A combustion chamber normally involves heat output but no heat input.
Then the energy balance for a typical steady-flow combustion process
becomes
(15–13)Energy in by mass Energy out by mass
per mole of fuel per mole of fuel
It expresses that the heat output during a combustion process is simply the
difference between the energy of the reactants entering and the energy of
the products leaving the combustion chamber.
Closed Systems
The general closed-system energy balance relation EinEoutEsystemcan
be expressed for a stationary chemically reacting closed systemas
(15–14)where Uprodrepresents the internal energy of the products and Ureactrepre-
sents the internal energy of the reactants. To avoid using another property—
the internal energy of formation u–f°—we utilize the definition of enthalpy
(u–h
- Pv–oru–f°u–u–°h
- °fh
- h
- °Pv) and express the above
equation as (Fig. 15–22)
- °Pv) and express the above
- h
- °fh
(15–15)where we have taken heat transfer tothe system and work done bythe sys-
tem to be positivequantities. The Pv–terms are negligible for solids and liq-
uids, and can be replaced by RuTfor gases that behave as an ideal gas. Also,
if desired, the terms in Eq. 15–15 can be replaced by u–.
The work term in Eq. 15–15 represents all forms of work, including the
boundary work. It was shown in Chap. 4 that UWbHfor nonreact-
ing closed systems undergoing a quasi-equilibrium Pconstant expansion or
compression process. This is also the case for chemically reacting systems.
There are several important considerations in the analysis of reacting sys-
tems. For example, we need to know whether the fuel is a solid, a liquid, or
a gas since the enthalpy of formation hf° of a fuel depends on the phase of
the fuel. We also need to know the state of the fuel when it enters the com-
bustion chamber in order to determine its enthalpy. For entropy calculations
it is especially important to know if the fuel and air enter the combustion
chamber premixed or separately. When the combustion products are cooled
to low temperatures, we need to consider the possibility of condensation of
some of the water vapor in the product gases.
EXAMPLE 15–6 First-Law Analysis of Steady-Flow CombustionLiquid propane (C 3 H 8 ) enters a combustion chamber at 25°C at a rate of
0.05 kg/min where it is mixed and burned with 50 percent excess air that
enters the combustion chamber at 7°C, as shown in Fig. 15–23. An analysis
of the combustion gases reveals that all the hydrogen in the fuel burns
to H 2 O but only 90 percent of the carbon burns to CO 2 , with the remaining
10 percent forming CO. If the exit temperature of the combustion gases ishPvQWaNp 1 h°fhh°Pv (^2) paNr 1 h°fhh°Pv (^2) r
1 QinQout 2 1 WinWout 2 UprodUreact¬¬ 1 kJ>kmol fuel 2
QoutaNr 1 h°fhh° (^2) raNp 1 h°fhh° (^2) p
Chapter 15 | 767
15555255553 15555255553
U = H – PV
= N(h°f + h – h°) – PV
= N(h°f + h – h° – Pv)
–––
––– –
FIGURE 15–22
An expression for the internal energy
of a chemical component in terms of
the enthalpy.
C 3 H 8 ()
CO 2
Q =?
1500 K
AIR
7 °C
25 °C, 0.05 kg/min
CO
N 2
Combustion
chamber
H 2 O
O 2
·
FIGURE 15–23
Schematic for Example 15–6.