Stoichiometry, Energy Balances, Heat ransfer, Chemical Equilibrium, and Adiabatic Flame emperatures Geof Silcox geoff@che.utah.edu (80)58-880 University of Utah Chemical Engineering Salt Lake City, Utah May 007 Stoichiometry
Outline SOICHIOMERY combustion reactions combustion with air equivalence ratio, stoichiometric ratio percent theoretical air, percent excess air percent excess oxygen in stack gases dry and wet basis for concentrations actual and theoretical concentrations the effects of mixing and in leakage SOICHIOMERY Combustion reactions CH O CO H O CO O CO H O HO In general nc no nco m m mh O HO m m CH n O nco HO or n m
SOICHIOMERY Combustion with air Assume air is % O and 79% N, by volume. hen each mole O brings with it 3.76 mole N. m m CH n m n O 3.76 n N m m nco HO 3.76 n N Equivalence ratio Φ= SOICHIOMERY Stoichiometric ratio SR = Φ mols fuel mols oxidant supplied by operator mols fuel mols oxidant reqd for complete combustion SR = /Φ = means that conditions are stoichiometric or chemically balanced. 3
SOICHIOMERY Φ and SR Fuel rich (reducing): SR < and Φ> SR > and Φ < Fuel lean (oxidizing): Alternate forms w/ constant amount of fuel Φ= SR mols oxidant reqd mols oxidant fed mols oxidant fed = mols oxidant reqd SOICHIOMERY Percent theoretical air 00 %A = 00( SR) = Φ Percent excess air %EA = %A 00 = 00 ( SR ) = 00 Φ
SOICHIOMERY Percent excess oxygen in stack gases Defined as actual mole (volume) percent in stack. Not to be confused with %EA or %A. Can be measured on a dry or wet basis. Dry or wet basis for stack measurement Dry basis gives higher value because water vapor acts as a diluent and is 0 5% of stack gases Flue gas mole fractions, dry basis SOICHIOMERY Actual and theoretical concentrations 0. 0.8 0.6 0. 0. 0. 0.08 0.06 0.0 0.0 0 CO theoretical effects of imperfect mixing effects of air leakage CO theoretical O theoretical 0.8...6.8 SR 5
SOICHIOMERY Controlling excess air ppm CO 00 50 he rise in CO is steep as you approach SR = and it is insensitive to air leakage. Air flow control by CO should lead to a close approach to SR =..0.. Stoichiometric ratio SOICHIOMERY Spreadsheets for furnace material balance calculations FURN_MB.xls: balance for liquid and solid fuels GAS_MB.xls: balance for gaseous fuels Author: Jost Wendt 6
Outline Conservation of mass and energy Example: drying salt with hot flue gas Energy balances on systems that involve reactions - the absolute enthalpy Example: adiabatic combustion of H in air Heats of combustion and higher and lower heating values Energy balance on a furnace available heat and the efficiency of combustion 7
Energy and mass are conserved Mass entering () Q Control volume or system W Mass leaving () e = u V h = u pv gz de dt CV V V = Q in, net W in, net m h gz m h gz dm dt CV = in out Changes in kinetic and potential energy are usually small compared to changes in internal energy Unsteady du dt CV = Q W mh h in, net in, net Steady Q W mh h in, net in, net = 0 8
Drying salt with combustion gases Flue gas at 000 F to be used to dry 0 tons/h salt that is 3 wt. % water. he inlet temperature of the salt is 0 F. he dry salt exits the dryer at 90-0 F. What flow rate of flue gas do we need if we assume the flue gas leaves the dryer at 00 F? Drying salt with combustion gases flue gas, 000 F m =? dry salt, 0 F 77,600 lb/h H O(l), 0 F 00 lb/h 3 Dryer 5 6 flue gas, 00 F = dry salt, 00 F 77,600 lb/h H O(g), 00 F 00 lb/h = = 5 = 3 6 9
Drying salt with combustion gases energy balance over adiabatic dryer flue gas, 000 F m =? = dry salt, 0 F 5 77,600 lb/h Dryer 3 6 H O(l), 0 F 00 lb/h ( ) ( ) ( ) h h h h h h 5 3 3 6 = 0 ( ) ( ) h h m h h = h h 5 3 6 3 flue gas, 00 F dry salt, 00 F 77,600 lb/h H O(g), 00 F 00 lb/h Drying salt with combustion gases energy balance over adiabatic dryer ( ) ( ) ( ) c ( ) c m c h c = salt 5 3 H O( l ) 5 3 fg ph O( g ) 6 5 h c c fg salt pgas HO Btu Btu = 978 cpgas = 0.5 lb lb F Btu Btu = 0. cph O( g ) = 0.7 lb F lb F Btu lb =.0 = 36,500 lb F h 0
Energy balances on reacting systems the absolute enthalpy he reference state for systems in which reactions are occurring is conveniently chosen to be the elements at ref = 5 C (98.5 K) and bar. o i( ) = f, i( ref) p, i ref Enthalpy of formation at bar, 98 K h h c d Absolute enthalpy Sensible enthalpy change Heat capacity as a function of (K) o h i( ) = hf, i( ref) cp, id ref c p 3 = a a a 3 a a 5 R h R a a a a a = a 3 5 3 3 5 6 where is in Kelvin and R is the universal gas constant, R = 8.35 kj/(kmol K). Spreadsheet: ther_coef.xls
Adiabatic combustion of H with air, no dissociation. What is exit temperature? Inlet (i) 0. kmol/s H 0. kmol/s O 0.79 kmol/s N 300 K, bar Control volume Q = 0 W = 0 in Q W nh n h Exit (e) 0. kmol/s H O 0.79 kmol/s N =?, bar in, net in, net i i e e = 0 i e in Spreadsheet: h_flame.xls Adiabatic combustion of H nh n h = 0 i i e e i e o i( ) = f, i( ref) p, i h h c d ref We have one equation in one unknown,. his is conveniently solved in Excel with the Goal Seek tool to give an exit temperature of 530K. If we allow for dissociation of the water (at equilibrium), e = 390K. Spreadsheet: h_flame.xls
Heats (enthalpies) of combustion and higher (gross) and lower (net) heating values (HHV and LHV) 98 hrxn = νp h fp [ νr hfr ] 98 f 98 f 98 f 98 f prod 98 kj CH hf = 7.600 mol O h = 0 CO h = 393.50 HOl ( ) h = 85.830 HOg ( ) h =.86 react CH O CO H O( liq) CH O CO H O( gas) Heats of reaction and HHV and LHV CH O CO H O( liq) 98 rxn ( ) ( ) h ( l) = 393.5 85.83 7.6 0 = 890.57 kj / mol CH CH O CO H O( gas) 98 rxn ( ) ( ) h ( g) = 393.5.86 7.6 0 = 80.56 kj / mol CH Gross heating value HHV = h () l = 98 rxn 890.57 kj / mol CH Net heating value LHV = h ( g) = 98 rxn 80.56 kj / mol CH 3
Energy balance over a furnace Inlet (i) in Control volume Exit (e) Q W = 0 Q nh n h in, net i i e e = 0 i e in Use absolute enthalpies for energy balances involving reactions. o i( ) f, i( ref) p, i h = h c d ref = 98.5 K ref Energy balance over a furnace available heat i e o o Q avail = n i hf, i ( ref ) cp, id ne hf, e( ref ) cp, ed i e ref ref Available heat = gross heat input latent heat loss unburned fuel loss sensible flue gas heat loss Available heat = net heat input unburned fuel loss sensible flue gas heat loss
hermal efficiency of a combustion process Definition in terms of HHV Q avail η gross = HHV fuel fuel ( ) Definition in terms of LLV Q avail η net = ( LHV) 5