Calculation

In regions using imperial units, fuel gas is often measured in cubic feet per hour (ft³/h). To determine the thermal power in kW, we first convert ft³/h to m³/h, because the calorific value we use here is in MJ/m³. The approximate conversion factor is:

1 ft³ ≈ 0.0283168 m³
                  

Hence,

Q(m³/h) = Q(ft³/h) × 0.0283168
                  

Once we have the flow in m³/h, the same formula from the previous calculation applies:

Power (kW) = [ Q(m³/h) × Efficiency × CV(MJ/m³) ] / 3.6
                  

This approach assumes you know or have an estimate of the calorific value in MJ/m³ for the type of gas, along with an efficiency factor representing how effectively the burner or boiler converts fuel energy into useful heat.

Example (Step by Step)

Suppose you have a flow of 1765 ft³/h. First, convert ft³/h to m³/h:

1765 ft³/h × 0.0283168 = 62.34 m³/h
                  

If the combustion efficiency is 0.94 and the calorific value is 39 MJ/m³, we calculate:

Power = (62.34 × 0.94 × 39) / 3.6
      ≈ 643 kW
                  

We arrive at roughly 643 kW of power, which is nearly identical to the result we would have gotten had we started directly with 62.34 m³/h. This example shows that the primary difference is simply the unit conversion from ft³/h to m³/h before applying the standard formula.

In practical terms, if your gas meter logs usage in ft³, you can still estimate your system’s heat release or required burner size by converting to SI units.

Calculation

If you have a target or required **output** (in kW), you can determine how much **input** power the burner must supply by considering the system efficiency. Efficiency here represents the fraction of the heat input that is effectively used as output. In formula form:

Burner Heat Input (kW) = Output (kW) / Efficiency
                  

Since no real-world system is 100% efficient, you always need more thermal energy input from the burner than the net heat output to the process or fluid (water, steam, air heating, etc.).

For instance, if you want 100 kW of usable heat (output) and the system efficiency is 85% (0.85), you would need:

Input = 100 / 0.85 ≈ 117.65 kW
                  

The difference between 117.65 kW and the 100 kW output is lost mostly as flue gas heat, radiant heat, and other inefficiencies in the system.

Example

For a common scenario: You have a hot water boiler that needs to deliver 200 kW of heat to the water. The boiler manufacturer or your on-site testing suggests that the boiler’s overall efficiency is about 90%.

Burner Input = 200 kW / 0.90 = 222.22 kW
                  

So, the burner must supply ~222 kW of fuel power (based on the fuel’s calorific value) to achieve the 200 kW net output.

In practice, checking actual flue gas temperature, oxygen levels, and other factors can help confirm if the real efficiency matches the theoretical value.

Calculation

When using oil (e.g., Diesel, Kerosene) as the fuel, you typically select a nozzle size (expressed in GPH—gallons per hour, or L/h) to match the required heat input. The fundamental steps are:

1. Determine the required fuel heat input based on the boiler's desired output and its efficiency:

Required Fuel Heat (kW) = Boiler Output (kW) / Boiler Efficiency
                  

2. Convert that required fuel heat from kW to mass flow (kg/h) or volume flow (L/h) using the oil’s calorific value (CV).
3. Match the resulting flow rate to nozzle manufacturer charts (e.g., for Riello, Delavan, Danfoss nozzles). These charts indicate the nozzle GPH or L/h based on the pump pressure you select (e.g., 7 bar, 10 bar, 12 bar).

If you need two nozzles (e.g., for a wide turndown range), you calculate the total capacity for maximum firing and a smaller capacity for lower turndown operation.

Example

Suppose you want a boiler output of 250 kW at 91% efficiency, and the chosen oil has a calorific value of 11.8 MJ/kg (typical diesel).

Step 1: Required fuel heat input
  = 250 / 0.91
  ≈ 274.73 kW (fuel side)

Step 2: Convert kW to MJ/h
  274.73 kW = 274.73 kJ/s
           = 274.73 × 3600 kJ/h
           = 989,028 kJ/h
  That is ≈ 989 MJ/h.

Step 3: Mass flow of oil
  If CV = 11.8 MJ/kg,
  Mass flow = 989 MJ/h ÷ 11.8 MJ/kg
            ≈ 83.9 kg/h

Step 4: Using an oil density of ~0.84 kg/L (diesel),
  Volume flow ≈ 83.9 / 0.84 ≈ 99.9 L/h

Step 5: Match nozzle
  Look at nozzle charts: you pick a nozzle (or multiple) 
  that can deliver ~100 L/h at your chosen pump pressure 
  (e.g., 10 bar). If two nozzles are needed, split the total 
  capacity accordingly (e.g., one nozzle at 30 L/h, another at 70 L/h).
                  

Remember that pump pressure changes the flow through the nozzle. Each manufacturer provides nominal flow at a reference pressure (often 7 bar or 10 bar). Adjust accordingly with Flow ∝ √Pressure relationships if not using official charts.

Calculation

Liquefied Petroleum Gas (LPG) can be measured in liters, kilograms, or occasionally in cubic meters of liquid. Some engineers or technicians express consumption in m³/h (liquid volume flow). The energy content of LPG in liquid form is quite high. Roughly, 1 m³ of liquid LPG is about 540 liters, but note that the exact factor can depend on composition (propane vs. butane mix).

A very rough rule of thumb for propane is around 25 kW per liter/hour of liquid consumption, although a more accurate approach is to use the actual calorific value in MJ/kg or MJ/l. If you only have an approximate consumption in m³/h, you can multiply by 540 to get liters/hour, then multiply by the approximate kW/l factor.

For more precise computations, you should factor in the exact composition, the net or gross calorific value, and the actual density at your storage temperature. This simplified approach helps give a ballpark figure for quick design or check calculations.

Example

If you have 2 m³/h of LPG in liquid form, that’s approximately 2 × 540 = 1080 liters per hour. Using a rough energy content estimate of 25 kW per liter/hour, you get:

1080 L/h × 25 kW/L/h = 27,000 kW = 27 MW
                  

Clearly, this is a very large consumption for a typical commercial or industrial scenario. If your real system is smaller, you might see correspondingly lower volumes of liquid LPG. Always confirm with actual product data sheets for best accuracy (some sources might use ~46 MJ/kg for propane, ~45.7 MJ/kg for butane, etc.).

Also, in real operation, an evaporator or vaporizer step is usually needed, since burners typically consume LPG in gaseous form. The calculation here focuses purely on the energy content in the liquid volume.

Calculation

This approach is more direct when you have a volume flow rate (m³/h) and you know the density (kg/m³) plus the fuel’s calorific value (MJ/kg). Steps:

1. Convert volume flow (m³/h) to mass flow (kg/h) using the density.
2. Use the calorific value (MJ/kg) to find total MJ/h. Then convert MJ/h to MW (since 1 MW = 1 MJ/s = 3600 MJ/h).

Mass Flow (kg/h) = Volume Flow (m³/h) × Density (kg/m³)

Power (MW) = [ Mass Flow (kg/h) × CV (MJ/kg) ] / 3,600
                  

This method is often used when dealing with process or industrial gases where the density and composition may be well-defined (e.g., a syngas, biogas, or specialized fuel gas).

Example

If volume flow = 500 m³/h,
   density = 1.2 kg/m³,
   CV = 45 MJ/kg,

Step 1: Mass Flow = 500 × 1.2 = 600 kg/h

Step 2: Total MJ/h = 600 × 45 = 27,000 MJ/h

Step 3: Convert to MW:
         27,000 MJ/h ÷ 3,600 = 7.5 MW
                  

Interpretation: The fuel stream has an energy release potential of 7.5 MW. If you have a 90% efficient boiler or burner, the net output might be ~6.75 MW (0.90 × 7.5).

Always ensure you’re using consistent units. Gas densities may change significantly with pressure and temperature, so use the correct density at operating conditions.

Calculation

Many industrial boilers are rated in tonnes of steam per hour (T/h) at a specified pressure and temperature. The “Boiler Load” is often expressed as a percentage of the maximum design capacity:

Boiler Load (%) = (Actual Steam Load / Design Steam Load) × 100

This percentage indicates how much of the boiler’s potential is being utilized. Running close to 100% indicates near-max capacity.

Example

If your boiler design capacity is 1000 T/h but you’re currently generating 800 T/h of steam:

Boiler Load = (800 ÷ 1000) × 100 = 80%
                  

This means you’re operating at 80% of the boiler’s rated design. If the steam load approaches or exceeds 100%, that may cause stress or reduce steam quality.

Monitoring boiler load helps in scheduling maintenance, anticipating equipment needs, and ensuring safe operation.

Calculation

Modern burners can modulate or operate at some fraction of their maximum rated capacity. The formula is:

Burner Load (MW) = Max Power (MW) × (Burner % / 100)
                  

For instance, if you have a 50 MW burner set to 60% firing rate, you’re currently firing at 30 MW of input power (fuel flow adjusted accordingly).

Example

If your burner’s max power is 50 MW and you operate at 60%:

Burner Load = 50 × (60 ÷ 100) = 30 MW
                  

You’d typically measure or confirm this by checking the fuel flow or the control signal. This concept is widely used in large industrial or commercial burners that have turn-down ratios (the ratio between max firing rate and minimum stable firing rate).

Operating at part load can enhance efficiency in some cases or reduce it in others, depending on the equipment design.

Flue Gas Recirculation (FGR)

Flue Gas Recirculation is used to reduce flame temperature and thus lower NOx emissions. It typically involves diverting a portion of the flue gas back into the combustion air or the windbox. By diluting the oxygen (and adding heat capacity), peak flame temperature goes down. A common approximate formula is:

FGR (%) = [ (Air O₂ - WB O₂) / (WB O₂ - Stack O₂) ] × (100 / 1.06)
                  

Air O₂ is typically around 21% for ambient air. WB O₂ is the measured oxygen in the windbox, and Stack O₂ is the measured oxygen in the flue gas at the stack. The factor 1.06 accounts for differences in specific heat and small measurement offsets. Different organizations might use slightly different correction factors.

Example

Given:

• Air O₂ = 21%
• WB O₂ = 15%
• Stack O₂ = 3%

Step 1: numerator = 21 - 15 = 6
        denominator = 15 - 3 = 12
        ratio = 6 / 12 = 0.5

Step 2: multiply by 100 → 50%
        then divide by 1.06 → 50 / 1.06 ≈ 47.17%

Hence FGR ≈ 47%
                  

This is a relatively high FGR rate, but in low-NOx burner designs or ultra-low NOx boiler configurations, FGR can indeed be significant. The result is an approximate gauge of how much flue gas is recirculated compared to the total airflow.

NOx Conversion & O₂ Correction

Regulations often specify NOx emissions in mg/Nm³ at a given reference oxygen level (e.g., 3% O₂ for boilers). If you measure NOx at some different O₂ content, you apply a correction formula. One common approach is:

NOx (corrected) = NOx (raw) × 17.9 / (20.9 - O₂ref)
                  

The factor 17.9 can vary depending on dryness, temperature, or if you are converting from NO to NO₂ mass basis. Another widely used formula is:

NOx(corr to X% O₂) = NOx(measured) × (21 - X) / (21 - O₂measured)
                  

The exact formula used depends on the jurisdiction and measurement standards. Some references incorporate dryness corrections, total wet or dry gas basis, and occasionally a factor to adjust from ppm to mg/m³.

Example

Let’s say:

• Measured NOx = 100 mg/Nm³
• O₂ reference we want is 5%
• And the flue gas O₂ measured was 7%

Using the formula NOx@5% = 100 × (21 - 5) / (21 - 7)
                         = 100 × 16 / 14
                         ≈ 114.3 mg/Nm³
                  

This simple ratio corrects the NOx from the measured O₂ to a standardized reference O₂. The concept is that if you had less excess air (lower O₂%), the NOx concentration changes proportionally. In reality, actual formation depends on many factors, but regulatory compliance typically demands such standardized reporting.

Air Flow Calculation

The stoichiometric ratio for complete combustion of methane (CH₄), for example, is about 9.52 volumes of air per 1 volume of methane under standard conditions. However, many burners operate with excess air (10-20% or more) to ensure stable, safe combustion and to reduce CO formation. The formula used here is:

Air Flow (m³/h) = Fuel Flow (m³/h) × Stoich Factor × Excess Air
                  

For natural gas (mostly methane), stoich factor ~9.5. If you have 10 m³/h of methane and you run at 20% excess air (factor 1.2):

Air Flow = 10 × 9.5 × 1.2 = 114 m³/h
                  

Real fuels often aren’t pure methane, so a more accurate stoich ratio might be 9.3 or 9.6, depending on composition. For other hydrocarbons, the stoich factor changes. The principle remains the same: stoichiometric air is the minimum needed for complete combustion, and excess air is added for safety and completeness.

Example

Given 10 m³/h CH₄,
stoich factor = 9.5,
excess air = 1.20,

Air Flow = 10 × 9.5 × 1.20 = 114 m³/h
                  

This is a straightforward approach that helps to size blowers, fans, or specify the design airflow for burners. In practice, you verify with flue gas analyzers measuring O₂ in the exhaust to confirm that the actual air supply matches your intended excess air level.

Calculation

The ratio of measured CO₂ in the flue gas to the “maximum possible” CO₂ (if combustion were perfectly stoichiometric) can be used as an indicator of “chemical combustion efficiency.” It looks at how thoroughly the carbon in the fuel was oxidized to CO₂, ignoring other heat losses:

Efficiency (%) = (Measured CO₂ / Max Theoretical CO₂) × 100
                  

A higher measured CO₂ means more complete conversion of carbon to CO₂ and typically less unburnt fuel or CO. However, real efficiency in a boiler or furnace also depends on stack temperatures and other losses.

Example

If max CO₂ for your fuel is 15%,
and you measure 12% in the flue,

Efficiency (%) = (12 ÷ 15) × 100 = 80%
                  

This implies about 80% “chemical efficiency.” If the measured CO₂ is very close to the maximum, the system has near-complete combustion, though you still need to check if you have excessive O₂ or flue gas temperature losses.

Many analyzers or “combustion efficiency” testers rely on this ratio plus corrections for flue gas temperature, O₂ or CO, and so on, to provide a more inclusive “combustion efficiency” measure.

Calculation

Orifice or nozzle sizing commonly uses the fundamental orifice equation:

Q = C × A × √(2∆P / ρ)
                  

Q is volumetric flow, C is the discharge coefficient (depends on shape), A is cross-sectional area, ∆P is the pressure drop, and ρ is the fluid density. Rewriting to solve for diameter (d) of a circular orifice:

A = Q / [C × √(2∆P / ρ)]

d = 2 × √(A / π)
                  

Because Q might be given as mass flow rate, you first convert to volumetric flow rate. The user must be careful about the units: Pressure must be in Pa (SI units), density in kg/m³, flow in m³/s, etc., for a consistent result in meters (m).

The “control valve opening (%)” and “divisor” allow for adjustments if you’re not operating at full design or if you intentionally want a fraction of the orifice capacity.

Example

Suppose you want 500 kg/h of a gas with density ~1.0 kg/m³ at 2 bar differential pressure, and you pick a discharge coefficient of 0.83. Converting 500 kg/h (≈ 0.1389 kg/s) to 0.1389 m³/s (since ρ=1.0, mass flow ÷ ρ = volume flow):

Q = 0.1389 m³/s
∆P = 2 bar = 200,000 Pa
C = 0.83
ρ = 1.0 kg/m³
                  

Then solve for area and diameter. If you incorporate partial valve opening or a divisor, the final diameter effectively becomes smaller. In practice, you’d do a step-by-step approach ensuring the correct units and factoring safety margins.

Real nozzles also experience potential multi-phase flow or velocity constraints near sonic speeds, which complicates advanced designs. This formula is a simplified incompressible or low Mach assumption.

Calculation

Oil flows differently than gases due to higher viscosity. Nonetheless, the standard orifice equation can still approximate the required diameter:

Q = C × A × √(2∆P / ρ)
                  

But “C” (discharge coefficient) may be lower or adjusted further because of oil viscosity. We do:

• Convert oil mass flow (kg/h) into volumetric flow (m³/s) using density.
• Apply the orifice formula to solve for cross-sectional area, then diameter.
• Viscosity might reduce the effective discharge coefficient further, so real systems rely on empirical data or orifice vendor charts.

The “control valve percentage” and “divisor” handle partial operation or multi-orifice scenarios.

Example

Suppose you have 700 kg/h oil flow, pressure drop of 3 bar, density ~850 kg/m³, viscosity ~2 mPa·s, and you estimate a discharge coefficient of 0.7.

1. Mass flow in kg/s: 700 ÷ 3600 ≈ 0.1944 kg/s.
2. Volumetric flow in m³/s: 0.1944 ÷ 850 ≈ 2.29 × 10⁻⁴ m³/s.
3. Apply Q = C × A × √(2∆P / ρ) → solve for A, then d = 2√(A/π).

Adjust final diameter for partial opening or a divisor factor if needed. Real nozzles for oil burners might also factor swirl, droplet size requirements, and swirl plates, making an empirical approach necessary. This formula simply provides a theoretical baseline diameter.

Calculation

Air orifice sizing again uses the orifice equation. You first figure out the volumetric flow in m³/s by converting mass flow (kg/s) with the density of air at your given pressure and temperature. The Ideal Gas Law or standard tables can provide density. Then:

Q = C × A × √(2∆P / ρ)
                  

You rearrange to find the orifice area A, and from that, the diameter d. The “discharge coefficient” depends on the orifice geometry. Real-life air flow might approach compressible flow conditions if the pressure drop is large, requiring more advanced compressible flow equations.

Example

You have 2000 kg/h air flow at a differential pressure of 2 bar, temperature ~25 °C, and an estimated discharge coefficient of 0.7:

1. Convert 2000 kg/h → 0.556 kg/s.
2. Find air density at 2 bar & 25 °C (roughly 2.4 kg/m³ if 1 bar & 25 °C is ~1.2 kg/m³). Then volumetric flow = 0.556 ÷ 2.4 ≈ 0.2317 m³/s.
3. Plug into the orifice formula to find area, then diameter.

If partial valve opening or a factor is used, the effective orifice diameter changes. In real applications, orifices or flow nozzles for air are often validated by direct measurement (e.g., flow benches).

Calculation

Gas density can be estimated using the Ideal Gas Law:

PV = nRT
                  

or in a rearranged form:

ρ = (P × M) / (R × T)
                  

where P is absolute pressure (Pa), M is molar mass (kg/mol), R is the universal gas constant (8.314 J/(mol·K)), and T is absolute temperature (K). In engineering practice, you convert your gauge or bar reading to absolute pressure and your °C reading to Kelvin by adding 273.15. The result is the density in kg/m³ under those conditions.

Example

For air at T=25 °C (298.15 K), P=1.2 bar (i.e., 120,000 Pa absolute), molar mass of air ~28.97 g/mol (0.02897 kg/mol):

ρ = (120,000 Pa × 0.02897 kg/mol) 
     ÷ (8.314 J/(mol·K) × 298.15 K)

  = 3,476.4 ÷ 2,476.8
  ≈ 1.4 kg/m³
                  

This is a rough calculation showing that if you slightly pressurize air (1.2 bar) at near-room temperature, the density is ~1.4 kg/m³. Actual measuring equipment or real gas effects (especially for heavier fuels) can lead to small deviations from ideal gas behavior.

Calculation

This is a straightforward conversion: if you burn a certain amount of oil (in kg/h), you can find the total MJ/h by multiplying by the oil’s calorific value (in MJ/kg). Then to convert MJ/h to MW, recall that 1 MW = 1 MJ/s = 3600 MJ/h. The formula:

Power (MW) = [ Mass (kg/h) × CV (MJ/kg) ] / 3,600
                  

This yields the theoretical heat input in MW, ignoring boiler/burner efficiency. In real usage, some fraction is lost (stack losses, incomplete combustion, etc.).

Example

If you burn 100 kg/h of oil
   with a CV = 25 MJ/kg,

Total MJ/h = 100 × 25 = 2,500 MJ/h
MW = 2,500 / 3,600 ≈ 0.694 MW (694 kW)
                  

This indicates that you’re supplying about 694 kW of thermal power in the oil stream. If your combustion system is 90% efficient, net output would be about 625 kW.

Pressure Changes with Temperature

When gas is confined at a constant volume, the relationship between absolute pressure and absolute temperature (in Kelvin) follows the law:

P₁ / T₁ = P₂ / T₂
                  

If you have a gas supply at a certain temperature and want to know how the pressure changes if temperature changes (still at constant volume), you rearrange:

P₂ = P₁ × (T₂ / T₁)
                  

T in Kelvin = °C + 273.15, and pressure must be absolute (e.g., gauge + atmospheric). For small changes or in real supply lines that are not strictly constant volume, actual values might differ from this ideal approach.

Example

Let’s say you have a supply cylinder at 2 bar gauge (roughly 3 bar absolute if atmospheric is 1 bar) at 0 °C (273 K). If the outside temperature rises to 25 °C (298 K):

P₂ ≈ 3 bar_abs × (298 / 273) ≈ 3.27 bar_abs
Gauge pressure = 3.27 - 1 = 2.27 bar_g
                  

This is a simplified estimate. Real cylinders and supply lines can also be affected by volume changes, consumption rates, and non-ideal gas behavior. However, the formula provides a basic reference for how pressure might shift with temperature.

Purge Volume & Time

When purging a pipeline (e.g., removing hydrocarbon gas before maintenance by pushing inert gas such as nitrogen through), you typically:

1. Calculate the pipeline volume:

   Volume = π × (ID/2)² × Length
                         

   where ID is the internal diameter of the pipe, and Length is how long the pipeline or flare header is.
2. Determine how many volume exchanges are needed (e.g., 3-5 times the pipeline volume) to ensure the residual concentration of flammable gas is below a safe threshold.
3. Compare that total needed purge volume to your nitrogen supply (bottle volumes, pressures, etc.) to see how many cylinders or how long the purge will take at a given flow rate (Nm³/h).

Example

Suppose ID=0.5 m, length=100 m, so pipeline volume:

π × (0.25)² × 100 ≈ 19.63 m³
                  

If your safety procedure demands purging 5 times that volume, total = 5 × 19.63 ≈ 98.15 m³ of N₂. If the flow rate is 10 Nm³/h, you need roughly 9.8 hours. Next, to supply 98 Nm³ at your bottle pressure and volume, you check how many cylinders or bundles are required (each cylinder might hold a certain fraction at a given bar).

Always confirm actual partial pressure, real gas behavior at high pressures, and plan for safe venting. Pipeline purge guidelines vary by region and hazard classification.

CO₂ vs O₂ Relationship

In combustion, carbon in the fuel combines with oxygen to form CO₂. The maximum CO₂ that can appear in the flue gas depends on stoichiometric conditions. If you measure a certain CO₂, you can infer how much O₂ is left (excess air). Usually, the sum of CO₂ + O₂ + other components is the composition of the dry flue gas. Each fuel has a theoretical maximum CO₂ yield:

• For natural gas (mostly CH₄), max CO₂ is ~11.7-12%.
• LPG might be around 13-14% max CO₂, depending on exact composition.
• Kerosene and diesel can have higher potential CO₂ if burned stoichiometrically.
• Hydrogen has no CO₂ from stoichiometric combustion (though real burners often see small amounts from air infiltration or NOx formation by-products).

By relating measured CO₂ to a maximum possible value, you can estimate the leftover O₂. Some analyzers include a chart or a function that given CO₂, calculates approximate O₂ for a specific fuel type.

Example

Natural gas max CO₂ is often ~11.7%. If you measure 10% CO₂ in flue gas, you might estimate the leftover O₂ is a few percent (and thus an excess air of around 15-20%). Another approach is using an approximate correlation or direct flue gas analyzers that read both O₂ and CO₂ simultaneously.

As a rule of thumb, for natural gas, each 1% O₂ in flue gas corresponds to a drop of about 0.6% in CO₂ from the stoichiometric maximum (though the exact ratio can vary).

Advanced NO_x Calculations: Step-by-Step

In many regulatory frameworks, NO_x emissions must be reported in mg/Nm³ at a specific O₂ reference (like 3% or 6%), on a dry basis, at standard temperature (0 °C), and occasionally referencing NO₂ mass if originally measured in NO. Below is a typical multi-step process:

1. Convert from ppm to mg/m³ (if measured in ppm): For NO₂, 1 ppm ≈ 2.05 mg/m³ (at standard conditions). For NO, the factor is different (~1.34 mg/m³ per ppm). If you measure total NO_x as “NO₂ equivalent,” use the relevant factor.
2. Correct for actual temperature to standard temperature (0 °C):

   mg/Nm³ @ 0°C = mg/m³ @ T × (273 / (273 + T[°C]))

   This step standardizes the volume to normal conditions (Nm³).
3. Convert wet to dry basis (if needed): If your measurement is wet and you have a known moisture fraction in the flue gas, you adjust:

   NOxdry = NOxwet / (1 − %H₂O)

4. O₂ reference correction:

   NOx@RefO₂ 
     = NOxdry × [(21 − RefO₂) / (21 − Measured O₂)]
                         

   This normalizes the NO_x reading to the same oxygen level. If you have less O₂ in your actual reading, the concentration of NO_x tends to be higher, so a correction ensures a fair comparison across different excess air levels.

Example Calculation

1. Measured NO_x = 100 ppm at 200 °C, with 10% H₂O, flue O₂=8%. We want to correct to 3% O₂, 0 °C, dry basis, and measure NO_x as NO₂.
2. 1 ppm NO₂ ≈ 2.05 mg/m³ at STP, so 100 ppm → 205 mg/m³ (wet, 200 °C).
3. Temperature normalization to 0 °C:

   mg/Nm³(0°C) = 205 × (273 / (273 + 200))
                                   = 205 × 273/473
                                   ≈ 205 × 0.577 = 118 mg/Nm³ (wet)

4. Convert wet to dry:

   NOxdry = 118 ÷ (1 − 0.10)
                                        = 118 ÷ 0.90
                                        ≈ 131 mg/Nm³

5. O₂ reference from 8% to 3%:

   NOx@3% O₂ 
    = 131 × [(21 − 3) ÷ (21 − 8)]
    = 131 × (18 ÷ 13)
    = 131 × 1.3846
    = ~181.4 mg/Nm³

Final reported NO_x would be ~181 mg/Nm³ @ 3% O₂, 0 °C, dry, as NO₂. The exact factors can differ by standard or local regulations, but this demonstrates the multi-step approach for advanced NO_x reporting.