I will explain the combustion calculation for the boiler engineer exam so that you can understand it visually.

This is one of the most common calculation problems on the exam that is easy to get wrong.

Even though there are many definitions and terms, it is difficult to understand immediately because we try to express it using only mathematical formulas.

Therefore, I will explain the definitions that appear in combustion calculations, which are factors that hinder understanding, so that you can visualize the amount of air and the amount of combustion gas.

## Combustion calculation definition

There are six important definitions when calculating boiler combustion.

If you don’t have these definitions in mind, you may end up making incorrect calculations.

## air volume

The following reactions occur during combustion:

Fuel + air → combustion gas

Let’s look at air and combustion gas respectively.

Let’s start with the air.

### Theoretical oxygen amount

Theoretical oxygen amount is the **amount of oxygen** derived from **the chemical reaction equation** , which is a **theoretical calculation .**

Calculation of the theoretical amount of oxygen begins with elemental analysis of the exhaust gas.

Fuel contains combustible components such as C, H, and S in certain proportions.

This ratio is difficult to know because the fuel is hard.

This is because they are contained in combinations of various compounds.

I thought it would be best to burn it to find out this easily.

In a combustion reaction, the following reactions take place from a macroscopic perspective.

- C+O
_{2}→CO_{2} - H
_{2}+1/2O_{2}→H_{2}O - S+O
_{2}→SO_{2}

Regardless of the form of C, H, and S compounds, O _{2} is used in the reaction in proportion to the number of moles of C, H, and S.

If we can analyze the amounts of C, H, and S, we can conversely calculate the amount of O _{2 .}

If the mass (kg) of carbon, hydrogen, and sulfur per 1 kg of liquid/solid fuel is c, h, s, the amount of oxygen (m3) for each is

- c*22.4/12
- h*22.4/4
- s*22.4/32

It can be expressed as

This is simply converted to the volume of oxygen according to the molar ratio from the reaction equation above.

Actually, we have to remove the oxygen component from hydrogen h, but we will omit that part.

### Theoretical air volume

Theoretical air volume is easy.

The theoretical oxygen amount is **multiplied by 1/0.21** .

This is because air contains nitrogen:oxygen in a ratio of **0.79:0.21 .**

### Actual air volume

The actual air volume is the theoretical air volume multiplied by **the air ratio .**

The amount of air actually required for combustion is usually greater than the amount determined by theoretical calculations.

This means that let’s generalize this using a coefficient called **air ratio .**

In general chemical reactions, the optimum reaction rate is obtained by slightly changing the molar ratio, and this is the same.

m=A/A _{0}

It is determined by the ratio.

The excess air volume is defined as AA _{0} =(m-1)A _{0 .}

It is calculated that this amount of air (i.e. nitrogen and oxygen) does not contribute to the reaction.

## Combustion gas amount

Next, let’s look at combustion gas compared to air.

### Theoretical dry combustion gas amount

Theoretical dry combustion gas amount is the amount of completely combusted combustion gas determined from theoretical calculations after removing the water vapor component from the combustion gas.

It’s hard to understand, isn’t it?

Like the theoretical amount of oxygen, it is **a theoretical formula** determined from **a chemical reaction formula** .

V _{do} =0.79A _{o} +c*22.4/12+s*22.4/32+n*22.4/28

This is the definition.

Looking at the formula, you can see that the theoretical combustion gas amount is made up of the following combinations.

- Theoretical air amount – Theoretical oxygen amount (approximately the theoretical nitrogen amount. N
_{2}) - Volume
_{of}CO2 - Volume
_{of}SO2 - Volume
_{of}N2

Combustion gas may seem complicated, but it actually appears in the form of CO _{2} , SO _{2} , and N _{2} , so it would be best to control its **source .**

** _{}**The key is not to forget about

_{N2}**in the air .**

### Dry combustion gas amount

The dry combustion gas volume is the same as the relationship between the theoretical air volume and the actual air volume.

The excess air volume contains gases that do not contribute to combustion.

This is also emitted as combustion gas, so it needs to be counted.

### Combustion gas amount

Finally, consider the combustion gas as a whole.

Total combustion gas for dry combustion gas.

What it contains is **water vapor** .

It comes from the hydrogen (h) and water (w) contained in the fuel.

I think it would be a good idea to call it **wet combustion gas .**

V _{w} =V _{d} +h*22.4/2+w*22.4/18

I think the h in fuel is easy to understand, but we also don’t want to forget that there is **water in it.**

## reference

The boiler engineer exam is not very difficult for plant engineers, and there are a large number of books available.

If you read just one of the articles on this site, you will be able to deepen your understanding considerably.

## lastly

We explained the definition of combustion calculation for the boiler engineer examination.

Theoretical oxygen amount, theoretical air amount, actual air amount, theoretical dry combustion gas amount, dry combustion gas amount, combustion gas amount

Each one is simple, but the overall structure is difficult to understand at first glance, which is probably why people avoid it.

First, let’s understand it with an image.

Please feel free to post any concerns, questions, or concerns you may have regarding the design, maintenance, and operation of chemical plants in the comments section. (The comment section is at the bottom of this article.)

*We will read all comments received and respond seriously.