# Calculate coal consumption of coal-fired boiler from air flow

It is not always easy to determine the coal consumption of a coal-fired boiler which is generally used at power plants; therefore, different methods are used which are compared to determine the most likely consumption.
One method to determine the coal consumption is to use the air flow and the oxygen content in the flue gas. If the coal composition is known, the flue gas composition can be determined and from the oxygen in the flue gas the coal flow can be calculated.
This method will be useful to:
* Confirm the measured coal consumption
* Calculate the carbon dioxide (CO$_2$) and sulphur dioxide (SO$_2$) emissions.

## Coal combustion

Coal consists of C, H$_2$, O$_2$, N$_2$, S, H$_2$O and air consists of O$_2$, N$_2$ and Ar.
The following combustion reactions take place in the boiler:
    $C+O_{2}\rightarrow CO_{2}$
    $H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O$
    $N_{2}+2O_{2}\rightarrow 2NO_{2}$
    $S+O_{2}\rightarrow SO_{2}$

## Symbols and constants

The following is a list of symbols which will be used:
* $c$, $h$, $o$, $n$, $s$ = mass fractions of C, H$_2$, O$_2$, N$_2$ and S in the coal according to the ultimate analysis based on air dry (ad) coal.
* $m=m^{air\ dry}$ = mass fraction of total coal moisture based on air dry (ad) coal. Total moisture is the sum of superficial moisture and inherent moisture. However, normally the mass fraction of superficial moisture ($m_{s}$) is based on received (ar) coal and the inherent moisture mass fraction ($m_{i}$) is based on an air dry (ad) coal basis. The total moisture can be written in terms of $m_{s}$ and $m_{i}$ as follows:
    $m^{air\ dry}={m_{s} \over 1-m_{s}}+m_{i}$ (This formula is derived later in the article.)
* $n_{C}$, $n_{H2}$, $n_{O2}$, $n_{N2}$, $n_{S}$, $n_{H2O}$ = number of moles of components in coal ending up in the boiler (kmol/h). Note that both inherent moisture and superficial moisture ends up in the boiler.
* $Coal=Coal_{100\%\ conversion}^{ad}$ = mass flow of air dry coal flow which is completely combusted (kg/h)
* $Air$ = air flow (kmol/h). Take note that this is the total air flow to the boiler. It includes both the measured air flow and the air leaking into the fire box.
* $wa$ = mol fraction of water vapour in air (due to air humidity)
* $X_{i}$ = conversion of component i during combustion. For example, about 90% of S will convert to SO$_2$. Thus, $X_{S}\approx 0.9$. Very little N$_2$ will convert to NO$_2$. Thus, $X_{N2}$ will be very small.

### Air composition

The following air composition is assumed:
| Component | Dry air (mol frac) | Humid air (mol frac) |
| N$_2$ | 0.781 | (1-wa) × 0.781 |
| O$_2$ | 0.21 | (1-wa) × 0.21 |
| Ar | 0.009 | (1-wa) × 0.009 |
| H$_2$O | 0 | wa |
| Total | 1.000 | 1.000 |
If another composition needs to be used, only change the 0.781, 0.21 and 0.009 constants throughout the document.

### Number of moles before combustion

The following table shows the number of moles of each component before combustion:
| Component | Name | MW (kg/kmol) | Moles of component before combustion |
| Air | Air (humid) | | $Air$ |
| Air$^{dry}$ | Air dry | | $(1-wa)Air$ |
| C | Carbon | 12 | $n_{C}={\frac {c\cdot Coal}{12}}$ |
| H$_2$ | Hydrogen | 2 | $n_{H2}={\frac {h\cdot Coal}{2}}$ |
| O$_2$ | Oxygen | 32 | $n_{O2}={\frac {o\cdot Coal}{32}}+0.21(1-wa)Air$ |
| N$_2$ | Nitrogen | 28 | $n_{N2}={\frac {n\cdot Coal}{28}}+0.781(1-wa)Air$ |
| S | Sulphur | 32 | $n_{S}={\frac {s\cdot Coal}{32}}$ |
| Ar | Argon | 40 | $n_{Ar}=0.009(1-wa)Air$ |
| H$_2$O | Water | 18 | $n_{H2O}={\frac {m\cdot Coal}{18}}+wa\cdot Air$ |

## Component balance before and after combustion

The following table shows the number of moles before and after combusion of the coal with air:
| Short | Component | MW kg/kmol | Reaction | Moles before | Change | Moles after combustion |
| C | carbon | 12 | $C+O_{2}\rightarrow CO_{2}$ | $n_{C}$ | $-n_{C}$ | $0$ |
| H$_2$ | hydrogen | 2 | $H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O$ | $n_{H2}$ | $-n_{H2}$ | $0$ |
| O$_2$ | oxygen | 32 | All combustion reactions | $n_{O2}$ | $-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)$ | $n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)$ |
| N$_2$ | nitrogen | 28 | $N_{2}+2O_{2}\rightarrow 2NO_{2}$ | $n_{N2}$ | $-X_{N2}n_{N2}$ | $(1-X_{N2})n_{N2}$ |
| S | sulphur | 32 | $S+O_{2}\rightarrow SO_{2}$ | $n_{S}$ | $-X_{S}n_{S}$ | $(1-X_{S})n_{S}$ |
| CO$_2$ | carbon dioxide | 44 | $C+O_{2}\rightarrow CO_{2}$ | $0$ | $+n_{C}$ | $n_{C}$ |
| H$_2$O | water | 18 | $H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O$ | $n_{H2O}$ | $n_{H2}$ | $n_{H2O}+n_{H2}$ |
| NO$_2$ | nitrogen dioxide | 46 | $N_{2}+2O_{2}\rightarrow 2NO_{2}$ | $0$ | $+2X_{N2}n_{N2}$ | $2X_{N2}n_{N2}$ |
| SO$_2$ | sulphur dioxide | 64 | $S+O_{2}\rightarrow SO_{2}$ | $0$ | $+X_{S}n_{S}$ | $X_{S}n_{S}$ |
| Ar | argon | 40 | n/a | $n_{Ar}$ | $0$ | $n_{Ar}$ |

## Develop formula
### Total flue gas

The total flue gas is the sum of all the products and all unreacted components, except sulphur (S) which will not leave with the flue gas but will leave with the ash.
$Flue\ gas=n_{O2}-(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S})+(1-X_{N2})n_{N2}+n_{c}+n_{H2O}+n_{H2}+2X_{N2}n_{N2}+X_{s}n_{s}+n_{Ar}$
       $=n_{O2}-n_{C}-{\frac {1}{2}}n_{H2}-2X_{N2}n_{N2}-X_{S}n_{S}+(1-X_{N2})n_{N2}+n_{c}+n_{H2O}+n_{H2}+2X_{N2}n_{N2}+X_{s}n_{s}+n_{Ar}$
       $=n_{O2}+{\frac {1}{2}}n_{H2}+(1-X_{N2})n_{N2}+n_{H2O}+n_{Ar}$
Substitute the mol fractions n$_i$ with equations in table above:
$Flue\ gas=\left({\frac {o.Coal}{32}}+0.21(1-wa)Air\right)+{\frac {1}{2}}\left({\frac {h.Coal}{2}}\right)+(1-X_{N2})\left({\frac {n.Coal}{28}}+0.781(1-wa)Air\right)+\left({\frac {m.Coal}{18}}+wa.Air\right)+(0.009(1-wa)Air)$
       $=Coal{\frac {o}{32}}+Air0.21(1-wa)+0.5Coal{\frac {h}{2}}+Coal{\frac {n}{28}}(1-X_{N2})+Air0.781(1-wa)(1-X_{N2})+Coal{\frac {m}{18}}+Air.wa+Air0.009(1-wa)$
       $=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)(1-X_{N2})+0.009(1-wa)+0.21(1-wa)\right]$
       $=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+0.781(1-wa)-0.781X_{N2}(1-wa)+0.009(1-wa)+0.21(1-wa)\right]$
       $=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[wa+(1-wa)-0.781X_{N2}(1-wa)\right]$
* $Flue\ gas=Coal\left({\frac {o}{32}}+0.5{\frac {h}{2}}+{\frac {n}{28}}(1-X_{N2})+{\frac {m}{18}}\right)+Air\left[1-0.781X_{N2}(1-wa)\right]$
Where $Flue\ gas$ and $Air$ is in kmol/h, $Coal$ in kg/h, $c,h,o,n,s$ in mass fraction, $wa$ in mol fraction, $X_{N2}$ in fraction

### Oxygen in flue gas

The oxygen in the flue gas can be written as follows:
$Oxygen\ in\ flue\ gas=n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)$
       $=n_{O2}-n_{C}-{\frac {1}{2}}n_{H2}-2X_{N2}n_{N2}-X_{S}n_{S}$
Substitute the mol fractions n$_i$ with equations in table above:
$Oxygen\ in\ flue\ gas=\left({\frac {o.Coal}{32}}+0.21(1-wa)Air\right)-\left({\frac {c.Coal}{12}}\right)-{\frac {1}{2}}\left({\frac {h.Coal}{2}}\right)-2X_{N2}\left({\frac {n.Coal}{28}}+0.781(1-wa)Air\right)-X_{S}\left({\frac {s.Coal}{32}}\right)$
       $=Coal{\frac {o}{32}}+Air0.21(1-wa)-Coal{\frac {c}{12}}-0.5Coal{\frac {h}{2}}-Coal.2X_{N2}{\frac {n}{28}}-Air2X_{N2}0.781(1-wa)-Coal.X_{S}{\frac {s}{32}}$
* $Oxygen\ in\ flue\ gas=Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]$
Where $Oxygen$ and $Air$ is in kmol/h, $Coal$ in kg/h, $c,h,o,n,s$ in mass fraction, $wa$ in mol fraction, $X_{s},X_{N2}$ in fraction

### % oxygen in flue gas

The percentage oxygen in flue gas is given by the following formula:
          $x_{O2}=\%\ oxygen\ in\ flue\ gas={\frac {Oxygen\ in\ flue\ gas}{Total\ flue\ gas}}$
If substituting above formulas into this, then:
$Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]$ $=Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]+Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]$
Get "Coal" on the left and "Air" on the right:
$Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]-Coal\left[{\frac {o}{32}}x_{O2}+0.5{\frac {h}{2}}x_{O2}+{\frac {n}{28}}(1-X_{N2})x_{O2}+{\frac {m}{18}}x_{O2}\right]$ $=Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}]-Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]$
$Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]$
$=Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)]$
$Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]$
$=Air[x_{O2}-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)]$
Taking the right hand side separately:
$RH=Air[-0.781X_{N2}(1-wa)x_{O2}-0.21(1-wa)+2X_{N2}0.781(1-wa)+x_{O2}]$
$RH=Air[(1-wa)[-0.781X_{N2}x_{O2}+2X_{N2}0.781-0.21]+x_{O2}]$
$RH=Air[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}]$
Combining the two equations again:

### Coal consumption (total combustion)

$Coal\left[{\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}(2X_{N2}+(1-X_{N2})x_{O2})-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}\right]$
$=Air\left[(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2}\right]$
Thus, the total air dry coal which underwent total combustion is:
* $Coal=Air\times {(1-wa)[0.781X_{N2}(2-x_{O2})-0.21]+x_{O2} \over {\frac {o}{32}}(1-x_{O2})-{\frac {c}{12}}-0.5{\frac {h}{2}}(1+x_{O2})-{\frac {n}{28}}[2X_{N2}+(1-X_{N2})x_{O2}]-X_{S}{\frac {s}{32}}-{\frac {m}{18}}x_{O2}}$
Where:
* $Coal$ is air-dry coal in kg/h. This is the coal which fully combusted and excludes the portion of the coal which leaves with the ash. If the CO$_2$ and SO$_2$ emissions are calculated, this value can use used as is. However, if one wants to compare this calculated value with the actual coal received, on should take into consideration the uncombusted coal in the ash.
* $Air$ is the humid air flow in kmol/h with a dry air composition of 21 mol% oxygen, 78.1 mol% nitrogen and 0.9% argon.
* $Air=air\ measured+air\ leakage$

### Individual components in flue gas

The individual components in the flue gas is then the following (kmol/h):
    $O_{2}\ in\ flue\ gas\quad =n_{O2}-\left(n_{C}+{\frac {1}{2}}n_{H2}+2X_{N2}n_{N2}+X_{S}n_{S}\right)$
             $=Coal\left[{\frac {o}{32}}-{\frac {c}{12}}-0.5{\frac {h}{2}}-2X_{N2}{\frac {n}{28}}-X_{S}{\frac {s}{32}}\right]+Air\left[0.21(1-wa)-2X_{N2}0.781(1-wa)\right]$
    $N_{2}\ in\ flue\ gas\quad =\quad (1-X_{N2})n_{N2}$
             $=(1-X_{N2})\left[{\frac {n\cdot Coal}{28}}+0.781(1-wa)Air\right]$
    $CO_{2}\ in\ flue\ gas\quad =\quad n_{C}$
             $=Coal{\frac {c}{12}}$
    $H_{2}O\ in\ flue\ gas\quad =\quad n_{H2O}+n_{H2}\quad =\quad \left(Coal{\frac {m}{18}}+wa.Air\right)+\left(Coal{\frac {h}{2}}\right)$
             $=Coal\left({\frac {m}{18}}+{\frac {h}{2}}\right)+wa.Air$
    $NO_{2}\ in\ flue\ gas\quad =\quad 2X_{N2}n_{N2}$
             $=2X_{N2}\left(Coal{\frac {n}{28}}+0.781(1-wa)Air\right)$
    $SO_{2}\ in\ flue\ gas\quad =\quad X_{s}n_{S}\quad =\quad X_{S}\left(Coal{\frac {s}{32}}\right)$
             $=Coal.X_{S}{\frac {s}{32}}$
    $Ar\ in\ flue\ gas\quad =\quad n_{Ar}$
             $=0.009(1-wa)Air$

### Emissions

The CO$_2$, SO$_2$ and NO$_2$ emissions can be calculated as follows:
    $CO_{2}\ emissions=c\times Coal\times {\frac {44}{12}}\ in\ kg/h$
    $SO_{2}\ emissions=X_{s}\times c\times Coal\times {\frac {64}{32}}\ in\ kg/h$
    $NO_{2}\ emissions=2X_{s}\times \left({\frac {n\cdot Coal}{28}}+0.781(1-wa)Air\right)\times 46\ in\ kg/h$

## Calculating water vapour in air from humidity

If:
* $T$ = Ambient temperature in °C
* $RH$ = Relative humidity as a fraction between 0 and 1
* $P$ = Atmospheric pressure in kPa
* $p_{sat}$ = Vapour pressure (at ambient temperature) in kPa
* $x_{water}$ = Mol (volume) fraction of water in air
The maximum amount of water vapour in air is:
    $x_{water,max}={\frac {p_{sat}}{P}}$
The actual mol fraction of water in air is:
$x_{water}=RH\times x_{water,max}$
The following Buck equation is a very good approximation of the vapour pressure of water below 100°C:
$p_{sat}=0.61121\times e^{\frac {17.368T}{T+238.88}}$
Thus, the mol fraction of water in air is:
* $x_{water}={\frac {RH}{P}}\times 0.61121\times e^{\frac {17.368T}{T+238.88}}$

## Total coal moisture in terms of superficial and inherent moisture

The superficial moisture fraction (m$_s$) is based on as received (AR) coal and the inherent moisture fraction (m$_i$) is based on air dry (AD) coal. However, the total moisture (m), can be expressed in terms of m$_s$ and m$_i$ on an air dry basis as follows.
    $m^{ad}={\frac {M}{Coal^{ad}}}={M_{s}+M_{i} \over Coal^{ad}}={Coal^{ar}m_{s}+Coal^{ad}m_{i} \over Coal^{ad}}$
If:
    $Coal^{ad}=(1-m_{s})Coal^{ar}\qquad \Rightarrow \qquad Coal^{ar}={Coal^{ad} \over 1-m_{s}}$
Then:
    $m^{ad}={{Coal^{ad} \over 1-m_{s}}m_{s}+Coal^{ad}m_{i} \over Coal^{ad}}$
* $m^{ad}={m_{s} \over 1-m_{s}}+m_{i}$

## Also see

* Mass flow rate of coal consumption and theoretical air mass
* Application of mass and energy balances to determine coal, air required and flue gas flow rates in a power plant], Katende, Landry Mbangu, 2019
* Calculation of combustion air required for burning solid fuels (coal / biomass / solid waste) and analysis of flue gas composition], Lizica Simona Paraschiv, Alexandru Serban, Spiru Paraschiv, 6 September 2019