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Enthalpy • Form of energy important in flow systems where temperature and chemical composition play an important role h = u + Pv h : specific enthalpy, (kJ/kg, J/mol, Btu/lb etc.) u : specific internal energy P : pressure v : specific volume • Units for u and Pv obviously need to be consistent with h • Total enthalpy H (kJ) = (m)(h) ; i.e. (kg) (kJ/kg) etc. Specific enthalpy for a pure substance • From phase rule, C=1, P=1, F=2 Thus, h=h(T,P) • For ideal gases: h=h(T) Holds for many other substances as well • Notable exception: STEAM (That’s why we need steam tables to look up properties of superheated steam and compressed liquid as a function of both T and P) Change of enthalpy with temperature dh C p heat capacity at constant pressure dT kJ Btu kg.C , lb - mol.F etc. ith Thus, if C p is constant w temperature : Δh C p ΔT ; i.e. hT2 hT1 C p (T2 T1 ) ith if C p is NOT constant w temperature : T2 hT2 hT1 C p dT T1 Change of enthalpy with phase The (latent) heat of vaporization at T: hvT hvapourT hliquid T Heat of fusion at T: h fT hliquid T hsolid T Heat of sublimation at T: hsT hvapourT hsolid T Note: we have h vs T, for water, used h for including specific phase enthalpy Changes Adapted from Fig. 23.5, Himmelblau, 7th ed. hv100C hE hD h f 0 C hC hB hs30C hG hA G . hF hE C p H O vapour (130 100) 2 hD hC C p H O liquid (100 0) 2 hB hA C p H2O ice (0 (30)) hvT hvapourT hliquid T h fT hliquid T hsolid T hsT hvapourT hsolid T • The above formulation implies that each of the phase changes (vaporization, fusion (melting) and sublimation) can take place at different temperatures and the enthalpy change associated with the phase change is a function of temperature. • In practice it is only the vaporization temperature and enthalpy that shows any variation which we will be interested in. • The enthalpy of vaporization is at its maximum value at 0 C (the normal melting point) and it is 0 at _____ • For water, the value of 40.65 kJ/mol is the value at the normal boiling temperature Absolute (relative ?) value of h choosing a DATUM • We can calculate changes in specific enthalpy by dh=CpdT • What about the absolute value? h=h(T) • Remember h is what we need, to find H=mh We have, T2 hT2 hT1 C p (T2 T1 ) or hT2 hT1 C p dT T1 Define hT1 0 ; this is called " choosing a datum" Then, T2 hT2 C p (T2 T1 ) or hT2 C p dT T1 relative to T1 Enthalpy of a mixture - ideal mixtures • Simply add the enthalpy of the components to get the total enthalpy mmixture m A mB mC ... m: moles or mass H mixture H A H B H C ... H : enthalpy (kJ, Btu etc.) H A m A hA kJ kJ kg or kg lb mol Btu etc. lb mol H mixture mi h i H mixture hmixture mmixture Enthalpy of a mixture - nonideal mixtures 1+12 ! • Examples: strong acid and base solutions (1 L of 2 M HCl @ 25C, 1atm) (1 L of H 2O @ 25C, 1atm) (2 L of 1 M HCl @ 25 C, 1 atm) heat • The resulting solution must have less energy than the sum of the energies of its constituents at the same temperature and pressure. Properties of humid air expressed per unit mass of BDA BDA H2O (vapour) Humid air Mass (Kg) 1 w 1+w Cp, kJ/kg.C, (assumed constant 1.00 1.88 1.00+1.88w over ambient temperature range) h, kJ/kg, at T (C) Relative to BDA and T H2O (liquid) at 0 C, 1 T hv , H 2O ,0 C 1.88T (hv, H 2O,0 C 1.88T ) w atm air hhumid Cp BDA (T 0) hv , H 2O ,0 C C p H 2Ovapour T 0 w kJ kJ kJ kJ kg H 2O C C kg BDA (kg BDA)(C ) kg H 2O (kg H 2O)(C ) kg BDA Properties of humid air expressed per unit mass of BDA BDA H2O (vapour) Humid air Mass (lb) 1 w 1+w Cp, Btu/lb.F, (assumed constant 0.25 0.45 0.25+0.45w over ambient temperature range) h, Btu/lb, at T (F) Relative to BDA at 0 F T 0.25T and H2O (liquid) at 32 hv, H 2O,32 F 0.45T F, 1 atm (1061 0.45T ) w hhumid Cp BDA (T 0) hv , H 2O ,32 F C p H 2Ovapour T 32 w air Btu Btu Btu Btu lb H 2O F F lb BDA (lb BDA)(F ) lb H 2O (lb H 2O)(F ) lb BDA Temperatures and temperature differences • Temperature: C -17.8 0 10 20 25 K 290.8 273 283 293 298 F 0 32 50 68 77 R 460 492 510 528 537 • Temperature difference: C = K F = R F = 1.8 C Energy balance Conservation of energy in a flow system dominated by thermal and composition effects (i.e. kinetic and potential energy differences are relatively small) Hin + Q = Hout + W Where: Hin, Hout : The enthalpy of the material entering and leaving the system Q : heat (thermal energy) added to the system by heat transfer (conduction, convection, radiation) Q = UA(Tsurroundings - Tsystem) W : Work done by the system on the surroundings (via some mechanical means like a shaft) • Each or these terms has units of energy, kJ, Btu, etc. Energy balance for a flow system dominated by thermal and composition effects (i.e. kinetic and potential energy differences are relatively small) Hin + Q = Hout + W W : Work done by the system on the surroundings Hin Hout (with material coming in) (with material leaving) Q : heat (thermal energy) added to the system by heat transfer Energy balance Hin + Q = Hout + W This is an expression of the first law of thermodynamics (conservation of energy). Usual form in MAAE2400: ΔU = Q – W What is different here? - flow vs open system; if the system is closed we are only dealing with the changes in the internal energy of the material in the system, when material enters and leaves the system the pv term in h = u + pv represents the the work that is involved in pushing the material into the system and pushing it out again. Another way of looking at the situation is to consider pv as the material’s ability to do work (hence energy) by virtue of its pressure. - The Δ in ΔU indicates change from initial condition to final condition. In a flow system we can have multiple streams and so we are not looking at the difference in energy of the same material. We need to look at all the energy in, and all the energy out instead of defining a difference between two streams Strategy for Analyzing Energy Balance problems mh in Q mhout (assuming W 0) 1) See strategy for analyzing material balance problems. - Usually the material balances have to be solved first before the energy balance is attempted, so that the m’s are known. - Sometimes they are simultaneous. - No problem that involves just an energy balance. - Never start with the energy balance 2) Choose a DATUM for all components going in and out of your system, so that the h’s can be calculated. 3) Determine the status of Q: - Unknown: the energy balance has to be solved for Q - Known : the energy balance can be used as one more equation to solve for an unknown (a material balance unknown, temperature, or state - Often we are interested in the adiabatic, Q=0, case • SYSTEM: Wet Bulb Large amount of air blown over the wet bulb (temperature and humidity of the air in and out are the same) System is really not at steady state since the water on the wick is evaporating. However, the wet bulb temperature is established pretty quickly and is maintained until all the water is evaporated. So we can analyze the flows at steady state. • MASS BALANCES: – air: IN = OUT – water: DEPLETION = OUT Wet bulb • ENERGY BALANCE: Hin + Q = Hout • Datum: BDA @ Tair, H2O (liquid) @ Twb Hair + Q = Hair + H evaporated water Rate of water evaporation : k g wTwb wT mass/time H evaporated water mh k g wTwb wT hv energy/time Q, Rate of heat transfer : hc (T Twb ) energy/time kg : mass transfer coefficient, hc : heat transfer coefficient Wet bulb energy balance Q H evaporated water hc (T Twb ) k g ( wTwb wT )hv wTwb wT hc 1 Twb T k g hv For the conditions of the wet bulb system: hc 0.25 Btu / lb.F ; 1.04 kJ / kg.K kg • SYSTEM: Adiabatic saturation tower • MASS BALANCES: • BDA: 1 kg BDA in = 1 kg BDA out (BASIS) • Water: w kg in with air + m kg make-up = ws kg out with air Adiabatic saturation • ENERGY BALANCE: Hin = Hout • Datum: BDA @ Ts, H2O (liquid) @ Ts H IN BDA : C pBDA (T Ts ) water vapour : [hv C pvapour (T Ts )]w make up water : mC p liquid (Ts Ts ) H OUT BDA : C pBDA (Ts Ts ) water vapour : [hv C pvapour (Ts Ts )]ws Adiabatic saturation energy balance C pBDA (T Ts ) [hv C pvapour (T Ts )]w hv ws (C pBDA C pvapour w)(T Ts ) hv ( ws w) (CS )(T Ts ) hv ( ws w) wTs wT Cs Ts T hv • Energy balance on wet bulb • Energy balance for adiabatic saturation wTwb wT hc 1 wTs wT Cs Twb T k g hv Ts T hv hc 1.04 kJ / kg.K (water air systemin wet bulb scenario) kg Cs C pBDA C pvapour w 1.0 1.88w kJ / kgBDA.K w 0.01 0.1 Cs 1.02 1.19 ; very close to the1.04 kJ / kg.K value Thus there is little error in taking the wet bulb lines to also represent adiabatic saturation lines on the w vs T coordinates of the water- air system, the PSYCHROMET RIC CHART. Energy balance for systems with chemical reaction • So far we have considered the change of enthalpy only with temperature and phase (solid, liguid, vapour) • Thus, in choosing a DATUM we have only specified a temperature and a state, e.g. H2O (liquid) at 0 C, BDA at 0 C etc. • What about the following system? 1 mol CH4 1 mol CO2 2 mol O2 2 mol H2O At 25 C, 1 atm At 25 C, 1 atm CH 4 2O2 CO2 2 H 2O 1 mol CH4 1 mol CO2 2 mol O2 2 mol H2O (g) Q At 25 C, 1 atm At 25 C, 1 atm • If we set DATUM: 25 C for all gases Then, Hin = Hout = 0, and Q = Hout – Hin = 0 • However, when we carry out this process, we observe that Q = - 802.32 kJ, i.e. the system gives off energy to the surroundings • It looks like there is “generation” of energy in the system. So we revise our energy balance to read: Hin + Q + heat generated = Hout • We know that energy is neither created nor destroyed but can change from one form to another • So what we are observing in a chemical reaction is the transformation of energy from one form to another • We are transforming the bond energies in the molecules to thermal energy • We cannot measure bond energies by measuring the temperature or phase of a substance, so it appears as if energy is being “created” • Other reactions may appear to “consume” energy. CH 4 2O2 CO2 2 H 2O H rxn 802 .32 kJ 0 • The negative sign indicates an exothermic reaction, i.e. the reaction creates energy • The opposite is an endothermic reaction, consuming energy • We calculate H0rxn by: H = ( mi h )products - ( mi h )reactants 0 rxn 0 fi 0 fi • Where the h0f are the standard heats of formation for each compound, tabulated in chemistry texts. • The superscript 0 refers to the standard state (25 C, 1 atm) • The bar over the H indicates that the value has been calculated for the indicated quantity of reactans/products, i.e. 1 mol of methane Energy balance with chemical reaction – “The Heat of reaction” approach We revise the energy balance to include a term to account for the heat effects of chemical reactions: Hin + Q + heat generated = Hout Using the convention for exothermic and endothermic reactions, we we can write this as: H IN Q H rxn H OUT When Hrxn is negative we end up with a positive value for “heat generated”, and vice versa The Hrxn term must account for the actual amount of reaction that took place. It must account for reactants going to products at the specified DATUM The DATUM does not have to be at 25 C but it is almost always more convenient to choose a DATUM of 25 C • The standard heats of formation are themselves the heats of reactions that form the compounds from the constituent elements in their stable forms at 25 C and 1 atm. • e.g. C O2 CO2 h 393.51 kJ 0 f 1 H 2 O2 H 2O ( g ) h 0 241.83 kJ f 2 1 H 2 O2 H 2O (l ) h f 285.84 kJ 0 2 C 2 H 2 CH 4 h 0 74.84 kJ f • The heat of formation of elemental species are 0! (by definition) Energy balance with chemical reaction – “The heat of formation” approach We retain the energy balance developed for non-reacting systems: Hin + Q = Hout However, we recognize that the enthalpy of any material entering or leaving the system must account for differences in enthalpy due to the chemical bond structure of the material. We therefore add a term to the specific enthalpy of substances: energy due to energy in chemical bonds temperature or phase, h i.e. " sensible" enthalpy T h h 0 f C p dT 25 C “The heat of formation” approach DATUM: Elements in their natural states at 25 C and 1 atm hO2 , 25C 0 hN 2 , 25C 0 hH 2 , 25C 0 hCO2 , 25C h 0 f CO 2 393.51 kJ / mole hH 2O ( g ), 25C h 0 f H 2O ( g ) 241.83 kJ / mole hH 2O (l ), 25C h 0 f H 2O ( l ) 285.84 kJ / mole

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enthalpy change, internal energy, chemical reaction, Standard enthalpy, heat content, thermodynamic system, enthalpy of reaction, reactants and products, specific enthalpy, open systems

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posted: | 3/25/2011 |

language: | English |

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