Thermodynamics

Q out a N r 1h° f h h°2 r15555255553 a N p 1h° f h h°2 p15555255553(15–13)Chapter 15 | 767A combustion chamber normally involves heat output but no heat input.Then the energy balance for a typical steady-flow combustion processbecomesEnergy in by massper mole of fuelEnergy out by massper mole of fuelIt expresses that the heat output during a combustion process is simply thedifference between the energy of the reactants entering and the energy ofthe products leaving the combustion chamber.Closed SystemsThe general closed-system energy balance relation E in E out E system canbe expressed for a stationary chemically reacting closed system as1Q in Q out 2 1W in W out 2 U prod U react 1kJ>kmol fuel2(15–14)where U prod represents the internal energy of the products and U react representsthe internal energy of the reactants. To avoid using another property—the internal energy of formation u – °—we f utilize the definition of enthalpy(u – h – Pv – or u – f ° u– u – ° h – ° f h – h – ° Pv) and express the aboveequation as (Fig. 15–22)Q W a N p 1h° f h h° Pv 2 p a N r 1h° f h h° Pv 2 r(15–15)where we have taken heat transfer to the system and work done by the systemto be positive quantities. The Pv – terms are negligible for solids and liquids,and can be replaced by R u T for gases that behave as an ideal gas. Also,if desired, the h Pv terms in Eq. 15–15 can be replaced by u – .The work term in Eq. 15–15 represents all forms of work, including theboundary work. It was shown in Chap. 4 that U W b H for nonreactingclosed systems undergoing a quasi-equilibrium P constant expansion orcompression process. This is also the case for chemically reacting systems.There are several important considerations in the analysis of reacting systems.For example, we need to know whether the fuel is a solid, a liquid, ora gas since the enthalpy of formation h° f of a fuel depends on the phase ofthe fuel. We also need to know the state of the fuel when it enters the combustionchamber in order to determine its enthalpy. For entropy calculationsit is especially important to know if the fuel and air enter the combustionchamber premixed or separately. When the combustion products are cooledto low temperatures, we need to consider the possibility of condensation ofsome of the water vapor in the product gases.EXAMPLE 15–6First-Law Analysis of Steady-Flow CombustionLiquid propane (C 3 H 8 ) enters a combustion chamber at 25°C at a rate of0.05 kg/min where it is mixed and burned with 50 percent excess air thatenters the combustion chamber at 7°C, as shown in Fig. 15–23. An analysisof the combustion gases reveals that all the hydrogen in the fuel burnsto H 2 O but only 90 percent of the carbon burns to CO 2 , with the remaining10 percent forming CO. If the exit temperature of the combustion gases isU = H – PV– – –= N(h°f + h – h°) – PV– – – –= N(h°f + h – h° – Pv)FIGURE 15–22An expression for the internal energyof a chemical component in terms ofthe enthalpy.C 3 H 8 ()25°C, 0.05 kg/minAIR7°C·Q = ?CombustionchamberFIGURE 15–23Schematic for Example 15–6.H 2 OCO1500 K 2COO 2N 2