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 (CO₂) and sulphur dioxide (SO₂) emissions.

Coal combustion

File:BoilerCombustion.png

Coal-fired boiler combustion

Coal consists of C, H₂, O₂, N₂, S, H₂O and air consists of O₂, N₂ and Ar.

The following combustion reactions take place in the boiler:

C + O_{2} \to C O_{2}

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

N_{2} + 2 O_{2} \to 2 N O_{2}

S + O_{2} \to S O_{2}

Symbols and constants

The following is a list of symbols which will be used:

$c$ , $h$ , $o$ , $n$ , $s$ = mass fractions of C, H₂, O₂, N₂ and S in the coal according to the ultimate analysis based on air dry (ad) coal.
$m = m^{a i r d r y}$ = 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^{a i r d r y} = \frac{m_{s}}{1 - m_{s}} + m_{i}

Template:Spaces (This formula is derived later in the article.)

$n_{C}$ , $n_{H 2}$ , $n_{O 2}$ , $n_{N 2}$ , $n_{S}$ , $n_{H 2 O}$ = 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.
$C o a l = C o a l_{100 % c o n v e r s i o n}^{a d}$ = mass flow of air dry coal flow which is completely combusted (kg/h)
$A i r$ = 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.
$w a$ = 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 SO₂. Thus, $X_{S} \approx 0.9$ . Very little N₂ will convert to NO₂. Thus, $X_{N 2}$ will be very small.

Air composition

The following air composition is assumed:

Component	Dry air (mol frac)	Humid air (mol frac)
N₂	0.781	(1-wa) × 0.781
O₂	0.21	(1-wa) × 0.21
Ar	0.009	(1-wa) × 0.009
H₂O	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)		$A i r$
Air^dry	Air dry		$(1 - w a) A i r$
C	Carbon	12	$n_{C} = \frac{c \cdot C o a l}{12}$
H₂	Hydrogen	2	$n_{H 2} = \frac{h \cdot C o a l}{2}$
O₂	Oxygen	32	$n_{O 2} = \frac{o \cdot C o a l}{32} + 0.21 (1 - w a) A i r$
N₂	Nitrogen	28	$n_{N 2} = \frac{n \cdot C o a l}{28} + 0.781 (1 - w a) A i r$
S	Sulphur	32	$n_{S} = \frac{s \cdot C o a l}{32}$
Ar	Argon	40	$n_{A r} = 0.009 (1 - w a) A i r$
H₂O	Water	18	$n_{H 2 O} = \frac{m \cdot C o a l}{18} + w a \cdot A i r$

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	$C + O_{2} \to C O_{2}$	$n_{C}$	$- n_{C}$	$0$
H₂	hydrogen	2	$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$	$n_{H 2}$	$- n_{H 2}$	$0$
O₂	oxygen	32	All combustion reactions	$n_{O 2}$	$- (n_{C} + \frac{1}{2} n_{H 2} + 2 X_{N 2} n_{N 2} + X_{S} n_{S})$	$n_{O 2} - (n_{C} + \frac{1}{2} n_{H 2} + 2 X_{N 2} n_{N 2} + X_{S} n_{S})$
N₂	nitrogen	28	$N_{2} + 2 O_{2} \to 2 N O_{2}$	$n_{N 2}$	$- X_{N 2} n_{N 2}$	$(1 - X_{N 2}) n_{N 2}$
S	sulphur	32	$S + O_{2} \to S O_{2}$	$n_{S}$	$- X_{S} n_{S}$	$(1 - X_{S}) n_{S}$
CO₂	carbon dioxide	44	$C + O_{2} \to C O_{2}$	$0$	$+ n_{C}$	$n_{C}$
H₂O	water	18	$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$	$n_{H 2 O}$	$n_{H 2}$	$n_{H 2 O} + n_{H 2}$
NO₂	nitrogen dioxide	46	$N_{2} + 2 O_{2} \to 2 N O_{2}$	$0$	$+ 2 X_{N 2} n_{N 2}$	$2 X_{N 2} n_{N 2}$
SO₂	sulphur dioxide	64	$S + O_{2} \to S O_{2}$	$0$	$+ X_{S} n_{S}$	$X_{S} n_{S}$
Ar	argon	40	n/a	$n_{A r}$	$0$	$n_{A r}$

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.

$F l u e g a s = n_{O 2} - (n_{C} + \frac{1}{2} n_{H 2} + 2 X_{N 2} n_{N 2} + X_{S} n_{S}) + (1 - X_{N 2}) n_{N 2} + n_{c} + n_{H 2 O} + n_{H 2} + 2 X_{N 2} n_{N 2} + X_{s} n_{s} + n_{A r}$

= n_{O 2} - n_{C} - \frac{1}{2} n_{H 2} - 2 X_{N 2} n_{N 2} - X_{S} n_{S} + (1 - X_{N 2}) n_{N 2} + n_{c} + n_{H 2 O} + n_{H 2} + 2 X_{N 2} n_{N 2} + X_{s} n_{s} + n_{A r}

= n_{O 2} + \frac{1}{2} n_{H 2} + (1 - X_{N 2}) n_{N 2} + n_{H 2 O} + n_{A r}

Substitute the mol fractions n_i with equations in table above:

$F l u e g a s = (\frac{o . C o a l}{32} + 0.21 (1 - w a) A i r) + \frac{1}{2} (\frac{h . C o a l}{2}) + (1 - X_{N 2}) (\frac{n . C o a l}{28} + 0.781 (1 - w a) A i r) + (\frac{m . C o a l}{18} + w a . A i r) + (0.009 (1 - w a) A i r)$

= C o a l \frac{o}{32} + A i r 0.21 (1 - w a) + 0.5 C o a l \frac{h}{2} + C o a l \frac{n}{28} (1 - X_{N 2}) + A i r 0.781 (1 - w a) (1 - X_{N 2}) + C o a l \frac{m}{18} + A i r . w a + A i r 0.009 (1 - w a)

= C o a l (\frac{o}{32} + 0.5 \frac{h}{2} + \frac{n}{28} (1 - X_{N 2}) + \frac{m}{18}) + A i r [w a + 0.781 (1 - w a) (1 - X_{N 2}) + 0.009 (1 - w a) + 0.21 (1 - w a)]

= C o a l (\frac{o}{32} + 0.5 \frac{h}{2} + \frac{n}{28} (1 - X_{N 2}) + \frac{m}{18}) + A i r [w a + 0.781 (1 - w a) - 0.781 X_{N 2} (1 - w a) + 0.009 (1 - w a) + 0.21 (1 - w a)]

= C o a l (\frac{o}{32} + 0.5 \frac{h}{2} + \frac{n}{28} (1 - X_{N 2}) + \frac{m}{18}) + A i r [w a + (1 - w a) - 0.781 X_{N 2} (1 - w a)]

$F l u e g a s = C o a l (\frac{o}{32} + 0.5 \frac{h}{2} + \frac{n}{28} (1 - X_{N 2}) + \frac{m}{18}) + A i r [1 - 0.781 X_{N 2} (1 - w a)]$

Where $F l u e g a s$ and $A i r$ is in kmol/h, $C o a l$ in kg/h, $c, h, o, n, s$ in mass fraction, $w a$ in mol fraction, $X_{N 2}$ in fraction

Oxygen in flue gas

The oxygen in the flue gas can be written as follows:

$O x y g e n i n f l u e g a s = n_{O 2} - (n_{C} + \frac{1}{2} n_{H 2} + 2 X_{N 2} n_{N 2} + X_{S} n_{S})$

= n_{O 2} - n_{C} - \frac{1}{2} n_{H 2} - 2 X_{N 2} n_{N 2} - X_{S} n_{S}

Substitute the mol fractions n_i with equations in table above:

$O x y g e n i n f l u e g a s = (\frac{o . C o a l}{32} + 0.21 (1 - w a) A i r) - (\frac{c . C o a l}{12}) - \frac{1}{2} (\frac{h . C o a l}{2}) - 2 X_{N 2} (\frac{n . C o a l}{28} + 0.781 (1 - w a) A i r) - X_{S} (\frac{s . C o a l}{32})$

= C o a l \frac{o}{32} + A i r 0.21 (1 - w a) - C o a l \frac{c}{12} - 0.5 C o a l \frac{h}{2} - C o a l . 2 X_{N 2} \frac{n}{28} - A i r 2 X_{N 2} 0.781 (1 - w a) - C o a l . X_{S} \frac{s}{32}

$O x y g e n i n f l u e g a s = C o a l [\frac{o}{32} - \frac{c}{12} - 0.5 \frac{h}{2} - 2 X_{N 2} \frac{n}{28} - X_{S} \frac{s}{32}] + A i r [0.21 (1 - w a) - 2 X_{N 2} 0.781 (1 - w a)]$

Where $O x y g e n$ and $A i r$ is in kmol/h, $C o a l$ in kg/h, $c, h, o, n, s$ in mass fraction, $w a$ in mol fraction, $X_{s}, X_{N 2}$ in fraction

% oxygen in flue gas

The percentage oxygen in flue gas is given by the following formula:

x_{O 2} = % o x y g e n i n f l u e g a s = \frac{O x y g e n i n f l u e g a s}{T o t a l f l u e g a s}

If substituting above formulas into this, then:

$C o a l [\frac{o}{32} - \frac{c}{12} - 0.5 \frac{h}{2} - 2 X_{N 2} \frac{n}{28} - X_{S} \frac{s}{32}] + A i r [0.21 (1 - w a) - 2 X_{N 2} 0.781 (1 - w a)]$ $= C o a l [\frac{o}{32} x_{O 2} + 0.5 \frac{h}{2} x_{O 2} + \frac{n}{28} (1 - X_{N 2}) x_{O 2} + \frac{m}{18} x_{O 2}] + A i r [x_{O 2} - 0.781 X_{N 2} (1 - w a) x_{O 2}]$

Get "Coal" on the left and "Air" on the right:

$C o a l [\frac{o}{32} - \frac{c}{12} - 0.5 \frac{h}{2} - 2 X_{N 2} \frac{n}{28} - X_{S} \frac{s}{32}] - C o a l [\frac{o}{32} x_{O 2} + 0.5 \frac{h}{2} x_{O 2} + \frac{n}{28} (1 - X_{N 2}) x_{O 2} + \frac{m}{18} x_{O 2}]$ $= A i r [x_{O 2} - 0.781 X_{N 2} (1 - w a) x_{O 2}] - A i r [0.21 (1 - w a) - 2 X_{N 2} 0.781 (1 - w a)]$

$C o a l [\frac{o}{32} (1 - x_{O 2}) - \frac{c}{12} - 0.5 \frac{h}{2} (1 + x_{O 2}) - \frac{n}{28} (2 X_{N 2} + (1 - X_{N 2}) x_{O 2}) - X_{S} \frac{s}{32} - \frac{m}{18} x_{O 2}]$

$= A i r [x_{O 2} - 0.781 X_{N 2} (1 - w a) x_{O 2} - 0.21 (1 - w a) + 2 X_{N 2} 0.781 (1 - w a)]$

$C o a l [\frac{o}{32} (1 - x_{O 2}) - \frac{c}{12} - 0.5 \frac{h}{2} (1 + x_{O 2}) - \frac{n}{28} (2 X_{N 2} + (1 - X_{N 2}) x_{O 2}) - X_{S} \frac{s}{32} - \frac{m}{18} x_{O 2}]$

$= A i r [x_{O 2} - 0.781 X_{N 2} (1 - w a) x_{O 2} - 0.21 (1 - w a) + 2 X_{N 2} 0.781 (1 - w a)]$

Taking the right hand side separately:

$R H = A i r [- 0.781 X_{N 2} (1 - w a) x_{O 2} - 0.21 (1 - w a) + 2 X_{N 2} 0.781 (1 - w a) + x_{O 2}]$

$R H = A i r [(1 - w a) [- 0.781 X_{N 2} x_{O 2} + 2 X_{N 2} 0.781 - 0.21] + x_{O 2}]$

$R H = A i r [(1 - w a) [0.781 X_{N 2} (2 - x_{O 2}) - 0.21] + x_{O 2}]$

Combining the two equations again:

Coal consumption (total combustion)

$C o a l [\frac{o}{32} (1 - x_{O 2}) - \frac{c}{12} - 0.5 \frac{h}{2} (1 + x_{O 2}) - \frac{n}{28} (2 X_{N 2} + (1 - X_{N 2}) x_{O 2}) - X_{S} \frac{s}{32} - \frac{m}{18} x_{O 2}]$

$= A i r [(1 - w a) [0.781 X_{N 2} (2 - x_{O 2}) - 0.21] + x_{O 2}]$

Thus, the total air dry coal which underwent total combustion is:

$C o a l = A i r \times \frac{(1 - w a) [0.781 X_{N 2} (2 - x_{O 2}) - 0.21] + x_{O 2}}{\frac{o}{32} (1 - x_{O 2}) - \frac{c}{12} - 0.5 \frac{h}{2} (1 + x_{O 2}) - \frac{n}{28} [2 X_{N 2} + (1 - X_{N 2}) x_{O 2}] - X_{S} \frac{s}{32} - \frac{m}{18} x_{O 2}}$

Where:

$C o a l$ 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 CO₂ and SO₂ 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.
$A i r$ 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.
$A i r = a i r m e a s u r e d + a i r l e a k a g e$

Individual components in flue gas

The individual components in the flue gas is then the following (kmol/h):

O_{2} i n f l u e g a s = n_{O 2} - (n_{C} + \frac{1}{2} n_{H 2} + 2 X_{N 2} n_{N 2} + X_{S} n_{S})

= C o a l [\frac{o}{32} - \frac{c}{12} - 0.5 \frac{h}{2} - 2 X_{N 2} \frac{n}{28} - X_{S} \frac{s}{32}] + A i r [0.21 (1 - w a) - 2 X_{N 2} 0.781 (1 - w a)]

N_{2} i n f l u e g a s = (1 - X_{N 2}) n_{N 2}

= (1 - X_{N 2}) [\frac{n \cdot C o a l}{28} + 0.781 (1 - w a) A i r]

C O_{2} i n f l u e g a s = n_{C}

= C o a l \frac{c}{12}

H_{2} O i n f l u e g a s = n_{H 2 O} + n_{H 2} = (C o a l \frac{m}{18} + w a . A i r) + (C o a l \frac{h}{2})

= C o a l (\frac{m}{18} + \frac{h}{2}) + w a . A i r

N O_{2} i n f l u e g a s = 2 X_{N 2} n_{N 2}

= 2 X_{N 2} (C o a l \frac{n}{28} + 0.781 (1 - w a) A i r)

S O_{2} i n f l u e g a s = X_{s} n_{S} = X_{S} (C o a l \frac{s}{32})

= C o a l . X_{S} \frac{s}{32}

A r i n f l u e g a s = n_{A r}

= 0.009 (1 - w a) A i r

Emissions

The CO₂, SO₂ and NO₂ emissions can be calculated as follows:

C O_{2} e m i s s i o n s = c \times C o a l \times \frac{44}{12} i n k g / h

S O_{2} e m i s s i o n s = X_{s} \times c \times C o a l \times \frac{64}{32} i n k g / h

N O_{2} e m i s s i o n s = 2 X_{s} \times (\frac{n \cdot C o a l}{28} + 0.781 (1 - w a) A i r) \times 46 i n k g / h

Calculating water vapour in air from humidity

If:

$T$ = Ambient temperature in °C
$R H$ = Relative humidity as a fraction between 0 and 1
$P$ = Atmospheric pressure in kPa
$p_{s a t}$ = Vapour pressure (at ambient temperature) in kPa
$x_{w a t e r}$ = Mol (volume) fraction of water in air

The maximum amount of water vapour in air is:

x_{w a t e r, m a x} = \frac{p_{s a t}}{P}

The actual mol fraction of water in air is:

$x_{w a t e r} = R H \times x_{w a t e r, m a x}$

The following Buck equation is a very good approximation of the vapour pressure of water below 100°C:

$p_{s a t} = 0.61121 \times e^{\frac{17.368 T}{T + 238.88}}$

Thus, the mol fraction of water in air is:

$x_{w a t e r} = \frac{R H}{P} \times 0.61121 \times e^{\frac{17.368 T}{T + 238.88}}$

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^{a d} = \frac{M}{C o a l^{a d}} = \frac{M_{s} + M_{i}}{C o a l^{a d}} = \frac{C o a l^{a r} m_{s} + C o a l^{a d} m_{i}}{C o a l^{a d}}

If:

C o a l^{a d} = (1 - m_{s}) C o a l^{a r} \Rightarrow C o a l^{a r} = \frac{C o a l^{a d}}{1 - m_{s}}

Then:

m^{a d} = \frac{\frac{C o a l^{a d}}{1 - m_{s}} m_{s} + C o a l^{a d} m_{i}}{C o a l^{a d}}

$m^{a d} = \frac{m_{s}}{1 - m_{s}} + m_{i}$

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

Calculate coal consumption of coal-fired boiler from air flow

Contents

Coal combustion

Symbols and constants

Air composition

Number of moles before combustion

Component balance before and after combustion

Develop formula

Total flue gas

Oxygen in flue gas

% oxygen in flue gas

Coal consumption (total combustion)

Individual components in flue gas

Emissions

Calculating water vapour in air from humidity

Total coal moisture in terms of superficial and inherent moisture

Also see

Navigation menu

Calculate coal consumption of coal-fired boiler from air flow

Coal combustion

Symbols and constants

Air composition

Number of moles before combustion

Component balance before and after combustion

Develop formula

Total flue gas

Oxygen in flue gas

% oxygen in flue gas

Coal consumption (total combustion)

Individual components in flue gas

Emissions

Calculating water vapour in air from humidity

Total coal moisture in terms of superficial and inherent moisture

Also see

Navigation menu

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