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Section summary |
---|

1. Process furnace
efficiency definition |

2. Process furnace
efficiency calculation step by step guide |

3. Process furnace efficiency calculation
example |

A process furnace efficiency is **defined as the % of heat
actually absorbed the process, compared to the heat introduced in
the furnace through the burners**. **The efficiency is
increasing if the furnace design and operation allows it to reduce
the temperature of the flue gas (sign that the heat is absorbed by
the process) and operates with very few excess air.**

It is possible to make a quick estimation of a furnace efficiency simply by measuring the oxygen in the flue gas and the temperature of the flue gas. However it is sometimes a bit imprecise, thus a more accurate method to calculate a furnace efficiency is given in this page.

No calculation can be done if data cannot be gathered

**Fluids properties**

- Fuel composition
- Air inlet composition (humidity)
- Calorific values

**Furnace process data**

- Key temperatures
- Flows
- Flue gas composition (% O2, %CO2)

The combustion air is taken from the atmosphere, it is therefore not dry. Temperature and relative humidity of the incoming air must be measured so that the amount of water carried can be calculated :

y_{water} = [%RH*P^{sat}] / Patm

The molar fraction is equal for a perfect gas, to the volume ratio, considering the other components of the air as :

Nitrogen | 78.09 % |

Oxygen | 20.95 % |

Inerts | 0.96 % |

The composition can then be calculated.

The stoichiometric **volume of air / kg of fuel** can be
determined thanks to the table 1. If the fuel is complex and data
are known about the atomic composition (C, H, S), then it is
possible to calculate the required air volume by summing the
requirement for the individual components, if the fuel is pure, such
as methane, the value is direct and tabulated.

**Table 1 : stoechiometric combustion
of fuel**

Fuel |
Combustion reaction |
C/H |
Heat of reactionkcal/mol (H2O gas) |
Calorific valuekcal/kg (H2O gas) |
StoichometricFuel / Air ratio Nm3 air / kg fuel |

Carbon | C+O2 -> CO2 | 94.05 | 7840 | 8.9 | |

Carbon | C+0.5*O2 -> CO | 26.45 | 2205 | 4.45 | |

Carbon monoxide | CO+0.5O2 -> CO2 | 67.6 | 2413 | 1.91 | |

Hydrogen | H2+1/2O2 -> H2O | 57.7 | 28640 | 26.5 | |

Sulfur | S+O2 -> SO2 | 70.7 | 2205 | 3.33 | |

Hydrogen sulfide | H2S+1.5O2 -> SO2 + H2O | 123.9 | 3635 | 4.7 | |

Methane | CH4+2O2 -> CO2+2H2O | 3 | 191.8 | 11960 | 13.32 |

Ethane | C2H6+3.5O2 -> 2CO2+3H2O | 4 | 341.4 | 11355 | 12.44 |

Propane | C3H8+5O2 -> 3CO2+4H2O | 4.5 | 488.7 | 11085 | 12.12 |

N-Butane | NCH4H10+6.5O2 -> 4CO2+5H2O | 4.8 | 635.6 | 10935 | 11.95 |

The Excess air is directly dependent on** the % of O2 analysed in
the exhaust fumes**.

It is possible to calculate the excess air **thanks to the
following graph** :

Note that similar abacus exist for CO2 in flue gas (not given here)

The heat introduced in the system must be calculated. One must note that the main component of the heat introduced in the furnace is of course the combustion heat, but the heat entered is actually made of the following contributors :

**Heat of combustion****Heat supplied by the fuel itself****Heat supplied by the air****Heat supplied by atomization steam (if present)**

It is very important that the reference of every enthalpy used in the calculation is the same (typically at 0c).

It is assumed that the water produced during the combustion process is not condensed in the furnace, thus the vaporization energy contained in the water vapor emitted is not recovered. The heat of combustion is thus equal to the Lower Heating Value (LHV) of the fuel burnt.

If the fuel is pure, for example only methane, the LHV is usually tabulated. For more complexed feeds, such as fuel oil, some abacus are available to calculate the LHV as a function of the density and sulfur content of the fuel.

The fuel is supplied at a given temperature, the fuel enthalpy can then be calculated either thanks to tables / graphs, or by using m.Cp.(T-Tref) if the specific heat is known.

The air contribution is calculated as : m_{air}.Cp_{air}.(T_{air}
- T_{ref})

With

m_{air} = mass flow of air (kg air / kg of fuel)

Cp_{air} = specific heat of air (kcal/kg/c)

T_{air} = air temperature (c)

The steam contribution is defined as the enthalpy of the steam used (can be determined thanks to steam tables).

The heat carried away by the flue gas can be calculated knowing the composition of the flue gas (CO2, H2O, N2, O2, SO2...), the enthalpy of each of these components, and the flow of each of the components.

The heat lost through the walls of the furnace is normally 1 to 2% of Qentered

It is then possible to calculate **Q _{absorbed} = Q_{entered}
- Q_{fluegas} - Q_{casing},**

and then the **efficiency Q _{absorbed} / Q_{entered}
* 100**

A factory is operating a process furnace used to heat up the feed to a distillation column. The furnace is run with some fuel oil.

- Fuel oil : d=0.98, S = 3.5% w, C/H=8, temperature = 134c, flowrate = 7000 kg/h
- Atomization steam : 1000 kg/h of steam at 180c
- The flue gas is used to pre-heat the combustion air
- %O2 of dry flue gas = 4%
- %CO2 of dry flue gas = 12.5%
- Exhaust Temperature = 270c
- Inlet air : 65% humidity, temperature = 15c, preheated at 200c by the flue gas

The relative humidity of air is 65%. It means that the actual pressure of water in the air is equal to 65% of the vapor pressure of water at the temperature of air (here 15c).

y_{water} = [%RH*P^{sat}] / P_{atm}

The saturation pressure of water at 15c is 0.017 bar

y_{water} = [0.65*0.017]/1.03125 = 0.01

The molar fraction of water in the incoming air** is 0.01, which
is also equal, for a perfect gas, to the volume fraction**. The
volume fraction of water in the combustion air is thus 1%.

**The composition of the air is thus :**

Dry air (% volume) | Actual air (% volume) | Actual air (% weight) | |

Nitrogen | 78.09 | 77.31 | 75.05 |

Oxygen | 20.95 | 20.74 | 23.01 |

Inerts | 0.96 | 0.95 | 1.32 |

H2O | 0 | 1 | 0.62 |

Total | 100% | 100% | 100% |

The 1st step is to decompose the fuel in C, H and S. It is already known that the % weight of the fuel in Sulfur is 3.5%. The ration C/H is also known = 8, which means C=8*H. The composition is thus

kg / kg of fuel | |

Carbon | 0.858 |

Hydrogen | 0.107 |

Sulfur | 0.035 |

**The Excess can be calculated thanks to the graph 1.**

The ratio C/H is equal to 8 and the %O2 in the flue gas is 4%.

Using the graph, the excess air is 22.5%.

It can then be determined what will be the quantity of air allowing to have a stoichimetric combustion thanks to table 1. From the air requirement, the products of combustion can then be determined :

1 mole of O2 -> 1 mole of CO2

1 mole of O2 -> 2 moles of H2O

1 mole of O2 -> 1 mole of SO2

kg / kg of fuel |
Stoichometric Fuel / Air ratio Nm3 air / kg fuel |
Air requirement Nm3 | O2 | N2 | H2O | CO2 | H2O | SO2 | N2 | |

Carbon |
0.858 |
8.9 | 7.63 | 1.58 | 5.9 | 0.076 | 1.58 |
0.076 |
0 |
5.9 |

Hydrogen |
0.107 |
26.5 | 2.84 | 0.59 | 2.2 | 0.028 | 0 |
1.18 |
0 |
2.2 |

Sulfur |
0.035 |
3.33 | 0.117 | 0.024 | 0.09 | 0.0011 | 0 |
0.0011 |
0.024 |
0.09 |

Total | 1 | - | 10.59 | 2.19 | 8.19 | 0.105 | 1.58 |
0.024 |
8.19 |

It is necessary to add to the H2O total the steam used for a better atomization of fuel oil. It is 1000 kg/h of steam for 7000 kg/h of fuel, thus 0.143 kg of steam / kg of fuel, which is 0.143/18*22.4 = 0.18 Nm3 of water / kg of fuel.

The flue has composition is thus :

**CO2 = 1.58 Nm3/kg fuel****H2O = 1.257+0.18 = 1.43 Nm3/kg fuel****SO2 = 0.024 Nm3/kg fuel****N2 = 8.19 Nm3/kg of fuel**

It is possible to convert it to mass by dividing by the molar volume and multiplying by the molar mass

- CO2 = (1.58 Nm3/kg fuel) / (0.0224 m3/mol) *( 0.042 kg/mol) = 2.96 kg/kg fuel
- H2O = (1.43 Nm3/kg fuel) / (0.0224 m3/mol) *( 0.018 kg/mol) = 1.15 kg/kg fuel
- SO2 = (0.024 Nm3/kg fuel) / (0.0224 m3/mol) *( 0.064 kg/mol) = 0.07 kg/kg fuel
- N2 = (8.19 Nm3/kg of fuel) / (0.0224 m3/mol) *( 0.028 kg/mol) = 10.23 kg/kg fuel

The O2 in the exhaust gas is 4% :

- O2 = (0.5 Nm3/kg of fuel) / (0.0224 m3/mol) *( 0.032 kg/mol) = 0.7 kg/kg fuel

The determination of the heat of combustion starts by the calculation of the LHV of the fuel used in the furnace. In this example, it is a fuel oil whose LHV can be calculated thanks to abacus. The abacus is not given here but can be found in the litterature.

For a Fuel oil with d=0.98, S = 3.5% w, **the LHV is 9575 kcal/kg**.

The heat supplied by the fuel oil can be determined through an enthalpy diagram, or by knowing the specific heat. Assuming a specific heat of 1.8 kJ/kg/K, or 0.43 kcal/kg/K.

The heat brought by the fuel is then Cp*(T-Tref) = 0.43*(134-0) = 57.7 kcal/kg

The air contribution is calculated as : Cpair.(Tair - Tref) = 0.24*(15-0)= 3.6 kcal/kg of air

It must then be converted to have the heat contribution per kg of fuel. The air flow for stoichometric combustion is 10.59 Nm3/kg of fuel, with an excess air of 22.5% which makes 10.59*(1+0.225) = 12.97 Nm3/kg of fuel, with a density of 1.24 kg/Nm3, the air weight is thus 12.97*1.24 = 16.08 kg of air / kg of fuel, thus the heat entered due to the air is 3.6*16.08 = 57.9 kcal/kg of fuel.

The steam enthalpy at 180c is 78 kcal /kg (with reference 0 at 0c vapor), as we inject 1000 kg/h of steam for 7000 kg/h of fuel, the heat contribution of the steam is 78*1000/7000 = 11.1 kcal / kg of fuel.

Qentered = 9575 + 57.7 + 57.9 + 11.1 = 9701.7 kcal / kg of fuel

The enthalpy that is "escaping" with the flue gas can be calculated, knowing the composition of the flue gas, the flow as a function of a kg of fuel burn, and the enthalpy of all the components. The tempature of the flue gas is 270 c

Enthalpy at flue gas temperature kcal / kg | kg / kg of fuel (see step 5) | kcal / kg of fuel | |

CO2 | 60 | 2.96 |
177.8 |

H2O | 120 | 1.15 |
137.9 |

N2 | 66 | 10.23 |
614.25 |

O2 | 60 | 0.7 |
42.9 |

SO2 | 40 | 0.07 |
2.74 |

The total heat carried out of the furnace by the flue gas is thus :
**975.5 kcal / kg of fuel**

The heat lost through the walls of the furnace is normally 1 to 2% of Qentered, which is here 9701.7*0.02 = 194 kcal / kg of fuel

It is then possible to calculate **Q _{absorbed} = Q_{entered}
- Q_{fluegas} - Q_{casing}, thus**

**Q _{absorbed} = 9701.7 - 975.5-194 = 8532.2 kcal/kg of
fuel_{}**

and then the **efficiency Q _{absorbed} / Q_{entered}
* 100**

**Efficiency = 8532.2/9701.7 * 100 = 87.9%
**