2. C 4 First Law Analysis of Reacting Systems The form of the first law for the control volume is(there is no shaft work) ∑nh1+Qcv=∑nh This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid(say air)with no reactions occurring. We need to specify one parameter as the basis of the solution 1 kmole of fuel. 1 kmole of air. 1 kmole total etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen. H2+202→2H20 The reactants and the products are both taken to be at o 1MPa and 25C, so the inlet and exit P and Tare specified. The control volume is the combustion chamber. There is no shaft work done and the sfee is in the form of Equation(C 1. 2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen(elements have enthalpies defined as zero at the reference state) If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by H2oneH2O-n H2o(g) 2x( -241, 827)kJ=-483, 654 k; gaseous state at exit If the water is in a liquid state at the exit of the process 2x(-285,783)kJ=-571,676 There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25C: h= 2442 kJ/kmole A more complex example is provided by the burning of methane(natural gas )in oxygen, CH4+2O2→CO2+2H2O) The components in this reaction equation are three ideal gases(methane, oxygen, and cO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that ∑n再-(列2m=-4873J n-()2+212)m =-393,522+2(-285,838)=-965,198kJ Qcv=-965,198kJ-(-74,873)=-890,325kJ 2C-52.C.4 First Law Analysis of Reacting Systems The form of the first law for the control volume is (there is no shaft work): ∑n h + Q ˙ ˙ ˙ CV = ∑nh . ii e e R P This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid (say air) with no reactions occurring. We need to specify one parameter as the basis of the solution; 1 kmole of fuel, 1 kmole of air, 1 kmole total, etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen. H2 + 20 → 2HO 2 2 The reactants and the products are both taken to be at 0.1MPa and 25o C, so the inlet and exit P and T are specified. The control volume is the combustion chamber. There is no shaft work done and the SFEE is in the form of Equation (C.1.2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen (elements have enthalpies defined as zero at the reference state). If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by: o hf n˙eH O he HO = n˙e ( ) 2 HO HOg ( ) 2 2 2 = 2 x (- 241,827)kJ = - 483,654 kJ; gaseous state at exit. If the water is in a liquid state at the exit of the process: o hf n˙eH O he HO = n˙e ( ) 2 HO HOl ( ) 2 2 2 = 2 x (- 285,783) kJ = - 571, 676. There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25o C: hfg = 2442 kJ/kmole. A more complex example is provided by the burning of methane (natural gas) in oxygen, producing . CH4 + 2O2 → CO2 + 2H O ( )l 2 The components in this reaction equation are three ideal gases (methane, oxygen, and CO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that: ∑nh hf o , ii = ( ) = −74 873 kJ R CH4 ∑nh h ee = ( ) f + 2( ) o CO2 hf o HOl P 2 ( ) Q =-393,522 + 2(-285,838) = -965,198 kJ CV = −965 198 , kJ – (-74,873) = -890,325 kJ. 2C-5