Mechanical Engineering - Energy Systems LM

First part of the course

Divided by topic

A.Y. 2021/2021 Curri Davide Master of Science in Mechanical EngineeringEnergy Systems Lecture Notes Contents 1 Thermodynamic Properties of Gases 21.1 Ideal gas model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 1.2 Perfect gas model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 1.3 Real gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 2 Internal energy and Mass balance equations 62.1 Mass balance equation . . . . . . . . . . . . . . . . . . . . . . . . . .62.1.1 Mass balance equation for open control volumes under steadystate flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 2.1.2 Mass balance equation for average flow under steady statecondition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 2.2 Balance equations of atomic species . . . . . . . . . . . . . . . . . . .102.2.1 Relation between atomic flow rates and molecular flow rates .11 2.2.2 Example — Combustion of ethanol in steady-flow reactor . . .12 2.3 Energy balance equation . . . . . . . . . . . . . . . . . . . . . . . . .132.3.1 Law of conservation of energy (1st principle of thermodynamics)14 2.4 Energy balance equation for closed systems . . . . . . . . . . . . . . .15 2.5 Energy balance equation for open systems . . . . . . . . . . . . . . .16 2.6 Energy balance equation for open systems without chemical reactions17 2.7 Energy balance equation with chemical reactions . . . . . . . . . . . .18 2.8 Chemical energy of molecules . . . . . . . . . . . . . . . . . . . . . .182.8.1 Determining the chemical energy of molecules . . . . . . . . .19 2.9 Energy balance equation for a chemical reactor (general case) . . . .21 2.10 Possible types of mixtures and their thermal enthalpy h . . . . . . . .232.10.1 Real mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . .23 2.10.2 Ideal mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . .23 2.10.3 Ideal mixtures of ideal gases . . . . . . . . . . . . . . . . . . .23 2.11 Heat of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24 2.12 Entropy balance equation . . . . . . . . . . . . . . . . . . . . . . . .25 3 Heat Exchangers 283.1 Type of heat exchangers . . . . . . . . . . . . . . . . . . . . . . . . .283.1.1 Direct transfer heat exchangers . . . . . . . . . . . . . . . . .28 3.1.2 Regenerative heat exchangers - The Ljungstrom air preheater29 0 3.2 Focus on HRSG and Boilers . . . . . . . . . . . . . . . . . . . . . . .31 3.3 Design procedure for direct transfer heat exchangers . . . . . . . . . .32 3.4 Heat transfer rate equation . . . . . . . . . . . . . . . . . . . . . . . .32 3.4.1 Note 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 3.4.2 Note 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 3.5 Cross flow heat exchangers . . . . . . . . . . . . . . . . . . . . . . . .34 3.6 Heat transfer rate equation for cylindrical geometries (tubes) . . . . .36 3.7 Extended surface heat exchangers (fins) . . . . . . . . . . . . . . . . .38 3.8 Rating and Off-Design calculation of heat exchangers . . . . . . . . .40 3.9 Rating and Off-design calculation of Hx with theε-NTU method . . .43 3.10 Heat exchangers with variable Cp streams . . . . . . . . . . . . . . .44 3.11 Condensers for steam cycles . . . . . . . . . . . . . . . . . . . . . . .443.11.1 Water cooled condensers . . . . . . . . . . . . . . . . . . . . .45 3.11.2 Cooling towers . . . . . . . . . . . . . . . . . . . . . . . . . .46 3.11.3 Energy balance of a cooling tower V . . . . . . . . . . . . . .48 3.11.4 Air cooled condenser . . . . . . . . . . . . . . . . . . . . . . .48 4 Fuel combustion 504.1 Combustion reactions and fuels . . . . . . . . . . . . . . . . . . . . .504.1.1 Classification of fuels . . . . . . . . . . . . . . . . . . . . . . .50 4.1.2 Identification of fuels . . . . . . . . . . . . . . . . . . . . . . .51 4.2 Evaluation of the combustion products and heat of combustion . . . .54 4.3 Complete combustion theory/assumption . . . . . . . . . . . . . . . .544.3.1 Example: combustion of CH4with O 2with complete combus- tion assumption . . . . . . . . . . . . . . . . . . . . . . . . . .55 4.3.2 Advantages of the complete combustion assumption . . . . . .55 4.3.3 Disadvantages of the complete combustion assumption . . . .56 4.3.4 Complete combustion with air . . . . . . . . . . . . . . . . . .56 1.Thermodynamic Properties of Gases 1.1Ideal gas model The kinetic theory of gases defines "ideal gas" a gas made of molecules of the same type (identical) and size in which collision between molecules are perfectly elastic (no intermoleculare attractive forces).Under such hypotesis it is possible to prove that theEquation of Stateis: pV=RT(1.1) And the following results can be derived: 1.Mayer relationCp− C v= R(1.2) 2.Internal energyu=u(T)(1.3) item Enthalphyh=h(T)(1.4) they both depend only on T. 3.Specific heat capacity at constant pressure Cp=∂ h∂ T = C p( T)(1.5) 4.u(T) =u(T ref) +Z T T rifC v( T)dT(1.6) h(T) =h(T ref) +Z T T rifC p( T)dT(1.7) 5.Quantum Thermodynamicsallows to determine the relation Cv( T)(1.8) Indeed it has established the link betweenu=internal energyande c=kinetic energy of the gas molecules. 2 A gas molecule has a certain number of degrees of freedom (traslational, rota- tional and vibrational) which are associated to the kinetic energy.GDL = degrees of freedom of a gas molecule ft= traslational →3 fr= rotational →possible rotations which cause a variation of the relative distances between atoms and contributing to the kinetic energy of the molecule.fv= vibrational →possible vibrations of the atoms around their average posi- tion. •Monoatomic gases (Ar):f T= 3 , f R= 0 , f V= 0 ; •Biatomic gases (O 2, N 2): f T= 3 , f R= 2 , f V= 1 ; •Triatomic linear gases (C O 2): f T= 3 , f R= 2 , f V= 4 ; •Monoatomic nonlinear gases (H 2O ):f T= 3 , f R= 3 , f V= 3 ; It is possible to write that Cv=f r+ f t2 R +f vR (1.9) howeverf rand f vare not constant but they depend on T, as they are activated only if T is greater than certain values.Figure 1.1: Example of C v( T)f or H 2 B: activation of the rotational degrees of freedom; D: activation of the vibrational degrees of freedom; 1.2Perfect gas model It is an ideal gas withC vconstant (it does not depend on T). Such a model is valid to describe the behaviour of: •Monoatomic gases(Ar) 3 • Ideal gases over a narrow range of temperatures in which you can assume that ∂ Cv∂ T ≈ 0 For perfect gases you can write: •∆h=C p∆ T(1.10) •∆u=C v∆ T(1.11) •s(T , p) =C pln(TT ref ) −Rln(PP ref ) (1.12) •pvγ =cost(1.13) along an isoentropic transformation •and define "politropic transformations" as those with: –ConstantC p –pvk =costor (equivalent)fT 2T 1= βk 1.3Real gas The ideal gas model cannot be used when:•Low temperatures (T≤4T critical): interaction forces between molecules are not negligible→Real gas effect •High pressures (p≤kp critical): interaction forces between molecules are not negligible→Real gas effect •High temperatures (T≫2000°C): molecules tend to dissociate into atoms and atoms tend to ionize (loosing electrons). Forces between molecules arise and the ideal model is not valid. For instance steam (H 2O vap) at temperatures of 0°Cup to700°Cand higher cannot be modeled as an ideal gas because its behaviour has a considerable real gas effect. z=compressibility f actor=V of real gas at T ,pV of ideal gas at T ,p= V of real gasRT P (1.14) 4 Figure 1.2: Compressibility factor for N 2 z gives a measure of the real gas effect, i.e. of the difference between the ideal gas behaviour and that of the real gas. Ifz= 1the real gas can be approximated as an ideal gasFor givenˆ p, the higher is T and the closer z is to 1. For given T, the lower is p and the closer z is to 1.It is wrong/mistaken to compute the properties of a real gas (e.g. steam) with the ideal gas relationspv=RT(1.15) ∆h=Z Cp( T)dT(1.16) For real gases we have to use: •Diagrams T-s, h-s, p-h obtained from experimental measures (e.g.Mollier Diagram); •Correlations obtained from measures like theIAPWSfunctions for Steam; •Equations of statev=v(p, T)and h, u, etc. 5 2.Internal energy and Mass bal- ance equations 2.1Mass balance equation It derives from thePrinciple of conservation of mass(valid only if particle velocities are≪than the speed of light= 299 792 458 m/s)→always valid in the study of energy systems The mass of the particles contained in a material control volumeΩ(t)is constant during time:M(t) =cost— Antoine Lavoisier 1789 Amaterial control volumeis a volume with time-depending shape and size, which always contains the same set of particles or fluid molecules. It is also called Lagrangian Control Volume.Figure 2.1 6 The Material control volumefollows the fluid particles according to the la- grangian approach and to do so it has to move, changing shape and size. Such material control volume is not convenient to study in Energy Systems. It is more convenient to have a control volume whose position is fixed in the space (Eulerian approach). We can convert theprinciple of conservation of massvalid for ma- terial control volumes into themass balance equationvalid for eulerian control volumes by applying thereynolds transport theorem: LetΩ(t)be a material control volume whose surface/border is∂Ω(t)and whose velocity field isv( x, t )andf(x, t )a vector or scalar field then: 1.Z Ω(t)f (x, t )dV=F(t) (its integral depends only on t)(2.1) 2.ddt Z Ω(t)f (x, t )dV=Z Ω(t)∂ f∂ t dV +Z Ω(t)f (v· n) dA(2.2) where nis outward pointing normal relative to the border ∂Ω(t)and vis the velocity vector on the border point. REMARKThe theorem states that: The variation of F(t) = variation due to the variation of f(x,t) + variation due to the change of shape, size and position ofΩ(t) Let’s derive now the mass balance equation for the following eulerian control volume V (eulerian)Figure 2.2 Let V denote theEuleriancontrol volume under study. Let’s consider a time instantt 0and let’s consider a material control volume Ω(t)equal to V int 0so that: Ω(t 0) ≡V(2.3) 7 then let’s apply the Reynolds transport theorem to the Mass conservation law ofΩ(t) dMf orΩ(t)dt = 0 (2.4) ddt Z Ω(t)ρ (x, t )dV= 0(2.5) dMΩ(t)dt =Z Ω(t)∂ ρ∂ t dV +Z Ω(t)ρ (x, t )(v· n) dA(2.6) if we now considert=t 0in which Ω(t 0) ≡Vand∂Ω(t 0) = S 1∪ S 2∪ S channel∪ Srotor∪ S piston∪ S out Z V∂ ρ∂ t dV +Z Stotρ (v· n) dA= 0(2.7) ZV∂ ρ∂ t dV +Z Sin1∪ S in2ρv· ndA +Z Soutρv· ndA Z Schannelρv· ndA + +Z Srotorρv· ndA +Z Spistonρv· ndA = 0(2.8) where the integral on the channel surface is =0 because the channel is rigid and impermeable, so(v· n) = 0 over its surface:Figure 2.3 ZV∂ ρ∂ t dV + ˙m out− ˙ m in1− ˙ m in2+ +Z Srotorρv· ndA +Z Spistonρv· ndA = 0(2.9) •R V∂ ρ∂ t dV is an unsteady flow term due to the time variation of fluid density at fixed volume V of the control system; •R Srotorρv· ndA +R Spistonρv· ndA are unsteady flow terms due to the change of shape/volume of the control volume caused by the displacement of the piston (which increases/decreases the volume) or the rotation of the rotor (which causes variations of the control volume shape). 8 2.1.1Mass balance equation for open control volumes under steady state flow Steady flow condition=∂ ϕ∂ t = 0 ∀parameterϕof the flow field in each point of the control volume V. Soρ,¯ v, p, Tdoes not change over time. Also the parameters/shape of the control volume cannot change over time. This means that under steady state flow the piston must be kept at a fixed position (it cannot be moved!). Its displacement would change the system volume making the flow field unsteady! Moreoveralso the Rotor must be kept stil l(fixed and not rotating) otherwise it would cause the fluctuations of the velocity/pressure field! Under the steady flow condition, the mass balance equation is: ˙ m out= ˙ m in1+ ˙ m in2(2.10) Or forN inand M outstreams:Figure 2.4 N X i=1˙ m ini=N X j=1˙ m outj(2.11) 2.1.2Mass balance equation for average flow under steadystate condition When dealing with a turbomachine the flow is never steady because of the fluctu- atiions caused by the rotating blades. For those systems it is necessary to define the condition ofSteady average flow: the average over time values of p, T,¯ v, ρetc. (parameters of the flow field) are constant: ∂¯ ϕ∂ t = 0 where¯ ϕ=1T Z T 0ϕ (t)dt(2.12) 9 Figure 2.5 Figure 2.6 for each point xof the system V, and ¯ Φis the average value over time: Under such condition:R V∂ ¯ ρ∂ t dV = 0because∂ ¯ ρ∂ t = 0 for each point xin V, andR Srotor¯ ρ(¯v· n) dA= 0because this is equal to the average flow rate across the rotor surface (˙ ¯ m rotor) which is zero because the rotor is impermeable. (fluid cannot exit/enter through the rotor surface)Then theMass balance equation is: N X i=1˙ ¯ m ini=N X j=1˙ ¯ m outj(2.13) 2.2Balance equations of atomic species The mass conservation law of Lavoisier can be applied to each single atomic species (Ar, H, O, N, etc.)‘if no molecular reaction (but only chemical reactions) take place in the control volume. Indeed the chemical reaction changes the arrangement of atoms by generatng new molecules but no atom is generated or destroyed. 10 Given a material control volume Ω(t)dA jdt = 0 whereA jis the number of moles of atomic species j inΩ(t) For an Eulerian control volume V:Figure 2.7 ∂ Aj∂ t =X in˙ Aini−X out˙ Aoutj(2.14) for each atomic species j.∂ A j∂ t is an unsteady flow term storage of atoms inside the volume V, whileP in˙ AiniandP out˙ Aoutjare molar flow rate of atom j Kmol/s 2.2.1Relation between atomic flow rates and molecular flowrates Typically we know the flow rates of molecular species which enters/exits the control volume. For instance, for the reactor below you know the inlet flow rate ofC H 4and O2and NOTthose of C, H and O atoms. Figure 2.8 Note that the flows of molecules MAY NOTbe equal at inlet and outlet because no law/principle guarantees this. Typically˙ Nmoleculesin̸ =˙ Nmoleculesoutbecause the chemical reaction may change the number of molecules. Only for the flow of atomic species you can write the equation: dAjdt = ˙ Ain,j−˙ Aout,j(2.15) In this example j=C, H, O so we can write 3 equations of balance of the atomic species. 11 • for C atoms:dA Cdt = ˙ Ain,C−˙ Aout,Cwhere˙ Ain,C=˙ NC H4· [number of C atoms in eachC H 4molecule ] =˙ NC H4and˙ Aout,C=˙ NC O2· [number of C atoms in eachC O 2molecule ] =˙ NC O2so the balance equation of carbon atom is: dACdt = ˙ NC H4−˙ NC O2(2.16) If the reactor is in steady state, thendA Cdt = 0 and˙ NC H4−˙ NC O2= 0 •for H atoms the equation is:dA Hdt = 4 ·˙ NC H4− 2·˙ NC O2 •for O atoms the equation is:dA Odt = 2 ·˙ NO2− 2·˙ NC O2−˙ NH2O Thus the general equation of balance of atomic species can be written as: dAjdt =N f low−in X i=1N molecularspecies X k=1˙ Ni,k· f k,j−N f low−out X i=1N molecularspecies X k=1˙ Ni,k· f k,j(2.17) (where k is the number of atoms of type j in each molecule k) and it can be written for each atomic species involved in the process/control volume. 2.2.2Example — Combustion of ethanol in steady-flow re-actorFigure 2.9 ˙ NO2,˙ NC O2and˙ NH2Omust satisfy the Atomic balance equations. Steady-flow conditiondA Cdt = 0 ,dA Hdt = 0 ,dA Odt = 0 . Atomic balance equations under steady flow condition: •C atoms:1 kmol/s·2 +˙ NO2· 0−˙ NC O2· 1−˙ NH2O· 0 = 0 •H atoms:1 kmol/s·6 +˙ NO2· 0−˙ NC O2· 0−˙ NH2O· 2 = 0 •O atoms:1 kmol/s·0 +˙ NO2· 2−˙ NC O2· 2−˙ NH2O· 1 = 0 System of 3 linear equations with 3 unknowns. The solution is: •˙ NO2=72 kmol/s •˙ NC O2= 2 kmol /s 12 • ˙ NH2O= 3 kmol /s Note that: •˙ Nreactants=˙ NC2H 6+˙ NO2=92 kmol/s •˙ Nreactants=˙ NH2O+˙ NC O2= 5 kmol /s •˙ Nreactants̸ =˙ Nproductsin general! Inlet molar flow rates̸ =Outlet molar flow rate: only the flow rates of atoms are constant! 2.3Energy balance equation The types of energy relevant to the study of energy systems are:•Internal energy (u) = it is the macroscopic result/energy (i.e. the capability of doing work) of the kinetic energy of the system’s particles/molecules; •Kinetic energyV 22 ; •Potential energy = energy stored by the system due its position in aconser- vative force field; –Gravitational energy (gz); –Electrical/Elastic/Electro-magnetic potential energy are not needed in the analysis of conventional energy systems like turbines and steam cycles; •Chemical energy (e C H) = energy stored in a molecule due its molecular struc- ture and bonds between atoms. It can be released by means of a chemical reaction which re-arrange the molecular structure and breaks the bonds be- tween atoms; •Nuclear energy (e N U C) = energy stored in the particles which can be released with a modification of the atoms molecules (nuclear reaction). In the energy systems of this course only the following forms of energy will be considered: •u •e kinetic=V 22 •gz •e C H 13 as the streams flowing accross the energy syste do not undergo any variations of nuclear energy (as no nuclear reaction occurs) or electric energy (as they are not subjet to electric field). Then the total energy of a stream (in/out) or of any particle can be written as: etot= u+V 22 + gz+e C H(2.18) 2.3.1Law of conservation of energy (1st principle of ther-modynamics) For a material control volumeΩ(t), the variation of total energy of its particles is equal to the power of the external forces (called Active forces) applied toΩ(t)plus the thermal power enteringΩ(t)Figure 2.10 dEtotdt = ˙ Qin+˙ Win(2.19) whereEtot=Z Ω(t)( ρe tot) dV(2.20) ˙ Win=Z ∂Ω(t)⃗ Fexternal· ⃗vdA=Z ∂Ω(t)( −p¯ n+ ¯τ)·⃗vdA(2.21) 14 2.4Energy balance equation for closed systems The law of conservation of energy for material control volumes can be applied directly to closed systems:Figure 2.11 dEtotdt = ˙ Qin+˙ Win(2.22) Zt2 t1dE totdt =Z t2 t1˙ Qin+˙ Win(2.23) Etot( t 2) −E tot( t 1) = Q in1→2+ L in1→2(2.24) We assume it is a simple closed system with quasi-static transformations, there- fore: •E c≈ 0(velocity of fluid≈0) •∆z≈0(negligible variation of potential energy) •¯ τ≈0(due to quasi static transformations⃗v≈0→⃗τ= 0) =⇒E tot= U+E c+ gz+E C H= U+E C H= U T C H"Thermochemical internal energy"=⇒R ∂Ω(t)( −p¯ n+ ¯τ)·⃗vdA=−R ∂Ω(t)p⃗n ·⃗vdA=−pR ∂Ω(t)⃗v ·⃗ndA=Applying the Reynolds theorem=−pdVdt = ⇒dU T C Hdt = ˙ Qin− pdVdt or dU T C H= DQ in− pdV or with the integral fromt 1→ t 2 UT C H( t 2) −U T C H( t 1) = Q in1→2−Z t2 t1p (t)dV(2.25) 15 Figure 2.12 2.5Energy balance equation for open systems Starting from the law of conservation of energy applied to a material control vol- umeΩ(t 0) such thatΩ(t 0) ≡V, it is necessary to use theReynolds’ transport theoremso as to convertdE totdt into: Z V∂ (ρe tot)∂ t dV +Z Sin 1∪ S in 2∪ S wall∪ S rotor∪ S piston∪ S out( ρe tot) ⃗v·⃗ndA(2.26) Then we can apply the first principle of thermodynamics:Z Ω(t)∂ (ρe tot)∂ t dV +Z ∂Ω(t)( ρe tot) ⃗v·⃗ndA=˙ Qin+Z ∂Ω(t)( −p⃗n+⃗τ)·⃗vdA(2.27) Whent=t 0, Ω(t 0) = Vand∂Ω(t 0) = ∂ Vthe eulerian control volume and the material control volume coincide. Then: Z Ω(t 0)= V∂ (ρe tot)∂ t dV +Z ∂Ω(t 0)= V( ρe tot) ⃗v·⃗ndA=˙ Qin+Z ∂Ω(t 0)= V( −p⃗n+⃗τ)·⃗vdA(2.28) ZV∂ (ρe tot)∂ t dV +Z Sin 1∪ S in 2∪ S out( ρe tot) ⃗v·⃗ndA +Z SP iston∪ S Rotor( ρe tot) ⃗v·⃗ndA+Z Swall( ρe tot) ⃗v·⃗ndA= =˙ Qin+Z SP iston( −p⃗n+⃗τ)·⃗vdA+Z SRotor( −p⃗n+⃗τ)·⃗vdA+ +Z SW all( −p⃗n+⃗τ)·⃗vdA+Z SS in1∪ S in2∪ S out( −p⃗n+⃗τ)·⃗vdA(2.29) However, considering the wall as still (fixed), impermeable and rigid; and con- sidering that there is no slip condition on the boundary, we can write that: 16 Z V∂ (ρe tot)∂ t dV + +Z SP iston∪ S Rotor( ρe tot) ⃗v·⃗ndA= =−Z Sin∪ S out( ρe tot+ p⃗n)⃗v·⃗ndA+˙ Qin−˙ WoutP iston− ˙ WoutRotor(2.30) We can say that the shear term is≈0becauseτ≈0on inlet and outlet surfaces if the fluid flow is uniform and orthogonal.Definition of total enthalpy of a stream:ρ·e tot+ p=ρ(e tot+ pv) =ρh tot The total enthalpy of the stream is given by the sum of the total energy and pv: u+e ch+ gz+V 22 + pv We can then define the mass flow rate of a stream across a section S as: ˙ m=−Z Sρ⃗v⃗ndA (2.31) and the average total enthalpy of a stream as:¯ htot=R Sh totρ⃗v⃗ndA˙ m(2.32) Then the energy balance equation for a generale open system (eulerian control volume V) can be written as: Z V∂ (ρe tot)∂ t dV +Z SP iston∪ S Rotor( ρe tot) ⃗v·⃗ndA= =X instreams˙ m in¯ htotin−X outstreams˙ m out¯ htotout+˙ Qin−˙ Woutrotor−˙ Woutpiston(2.33) If the average flow is steady∂ ¯ ϕ∂ t = 0 then the unsteady flow terms are zero even if there is a Rotor:X instreams˙ ¯ m in¯ ¯ htotin−X outstreams˙ ¯ m out¯ ¯ htotout+¯ ˙ Qin−¯ ˙ Woutrotor−¯ ˙ Woutpiston(2.34) Therefore the piston must be kept still/fixed to have a steady flow, so˙ Woutpiston= 0 2.6Energy balance equation for open systems with-out chemical reactions Let’s consider a stedy state flow:X instreamsm inh totin−X outstreamsm outh totout+˙ Qin−˙ Woutrotor(2.35) whereh tot= u+e ch+ gz+V 22 + pvand we can isolatee chas no chemical reaction occurs in the control volume V, so the in and out chemical energies are balanced, since all the inlet molecules are transferred to the exit without any modification of their energy, finally writing: X instreamsm i( h i+ gz i+V 2 i2 ) −X outstreamsm i( h i+ gz i+V 2 i2 ) + ˙ Qin−˙ Woutrotor= 0 (2.36) 17 2.7Energy balance equation with chemical reac- tions If in the control volume V a chemical reaction takes place, then the chemical energy of the output molecules is different from that of the inlet molecules:e C Hin̸ =e C Hout 2.8Chemical energy of molecules The chemical energy is astate function: it depends only on the current state of the system/particle/molecule and its variation across a transformation depends only on the initial and final stages and NOT on the path of the transformation.More in detail the chemical energy depends on: •Molecular structure •Pressure p •Temperature T At fixed p and T (for example atT=T ref= 25 °C, p=p ref= 1 atm) it is possible to associate a value ofe C Hto each molecule. We can measure the variation of chemical energy/difference of chemical energy between two molecules (reactants and products) by means of the following special reactor:Figure 2.13: Isothermal T=Tref=constant; Isobaric p=pref=constant The reactor is at constant p and T equal to the reference values. Reactants also enter atT refand p ref The energy balance equation in steady flow condition can be written as: ˙ Qin= ˙ m p( eref C H,p+ href p+V 2 p2 + gz p) −˙ m r( eref C H,r+ href r+V 2 r2 + gz r) (2.37) The reactor is designed so as to havez r= z p, V r= V pso we can simplify the kinetic and gravitational terms. Moreover we can assume thatT ref, p refare reference 18 values for enthalpies. Hence: href p= 0 , href r= 0 . Thus: ˙ Qin= ˙ m r( eref C H,p− eref C H,r) (2.38) eref C H,p− eref C H,r=˙ Qin˙ m r(2.39) The variation of chemical energy at reference p and T is equal to the thermal power absorbed/released by a chemical reactor operating at constant pressure and temperature, divided by the mass flow rate. This result can be exploited to deter- mine the chemical energy of all the molecules in nature. 2.8.1Determining the chemical energy of molecules As done for other state functions (h, u, etc.) you have to define a reference sys- tem/state at whiche C H= 0 : •T=T ref= 25 °C •p=p ref= 1 atm) •Molecule structure? It is NOT possible to define only one reference molecule, because it is impossible to derive all the other molecules from just one. If we assume H2as reference molecule, how can you measure the chemical energy of CO2? We can’t use the special isobaric- isothermal reactor as it is impossible to obtain CO2from H 2because no C and O atoms enter the reactor!As reference state for the molecular structure you need to define one molecule for each atomic species so as to have aspanning set:Atom Molecule (elemental species) C C graphite solid H H2 N N2 O O2 Ar ArSuch species are called elemental species and their chemical energies at T= Tref= 25 °C,p=p ref= 1 atm) arezero. Defining one elemental species for each atomic species allows to generate any possible molecule by properly combining elemental species. For example, you can obtain H2O and H 2and measure its chemical energy with the special reactor in reference condition:Reaction:2H 2O 22 H 2O ˙ Qin= ˙ m H 2O eref C H,H 2O− ˙ m H 2eref C H,H 2− ˙ m O 2eref C H,O 2= ˙ m H 2O eref C H,H 2O(2.40) Since O2and H 2are elemental species. We can then write that: eref C H,H 2O=˙ Qin˙ m H 2O and we can defineeref C H,H 2Oas the Enthalpy of formation∆href f. The enthalpy 19 Figure 2.14 of formation of the most common species have been measured and are reported in several chemical handbooks.Figure 2.15 20 Figure 2.16 Thus we know the chemical energy at reference condition of any molecule:eC H( T ref, p ref) = ∆ href f(2.41) What about the chemical energy at T and p different from the reference ones? It is not necessary to know it at any T and p, as we can write the total enthalpy of a stream of molecules of typep kas: htotp k= ∆ href fp k+ [ h pk( T , p)−h pk( T ref, p ref)] + gz+V 22 (2.42) where the term in square brackets is the enthalpy variation of the transformation which takesp kfrom the reference state to T,p. The basic idea behind this formula is to consider a sequence of transformations from the reference state to p, T,p k: we start from the reference state where we have the elemental species at(T ref, p ref) , then we add the reaction of formation of the speciesp k, ∆href fp k. Now we have the molecule p kat T ref, p refand we can add the reactions of: heating, cooling, pressurizing and expandingh pk( T , p)−h pk( T ref, p ref) and we finally obtainp kat (T,p). For the sake of brevity since h pk( T ref, p ref) = 0 the total enthalpy can be written as: htotp k= ∆ href fp k+ h pk( T , p) +gz+V 22 (2.43) NOTEYou cannot refer h(T,p) at a temperature/pressure different from the reference one, because you have to use the same reference state as the enthalpy of formation. You can use any arbitrary reference state only if the energy balance equation does not involve chemical reactions (hence the enthalpy of formation are not used/do not appear in the energy equations). If the stream is a perfect gas you can write that: htotp k= ∆ href fp k+ C pp k( T−T ref) + gz+V 22 (2.44) 2.9Energy balance equation for a chemical reac-tor (general case) Consider a chemical reactor with "rigid walls" (no piston, no work can be extracted by changing its shape) operating at p and T not necesarily equalto the reference values ofp refand T ref. 21 Figure 2.17 The reactor performs the following chemical reaction:n X k=1˙ Nr,kR k→m X k=1˙ Np,kP k(2.45) where˙ Nr,kis the molar flow of molecule R kin [kmol/s]andR k, P kare the molecule type of the reactants and of the products. If we divide both sides of the equation by˙ Nr,1(one of the mole flow rates of molecules), you obtain: n X k=1˙ Nr,k˙ Nr,1R k→m X k=1˙ Np,k˙ Nr,1P k(2.46) ˙ Nr,k˙ Nr,1= a r,kis the stoichiometric coefficient of Rk in the chemical reaction. ˙ Np,k˙ Nr,1= a p,kis the stoichiometric coefficient of Pk in the chemical reaction. So the chemical reaction can be written as:n X k=1a r,kR k→m X k=1a p,kP k(2.47) like for example: H4+ 2 O 2→ CO 2+ 2 H 2O Let’s assume thata r,kand a p,kas well as˙ Nr,kand˙ Np,kare known. We want to write the energy balance equation for the reactor under steady state condition: ˙ Qin+ ˙ m Rh tot,R= ˙ m Ph tot,P+˙ Woutrotor(2.48) The flows of reactants and products may in general be a mixture of different molecular species. For instance the flow of reactants may be made of O2and CH 4, while the one of the products is made of CO2and H 2O. Then we must be able to compute theh totof mixtures. 22 2.10Possible types of mixtures and their thermal enthalpy h 2.10.1Real mixtures Molecules of different type tend to interact each other with attraction/repulsion forces. The thermal enthalpy of each molecule species depends also on the concen- tration of the other species.h R,k( T , p, xR ) = enthalpy of molecular speciesR kin the mixture of reactants, wherexR,k is the molar composition of the stream of reactants, defined as a vector where: xR,k=˙ NR,k˙ NR,tot(2.49) hR( T , p) =n X k=1y R,k[ h R,k( T , p, xR ) −h R,k( T ref, p ref)] (2.50) wherey R,kis the mass fraction˙ m R,k˙ m R,tot. 2.10.2Ideal mixtures There is no interaction between molecules of different type.h R,k( T , p R,k) →where pR,k= p·x R,kis the partial pressure of R kin the mixture of reactants. hR( T , p) =n X k=1y R,k[ h R,k( T , p·x R,k) −h R,k( T ref, p ref)] (2.51) 2.10.3Ideal mixtures of ideal gases The enthalpy of gases depends only on T: hR( T , p) =n X k=1y R,k[ h R,k( T)−h R,k( T ref)] =n X k=1y R,k¯ Cp R,k( T−T ref) (2.52) hR( T , p) =¯ Cp R( T−T ref) (2.53) where¯ Cp Ris the average Cp of the mixture:P n k=1y R,k¯ Cp R,k Enthalpy of formation of mixtures: ∆href fR=n X k=1y R,k· ∆href fR,k(2.54) Total enthalpy of a mixture:htotR( T , p) = ∆href fR+ h R( T , p) +V 2 R2 + gz R(2.55) 23 where h R( T , p)must be evaluated according to Real mixtures, Ideal mixtures or Ideal mixtures of ideal gases depending on the type of mixture.Thus the energy balance equation for a general reactor is: ˙ Qin+ ˙ m R[∆ href fR+ h R( T , p)] = ˙m P[∆ href fP+ h P( T , p)] +˙ Woutrotor(2.56) where we can assume thatV 2 P2 ≊V 2 R2 and z P≊ z R. If the reactants and products are ideal mixtures of ideal gases, then: ˙ Qin+ ˙ m R[∆ href fR+ ¯ Cp R( T−T ref)] = ˙ m P[∆ href fP+ ¯ Cp P( T−T ref)] +˙ Woutrotor(2.57) 2.11Heat of reactionFigure 2.18 The special reactor has the same feature as that used to compute the enthalpy of formation of species. We can use it to determine the heat generated/absorbed by a chemical reaction. Let’s denote the chemical reaction as seen in (2.45) or as(2.47).For example: [1kmol/s]CH 4+ [2 kmol]O 2→ [1kmol/s]CO 2+ [2 kmol/s]H 2O (2.58) Let’s assume that the reaction is operating at steady-flow, then the energy bal- ance equation is: ˙ Qref in= ˙ m P[ h Ptot( T ref, p ref) + h Rtot( T ref, p ref)] (2.59) ˙ Qref in=m X ˙ NP,k∆˜ href fP,k−n X ˙ NR,k∆˜ href fR,k(2.60) ˙ Qinis the heat generated by the reaction at reference condition (T ref, p ref) . Thus the heat of reaction∆˜ hreactioncan be defined as the thermal power gener- ated/absorbed at reference condition over the molar flow rate of the first reactant species: ∆˜ hreaction=˙ Qref in˙ NR,1= [ kJkmol ] (2.61) 24 • if∆˜ hreaction> 0⇒the reaction isEndothermic˙ Qin> 0 •if∆˜ hreaction< 0⇒the reaction isExothermic˙ Qin< 0 2.12Entropy balance equation The second principle of thermodynamics states that the entropy variation dS/dt of a material control volumeΩ(t)is equal to: dSdt =Z ∂Ω(t)˙ q inT dA +Z Ω(t)˙ Sgen,int· dV(2.62) where the first term is the entropy associated to the inlet/outlet heat flow, while the second one in the entropy generated by internal irreversible processes: •friction (mechanical) •heat transfer with∆T >0 •irreversible reactions •mixing gases with different compositions where˙ Sgen,int≥ 0 25 Figure 2.19 In order to derive the Entropy balance equation for an eulerian control volume V, you need to use the Reynolds’ transport theorem:Figure 2.20 Z V∂ (ρs)∂ t dV +Z SP iston∪ S Rotor( ρs)vndA =X in˙ m is i−X out˙ m is i+˙ QinT in+Z V˙ Sgen,intdV (2.63) The entropy of the streamss imust be evaluated with respect to a unique ref- erence state for all the i streams. It is universally adopted the following reference 26 state: T∗ = 0Kbecause at such low temperature all pure substances form a perfect crystal, whose entropy ("disorder") is zero for the third principle of thermodynamics. thus we can write s usingT∗ as a reference. More in detail for a pure speciesp kyou can write: spk( T , p) =s pk( T , p)−s pk( T= 0K, p ref) spk( T , p) =s pk( T , p)−s pk( T ref, p ref) + s pk( T ref, p ref) −s pk( T= 0K, p ref) spk( T , p) =s pk( T , p)−s pk( T ref= 25 C, p ref= 1 atm) + ∆sref fp k(2.64) where we can divide two terms: •s pk( T , p)−s pk( T ref= 25 C, p ref= 1 atm)is the variation of thermal entropy of pkalong the transformation between T ref, p refand T, p. •∆sref fp kis the standard entropy of substance/molecular species p k. The standard entropy for essentially all the available molecular species and sub- stances are known and reported in chemical handbooks.If the stream i is an ideal mixture of perfect gasesyou can write: smix=n X k=1y pk[( s pk( T , p·x pk) −s pk( T , p ref)) + ∆ sref fp k] smix=n X k=1y pk[( C p pkln(TT ref) −Rln(p ·x pkp ref) + ∆ sref fp k] (2.65) where(s pk( T , p·x pk) −s pk( T , p ref)) is the thermal entropy of a perfect gasp k; ypkis the mass fraction of p kin the mixture and x pkis the molar fraction of p kin the mixture. 27 3.Heat Exchangers 3.1Type of heat exchangers HXs can be classified according to the type of heat-transfer process:1.Indirect-contact heat exchangers in which the two fluids are not mixeddirect-transfer HXs (separation surface between hot and cold streams) regenerative type: a heat storage receives heat from the hot stream, stores it for a certain time and then releases heat to the cold stream 2.Direct-contact HXs with no separation surface between the two streamsgas-liquid contact towers (cooling towers) gas-gas and liquid-liquid mixers 3.1.1Direct transfer heat exchangers Direct transfer HXs can be classified according to the directions of the two streams: 1.Parallel-flow where the two fluids have the same directions; 2.Counter-flow where the fluids have opposite directions; 3.Cross-flow in which the fluids have orthogonal directions.Figure 3.1 The other classification can be done according to the construction: 28 1.Tubular type: double-pipe; shell and tube (liquid in the tube and gas/vapor in the shell side) spiral-tube This type of HXs are recommended if the streams have similar heat transfer coefficients, high pressures or large pressure differences between streams. The first two types have more problem with thermal stresses, while the spiral ones can vary the diameter according to the different temperatures. 2.Plate type: more compact then the shell and tube HXs because of the improvedheat transfer process. It is recommended for liquid streams of small volume and suitable for limited pressure differences between streams (the plates must be kept tight by an external force). There are some advantages and some cons: (a)Advantages:It is modular, so if we want to increase the contact area we can simply add more plates. The plates are easy to disassemble, so it is an easier solution to clean. (b)Problems:This configuration is not suitable for gases. It is hold together with screws, so we can’t have high pressures as it is difficult to prevent leakages. 3.Extended Surface, either finned tubes or plates, to increase the heat transferarea of the side which has the lowest heat transfer coefficient of the two. Typical examples are the radiators of vehicles or refrigerators. 3.1.2Regenerative heat exchangers - The Ljungstrom airpreheater The “Ljungstrom air-preheater” is a regenerative heat exchanger: heat from the hot fluid is intermittently stored in a thermal storage medium before it is transferred to the cold fluid. The Ljungstrom con- sists of a central rotating-plate element installed within a casing that is divided into two sectors containing seals around the element. The seals allow the element to rotate through all the sectors, but keep gas leakage between sectors to a minimum while providing separate gas air and flue gas paths through each sector. One sector (usually spanning about half the cross-section of the casing) is connected to the boiler hot gas outlet. The hot exhaust gas flows over the central element, transferring some of its heat to the element, and is then ducted away to the stack. The second sector is fed with ambient air by a fan, which passes over the heated element as it rotates into the sector, and is heated before being carried to the boiler furnace for combustion. The rotor is the medium of 29 hea transfer and is usually composed of baskets containing steel sheets. It rotates quite slowly (around 3-5 rpm).Figure 3.2 Figure 3.3 30 3.2Focus on HRSG and Boilers Heat recovery steam generators or HRSG can be classified asdirect transfer heat exchangers, usually with finned tubes and cross flow directions. Fins are used toFigure 3.4 increase the heat transfer area of the flue gas side.InsteadSuperheatersandReheaters of combustion boilersdo not have finned tubes, but bare tubes, because at those high temperatures, fins would melt. Economizersinstead may have fins because flue gases are at lower temperatures.Figure 3.5 The membrane wall of theevaporatoris a special type of heat exchanger since its geometry is unique and most of the heat transfer takes place through irradiation. The tools required to size the membrane wall are not analyzed here. 31 3.3Design procedure for direct transfer heat ex- changers Problem statement:for a given (input data) [˙ m hot, ˙ m cold, T inhot, T incold, T outcold, T outhot] determine the heat exchanger designDesign procedure: 1.Determine˙ Q= ˙m coldC p cold( T outcold− T incold) (3.1) or alternatively in an equivalent way: ˙ Q= ˙m hotC p hot( T inhot− T outhot) (3.2) 2.Choose the heat exchanger construction type on the basis of:pressure difference of the two fluids; required cross sectional area; type of streams (gas or liquid or two phase stream). 3.Set all the velocities of the two streams, taking into account of heat transfercoefficient and pressure drops→see the typical ranges for liquids and gases. 4.Evaluate the required heat exchanger area with theheat transfer rate equa- tion 5.Workout the geometrical parameters of the heat exchanger: diameter andthickness of tubes, length and width of the heat exchanger, number of tubes in parallel, number of rows etc. Such geometrical parameters must satisfy two mean conditions: heat transfer area must be equal to the required value; cross sectional area must be equal to:˙ m hotρ hotV hotand ˙ m coldρ coldV coldfor the two streams. 6.For the given geometrical parameters carry out an accurate evaluation of pres-sure drops and heat transfer coefficients (thermo-hydraulical design) 3.4Heat transfer rate equation Under the assumption of no internal losses, steady flow,V in≊ V out(no velocity difference in/out) and that the streams have constant Cp (perfect gas or ideal liquid), we can prove that for a direct-transfer heat exchanger: ˙ Q=U A∆T mlog(3.3) where U is the global heat transfer coefficient, A is the heat transfer area and ∆T mlogis the mean logarithmic difference. 32 ( U A) =1R tot= 1P iR i(3.4) ∆T mlog=∆ T sideA− ∆T sideBln( ∆ T sideA∆ T sideB)(3.5) The formula (3.5) is valid for both parallel and counter flow heat exchangers:Figure 3.6 Hence for parallel and counter flow:∆T mlog=( T outhot− T outcold) −(T inhot− T incold)ln[( ( T out hot− T out cold)( T in hot− T in cold))](3.6) ∆T mlog=( T inhot− T outcold) −(T outhot− T incold)ln[( ( T in hot− T out cold)( T out hot− T in cold))](3.7) 3.4.1Note 1 For fixed in and out Temperatures, it can be proven that∆T mlog parallelf low≤ ∆T mlog counterf low. Then counter flow always require less heat transfer area than the parallel flow ar- rangement. 3.4.2Note 2 If one of two streams (hot or cold) has a constant temperature (phase change either evaporation or condensation) then∆T mlog parallelf low= ∆ T mlog counterf low. The direction of the two streams does not affect the heat transfer rate because one stream is at constant temperature, therefore for the heat transfer rate the two cases are equal. 33 Figure 3.7 3.5Cross flow heat exchangers The heat transfer rate equation for cross flow heat exchangers needs a correction fac- torF=F(p, R, Layout(numberof passes))which takes into account of the relative direction of the two streams: 34 Figure 3.8: Cross flow with exactly orthogonal direction (single pass design) Figure 3.9: Cross flow with orthogonal counter flow direction (multiple passes de- sign) So we can modify the equation (3.3) into:˙ Q=F U A∆T mlog counterf low= F U A[( T inhot− T outcold) −(T outhot− T incold)ln[( ( T in hot− T out cold)( T out hot− T in cold))]] (3.8) The two terms in F are defined as:p=| T tubeout− T tubein|| T shellin− T tubein|(3.9) R=( ˙ mC p) tubeside( ˙ mC p) shellside(3.10) 35 And ( ˙mC p)is calledheat capacity rate.Figure 3.10 Note that F=1 if p tends to zero, so a constant temperature phase change occurs in the tubes, and/or R tends to zero, so a constant temperature phase change occurs in the shell side. Hence if one of the two streams is evaporation/condensing at constant temperature then F=1. Each heat exchanger has its own plot F=F(p,R) as in figure 3.10 which depends on its layout (mainly on the number of passes). We can find such plots on heat transfer textbooks such as the "TEMA" Standards issued by theTubular exchanger manufacturer’s association. 3.6Heat transfer rate equation for cylindrical ge-ometries (tubes) Heat transfer rate equation:˙ Q=F U A∆T mlog counterf low(3.11) whereU A=1R tot. We need to distinguish between the internal Ai and external surface Ae of the tube, so we can write the previous equation for both surfaces, and for both cases we need to state that: UiA i= U eA e=1R tot(3.12) ˙ Q=F U iA i∆ T mlog counterf low(3.13) ˙ Q=F U eA e∆ T mlog counterf low(3.14) 36 Figure 3.11 If we choose Ai, we need to define Ui accordingly:AiU i=11 h iA i+ln D eD i2 πkL tube+ 1h eA e(3.15) Ui=11 h i+A ilnD eD i2 πkL tube+ A ih eA e(3.16) 37 Figure 3.12 Instead if we choose Ae, we can compute Ue as follows:AeU e=11 h iA i+ln D eD i2 πkL tube+ 1h eA e(3.17) Ue=A e1 h iA i+A elnD eD i2 πkL tube+ 1h e(3.18) Therefore we can define the heat transfer rate equation as defined in equations (3.13) and (3.14). 3.7Extended surface heat exchangers (fins) If one of the two streams has a poor heat transfer coefficient (e.g. air or gas) it is advantageous to extend the heat transfer area of that side by addingfins.Figure 3.13: Finned tube of an economizer Starting from the equation (3.16) we can see that if hi is very large com- pared to he and the tube is made of steel (high k) then:U i=A eh eAi ⇒ ˙ Q= 38 F U iA i∆ T mlog counterf low≊ A eh eF ∆T mlog counterf lowHence there is no advantage in in- creasing Ai and it is advantageous to increase only Ae by adding fins.However it is important to note that the heat exchanged does not increase linearly with the extension of the fins:A e= external tube surf ace=A bare tube+ A f ins area. So we cannot write that: Ui=11 h i+A ilnD eD i2 πkL tube+ A ih e( A bare tube+ A f ins area)(3.19) because such formula is valid only if the fins and the bare tube wall are at the same temperatureT w,ewhich in general is not true as we can see in this example:Figure 3.14 Figure 3.15 39 The fin temperature profile is between the flue gas temperature and the bare wall tube temperatureT w,e. This reduces the convective heat transfer rate between flue gases and fins: ˙ Q=h e( T e−¯ Tf ins) A f ins+ h e( T e− T w,e) A bare wall(3.20) where Te is the flue gases temperature,¯ Tf insis the average fins temperature, in our example around90℃andT w,eis the bare wall temperature. So we can note that:˙ Q < he( A F+ A B)( T e− T w,e) =˙ Qiso−f in(3.21) And the iso-fin heat is the thermal power exchanged by the finned surface if fins were at the bare wall temperatureT w,e. We can then define the overall finned surface effectivenessη 0as: η0=˙ Q˙ Qiso−f in(3.22) Hence the thermal resistance R of the convective process on the external fluid side can be written as: Rconvective external=T e, T w,e˙ Q= T e, T w,eη 0˙ Qiso−f in= 1h e( A F+ A B) η 0(3.23) This internal global heat transfer coefficient can be written as:Ui=11 h i+A ilnD eD i2 πkL tube+ A ih eA eη 0(3.24) NOTE THAT:η 0decreases as the height/extension of the fins increases (because the added extension is at higher temperature). As a consequence, there is a limit value for the length of the fins which should not be crossed, because the increase of cost, material and weight is larger than the benificial effect an increase of the heat transfer rate˙ Qgives. At that limit, instead of increasing the length of the fins it may be better: •to have more fins per meter of tube (more dense); •to increase the number of tubes in the heat exchanger. 3.8Rating and Off-Design calculation of heat ex-changers Problem statement:Given the geometry (Area, number of tubes, diameter of the tubes etc.) and the mass flow rates of the hot and cold streams, and the input tem- peraturesT hotin, T coldindetermine their output temperatures and the heat exchanged Thotout, T coldout,˙ Q. We could in principle apply theheat transfer rate equationin this way: 40 • Given˙ m hot, ˙ m coldand the cross sectional area →determine the velocities of the two streams; then determineh e, h iand then find U eor U i. •Solve the system of nonlinear equation:       ˙ Q= ˙m hotC p hot( T inhot− T outhot) ˙ Q= ˙m coldC p cold( T outcold− T incold) ˙ Q=U iA iF ∆T mlog(3.25) A numerical method and a computer are needed since the∆T mlogcontains the two variablesT hotout, T coldout. However there is another way to solve the problem without the need of a computer and of numerical methods: it is called theε-NTU method.εis defined as the effectiveness of the heat exchanger as the ratio between the actual heat exchanged and the heat that a counter flow HX with infinite area could exchange. ε=˙ Q˙ Qc∞(3.26) ˙ Qc∞= ( ˙ mC p) min( T hotin− T coldin) (3.27) Where( ˙mC p) minis the minimum heat capacity rate among those of the two streams:(a) Case A:( ˙mC p) hot> ( ˙mC p) cold(b) Case A:( ˙mC p) hot< ( ˙mC p) cold for both cases we can write that:˙ Qc∞= ( ˙ mC p) min( T hotin− T coldin) It is important to note that if: •both fluids have constant Cp; •there are no thermal losses; •we are in steady flow condition; •V in≊ V out 41 then the effectiveness εdepends only on Cr, NTU and the layout: ε=ε(C r, N T U, Layout )(3.28) This result comes from theBuckingham’s Theoremapplied to the heat ex- changer, as Cr and NTU (number of transfer units) are adimensional parameters defined as: N T U=U A( ˙ mC p) min(3.29) Cr=( ˙ mC p) min( ˙ mC p) max(3.30) This means that: •if two heat exchangers have the same NTU, Cr and geometry (not necessarily same area A but just same "shape": same directions of hot and cold streams, same arrangement of tubes, etc.) then they have the sameϵ •if the operating conditions of a given heat exchanger happen to have same NTU and Cr, then they have the sameε. We can now give some results which can be proven: •For parallel flow heat exchangers: ϵ=1 −e− N T U(1+C r)1 + C r(3.31) •For a counter flow heat exchangesr: ϵ=1 −e− N T U(1−C r)1 −C re− N T U(1−C r)(3.32) •For a heat exchanger with constant temperature phase change: ϵ= 1−e− N T U (3.33) because Cr = 0, and this formula is valid for all the arrangements: parallel, counter and cross flow. 42 Figure 3.16: Comparison between parallel, counter and cross flow The plot shows that for any fixed NTU, the heat exchanger has the highest effectivenessϵ. This means that it exchanges the greatest amount of heat for a given inlet temperature and area. Thus the counter flow layout is the most advantageous, the second best is the cross flow and the worst one is the parallel flow. Note that for NTU≥3 we have thatdεdN T U ≊ 0, hence an increase of heat transfer area A does not lead to an appreciable increase of the heat transferred. 3.9Rating and Off-design calculation of Hx withtheε-NTU method GivenT coldin, T hotin, ˙ m c, C p c, ˙ m h, C p h, U, A and HX layout/geometry, determine: ˙ Q, Thotout, T coldout. Solution procedure: 1.Determine Cr and NTU 2.Determineε(Cr, NTU, Layout) from formula or plots (plots are available in heat transfer handbooks) 3.Determine˙ Q ˙ Q=ε˙ Qc∞= ε( ˙mC p) min( T inhot− T incold) (3.34) 4.Determine: Touthot= T inhot−˙ (Q)( ˙ mC p) hot(3.35) Toutcold= T incold+˙ (Q)( ˙ mC p) cold(3.36) Note: such calculation does not require interactions. Instead the heat transfer rate equation (3.3) would require an alternative numerical method. 43 (a)(b) 3.10Heat exchangers with variable Cp streams Let’s consider a HRSG, since the Cp of liquid water is different from that of boiling water (Cp = +∞) and from that of superheated steam, then it is not possible to apply the heat transfer rate equation or theε-NTU method to the whole HRSG. It is necessary to divide the HRSG into 3 sections in which the stream Cp can be reasonably assumed constant. These 3 sections are:economizer, evaporator and superheaterFor each section we can write both the equation (3.11) and (3.28). As an example for the economizer we can write: ˙ Qeco= F ecoU ecoA eco( T D− T P) −(T B− T 2)ln T D− T PT B− T 2(3.37) εeco= ε eco(U ecoA eco( ˙ mC p) mineco, ( ˙ mC p) mineco( ˙ mC p) maxeco, Layout E C O )(3.38) 3.11Condensers for steam cycles There are three types of condensers:1.water-cooled condensers 2.cooling towers 3.air cooled condensers 44 3.11.1Water cooled condensers Figure 3.17 Cooling water is taken from a big reservoir like a lake, a big river or the sea. The condenser is a shell and tube heat exchanger, where steam flows on the shell side while cooling water flows in the tubes. The steam flows in the shell side because it has the larger volumetric flow rate and then it requires larger cross sectional areas.Figure 3.18 The water cooled condenser is the most advantageous system of condensation because water has a high transfer coefficient and the water from sea, lakes and rivers is typically at temperatures lower than that of air. As a consequence water cooled condensers allow to adopt lowerp condthan other types of condensers and require smaller heat transfer area (leading to higherη el,netand lower capital cost of the power plant). However it requires the availability of large quantities of water (sea, lake, large rivers) near the power plant. 45 3.11.2Cooling towers If the reservoir of water is limited (not sufficient to absorb the thermal power rejected by the condenser without an appreciable increase of temperature), it can be used to reintegrate the water of a cooling tower.Figure 3.19 The condenser system is made of: •a shell and tube condenser with water in the tubes and steam in the shell •a cooling tower which cools down the cooling water of the condenser from T2 to T1 with air. Air enters the cooling tower from the bottom and exits at the top. The drought is ensured by natural circulation or fans. If fans are used, the tower is not very tall (no need of high height to have natural drought).The tower is a direct contact, gas liquid heat exchanger. Water droplets fall from the top to the bottom exchanging heat with cold air through two mechanisms: 1.convection 2.evaporation (mass-transfer) 46 Figure 3.20 If air is not saturated, H2O molecules tend to evaporate from the water droplet and enter the air flow. Each H2O molecule which evaporates absorbs the evapora- tion enthalpy∆h eva( ≈2500[kJ/kg]) from the liquid droplet causing a considerable decrease of the droplet temperature. Since∆h evais large, most of the heat is trans- ferred from the liquid droplets to the air flow by the evaporation/mass-transfer mechanism. The exchange surface in the cooling tower is used to increase the con- tact area between air and liquid water so as to enhance evaporation.Make-up water is required to reintegrate: •water which evaporates and exits the tower with the air stream •water which is entrained by the air stream •water which is discharged from the bottom of the tower so as to keep the con- centration of dissolved solids (minerals and salts). Indeed, while H2O evapo- rates and leaves the tower with the air flow, dissolved solids remain in the water droplet and so then concentration tends to increase. The discharge stream of water at the bottom of the tower is reach of dissolved solids and it is replaced by make up water. Thus the make up water flow is given by the sum of these 3 terms. the water flow discharged is set so as to keep the concentration x of solids ad a certain level.In a cooling tower water can be cooled down to the wet bulb temperature of the inlet air flow. Indeed the wet bulb temperature is the lowest temperature which can be achieved by evaporative cooling of a water melted surface (that of the water droplets). The approach point temperature difference∆T apof the cooling tower (typically around 5-10℃) is the difference between T1 (cold water temperature) andT wet bulbof inlet air flow: ∆T ap= T 1− T wet bulb inlet air f low(3.39) ∆T apdepends on velocity, extension of the exchange surface and mass flow rate of air. 47 Note: For cooling towers the heat transfer rate equation and the ε-NTU method are not valid. Different (more sofisticated) equations must be used to determine the size of cooling towers. 3.11.3Energy balance of a cooling tower VFigure 3.21 ˙ m 2h 2+ ˙ m dry airi a+ ˙ m make uph make up= ˙ m dischargeh discharge+ ˙ m 1h 1+ ˙ m dry airi b(3.40) i=[kJkg of dry air ] = enthalpy of water vapor in the saturated/wet air. Note that i is referred to Tref and pref. Since i is referred to the reference condition, also the enthalpies of water (h2, h1, etch) must be referred to temperature: ˙ m 2c wT 2+ ˙ m dry airi a+ ˙ m make upc wT make up= ˙ m dischargec wT discharge+ ˙ m 1c wT 1+ ˙ m dry airi b (3.41) 3.11.4Air cooled condenser This type is adopted when at the plant site there isn’t enough water to feed a cooling tower. 48 Figure 3.22 Compared to a water cooled condenser, this type has the following disadvantages: 1.Air has poor heat transfer coefficient⇒much larger heat transfer area required 2.Air temperature may have significant oscillations during the days (hot in sum-mer and cold in winter) causing an increase of the condensation pressure and thus a decrease of theη el,net 3.The density of air is much lower than that of water causing the need of largercross-sectional areas and higher electric consumption of the circulation fan. 49 4.Fuel combustion 4.1Combustion reactions and fuels Combustion reactions:Reactions causing the oxidation of substances (called "chemical fuels") with considerable release of heat (Exothermic).Chemical fuel: Substance which can be converted into power (heat or work) by means of a chemical reaction.A chemical fuel can be both in a combustion based system generating heat, and in a fuel cell system generating electricity (work). 4.1.1Classification of fuels We can classify fuels on the basis of their state: liquid, solid or gas; or on the basis of their occurrence: natural (primary) or secondary (artificial).Classification Primary Secondary Solid Wood, coal, peat Coke, charcoal, petcoke Liquid Petroleum Diesel, gasoline, LPG, ethanol, etc. Gas Natural gas H 2, propane, coal gas, etc.Moreover, on the basis of their formation process they can be classified in re- newable and fossil fuels. Fossil fuels derive from a fossilization process, an anaerobic decomposition process of buried dead organisms (vegetables and animals). The fossilization process causes the progressive loss of N and O atoms and then a con- centration of C and H atoms. The fossilization process is very slow, taking millions of years. Hence fossil fuels cannot be considered as renewable energy sources as the consumption rate is much faster than the production rate. Instead Biomass fuels (wood, straw, etc.) can be considered renewable energy sources as their growth rate is of the same order of magnitude as the consumption rate. Note: PEAT is made of partially decomposed vegetation deposited at an average depth of 1,5-2 meters. 50 4.1.2Identification of fuels Identification of gases fuels The properties of gaseous fuels can be easily identified by providing their composition (molar or mass basis). Table 4.1: For natural gas:Molar composition % CH 495.2C 2H 62.5C 3H 80.2C 4hH 00.06C 5H 120.02CO 20.7O 20.02Table 4.2: For LPG: Molar composition % C 2H 60.2C 3H 857.3C 4hH 041.1C 5H 121.4Identification of liquid fuels Liquid fuels can be identified in terms of properties in two different ways: •proving the fraction of each molecular species of the mixture. For gasoline you could have the mass fraction of each compound:             isobuthane 0.0122 isopenthane0.1049 n−penthane0.0509 etc.(4.1) with a very long list of hydrocarbons; •providing the atomic composition mass basis, also called ultimate or elemental analysis. 51 For gasoline we can use the atomic composition: yH=m Hm tot= 14% (4.2) yC=m Cm tot= 86% (4.3) For fuel oil (residue of the distillation process) we may have a composition like below: yC= 84 .5%(4.4) yH= 11 .9%(4.5) yO= 0 .8%(4.6) yN= 0 .3%(4.7) yS= 2 .3%(4.8) yash= 0 .2%(4.9) Ash is a fraction of fuel which is not oxidized and does not react. Typically it is made of oxides coming from the soil (being already oxidized, they do not react with O2) such as:                   SiO 2= silica; Al2O 3= allumina; CaO=calcium oxide; Fe2O 3= iron oxide; etc.(4.10) Identification of solid fuels Solid fuels can be identified only with the ultimate analysis or atomic composition. Indeed it is not possible to identify the single molecules which make up the piece of coal or wood. The coal structure is not uniform like that of graphite but coal has micro-pores (cavities) containing gas/water, oxygen, nitrogen and hydrogen atoms that in some cases replace C atoms in a random way: 52 Figure 4.1 So it is necessary to perform first a proximate analysis which determines: •Moisture content: yH 2O =m H 2Om tot(4.11) •Volatile matter = gases driven off when the coal is heated up to a certaing high temperature for a short time •Solid residue: "fixed carbon" or ash These parameters determine the coal quality. A sample of coal has a high quality if it has a low moisture content, a high fixed carbon fraction and a low ash content.After the proximate analysis, the ultimate analysis is performed to identify the atomic composition of the sample of coal:                         y C= 73 .6% yH= 4 .6% yO= 1% yS= 2 .5% yash= 8 .4% ymoisture= 2 .7%(4.12) 53 4.2Evaluation of the combustion products and heat of combustion It is worth noting that the heat generated by a reaction depends on the products which are generated by the reaction: ˙ Qin= ˙ m P[∆ href fP− h P( T , p)]−˙ m R[∆ href fR− h R( T , p)](4.13) ∆href fp=X yp,k· ∆href fp,k(4.14) both the reference enthalpy and the enthalpy of mixture depends ony p,k, so in other words the heat generated depends on¯ y por ¯ x p. So in order to determine the heat we need to determine the species generated by the combustion process. There are three approaches (in order of the degree of complexity): •Complete combustion assumption; •Chemical equilibrium assumption; •Kinetic model of the reactor. 4.3Complete combustion theory/assumption All the atomic species which can be oxidized (i.e. C,H,S) are converted into the most chemically stable oxidized compound:                         C →CO 2 H→H 2O S→SO 2 N→N 2 O→O 2 Ar→Ar(4.15) N, O, Ar and all other atoms which cannot be oxidized are converted into the most stable compound at atmospheric conditions. The intent is to associate a single product species to each atom. 54 4.3.1Example: combustion of CH 4with O 2with complete combustion assumptionFigure 4.2 ˙ NRCH 4CH 4+˙ NRO 2O 2→˙ NPCO 2CO 2+˙ NPH 2OH 2O +˙ NPO 2O 2(4.16) How much are˙ NPCO 2, ˙ NPH 2O, ˙ NPO 2? The assumption of complete combustion has limited to CO2, O 2, H 2O the set of possible products excluding other species like CO, OH, etc. For the reactor I can write the equation balance of atomic species:       C :˙ NRCH 4= ˙ NPCO 2⇒ ˙ NPCO 2= 1 kmol/s H:˙ NRCH 4· 4 =˙ NPH 2O· 2⇒˙ NPH 2O= 2 kmol/s O:˙ NRO 2· 2 =˙ NPCO 2· 2 +˙ NPH 2O+ ˙ NPO 2· 2⇒˙ NPO 2= 0 kmol/s(4.17) It is a linear system with˙ NPkunknowns. Note that, thanks to the assumption of complete combustion, the number of unknowns is equal to the number of equations. The trick is to limit the number of possible products and set it equal to the number of atomic species so, in this way, the balance equation f