Section 6: F&R, Ch. 8 (Energy Balances – II)

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Section 6: F&R, Ch. 8 (Energy Balances – II)ENERGY BALANCES ON NONREACTIVE PROCESSESWe solved problems in Chapter 4 and ignored the fact that heat was added and that temperatures mighthave changed. In Chapter 8 we will start to include the energy effects. Here’s a brief review of whatwe’ve learned so far from Chapter 7:Q or Q W or W sm inm out(n A , Uˆ A , Hˆ A )in(n , Uˆ , Hˆ )B in( n A , Uˆ A , Hˆ A )out( n , Uˆ , Hˆ )(n C , Uˆ C , Hˆ C )in( n C , Uˆ C , Hˆ C )outBBBBB out (kg/s) (open)m (kg) (closed) or mni (mol)(closed) or n i (mol/s)(open)Uˆ (kJ/mol), Hˆ (kJ/mol)Each species in a feed or product stream is in a particular state (phase, T, P), and the same species mayenter and leave in several different streams and states.In Chapter 7, we saw that the first law of thermodynamics (energy balance equation) isQ W U Ek Ep[Closed (batch) system]Q W s H E k E p [Open (continuous), steady-state system]where U m finalUˆ final m out Hˆ out 11m out (uout )2 m in (uin )2 [where u(m/s) velocity]2input 2species ,states H outputstreams E k outputstreams E p minitialUˆ initialspecies ,statesoutputstreams m in Hˆ in [where Hˆ Uˆ PVˆ ]inputstreamsstreamsm out g zout m in g zin [where z height]inputstreamsEach of these terms has units of kJ(closed) or kJ/s(open).6-1Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Most problems in Ch. 8 have the following form: Given:––Feed and product states (phase, T, P), some flow rates (n’s) & compositions (y’s) (open) zero (no moving parts or electric currents)W (closed) or Wss Calculate other flow rates (Ch. 4–6 methods) Determine U (closed) or H (open), E k , E p (usually neglect the latter two). Substitute into energy balance to determine Q or Q .If we have tables of specific internal energies and enthalpies (as we do for water in the steam tables),calculating U, H is straightforward. Chapter 8 discusses how to do it if we don’t have those tables.Example:Q (kJ/s)W s (kJ/s)4 mol/s H2O(s, –5oC, 1 atm)4 mol/s H2O (v, 300oC, 5 atm)Energy Balance: Q W s H E k E p E p 0 (Why? )Neglect E k (Why? ) mol ˆ kJ ˆQ W s H n out Hˆ out - n in Hˆ in 4 H2 4 H1 s Q Ws 4 Hˆ mol The problem is now to determine Hˆ for the given process.Can’t use steam tables. (Why not? )General procedure for calculating Ĥ for a specified change in state Construct a process path out of steps of 5 types:1. Change P at constant T and phase (Section 8.2 for calculation of H and U)2. Change T at constant P and phase (Section 8.3)3. Change phase at constant T,P (Section 8.4)4. Mix dissimilar liquid species (e.g. acid & water), absorb gas in liquid at constant T,P andphase (Hmix) (Section 8.5)5. React at constant T,P (Hr) (Chapter 9)6-2Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II) Determine Hˆ for each step in the process path (formulas will be given for each of the 5 types) Hˆ (follows since Hˆ is a state variable).Calculate ( Hˆ ) overallprocess isteps Substitute in energy balance to calculate Q W sConstructing a hypothetical process path relies on the concept of state variables (Section 8.1).–State of a system: Set of all intensive variables [variables that don’t change with changing mass] thatdefine the system (phase, T, P, height, velocity, )State variable or state property: If a system changes from State 1 to State 2 in a process and X is some–property of the system (height, internal energy, )State 1State 2X1X2then X is a state variable if X X2 – X1 depends only on the initial and final states, and not how thesystem got from State 1 to State 2.Example: Travel from Denver, CO (elevation 5280 ft) to Raleigh, NC (elevation 300 ft). Path A vs.Path BDenverDenverZ 5280Path AZ 5280Path BRaleighRaleighZ 300 ftZ 300 ftLet:D reading on odometer (miles traveled), D D2 – D1Z reading on altimeter (ft), Z Z2 – Z1( D)Path A ( , ) ( D)Path B, so D (is, is not) a state variable.( Z)Path A ( , )( Z)Path B so Z (is, is not) a state variable.Most important concept in Chapter 8: Uˆ and Hˆ are state variables.6-3Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Exercise: For the water process shown on 6-2, construct a process path consisting of steps of some of thefive types.H2O (s, –5oC, 1 atm) Exercise: Look at the two process paths on p. 405. Identify Type (1–5) of each step on the hypothetical(bottom) path. Hˆ 1 : Hˆ 2 : Hˆ 3 : Hˆ 4 : Hˆ 5 : Hˆ 6 :Exercise: Construct a path that utilizes as many known enthalpy changes as possible.(1) Cyclohexane vapor at 180oC and 5 atm is cooled and condensed to liquid cyclohexane at 25oCand 5 atm. We know the specific enthalpy change for the condensation of cyclohexane at80.7oC and 1 atm.(2) Water at 30oC and 1 atm and NaOH at 25oC and 1 atm are mixed to form an aqueous NaOHsolution at 50oC and 1 atm. We know the enthalpy change for the dissolution of NaOH inwater at 25oC and 1 atm.6-4Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)(3) O2 at 170oC and 3 atm and CH4 at 25oC at 3 atm are combined and react completely to formCO2 and H2O at 300oC and 3 atm. The enthalpy change for the reaction occurring at 25oC and1 atm is known.6-5Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)ˆ and ΔHˆ for Process Types 1 and 2Calculating ΔUProcess Type 1: Change P at constant T and phase. (Section 8.2) Uˆ 0 (Exact for ideal gas, approximate for solids and liquids) Hˆ 0 (Ideal gas) Vˆ P (Solid or liquid - Vˆ is constant)Find Vˆ for solids and liquids from specific gravity in Table B.1 (convert to density, Vˆ 1 / ).Calculations for real gases are complex (topic in later thermodynamics class)Process Type 2: Change T at constant P and phase. (Section 8.3)Sensible heat: heat transferred to raise or lower the temperature depends strongly on T (molecules movefaster at higher temperatures), and therefore so doesHˆ ( Uˆ PVˆ ).Define kJ dHˆ Hˆ o mol C dT T PC p (T ) Constant pressure heat capacityCP is the rate of increase of specific enthalpy with temperature for a constant pressure process.Graphically, it is the slope of the tangent to the plot of Hˆ vs. T . To find the change in enthalpy from Cp: Hˆ If we combine this result with the previous expressions for Type 1 processes (change P at constant T), weget Hˆ C p (T )dTT2T1[exact for ideal gases with varying P, any gas at constant P ] C p (T )dT Vˆ PT2T1 [liquids and solids](8.3 -10a)(8.3 -10b)Look up polynomial expressions for Cp at 1 atm in Table B.2. For exampleAcetone (liquid): Cp [kJ/(mol oC)] Acetone (vapor): Cp [kJ/(mol oC)] TfSubstitute into Hˆ C p (T )dT.To6-6Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)We can also use APEx’s Enthalpy function to calculate the enthalpy change as a result of a temperaturechange: Enthalpy (“acetone”,150,250,”C”,”g”) 10.7237 kJ/mol Estimation of Cp for solids, liquids, mixtures: Kopp’s Rule (Section 8.3c and Table B.10 or theKopps function in APEx). Use when you can’t find the data in Table B.2. See Example 8.3-4.Kopps’s Rule: Cp for a molecular compound is the sum of contributions (given in Table B.10) for eachelement of the compound.Example: Cp [Ca(OH)2] Cpa [Ca] 2 Cpa [O] 2 Cpa [H]Note on Kopp’s rule units: Note that the units for the atomic heat capacities in Table B.10 are listed asJ/g-atom C. So what’s a g-atom? A gram-mole (or mol) is the amount of a molecular species whosemass in grams equals the molecular weight of the species. Similarly, a gram-atom (g-atom) is theamount of an atomic species (C, H, S ) whose mass in grams equals the atomic weight of the species.The heat capacity terms in Table B.10 have units of J/g-atom C because they apply to atomic speciesinstead of molecular species.1 mole of water contains 2 g-atoms of H and 1 g-atom of O. So to estimate the heat capacity of liquidwater, if you multiply the heat capacity term for H (18) by 2 g-atoms and add the term for O (25)multiplied by 1 g-atom, you get the heat capacity for a g-mole (mol) of H2O and its Cp units would beJ/mol C. Heat capacity of a mixture(CP ) mix (T ) yCipi(T )Note enthalpy values available for common gases (O2, N2, H2, air) in Tables B.8 and B.96-7Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)How about the dependence of Û on T? To find this, we do it the same way, starting by plotting Û vs. Tfor a process in which the temperature of a substance is raised holding the volume constant.Constant VDefine kJ dUˆ Uˆ o mol C dT T VCv (T ) Constant volume heat capacityProceeding exactly as before, we can show that, for a change in T at constant V:T2 Uˆ Cv (T )dTT1(8.3 - 6)That expression strictly applies to processes that take place at constant volume; however, since Û isalmost independent of P for every species but real gases at high pressures, it can be applied to mostprocesses in which volume changes occur as well. In short, this equation applies well for processes withtemperature changes but no phase changes.Recap: Eq. (8.3-6) is exact for ideal gases and an excellent approximation for liquids and solids, even if volume andpressure change. exact for constant volume processes. For real gases, Cv depends on the value of Vˆ at which theprocess takes place. inapplicable for real gases in which significant volume changes occur.So how do you determine CV? Determination of Cv from CpCv C p R (ideal gases)where R 8.314 10 -3 kJ/(mol K) C p (liquids and solids)(8.3-11)(8.3-12)Substitute expression for Cv in integral expression for Û (Eq. 8.3-6). (See Example 8.3-2)6-8Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Example : Calculate Hˆ (kJ/mol) for H 2 O(v, 30 C, 0.0424 bar) H 2 O(v, 350 C, 1.5 bar)(a) using the steam tables(b) using Eq. (8.3-10a)(c) using APEx’s SteamSatT (B.5) and SteamSH (B.7) functions(d) Which calculated value is most accurate, and why?Solution(a) From Tables B.5 and B.7, Hˆ ( )kJ0.01802 kgkgmol 11.13 kJ/mol(b.) From Table B.2, Cp [kJ/(mol oC)] 33.46 10–3 0.6880 10–5T 0.7604 10–8T2 – 3.593 10–12T3Pathway: H2O (v, 30 C, 0.0424 bar) H2O (v, 350 C, 0.0424 bar) H2O (v, 350 C, 1.5 bar)Assume ideal-gas behavior. From Eq. (8.3-10a),350 Hˆ [33.46 10 3 0.6880 10 5T 0.7604 10 8 T 2 3.593 10 12 T 3 ]dT30 11.2 kJ/molQ: Where did we use the assumption of ideal-gas behavior?A:(c) Using APEx:6-9Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)(d) Which estimate is more accurate, and why?Process Type 3: Change phase at constant T,P (Section 8.4)Latent Heat: enthalphy change associated with a phase transition (melting, vaporization, sublimation)(a)Solid to liquid: A(s, Tm ,1 atm) A(l, Tm ,1 atm), Hˆ Hˆ m where:Tm normal melting point (Table B.1) Hˆ heat of fusion (melting) at T (Table B.1 or APEx function Hm)m Uˆ m Hˆ mm(8.4-1)What about liquid solid transition (freezing)? Hˆ Hˆ m(b.) Liquid to vapor: A(l, Tb ,1 atm) A(v, Tb ,1 atm), Hˆ Hˆ v where:Tb normal boiling point (Table B.1 or APEx function Tb) Hˆ heat of vaporization at T (Table B.1 or APEx function Hv)vb Uˆ v Hˆ v PVˆ Hˆ v RTb(8.4-2)What about vapor liquid transition (condensation)? Hˆ Correlations for estimation of latent heats Hˆ v , Hˆ m : Section 8.4b. Use when you can’t findthe data in Table B.1. (See Example 8.4-3.)6-10Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)In most problems, we neglect kinetic energy and potential energy. Let’s see why:(a) Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid waterfrom room temperature (20 C) to its normal boiling point (100 C), while still remaining a liquid.(b) Calculate how much energy is required to heat a bottle containing 300 mL of liquid water at 100 C, changing it from liquid to vapor.(c) Adding the two values calculated in parts (a) and (b), if that amount of energy was insteaddedicated to accelerating the bottle of water, what would the final velocity be (miles per hour)?(d) Adding the two values calculated in parts (a) and (b), if that amount of energy was insteaddedicated to lifting the bottle of water, what would the final height (meters)?You can see in this example that compared to the sensible heat and the latent heat terms, kinetic andpotential energy effects are much smaller and can usually be neglected.6-11Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)HEATS OF SOLUTION & MIXING (Sect. 8.5)Ideal mixture: the heat of mixing or solution equals zero (gas mixtures, mixtures of similar liquids).When mixing acids and bases or dissolving certain gases or solids in a liquid solvent, the heat ofmixing/solution may be far from zero.Heat of Solution (dissolving a solid or gas in a liquid or mixing two liquids):A(s, l, or g, 25oC, 1 atm) rB(l, 25oC, 1 atm) A(soln, 25oC, 1 atm), Hˆ Hˆ s (r ) kJmol AA is the solute and B is the solvent. r moles of solvent per mole of solute.Q: What is the mole fraction of solute in terms of r? A: ysolute Values of Hˆ s for aqueous solutions of HCl(g) (hydrochloric acid), NaOH (caustic soda), and H2SO4(sulfuric acid) are given in Table B.11, p. 653. Note that the unit of Hˆ is kJ/(mol of solute), so toscalculate H (kJ) or H (kJ/s) for formation of a solution at 25 C from a solute and solvent at 25oC,multiply Hˆ by # moles of solute, not total solution.os Example: Calculate H for a process in which 2.0 mole of NaOH is dissolved in 400 mol H2O at25C. r 400/2 200 mol H2O/mol NaOH Hs (r 200, 25C from Table B.11) -42.26 kJ/mol NaOH dissolved H n Hs 2.0 mol NaOH (-42.26 kJ/mol NaOH) -84.52 kJWhat does Hˆ s represent physically? It’s the energy required to break solute-solute molecular bonds(strong for solids, moderate for liquids, negligible for gases) & solvent-solvent bonds minus energyreleased when solute-solvent bonds are formed (may be strong, moderate, or negligible) Values of Hˆ s in Table B.11 are negative (the solution is at a lower energy level than the pure solute& solvent) mixing & solution for the given solutes are exothermic energy is released by thesolution process. Unless you cool the mixer, the solution gets hot. Think about what happens whenyou mix acid with water. Another representation of the heat of mixing is the enthalpy-concentration diagram; two examplesare given in Figure 8.5-1 (for aqueous sulfuric acid solutions) and Figure 8.5-2 (for solutions ofammonia in water). Example: Find the specific enthalpy of a 40 wt% solution of H2SO4 at 120 Frelative to pure sulfuric acid at 77oF and pure water at 32oF (the reference conditions for Fig. 8.5-1).6-12Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II) Energy balances on processes involving mixing and solution are done in the same way used for othertypes of systems. References for enthalpy calculations are usually the pure solvent and solute at theconditions for which the heat of solution is known. If a reactant or product solution is at a differenttemperature, the process path for the enthalpy calculation is first to form the solution from the soluteand solvent at 25oC and 1 atm (the usual reference conditions), and then heat the solution to its statein the process. See Example 8.5-1.6-13Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Strategy for Solving the General Nonreactive System Energy BalanceQ or Q W(closed) or W s (open)Feeds(nA ,Uˆ A , Hˆ A )in(n ,Uˆ , Hˆ )BBProducts( n ,Uˆ , Hˆ )AAA out( nB ,Uˆ B , Hˆ B )out( n ,Uˆ , Hˆ )B in(nC ,Uˆ C , Hˆ C )inCCC outni (mol)(closed) or n i (mol/s)(open)Uˆ (kJ/mol), Hˆ (kJ/mol)Given information about(i)states (phase, T, P)(ii) amounts or flow rates (n) of feed and product components (A,B,C) (or equivalently, totalamounts or flow rates and component fractions)(iii) Q(iv) W or W scalculate the remaining variables.Procedure: Draw and label the flowchart. Include state information in the labeling, and label Q and W or Wsunless you know they equal zero. Perform a degree-of-freedom analysis. Include one energy balance along with material balances andother relations when counting equations. If you don’t get 0 degrees of freedom, try to figure outwhat’s missing or if go to another subsystem if there are multiple units in the process. Write all equations but the energy balance.–If Q or W is unknown, you should be able to solve the equations for the unknown variables inthem. If you plan to use Solver, you don’t need to do anything at this point beyond writing theequations.–If Q and W or W s are given in the problem statement or known to be zero, write the energybalance equation and solve it simultaneously with the other equations. Choose reference states for each species in the system for Û and Ĥ calculations.–For water, if you plan to use steam tables, choose H2O(l, triple point).6-14Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)–For a gaseous species listed in Table B.8, choose the species at 25oC and 1 atm (the referencestate used in the table). For Table B.9 (American engineering units), choose the species at 77oFand 1 atm.–Otherwise, choose the species in one of its feed or product states, so you can set at least onespecific internal energy or enthalpy equal to zero. Prepare an inlet-outlet internal energy (closed) or enthalpy (open) table. The enthalpy table is shownbelow as an illustration.References: A(state), B(state), C(state) (Fill in the reference states)–Speciesn inHˆ inn outHˆ outAn aHˆ an dHˆ dn (mol/s)Bn bHˆ bn eHˆ eHˆ (kJ/mol)Cn cHˆ cn fHˆ fIf a species exists in more than one inlet or outlet state in the process (e.g., in liquid and vaporphases), add extra rows to the table to make room for the flow rate and enthalpy in each state.–Cross out cells that don’t correspond to any inlet or outlet states, and replace the component flowrates in the table (n A1 , n B 2 , etc.) with known values and variable names from the flow chart.–Set Hˆ 0 for inlet and outlet species at their reference states, and label all unknown specificenthalpies.–For a closed system, use amounts rather than flow rates and internal energies rather thanenthalpies. Write and simplify the energy balance equation and substitute series expressions for H or U,filling in values and labeled quantities from the enthalpy table. Write expressions for all of the unknown specific enthalpies in the table. CalculateHˆ a as Hˆ for the process A(reference state) A(inlet state)Hˆ d as Hˆ for the process A(reference state) A(outlet state)and similarly for B, C, . (For closed systems, replace H with U.) Substitute the expressions for the enthalpies in the energy balance, which may now be solvedsimultaneously with the other system equations.6-15Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Example: Inlet-outlet enthalpy (or internal energy) table (solve for Q)Q (kJ/s)38.0 mol/s0.663 C6H6(v)0.337 C7H8(v)50oC, 163.7 mm Hg100 mol/s0.500 mol C6H6(l)/mol0.500 mol C7H8(l)/mol10oC, 2 atm62.0 mol/s0.400 C6H6 (l)0.600 C7H8 (l)50oC, 163.7 mm HgReferences: C6H6( l, 10oC, 2 atm), C7H8( v, 50oC, 163.7 mm Hg )SpeciesninHˆ innoutHˆ outC6H6(l)50Hˆ a24.8Hˆ cn(mol/s)C6H6(v)——25.2Hˆ dHˆ (kJ/mol)C7H8(l)50Hˆ b37.2ˆ 0HC7H8(v)——12.8Hˆ fHˆ a path: at reference, 0Hˆ b path: T (v, 50C, 163.7 mm Hg) T (v, 50C, 1 atm) T (v, 110C, 1 atm) T (l, 110C, 1 atm) T (l, 10C, 1 atm) T (l, 10C, 2 atm)ˆH c path: B (l, 10C, 2 atm) B (l, 50C, 2 atm) B (l, 50C, 163.7 mm Hg)Hˆ d path: B (l, 10C, 2 atm) B (1, 80.1C, 2 atm) B (1, 80.1C, 1 atm) B (v, 80.1C, 1 atm) B (v, 50C, 1 atm) B (v, 50C, 163.7 mm Hg)Hˆ e path: T (v, 50C, 163.7 mm Hg) T (v, 50C, 1 atm) T (v, 110C, 1 atm) T (l, 110C, 1 atm) T (l, 50C, 1 atm) T (l, 50C, 163.7 mm Hg)Hˆ f path: at reference, 06-16Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)zWrite expressions for all the specific enthalpies in the table and then the energy balance. State all yourassumptions.Ha Q H 25.9Hc 24.1Hd –50.0Hb Now let’s see what we’re doing. Ignore pressure effects on enthalpy.What is Ha? It’s H for 1 mol B(ref) 1 mol B(l, 10 C)oWhat is the process?25.9 mol/s B(l, 10 C) 25.9 mol/s B(ref.) 25.9 mol/s B(l, 50 C)oo6-17Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Example: Continuous air conditioning process. Take basis of 100 mol feed. Calculate Q.Q (kJ)n 2 (mol/s)y (mol H2O(v)/mol), saturated(1– y) (mol DA/mol)15oC, 1 atm100 mol/s0.100 mol H2O(v)/mol0.900 mol DA/mol110oC, 2 atmn 1 (mol/s) H 2 O (l)15oC, 1 atmQuestion: What sign should the value of Q have?Solution: Assume enthalpies are independent of pressure.(a) DOF Analysis:unknowns:4 (Q, n2, n1, y)equations:2 balances (water, DA)1 energy balance(b) Write all equations but energy balance. Circle unknowns for which you would solve.Raoult’s Law: y (5260) pw*(80oC)DA: 0.1(100) (1–y) n2Moles: 100 n1 n2(c) Choose references and prepare an inlet-outlet enthalpy table.References: H2O( l, 15oC, 1 atm), DA(g, 25oC, 1 atm ) (Why?)Speciesn inHˆ inn outHˆ outH2O(l)——n1Hˆ cn (mol/s)H2O(v)10Hˆ an2 yHˆ dHˆ (kJ/mol)DA90Hˆ bn2(1–y)ˆ 0H(d) Write and simplify the energy balance equation, substituting values and labeled variables from theenthalpy table.6-18Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Q W s H E k E p (e) Write expressions for each labeled specific enthalpy, using data in Tables B.2 and B.8. The first stepis partially shown below as an illustration.Hˆ a : H 2 O(l,15o C) H 2 O(l,100 o C) H 2 O(v,100 o C) H 2 O(v,110 o C) Hˆ a Q: Why 100oC? A:Continue with the remaining enthalpies in the table.The values of the specific enthalpies may be substituted into the energy balance, which may then besolved together with the other system equations to determine Q(kJ).6-19Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Psychrometric Chart (Section 8.4d)6-20Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)6-21Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Sectionn 6: F&R, Ch. 8 (Energyy Balances – III)6-22NotesNwith Gapss to accompanyy Felder, Rousseau, & Bullarrd, Elementaryy Principles of Chemical Proccesses.Copyright Johhn Wiley & Sonns, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Why does it work? Gibbs phase rule for humid air givesF 2 components – phases 2 2 – 1 3If we fix the system pressure at 1 atm, then we can specify any 2 intensive variables and determine all theother intensive system properties. These are the properties you can look up on the chart:a. Dry bulb temperature: Temperature as measured by a thermometer, thermocouple, etc.b. Absolute humidity: hA (lbm H2O/lbm DA) (also called moisture content). In terms of this quantity,the mass fraction of water is y ic. Relative humidity: h 100rha1 hapH 2OpH 2O* (T )d. Dew point temperature: Temperature at which humid air becomes saturated if cooled at constantpressure. Follow any point horizontally to the left until you reach the saturation curve.3e. Humid volume: VˆH (m / kg DA) - volume accompanied by 1 lbm DA plus the water vapor thataccompanies itf.Wet bulb temperature: Twb – temperature reading on a thermometer with a water-saturated wickaround the bulb immersed in a flowing stream of humid air. (Why do you feel cold when you step outof the shower or pool?)g. Specific enthalpy of saturated air: Btu/lbm DAh. Enthalpy deviation: Used to determine the enthalpy of humid air that is not saturated.Subtractenthalpy deviation from specific enthalpy of saturated air, which you find by following the wet bulbtemperature line to the saturation line.Example: Use the psychrometric chart to estimate the following properties of humid air at 41oC and 10%relative humidity: Absolute humidity Wet-bulb temperature Humid volume Dew point Specific enthalpyWhat is the amount of water in 150 m3 of air at these conditions?6-23Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)REVIEW OF KEY FORMULAS AND PROCEDURES IN CHAPTER 8First law of thermodynamics (energy balance equation):Q W U Ek Ep[Closed (batch) system]Q W s H E k E p[Open (continuous) steady-state system]where U m finalUˆ final m out Hˆ out 11m out (uout )2 m in (uin )2 [where u(m/s) velocity]2input 2species ,states H outputstreams E k outputstreams E p minitialUˆ initialspecies ,statesoutputstreams m in Hˆ in [where Hˆ Uˆ PVˆ ]inputstreamsstreamsm out g zout m in g zin [where z height]inputstreamsDetermining kinetic and potential energy changes is straightforward (usually neglect both unless there’sa big velocity change from inlet to outlet and no reactions, phase changes, or large temperature changes).Most problems in this course come down to determining values of U or H for a given process &substituting it in the energy balance equation to determine heat. To do that, need to evaluate changes inUˆ or Hˆ for each species in the process inlet and outlet streams.6-24Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Process Type 1: Change P at constant T in a single phase (Section 8.2) Uˆ 0 (Exact for ideal gas, approximate for solids and liquids) Hˆ 0 (Ideal gas) Vˆ P (Solid or liquid - Vˆ is constant) [Be able to derive this formula from the definition of Hˆ ]Process Type 2: Change T at constant P in a single phase (Section 8.3) Hˆ Hˆ 2 Hˆ 1[if you have tabulated enthalpies, as you do for water & species in Tables B.8 & B.9]T2 C dT [exact for ideal gases with varying P, any gas at constant P.] C dT Vˆ P [liquids and solids] T1T2ppT1 Uˆ Hˆ PVˆ [by definition] Hˆ R T (ideal gases--be able to derive) T2T1Cv dT[exact for ideal gases or constant Vˆ , otherwise approximate]whereCv C p R (ideal gases) C p (liquids and solids)To evaluate T2T1C p dT , integrate formula in Table B.2 term by term or (faster) use ICPP property database.Estimation of Cp for solids, liquids, mixtures—Section 8.3c.Process Type 3: Change phase at constant T,P (Section 8.4)A(s, Tm , 1 atm) A(l, Tm , 1 atm), H H mA(l, T , 1 atm) A(v, T , 1 atm), H H bbvwhere Tm normal melting point , H m heat of fusion (melting) at Tm (Table B.1) U m H m T normal boiling point, H heat of vaporization at T (Table B.1)bv U v H v RTb b(Be able to derive from ideal gas EOS)Estimation of latent heats—Section 8.4b.6-25Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.Copyright John Wiley & Sons, Inc.

Section 6: F&R, Ch. 8 (Energy Balances – II)Process Type 4: Mix two liquids or dissolve a solid or gas in a liquid at constant T,P (Section 8.5)A(s, l, or g, 25oC, 1 atm) B(l, 25oC, 1 atm)kJA(solution, 25oC, 1 atm) Hˆ Hˆ s (r ) Found in Table B.11 or APEx functions HCL, NaOH, and H2SO4mol solute ( A)heat of mixing or solutionProcess Type 5: Reaction at constant T, P (Chapter 9)Procedure for energy balance problems1. D

1. Change P at constant T and phase (Section 8.2 for calculation of H and U) 2. Change T at constant P and phase (Section 8.3) 3. Change phase at constant T,P (Section 8.4) 4. Mix dissimilar liquid species (e.g. acid & water), absorb gas in liquid at constant T,P and phase (H mix) (Section 8.5) 5. React at constant T,P (H r) (Chapter 9)

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