Conservation of mass and energy is the basis for any considerations in combustion engineering. The formulation presented in this lecture is based on the work of
2 Mass and Energy Balance
[9, 10, 11]. In formulating mass and energy balances we need first of all to identify the control volume. Control volume, Fig. 2.1, is a region of space bounded by a control surface (boundary) through which energy and matter may be transferred.
Fig. 2.1: Control volume for mass and energy balance
Once the control volume is identified an appropriate time basis must be specified.
There are two options for formulating the balances – a formulation at an instant and a formulation over a time interval.
2.1.1 Mass and Energy Balance at an Instant
Since the mass and energy balances must be satisfied at each and every instant of time t, one option involves formulating the law on a rate basis. That is, at any instant there must be a balance between all mass rates expressed in kg/s and simultaneously, there must be a balance between all energy rates expressed in J/s (=W).
The mass balance at an instant reads, Fig. 2.1:
˙
min+ ˙mg− ˙mout= ˙mstored≡ d(m)
dt (2.1)
where ˙min, ˙mg, ˙mout, ˙mstored are the rates with which the mass enters the control volume, the rate of mass generation inside the control volume, the rate with which matter leaves the control volume and the rate of accumulation (storage) of matter within the volume, respectively. All these terms are expressed in kg/s. Symbol m stands for the mass of the system expressed in kg.
Eq. (2.1) is the most general form of the mass balance at an instant. It contains the mass generation term ( ˙mg) that is non-zero only if nuclear reactions are considered
2.1 General Formulation of Mass and Energy Balance
during which matter can be converted into energy. In any other systems with non-nuclear reactions present, the mass generation term is zero and therefore during this combustion course we will use the following equation for the mass balance at an instance:
˙
min− ˙mout = dm
dt (2.2)
The energy balance at an instant reads:
E˙in+ ˙Eg− ˙Eout = ˙Estored ≡ dE
dt (2.3)
where ˙Ein, ˙Eg, ˙Eout, ˙Estored stand for the rates with which energy enters the con-trol volume, the rate of energy generation in the volume, the rate with which the energy leaves the volume and the rate of accumulation (storage) within the con-trol volume, respectively. All these terms are expressed in J/s (W). The symbol E represents the system energy expressed in joules.
2.1.2 Mass and Energy Balance over a Time Interval
The second option is to formulate balances over a time interval, thus the mass balance is formulated in amount of mass (kilograms) whilst the energy balance is formulated in amount of energy (joules). The mass balance over a time interval
∆t = t2− t1 reads:
The terms on the left hand side of Eqs. (2.4) and (2.5) denote the amount of mass entering the control volume over the time period ∆t, and the amount of mass that leaves the control volume over this interval, respectively. The right hand side of Eqs. (2.4) and (2.5) represents the amount of mass stored (accumulated) within the system in this time period.
The energy balance over a time interval reads:
t2
2 Mass and Energy Balance
or
Ein+ Eg− Eout = ∆E (2.7)
where the terms on the left hand side of Eqs. (2.6) and (2.7) stand for the amount of energy entering the control volume, the amount of energy generated in the volume, and the amount of energy leaving the volume in the time period ∆t = t2− t1. The right hand side of Eqs. (2.6) and (2.7) represents the energy stored (accumulated) in the control volume over the time interval ∆t = t2− t1.
2.1.3 Mass and Energy Balance under Steady-State Conditions Under steady-state conditions when there is neither mass nor energy accumulation (storage) in the system we obtain:
from Eq. (2.2) for mass balance:
˙
min = ˙mout (2.8)
from Eq. (2.3) for energy balance:
E˙in+ ˙Eg= ˙Eout (2.9)
2.1.4 Example of a Mass Balance of a Furnace
The purpose of making a mass balance under steady state conditions is to calculate the out-coming amount of mass and to calculate its make up (composition). A correct mass balance is a prerequisite to a subsequent energy balance.
There are two basic ways of formulating a mass balance; by expressing the in-coming and outin-coming flows in kmol/h (kmol/s) and their composition in mol (volume) fractions or by expressing the streams in kg/h (kg/s) and their compo-sition in mass fractions. The first way is popular among chemical engineers since it deals with kmol and the rates of chemical reactions are typically expressed using concentrations (kmol/m3/s). The latter way is common in combustion engineer-ing. Both methods have advantages and disadvantages and they are equivalent.
They both lead to the same results. The combustion stoichiometry calculations (see Chapter 1) are required to calculate composition of combustion products leaving the system.
2.1 General Formulation of Mass and Energy Balance
Example 2.1
A boiler fired with methane is used to generate hot water. The burner of the boiler is fed with 200 kg/h of methane and 4000 kg/h of dry air and it is operated under steady state conditions. Make the mass balance for the boiler assuming that the combustion is complete. Carry out the calculations using the molar flow rates (kmol/h). Repeat the calculations using mass flow rates (kg/h).
200 kg/h CH 4000 kg/h air
4 Boiler
Combustion products
control volume boundary
Fig. 2.2: Mass balance for the boiler
Assumptions:
(a) the methane combustion proceeds to completion resulting in CO2 and H2O, (b) we place the control volume over the boiler (see Fig. 2.2),
(c) the boiler operates under steady state and therefore we formulate the mass balance at an instance with the mass accumulation term equal to zero.
Mass balance in kmol/h
The starting point to any mass balance is an estimation of the incoming (in-put) streams and writing down proper chemical reactions. Thus, the incom-ing streams are: ˙mCH4 = 12.5 kmolCH4/h and since the molar mass of air is Mair= 0.21 · 32 + 0.79 · 28 = 28.84 kg/kmol we obtain for the incoming stream of air ˙mair= 28.84kg/kmol4000kg/h = 138.696(kmol air )/h. The oxidation reaction is:
CH4+ 2 O2 −−→ CO2+ 2 H2O
and therefore the minimum air requirement is 0.212 = 9.5238 kmol of air
kmol of fuel. The excess air ratio is
λ =
138.696kmol of air/h 12.5kmol of CH4/h
9.5238 kmol of air/kmol of CH4 = 1.165 Since λ > 1 the combustion products contain CO2, H2O, N2 and O2.
The amount of the wet combustion products is 12.5 · (3 + 0.79 · 9.5238 + (1.165 − 1) · 9.5238) = 151.20kmol/h while for the dry combustion products the figure of
2 Mass and Energy Balance
12.5 · (1 + 0.79 · 9.5238 + (1.165 − 1) · 9.5238) = 126.20kmol/h is applicable.
The composition of the combustion products is as follows:
xCO2,wet = 12.5
151.20 = 0.0827
xH2O = 2 · 12.5
151.20 = 0.1653 xN2,wet = 12.5 ·0.79 · 9.5238 + 0.79 · (1.165 − 1) · 9.5238
151.20 = 109.5654
151.20 = 0.7246 xO2,wet= 12.5 ·(1.165 − 1) · 9.5238 · 0.21
151.20 = 4.1250
151.20 = 0.0273
and the molar mass of the wet combustion products is:
Mwet,products= 0.0827 · 44 + 0.1653 · 18 + 0.7246 · 28 + 0.0273 · 32 = 27.7766kmolkg
The composition of the dry combustion products is as follows:
xCO
2,dry = 12.5
126.20 = 0.099 xN2,dry = 109.5654
126.20 = 0.8682 xO2,dry = 4.1250
126.20 = 0.0327
and the molar mass of the dry products is:
Mdry,products = 0.099 · 44 + 0.8682 · 28 + 0.0327 · 32 = 29.712kg/kmol
Knowing the molar fractions of the species, one may easily calculate the mass fractions since:
wCO2,wet= 0.0827 · 44
27.7766 = 0.1310 wCO2,dry = 0.099 · 44
29.712 = 0.1466 wH2O= 0.1653 · 18
27.7766 = 0.1071 wN2,wet = 0.7246 · 28
27.7766 = 0.7304 wN2,dry = 0.8682 · 28
29.712 = 0.8182
2.1 General Formulation of Mass and Energy Balance
and
wO2,wet = 0.0273 · 32
27.7766 = 0.0315 wO2,dry = 0.0327 · 32
29.712 = 0.0352
The tables below summarise the mass balance:
Table 2.1: The incoming and out-coming streams in kmol/h
IN OUT
Species kmol/h Species WET DRY
kmol/h kmol/h Methane 12.50 Carbon dioxide 12.50 12.50 Nitrogen 109.57 Water vapour 25.00 −
Oxygen 29.13 Nitrogen 109.57 109.57
Oxygen 4.13 4.13
P 151.20 P
151.20 126.20
Table 2.2: Composition of the combustion products
Species Molar fraction Mass fraction
Wet Dry Wet Dry
Carbon dioxide 0.0827 0.0990 0.1310 0.1466
Water vapour 0.1653 − 0.1071 −
Nitrogen 0.7246 0.8682 0.7304 0.8182 Oxygen 0.0273 0.0327 0.0315 0.0352
P 0.9999 0.9999 1.0000 1.0000
Mass balance in kg/h The incoming streams are:
˙
mCH4 = 200 kg/h and m˙air = 4000 kg/h.
The outcoming flow rate (in kg/h) is easily obtained since:
˙
mproducts= 200 kg/h + 4000 kg/h = 4200 kg/h.
While making the mass balance in kilograms it is convenient to write the oxidation reaction
CH4+ 2 O2 −−→ CO2+ 2 H2O
2 Mass and Energy Balance
using kilogram rather than kmol:
16 kg CH4+ 64 kg O2= 44 kg CO2+ 36 kg H2O
The minimum air requirement is 64/16/0.233 = 17.1674 kg of air/kg of methane and the excess air ratio is:
λ =
4000kg of air/h 200kg of CH4/h
17.1674 kg of air/kg of CH4 = 1.165
Since λ > 1 the combustion products will contain CO2, H2O, N2 and O2. The combustion products stream contains:
Methane none
Carbon dioxide 200 ·4416 = 550 kg/h
Water vapour 200 ·3616 = 450 kg/h
Nitrogen 0.767 · 4000 = 3068 kg/h
Oxygen 0.233 · (4000 − 200 · 17.1674) = 132 kg/h
TOTAL (wet) 4200 kg/h
and the amount of combustion products dry is then 3750 kg/h.
The species mass fractions can be easily calculated:
wCO2,wet = 550
4200 = 0.131 wCO2,dry = 0.1467 wH2O= 450
4200 = 0.1071 wN2,wet = 3068
4200 = 0.7305 wN2,dry = 0.8181 wO2,wet= 132
4200 = 0.0314 wO2,dry = 0.0352
The conversion of the mass fractions into the molar fractions (if required) is also rather straightforward:
1 Mproducts,wet
= 0.131
44 + 0.1071
18 + 0.7305
28 + 0.0314
32 = 0.0360
and Mproducts,wet= 27.7795 kg/kmol