Boiler Efficiency – Stack Losses for No.2 Oil
Calculating stack losses for boilers using Regular No. 2 Fuel Oil involves six steps and is based on the following:
Ultimate Analysis values for Regular No. 2 Fuel Oil‚ lb/lb fuel
Higher heating value (HHV) 19‚480 Btu/lb
- Loss due to CO in the flue gas is negligible. CO concentrations in the flue gas should normally be under 50 ppm by volume. A concentration of about 350 ppm CO is on the brink of violation of CCME emissions guidelines, and represents a loss of efficiency of about 0.035 % of fuel input.
- Loss due to unburned hydrocarbons in the flue gas is negligible. These should normally be under 50 ppm by volume. 100 ppm would represent a loss of efficiency of about 0.03 %of fuel input.
Step 1. Obtain and convert measurements
Obtain the maximum capacity rating (MCR) of the boiler unit. This is nameplate information.
- For steam boilers the units are lb/h
- For high temperature water generators the units are millions of Btu/h. 1 lb of steam = 1000 Btu is an approximation sufficiently accurate for this analysis.
- Flue gas temperature (FGT)‚ °F or °C
- Combustion air temperature (CAT)‚ °F or °C
- O2 in flue gas‚ % by volume
It is important that measurements of flue gas temperature and composition be representative. There may be stratification in the gas duct‚ particularly at low loads. Sampling at a few different locations in the gas duct is desirable.
Convert to °F: °F = [(°C x 1.8) + 32] if FGT and CAT are measured in °C.
Permanently installed (on-line) flue gas analyzers almost always have a moisture trap to condense water vapor and remove it from the sample upstream from the analyzer cell. The sample is therefore analyzed on a dry gas basis. Sometimes portable flue gas analyzers are used to obtain a spot measurement without employing a moisture trap. Then the sample is analyzed on a wet gas basis‚ and the result must be converted to the dry gas basis.
Convert to a dry gas basis using the following formula for Regular No. 2 Fuel Oil:
O2dry = O2wet x (1.139 - 0.007O2wet )‚ %
Obtain the CO2 ‚ % by volume‚ dry gas basis. When firing any given fuel the amount of CO2 in the flue gas is related to the amount of O2 in the flue gas. For regular No. 2 Fuel Oil‚ the relationship is defined by the following equation:
CO2dry ‚ % by volume‚ = 15.308-(0.732xO2dry )
Some portable analyzers will print out a value for CO2 in the flue gas‚ perhaps based on a calculation using the oxygen measurement. It is important to ensure that the O2 and CO2 values used in subsequent calculations are dry-basis values.
Determine the actual unit output at the time of test. This is usually available from the control panel display. It may be displayed in lb/h of steam or millions of Btu/h. Convert it to a percentage of the maximum capacity rating.
Step 2. Calculate dry flue gas (DG)‚ lb/lb fuel
For regular No. 2 Fuel Oil containing 0.2 % sulphur and fired with negligible CO and hydrocarbons, the ASME formula for DG simplifies to:
DG‚ lb/lb fuel‚ = [( 11CO2dry + 8O2dry + 7N2 ) / 3CO2dry] x (C + 0.375S )
|Where:||O2dry from Step 1|
|CO2dry from Step 1|
|N2 = 100 - O2dry - CO2dry|
|C in fuel, = 0.865, from ultimate analysis|
|S in fuel = 0.002, from ultimate analysis|
Step 3. Calculate excess air
The value of excess air is not needed for calculation of stack losses, but it is the most common parameter of burner performance, so it is useful to calculate it for the likely range of O2 in the flue gas.
The ASME formula for excess air, (EA), % by volume, when CO is negligible, is:
EA, % by volume, = ( 100 x O2dry ) / ( 0.2682N2 - O2dry )
For No. 2 fuel oil of the given composition, the table below gives the relationship of excess air to flue gas composition.
|O2dry‚ %||CO2dry ‚ %||N2dry ‚ %||Excess air‚ %|
Step 4. Calculate dry flue gas loss (LDG )‚ % of fuel input
The ASME formula for LDG‚ % of fuel input‚ is:
LDG‚ % of fuel input‚ = [DG x Cp x (FGT - CAT)] x (100 / HHV)
|Where:||DG‚ from Step 2|
|Cp ‚ the specific heat of flue gas‚ = 0.24|
|FGT and CAT‚ °F‚ from Step 1|
|HHV‚ from ultimate analysis‚ = 19‚480 Btu/lb|
Therefore‚ for No. 2 fuel oil‚ LDG‚ % of fuel input‚ = 0.00123 x DG x (FGT - CAT)
Commentary on dry flue gas loss
The dry flue gas loss (LDG ) can be reduced by reducing the amount of dry flue gas (DG) and by reducing the difference between flue gas temperature (FGT) and combustion air temperature (CAT).
DG can be reduced by reducing the excess air‚ but only up to a point; insufficient excess air leads to increased emissions of CO and hydrocarbons‚ even a risk of furnace explosion. The lower limits of excess air are determined by the ability of the burner to mix the fuel and air‚ and by the accuracy of the burner controls in metering the two fluids. Good burners with good controls should be able to operate with about 10 % excess air at full load‚ perhaps 30 % excess air at 20 % of full load.
Reducing excess air has a moderate effect on reducing the value of DG. This can be explored by substituting different values for CO2 ‚ O2 and N2 into the formula for DG. It is necessary to maintain the correct relationship in composition‚ so use the values from the table in Step 3‚ or interpolate between them.
While reducing excess air only moderately reduces DG‚ the effect on reducing LDG is greater‚ because reducing DG also reduces the flue gas temperature (FGT). The reduced volume of flue gas passing over the same heat exchange surface results in a lower end temperature‚ but the reduction in FGT is not easy to predict‚ and is best determined by experiment.
FGT can also be reduced by exposing more heat exchange surface to the flue gas‚ such as an economizer‚ which heats the feedwater‚ or an air heater which heats the combustion air. One can assume a lower value for FGT‚ calculate its effect on LDG ‚ and then determine whether the improvement in efficiency‚ and corresponding fuel savings‚ justify the cost of the additional heat exchanger.
It should be noted that when an air heater is employed‚ the combustion air temperature (CAT) is measured where the air enters the air heater‚ not at the burners.
When adding heat exchange surface it is important that the final FGT be kept above the dewpoint, to avoid condensation, unless the heat exchanger is designed to accommodate it. Condensing flue heat exchangers can offer substantial gains in efficiency, but are beyond the scope of this discussion.
When fuels containing sulphur are fired‚ small amounts of sulphuric acid may form‚ which raise the temperature at which condensation occurs‚ compared to the dewpoint that would prevail if only water vapor were present. If flue gas temperature is reduced‚ even locally‚ to this "acid dewpoint"‚ corrosion of metal heat exchangers can be expected. Thus‚ if the flue gas temperature is above the acid dewpoint but some parts of the heat exchanger are not‚ acid is likely to condense. The factors affecting acid dewpoint are beyond the scope of this discussion‚ but operating with minimal excess air is helpful. For fuel oil with 0.2 % sulphur‚ about 200 °F might be a minimum safe flue gas temperature.
Step 5. Calculate the latent heat loss (LH )‚ % of fuel input
The ASME formula for LH‚ % of fuel input is:
LH‚ % of fuel input = 9H2 x [(enthalpy of vapor at 1 psia & FGT) - (enthalpy of liquid at CAT)] x (100 / HHV)
- H2 in fuel‚ = 0.133‚ from ultimate analysis
- Enthalpy of vapor at 1 psia & FGT (hg‚ Btu/lb)
- Enthalpy of liquid at CAT (hf ‚ Btu/lb)
- FGT and CAT‚ °F‚ from Step 1
- HHV‚ from ultimate analysis‚ = 19‚480 Btu/lb
hg can be calculated as follows: hg ‚ Btu/lb = 1055 + (0.467 x FGT)
hf can be calculated as follows: hf ‚ Btu/lb = CAT - 32
Therefore‚ for No. 2 fuel oil‚ the ASME formula simplifies to:
LH ‚ % of fuel input = 0.00614 x (hg - hf )
Commentary on loss due to moisture from combustion of hydrogen
In the process of combustion the hydrogen content of the fuel is converted to H2O‚ which normally leaves the stack as water vapor‚ carrying with it the heat required to convert it from liquid to vapor. With fuel of the given composition LH is likely to be about 7 to 8 % of fuel input.
Much of this loss can be recovered by employing condensing flue heat exchangers‚ but as already stated‚ these are outside the scope of this discussion.
In conventional systems‚ minimizing LH is a matter of reducing FGT but staying above the dewpoint to avoid condensation. To avoid heat exchanger corrosion when firing fuels containing sulphur it is important to keep those parts of the heat exchangers exposed to flue gas above the acid dewpoint. As already stated, for the fuel in question this might be about 200 °F
Step 6. Calculate stack losses (LS )‚ % of fuel input
LS ‚ % of fuel input = LDG + LH
|Where:||LDG from Step 4|
|LH from Step 5|
Commentary on stack losses
Stack losses can be minimized by reducing excess air‚ which reduces the quantity of flue gas that is heated to exhaust temperature (FGT)‚ and by reducing the exhaust temperature. The latter can be accomplished by adding heat exchange surface‚ such as economizers and air heaters‚ within the constraints already discussed. The former can be accomplished by improving burner performance‚ through burner design or precise combustion controls. In heating plants with more than one generating unit‚ gains can also be made by adjusting the number of units in operation so those operating are close to peak efficiency‚ which is usually at about 75 % of capacity rating.
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