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-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 LB
Equation 17 illustrates the equation used to calculate dry gas per pound of as-fired fuel burned (item 25.)
(Item 34)
(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 for Combustion
(Item 11) ºF Heat Loss due to Dry
Flue Gas (Item 65),
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)
* Thes items come from Steam Tables from Combustion Engineering Fuel Burning and Steam Generation Handbook
BTU LB
Enthalpy of Liquid Water at Temperature Air Inlet Temperature (Item 11*),
BTU LB BTU
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
BTU LB
Enthalpy of Liquid Water at Temperature (Item 11), BTU LB
(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),
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
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.
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
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
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
Items 15 to 17 are computed from the Combustion Engineering Fuel Burning and Steam Generation Handbook (Course
Handout 7):
Unit Quantities
Item # Description Value 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 (from the steam tables, page 43, for 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
Item # Description Value
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 %
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.)
Boiler
Efficiencies Limits and Solution Implement
Constraints
Boiler 1 Boiler 2 Boiler 3
$/lb $/lb $/lb Steam to
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.
Figure 15. Typical Boiler Operating Line
This would be represented mathematically with a polynomial
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
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.
100
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.
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
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
Local Maxima
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?"
100
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
50 K LB/HR 50 K LB/HR $600/HR
80 K LB/HR 20 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.
100
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
20 K LB/HR 80 K LB/HR $440/HR
50 K LB/HR 50 K LB/HR $500/HR
80 K LB/HR 20 K LB/HR $560/HR
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
Water and Steam Data Fuel Data Gas
1. Feedwater - 206 degF 1. Flow - 2756 LB/Hr
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%
h. Heat Value 22,658 BTU/lb
Air and Flue Gas Data
1. Combustion Air - 48.4 degF
2. Flue Gas - 302.4 degF
4. Carbon Dioxide - 7.26%
Boiler Data
1. Rated Output - 50 MBTU/Hr 2. No. of Waterwalls - 2
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
Water and Steam Data Fuel Data Gas
1. Feedwater - 206 degF 1. Flow - 2756 LB/Hr
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%
h. Heat Value 22,658 BTU/lb
Air and Flue Gas Data
1. Combustion Air - 48.4 degF
2. Flue Gas - 302.4 degF
3. Oxygen (dry) - 8.7%
4. Carbon Dioxide - 7.26%
Boiler Data
1. Rated Output - 50 MBTU/Hr 2. No. of Waterwalls - 2
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)
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
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
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