10 Easy Hacks for Heater & Combustion Engineers

Since you can’t run a heater simulation every time you’re out in the field, here are some easy hacks, rules of thumb, and formulas to help fired heater engineers in their daily routine, especially when it comes to estimating the effect of fuel and air changes on burner capacity and heater efficiency.


1. Convert percent excess air to percent excess O2


Process heaters do not operate at exactly the right amount of air, so we need to provide “excess” air to the system to ensure complete combustion of the fuel. The recommended excess air level for a gas fired process furnace is 15%, according to industry recommended practices, like API 535. In certain process plants, such as ethylene and hydrogen production, furnaces operate steadily at high temperature. Here, the industry norm is an excess air level of 8 – 10%. Combustion of liquid fuels, on the other hand, requires excess air levels of 20 – 25% to prevent soot formation. Since the operator of the furnace typically only knows the firebox oxygen level, use the following formula to convert to excess air (EA) percentage.


EA (%)=  (92 O_2)/(21-O_2 )

with O2 expressed in vol% (dry). The other way around:

O2 (%)=  (21 EA)/(92+EA)

The equations work well for typical refinery fuel gas mixtures but deviate for fuels that are very high in inerts, hydrogen or carbon monoxide.

Example:

arch O2 = 5 vol%(dry) à  Excess Air = 92 x 5 / (21 – 5) = 29%  Excess Air = 15% à Excess O2 = 21 x 15 / (92 + 15) = 3 vol% (dry)

2. Fired heater efficiency

Fuel efficiency of a fired heater is an important indicator. It tells you how close the heater runs to the design sheet conditions, if there is fouling or damage causing excess fuel use, and ways to improve capacity or fuel consumption. The precise calculation of the fuel efficiency is a bit of an undertaking, as shown in Annex G of API 560, which requires the fuel composition, excess air, stack temperature, fuel temperature, combustion air temperature, etcetera.


However, it is possible to get a good estimate of efficiency just from excess air and stack temperature. See Figure 1 for the heater efficiency when operating on natural gas and ambient air. The graph can be used for other fuels as well, but accuracy will drop if the amount of hydrogen or inert components is high. Also note that the graph does not account for external heat sources like preheating the combustion air or fuel with waste steam.

Figure 1 Heater efficiency v Excess Air

Figure 1

If you do not know the excess air but only the stack O2 content, an even faster method is available:

Metric:

η=100-T_(stack, °C)/20-(O_2 (vol%))/2

Imperial:

η=100-[T_(stack, °F)-32]/36-(O_2 (vol%))/2

These linear equations are only valid for O2 < 5%; above 5% the efficiency drops exponentially.

Example: Stack temperature = 300°C and stack oxygen is 2%: Efficiency = 100 – 300/20 – 2/2 = 84%

3. Firebox draft

Fired heaters are typically controlled to a draft of around 0.05 – 0.15 in.H2O at the exit of the firebox. Sometimes we need to know what the draft is in other locations, for example if we want to estimate the pressure drop over a natural draft burner. Since draft varies linearly with height inside the firebox, this can be easily estimated with the vertical distance (in feet) between the location of the draft measurement and the point of interest. By multiplying this number by 0.01 we get the draft difference expressed in inches of water column:

Draft=0.01∙H_FB

For example, if a firebox is 45 ft. high and the arch draft is 0.15 in.H2O, the draft at the firebox floor is:

0.15 + 45 x 0.01 = 0.60 in.H2O

For natural draft burners, this is the maximum available pressure drop for the combustion air.

4. Estimate burner air capacity

If the performance of a burner is known at one condition (i.e. the design conditions on the data sheet), you can estimate the effect of changes, like air temperature, on capacity.


Change in burner duty as a function of air temperature (with constant pressure drop):

Q_new=Q_design √(T_design/T_new )

Change in burner pressure drop as a function of air temperature (with constant duty):

〖∆P〗_new=〖∆P〗_design  (150+460)/(250+460)= 〖∆P〗_design  ∙0.86

With T the absolute temperatures (in K or °R). So for example, if a burner was designed for a certain pressure drop with 250°F air temperature, and the new air temperature drops to 150°F we gain 8% extra capacity:


Q_new=Q_design √((250+460)/(150+460))= Q_design  ∙1.08

If we use the same example but keep the duty constant, we get 14% lower pressure drop:


〖∆P〗_new=〖∆P〗_design  (150+460)/(250+460)= 〖∆P〗_design  ∙0.86

5. Volumetric air to fuel ratio for any hydrocarbon

The volumetric air requirement for any hydrocarbon can be calculated from the ratio of carbon to hydrogen in the molecule (CxHy) and the fraction of excess air (EA):

(A/F)_V=4.76 EA(x+y/4)
"with" EA=  (Excess Air (%))/100+1

For example, methane is CH4, so x=1 and y=4. For an excess air of 20%, the air to fuel volume ratio is:

(A/F)_V=4.76∙1.2(1+4/4)=11.424  〖ft〗^3/〖ft〗^3

Which means that we need 11.424 ft3 (or m3) of air to combust 1 ft3 (or m3) of methane.

For hydrogen (H2), x=0 and y=2. So, hydrogen combustion with 10% excess air yields:


(A/F)_V=4.76∙1.1(0+2/4)=2.618 〖ft〗^3/〖ft〗^3

6. Air to fuel mass ratio

In many cases we need to know the mass ratio of air to fuel. This can be calculated from the volumetric ratio, the respective mole weights of air (which is 28.96), and fuel:

(A/F)_v  〖MW〗_air/〖MW〗_fuel

Methane mole weight is 16.04, so the air to fuel mass ratio for methane combustion at 20% excess air is:

(A/F)_w=11.424 28.96/16.04=20.62   lb/lb

In other words, it takes 20.62 lb (or kg) to combust 1 lb (or kg) of methane with 20% excess air. For hydrogen with 10% excess air:

(A/F)_w=2.618 28.96/2.016=37.61  lb/lb

7. Calculation of the amount of flue gas

Now that we know how much air is needed for combustion, we can calculate the amount of flue gas that is produced by adding 1.0 to the air to fuel mass ratio.

So, the flue gas to fuel mass ratio for methane combustion at 20% excess air is:

(A/F)_w=20.62+1=21.62  kg/kg

And for hydrogen with 10% excess air:

(A/F)_w=37.61+1=38.61 kg/kg

8. Estimate burner fuel capacity

Being able to estimate the heat release as a function of fuel pressure can be very useful to determine if a burner is fouled, check its maximum capacity for a new fuel, and/or verify maximum heat release. The actual equations to determine the fuel mass flow through orifices are a bit cumbersome, but the general trend of fuel capacity curves helps make quick estimates.


Fuel is injected through nozzles, also known as fuel ports, into the flame. If the fuel pressure is below a critical value, the flow through the nozzle will be subsonic. Above this critical pressure the flow turns sonic. The absolute critical pressure ratio can be calculated using:

P_crit/P_0 = (2/(γ+1))^(γ/(γ-1))

where y is the heat capacity ratio Cp/Cv. The heat capacity ratio and critical pressure for common fuel gases is shown below, assuming the atmospheric pressure P0 = 14.7 psia (1.013 bara):


Table 1 - heat capacity ratio and critical fuel pressure

Table 1 - heat capacity ratio and critical fuel pressure

See Figure 2 for an example capacity curve. Below the critical pressure, duty varies approximately with the square of the fuel pressure. In the choked flow region, it varies linearly with fuel pressure. We can use this behavior to estimate burner duty.


As long as the flow is choked, the burner capacity can be estimated linearly. If we know that the burner duty is 10 MMBtu/h at 16 psig, the duty at 25 psig will be:

10  MMBtu/h  ((25+14.7)/(16+14.7))=13  MMBtu/h

As long as the flow is subsonic, the duty can be estimated assuming a square root dependency on fuel pressure. For example, if the duty is 8 MMBtu/h at 10 psig, the duty at 1 psig is approximately:

8  MMBtu/h ((1+14.7)/(10+14.7))^2= 3  MMBtu/h<