen 8.06

e. Sulfur -

f. Ash -

g. Moisture -

Total 100%



1. Combustion Air - 48.4 degF

2. Flue Gas - 302.4 degF

3. Oxygen (dry) - 8.7%





Boiler Data

1. Rated Output - 50 MBTU/Hr


Work Aid 1B: ASME PTC 4.1

American Society of Mechanical Engineers Power Test Code 4.1 (Course Handout 10)

Work Aid 1C: Combustion Engineering Fuel Burning and Steam Generation Handbook

Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7).




WORK AID 2: RESOURCES REQUIRED TO CALCULATE BOILER EFFICIENCY BY THE HEAT LOSS METHOD FOR A GIVEN SET OF BOILER OPERATING CONDITIONS

Work Aid 2A: Example of Boiler Operating Parameters



2. Water Pressure - 160.3 psig 2. Analysis:

3A. Water Temp at S.H. Out - 356 degF a. Carbon 69.26

3B. Output Flow - 180.4 MLB/Hr b. Hydrogen 22.68

4. Blowdown - 0.0 MLB/Hr c. Oxygen -

d. Nitrogen 8.06

e. Sulfur -

f. Ash -

g. Moisture -

Total 100%



1. Combustion Air - 48.4 degF

2. Flue Gas - 302.4 degF




3. Oxygen (dry) - 8.7%


Boiler Data

1. Rated Output - 50 MBTU/Hr


Work Aid 2B: ASME PTC 4.1

American Society of Mechanical Engineers Power Test Code 4.1 (Course Handout 10)

Work Aid 2C: Combustion Engineering Fuel Burning and Steam Generation Handbook

Combustion Engineering Fuel Burning and Steam Generation Handbook (Course Handout 7).




WORK AID 3. RESOURCES REQUIRED TO CALCULATE TOTAL REQUIRED AIR FLOW TO MEET STOICHIOMETRIC CONDITIONS AND EXCESS AIR REQUIREMENTS FOR A GIVEN FUEL OF KNOWN COMPOSITION

Work Aid 3A: Combustion Engineering Fuel Burning and Steam Generation Handbook

Refer to The Combustion Engineering Fuel Burning and Steam Generation Handbook for fuel analysis data.

Work Aid 3B: Procedure to Calculate Excess Air

Excess air equals excess oxygen. Excess air equals the %O2 by volume in the flue gas divided by the total air required for stoichiometric combustion. The procedure to calculate excess air:

1. Obtain flue gas analyses CO2, CO, O2, N2.

2. From the percent N2, calculate the total O2 into the furnace.

3. Reduce the free O2 by the amount required to burn the CO to CO2. The remaining free O2 is excess. (CO is usually negligible)

4. O2 required = (total in) less (excess)

( )( )

( ) x100excess-total

excess=x100O requiredO excess

=O excess Percent 5.2

22




GLOSSARY

ASME PTC American Society of Mechanical Engineers Power Test Codes.

A.F. As-fired fuel. Fuel in the condition as it is fed to the fuel burning equipment.

absolute humidity The weight of water vapor in a gas water-vapor mixture per unit volume of space occupied.

blowdown Removal of a portion of boiler water to reduce chemical concentration, or to discharge sludge.

CO Carbon Monoxide.

CO2 Carbon Dioxide.

economic load allocation The minimization of cost by the proper allocation of steam demand to a set of boilers.

efficiency The ratio of the output to the input. The efficiency of a steam generating unit is the ratio of the heat absorbed by water and steam to the heat in the fuel fired.

enthalpy The amount of heat energy that is contained in a fluid or gas in BTU/lb.

H2 Hydrogen.

higher-heat value Amount of heat liberated by the fuel per unit quantity of the fuel.

O2 Oxygen.




relative humidity The ratio of the weight of water vapor present in a unit volume of gas to the maximum possible weight of water vapor in unit volume of the same gas at the same temperature and pressure.

sensible heat losses Dry gas losses at (Temperature of gas exiting boiler) - Ambient Temperature.




ADDENDUM

ADDENDUM TABLE OF CONTENTS

Addendum A: ASME Test Form for Example Boiler Efficiency Calculation - Input/Output Method

Addendum B: ASME Test Form for Example Boiler Efficiency Calculation - Heat Loss Method

Addendum C: ASME Test Form for Evaluation Boiler Efficiency Calculation - Boiler 1

Addendum D: ASME Test Form for Evaluation Boiler Efficiency Calculation - Boiler 2




Addendum A: ASME Test Form for Example Boiler Efficiency Calculation - Input/Output Method




Addendum B: ASME Test Form for Example Boiler Efficiency Calculation - Heat Loss Method




Addendum C: ASME Test Form for Evaluation Boiler Efficiency Calculation - Boiler 1




Addendum D: ASME Test Form for Evaluation Boiler Efficiency Calculation - Boiler 2

boiler combustion theory and efficiency

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