An Isentropic Process for an Ideal Gas

A General FormulationSteady State, Steady Flow in a Flow Channel of Arbitrary Cross-section with Work and HeatTransferdĖ = Ė final − Ė initial= Ė x+dx − Ė xwhereĖ = ṁ(e + Pv)= ṁ(u + (v∗ ) 22+ gz + Pv)From the 1st lawrate of energystorage=rate ofwork+rate ofheat transfer +net rate of energyleaving the systemdE CVdt= dẆ − d ˙Q − dĖ (1)5

where dE CV=0for steady state.dtEquation (1) becomes0=dẆ − d ˙Q − ṁd[u + Pv + (v∗ ) 22+ gz](2)From the 2nd lawrate ofentropystorage⎧⎪⎨ rate of= entropy⎪⎩inflow−rate ofentropyoutflow⎫⎪⎬ rate of⎪⎭ +entropyproductiondS CVdt=[ṁs] x − [ṁs] x+dx −d ˙QT TER+ Ṗ Swhere dS CVdt=0for steady state.0=−ṁds −d ˙QT TER+ Ṗ Sord ˙Q = T TER Ṗ S − T TER ṁds (3)Combining (2) and (3) through d ˙QT TER Ṗ S − T TER ṁds = dẆ − ṁd(u + Pv + (v∗ ) 22+ gz)(4)Equation (4) can be used for any SS-SF process.6

Special CasesReversible, SS-SF ProcessReversible implies ⇒ Ṗ S =0• frictionless process• heat transfer is allowed but must be across ΔT → 0• which means T TER ≈ T CV= TEquation 4 becomesdẆṁ= −Tds+ du + d(Pv)} {{ }+d=du + Pdv+ vdP} {{ } } {{ }=Tds( (v ∗ ) 2 )+ d(gz) (5)2Therefore(dẆ(v ∗ṁ = vdP + d ) 2 )+ d(gz) (6)2Integrating Eq. (6) between the inlet and the outlet∫Ẇ outṁ = vdP + (v∗ ) 2outout+ gzin2 ∣ ∣inin} {{ } } {{ }ΔKE ΔPE(7)but ΔKE and ΔPE are usually negligible.If ΔKE + ΔPE =0∫Ẇ outṁ = vdP (8)inEquation can be used for a reversible, SS-SF flow in a liquid or a gas.7

If we keep in mindρ liq >> ρ gas ⇒ v liq

d( Pρ + (v∗ ) 22+ gz)= 0 (11)Integrating givesPρ + (v∗ ) 22+ gz = constant (12)Equation (12) is Bernoulli’s equation for frictionless flow with constant density. The constant isBernoulli’s constant, which remains constant along a streamline for steady, frictionless, incompressibleflow.Isothermal Ideal Gas, Compression/ExpansionThis is a special case of Eq. (8) for an ideal gas where Pv = RTPv = constant =(Pv) in =(Pv) out∫Ẇ outṁ = vdP =in∫ outin(Pv) indPPTherefore( )Ẇṁ = P Poutinv in lnP in(13)Isentropic Ideal Gas, Compression/ExpansionIsentropic implies a reversible and adiabatic process where s = constant. With an ideal gas,Pv k = constant and (Pv k ) in =(Pv k ) out .Equation (8) becomes∫Ẇ outṁ = vdP =in∫ outin[ (Pv k ] 1/k) indPP9

( ) ⎡Ẇ kṁ = (Pv) in⎣k − 1(Pout⎤) (k−1)/k− 1⎦ = c p (T out − T in ) (14)P inThe right side of Eq. (14) is based ∫on the fact that ΔKE +ΔPE =0and dh = du + dP vand du =0. Which leads to h = vdP.Note: for the same inlet state and pressure ratio⎛ ⎞⇒ ⎝Ẇ ⎠ṁrev.,isothermal⎛ ⎞< ⎝Ẇ ⎠ṁrev.,adiabatic10

Review of First and Second Law of ThermodynamicsReading Problems2-6, 4-1, 4-2 4-27, 4-40, 4-415-1 → 5-36-1, 6-2, 7-13DefinitionsSYSTEM:• any specified collection of matter under study.• all systems possess properties like mass, energy, entropy, volume, pressure, temperature,etc.WORK & HEAT TRANSFER:• thermodynamics deals with these properties of matter as a system interacts with itssurroundings through work and heat transfer• work and heat transfer are NOT properties → they are the forms that energy takes tocross the system boundary1

First Law of ThermodynamicsControl Mass (Closed System)CONSERVATION OF ENERGY:• the energy content of an isolated system is constantenergy entering − energy leaving = change of energy within the systemExample: A Gas CompressorPerforming a 1st law energy balance:⎧⎪⎨⎪⎩InitialEnergyE 1⎫⎪⎬⎪ ⎭+−⎧{ }Energy gain W1−2⎪⎨=Energy loss Q 1−2 ⎪⎩F inalEnergyE 2⎫⎪⎬⎪ ⎭E 1 + W 1−2 − Q 1−2 = E 22

Control Volume Analysis (Open System)CONSERVATION OF MASS:⎧⎪⎨⎪⎩rate of increaseof mass withinthe CV⎫ ⎧⎪⎬ ⎪⎨ net rate of⎪⎭ = mass flow⎪ ⎩IN⎫ ⎧⎪⎬ ⎪⎨ net rate of⎪⎭ − mass flow⎪ ⎩OUT⎫⎪⎬⎪⎭ddt (m CV )=ṁ IN − ṁ OUTwhere:m CV =∫VρdVṁ IN = (ρv ∗ A) INṁ OUT = (ρv ∗ A) OUTwith v ∗ = average velocity3

CONSERVATION OF ENERGY:The 1st law states:E CV (t) +ΔQ +ΔW shaft +(ΔE IN − ΔE OUT )+(ΔW IN − ΔW OUT ) = E CV (t +Δt) (1)where:ΔE IN = e IN Δm INΔE OUT = e OUT Δm OUTΔW = flow worke = E m = u }{{}internal+ (v∗ ) 22} {{ }kinetic+ gz}{{}potential4

Second Law of ThermodynamicsFundamentals1. Like mass and energy, every system has entropy.Entropy is a measure of the degree of microscopic disorder and represents our uncertaintyabout the microscopic state.2. Unlike mass and energy, entropy can be produced but it can never be destroyed. That is, theentropy of a system plus its surroundings (i.e. an isolated system) can never decrease (2ndlaw).P m = m 2 − m 1 = 0 (conservation of mass)P E = E 2 − E 1 = 0 (conservation of energy) → 1st lawP S = S 2 − S 1 ≥ 0 → 2nd lawThe second law states:(ΔS) system +(ΔS) surr. ≥ 0where Δ ≡ final − initialExample: A freezing process5

3. Reference: In a prefect crystal of a pure substance at T =0K, the molecules are completelymotionless and are stacked precisely in accordance with the crystal structure. Sinceentropy is a measure of microscopic disorder, then in this case S =0. That is, there is nouncertainty about the microscopic state.4. Relationship to Work: For a given system, an increase in the microscopic disorder (that isan increase in entropy) results in a loss of ability to do useful work.5. Heat: Energy transfer as heat takes place as work at the microscopic level but in a random,disorganized way. This type of energy transfer carries with it some chaos and thus results inentropy flow in or out of the system.6. Work: Energy transfer by work is microscopically organized and therefore entropy-free.Example: Slow adiabatic compression of a gasA process 1 → 2 is said to be reversible if the reverse process 2 → 1 restores the systemto its original state without leaving any change in either the system or its surroundings.→ idealization where S 2 = S 1 ⇒P S =0T 2 >T 1 ⇒ increased microscopic disorderV 2

where:> 0 irreversible (real world)=0 reversible (frictionless, perfectly elastic, inviscid fluid)But does:Isentropic P rocess ⇒ Reversible + AdiabaticNOT ALWAYS - the entropy increase of a substance during a process as a result of irreversibilitiesmay be offset by a decrease in entropy as a result of heat losses.General Derivation of Gibb’s EquationFrom a 1st law energy balance when KE and PE are neglectedEnergy Input = Energy Output + Increase in Energy StoragedQ}{{}amount= dW + dU }{{}differential(1)We know that the differential form of entropy isdS = dQ T(2) dW = PdV (3)Combining Eqs. 1, 2 and 3dS = dU T+ PdVT⇒ds = duT+ Pdv} {{ T }per unit mass7

Second Law Analysis for a Control Mass• control mass is uniformly at T TER at all times• control mass has a fixed size (V = constant)From Gibb’s equationT TER dS = dU + PdV↗ 0From the 1st lawdU = dQTherefore for a reversible processdS =dQT TERWe integrate to giveS 2 − S 1 = Q 1−2T TERand for a non-reversible processdS =We integrate to givedQT TER+ dP SS 2 − S 1 = Q 1−2T TER+ P S1−28

Second Law Analysis for a Control Volumewhere:FR - fluid reservoirTER - thermal energy reservoirMER - mechanical energy reservoir9

For the isolated system(ΔS) sys +(ΔS) sur = P S1−2 ≥ 0ΔS CV − s A m A 1−2 + s Bm B 1−2 − QA 1−2T A TER+ QB 1−2T B TER= P S1−2or as a rate equation( ) dSdtCV⎛= ⎝sṁ +˙Q⎞⎠T TERIN⎛− ⎝sṁ +˙Q⎞⎠T TEROUT+ Ṗ SThis can be thought of asaccumulation = IN − OUT + generation10

AvailabilityReading Problems8-1 → 8-8 8-29, 8-34, 8-50, 8-53, 8-638-71, 8-93, 8-103, 8-118, 8-137Second Law Analysis of SystemsAVAILABILITY:• the theoretical maximum amount of work that can be obtained from a system at a givenstate P 1 and T 1 when interacting with a reference atmosphere at the constant pressureand temperature P 0 and T 0 .• describes the work potential of a given system.• also referred to as “exergy”.The following observations can be made about availability:1. Availability is a property - since any quantity that is fixed when the state is fixed is a property.For a system at state 1 and specified values of the atmosphere of T 0 and P 0 ,themaximum useful work that can be produced is fixed.2. Availability is a composite property - since its value depends upon an external datum - thetemperature and pressure of the dead state.3. Availability of a system is 0 at its dead state when T = T 0 and P = P 0 . It is not possiblefor the system to interact with the reference atmosphere at the dead state. The system is saidto be in thermodynamic equilibrium with its surroundings.4. Unless otherwise stated, assume the dead state to be:P 0 = 1atmT 0 = 25 ◦ C1

5. The maximum work is obtained through a reversible process to the dead state.REV ERSIBLE W ORK} {{ }W rev= USEFUL } {{ WORK}W usef ul+ IRREV ERSIBILIT Y} {{ }IControl Mass Analysis• we knowW rev = W usef ul+ Ibut as shown in the figure, the actual work of the process is divided into two componentsW actual = W usef ul + W sur• where W sur is the part of the work done against the surroundings to displace the ambientairW sur = P 0 (V 2 − V 1 )=−P 0 (V 1 − V 2 )2

• this is unavoidable → this is not useful work. Nothing is gained by pushing the atmosphereaway.To find W actual , from the 1st lawE 1 − Q − W actual = E 2 → Q = E 1 − E 2 − W actualFrom the 2nd lawP s = ΔS system +ΔS sur ≥ 0= S 2 − S 1 + Q T 0But from the 1st law balance we knowQT 0= E 1 − E 2 − W actualT 0and when we combine this with the 2nd lawP s = S 2 − S 1 + E 1 − E 2 − W actualT 0which leads toW actual =(E 1 − E 2 )+T 0 (S 2 − S 1 ) − T 0 P sor by reversing the order of S 2 and S 1W actual =(E 1 − E 2 ) − T 0 (S 1 − S 2 ) − T 0 P sBut we also know thatW usef ul = W actual − W sur3

thereforeandW usef ul =(E 1 − E 2 ) − T 0 (S 1 − S 2 )+P 0 (V 1 − V 2 ) −T 0 P s} {{ }−W surW rev = W usef ul + I= W actual − W sur + IwhereI = T 0 P SThereforeW rev =(E 1 − E 2 ) − T 0 (S 1 − S 2 )+P 0 (V 1 − V 2 )In summaryW actual = (E 1 − E 2 ) − T 0 (S 1 − S 2 ) − T 0 P sW usef ul = (E 1 − E 2 ) − T 0 (S 1 − S 2 )+P 0 (V 1 − V 2 ) − T 0 P sX =Φ=W rev = (E 1 − E 2 ) − T 0 (S 1 − S 2 )+P 0 (V 1 − V 2 )DefineX =Φ = CONT ROL MASS AV AILABILIT Y= W rev (in going to the dead state)= (E − E 0 ) − T 0 (S − S 0 )+P 0 (V − V 0 )where the specific availability is defined asφ = Φ m4

What is the availability in going from one state to another?The reversible work isW rev =(Φ 1 − Φ 0 ) − (Φ 2 − Φ 0 )=Φ 1 − Φ 2but we also knowΦ 1 = (E 1 − E 0 ) − T 0 (S 1 − S 0 )+P 0 (V 1 − V 0 )Φ 2 = (E 2 − E 0 ) − T 0 (S 2 − S 0 )+P 0 (V 2 − V 0 )Φ 1 − Φ 2 = (E 1 − E 2 ) − T 0 (S 1 − S 2 )+P 0 (V 1 − V 2 )The availability destroyed isX des = I = W rev − W usef ul = T 0 P s = T 0 S genThis can be referred to as: irreversibilities, availability destruction or loss of availability.Control Volume AnalysisConsider a steady state, steady flow (SS-SF) processFrom the 1st lawdE cvdt ↗0 = −Ẇ actual − ˙Q +[ṁ(h + (v∗ ) 22+ gz)]in−[ṁ(h + (v∗ ) 22+ gz)]out(1)From the 2nd lawdS cvdt ↗0 =⎛⎝ṁs + ˙Q↗⎞0⎠T TERin⎛− ⎝ṁs +5˙Q⎞⎠T 0out+ Ṗ s (2)

Combining (1) and (2) through the ˙Q term, leads to the actual work output of the turbine, given asẆ actual =[ṁ(h + (v∗ ) 22)] [+ gz − T 0 s − ṁ(h + (v∗ ) 22in)]+ gz − T 0 s − T 0 Ṗ Sout= ṁ [−T 0 Δs +Δh +ΔKE +ΔPE] − (T 0 Ṗ s ) (3)Ẇ actual is the actual work output of the turbine.The specific flow availability, ψ,isgivenasψ = −T 0 (s − s 0 )+(h − h 0 )+( (v ∗ ) 22− (v∗ ↗0 0) 2 )+ g(z − z↗ 0 0 ) (4)2For a steady state, steady flow process where we assume KE=PE=0Ẇ rev = (ṁψ) in − (ṁψ) out (5)Ẋ des = ˙ I = Ẇ rev − Ẇ actual = T 0 Ṗ s = T 0 Ṡ gen (6)ψ = (h − h 0 ) − T 0 (s − s 0 ) (7)6

The General Exergy EquationFrom the 1st lawdE cvdt= Ẇ in − Ẇ out − ˙Q 0 + ˙Q 1 − ˙Q 2 +[ṁ(e + Pv)] in − [ṁ(e + Pv)] out (1)From the 2nd lawdS cvdt⎛= ⎝ṁs −Q ˙ 0+ ˙T 0⎞Q 1⎠T 1in⎛− ⎝ṁs +Q ˙ ⎞2⎠T 2out+ Ṗ s (2)Multiply (2) by T 0 and subtract from (1) to eliminate Q 0 , which leads to the generalized exergyequationddt (E − T 0S) CV = Ẇ in − Ẇ out +[ṁ(e + Pv − T 0 s)] in⎛[ṁ(e + Pv − T 0 s)] out + ⎝ ˙Q 1 − T ˙Q⎞0 1⎠T 1in⎛− ⎝ ˙Q 2 − T ˙Q⎞0 2⎠ − T 0 Ṗ S (3)T 2out7

We can rewrite Eq. (3) in a generalized form by introducing the definitions of Φ and ψ.dΦdt= P 0dV CVdt−[+[Ẇ + ṁψ + ˙QẆ + ṁψ + ˙Q((1 − T 0T TER)]in1 − T 0− I˙T TER)]outwhere˙ I = Ẋ des = T 0 Ṗ s= exergy destruction rateΦ = [(E − E 0 )+P 0 (V − V 0 ) − T 0 (S − S 0 )]= non − flow exergyψ = (h − h 0 ) − T 0 (s − s 0 )+ 1 [(v ∗ ) 2 − (v ∗ 02)2] + g(z − z 0 )= flow exergyẆ usef ul =( )dV CV(Ẇ in − Ẇ} {{ out ) − P} 0dtẆ actual} {{ }W surAside: The left side of the above equation isdΦdt = dE + P 0dV − T 0 dSdtbut the left side of Eq. (3) does not contain the term P 0 dV . Therefore, we must add a term P 0 dVto both the left and right side of the above equation in order to preserve a balance. Hence the termP 0dVdtappears on the right side of the above equation to preserve this balance, while the left side isdΦdt = (dE + P 0dV − T 0 dS)dt8

Efficiency and Effectiveness1. First law efficiency (thermal efficiency)η =Carnot cyclenet work outputgross heat input = W netQ inη = Q H − Q LQ H=1− T LT H2. Second Law Efficiency (effectiveness)η 2nd =net work outputmaximum reversible work=net work outputavailabilityTurbine → η 2nd = Ẇ/ṁψ e − ψ iCompressor →η 2nd = ψ e − ψ iẆ/ṁHeat Source → η 2nd =Ẇ/ṁ[˙Q/ṁ 1 − T ]0T TER3. Isentropic efficiency (process efficiency)(a) adiabatic turbine efficiencyη T =work of actual adiabatic expansionwork of reversible adiabatic expansion = W actW S(b) adiabatic compressor efficiencyη C =work of reversible adiabatic compressionwork of actual adiabatic compression= W SW act9

Carnot CycleReading6-7, 6-8, 6-10, 9-2, 10-1Problems• an ideal theoretical cycle that is the most efficient conceivable• based on a fully reversible heat engine - it does not include any of the irreversibilities associatedwith friction, viscous flow, etc.• in practice the thermal efficiency of real world heat engines are about half that of the ideal,Carnot cycleProcess State Points DescriptionPump 1 → 2 isentropic compression from T L → T Hto return vapour to a liquid stateHeat Supply 2 → 3 heat is supplied at constanttemperature and pressureWork Output 3 → 4 the vapour expands isentropicallyfrom the high pressure and temperatureto the low pressureCondenser 4 → 1 the vapour which is wet at 4 has to becooled to state point 11

Cycle Efficiency• defined as the net work output divided by the gross heat suppliedη = W netQ H= Q H − Q LQ H= 1 − T LT HFrom the figure the gross heat supplied isQ H = area(s 1 → s 4 → 3 → 2 → s 1 )=T H (s 4 − s 1 )The net work output isQ H − Q L = area(1 → 4 → 3 → 2) = (T H − T L )(s 4 − s 1 )Therefore the Carnot efficiency isη = (T H − T L )(s 4 − s 1 )T H (s 4 − s 1 )=1− T LT H2

Practical Problems• at state point 1 the steam is wet at T L and it is difficult to pump water/steam (two phase) tostate point 2• the pump can be sized smaller if the fluid is 100% liquid water• the pump is smaller, cheaper and more efficient• can we devise a Carnot cycle to operate outside the wet vapour region– between state points 2 and 3 the vapour must be isothermal and at different pressures -this is difficult to achieve– the high temperature and pressure at 2 and 3 present metallurgical limitationsThe net effect is that the Carnot cycle is not feasible for steam power plants.3

Rankine CycleReading Problems10-2 → 10-7 10-16, 10-34, 10-37, 10-44, 10-47, 10-59Definitions• working fluid is alternately vaporized and condensed as it recirculates in a closed cycle• water is typically used as the working fluid because of its low cost and relatively large valueof enthalpy of vaporization• the standard vapour cycle that excludes internal irreversibilities is called the Ideal RankineCycle• the condensation process is allowed to proceed to completion betweenstate points 4 → 1– provides a saturated liquid at 1• the water at state point 1 can be conveniently pumped to the boiler pressure at state point 2• but the water is not at the saturation temperature corresponding to the boiler pressure• heat must be added to change the water at 2 to saturated water at ‘a’• when heat is added at non-constant temperature (between 2 − a), the cycle efficiency willdecrease1

Analyze the ProcessAssume steady flow, KE = PE =0.From a 1st law balance, we knowenergy in = energy outDevice1st Law BalanceBoiler h 2 + q H = h 3 ⇒ q H = h 3 − h 2 (in)Turbine h 3 = h 4 + w T ⇒ w T = h 3 − h 4 (out)Condenser h 4 = h 1 + q L ⇒ q L = h 4 − h 1 (out)Pump h 1 + w P = h 2 ⇒ w P = h 2 − h 1 (in)The net work output is given asw T − w p = (h 3 − h 4 ) − (h 2 − h 1 )= (h 3 − h 4 )+(h 1 − h 2 )The net heat supplied to the boiler isq H =(h 3 − h 2 )The Rankine efficiency isη R =net work outputheat supplied to the boiler= (h 3 − h 4 )+(h 1 − h 2 )(h 3 − h 2 )2

If we consider the fluid to be incompressible(h 2 − h 1 )=v(P 2 − P 1 )Since the actual process is irreversible, an isentropic efficiency can be defined such thatExpansion process ⇒ Isentropic efficiency =actual workisentropic workisentropic workCompression process ⇒ Isentropic efficiency =actual workBoth isentropic efficiencies will have a numerical value between 0 and 1.Effects of Boiler and Condenser PressureWe know the efficiency is proportional toη ∝ 1 − T LT HThe question is → how do we increase efficiency ⇒ T L ↓ and/or T H ↑.1. INCREASED BOILER PRESSURE:• an increase in boiler pressure results in a higher T H for the same T L , therefore η ↑.• but 4 ′ has a lower quality than 4– wetter steam at the turbine exhaust3

2. LOWER T L :– results in cavitation of the turbine blades– η ↓ plus ↑ maintenance• quality should be > 80% at the turbine exhaust• we are generally limited by the TER(lake, river, etc.)eg. lake @ 15 ◦ C +ΔT } =10 {{ ◦ C}=25 ◦ Cresistance to HT⇒ P sat =3.2 kP a.• this is why we have a condenser– the pressure at the exit of the turbine can be less than atmospheric pressure– the closed loop of the condenser allows us to use treated water on the cycle side– but if the pressure is less that atmospheric pressure, air can leak into the condenser,preventing condensation3. INCREASED T H BY ADDING SUPERHEAT:• the average temperature at which heat is supplied in the boiler can be increased bysuperheating the steam– dry saturated steam from the boiler is passed through a second bank of smallerbore tubes within the boiler until the steam reaches the required temperatureThe advantage isη = W net ↑Q H ↑overall ↑The value of T H , the mean temperature at which heat is added, increases, whileT L remains constant. Therefore the efficiency increases.– the quality of the turbine exhaust increases, hopefully where x>0.9– with sufficient superheating the turbine exhaust can fall in the superheated region.4

Rankine Cycle with Reheat• the wetness at the exhaust of the turbine should be no greater that 10% - this can result inphysical erosion of the turbine blades• but high boiler pressures are required for high efficiency - tends to lead to a high wetnessratio• to improve the exhaust steam conditions, the steam can be reheated with the expansion carriedout in two stepsTQHQRHSteamGeneratorQHHigh PressureTurbineQRHLow PressureTurbineWTsWPQCondenserPumpL• the temperature of the steam entering the turbine is limited by metallurgical constraints• modern boilers can handle up to 30 MPa and a maximum temperature ofT max ≈ 650 ◦ C.• newer materials, such as ceramic blades can handle temperatures up to 750 ◦ C.5

Rankine Cycle with Regeneration• Carnot cycle has efficiency:η =1− T L /T H– add Q H at as high a T H as possible– reject Q L at as low a T L as possible• the Rankine cycle can be used with a Feedwater Heater to heat the high pressure sub-cooledwater at the pump exit to the saturation temperature– most of the heat addition (Q H ) is done at high temperatureFeedwater HeatersThere are two different types of feedwater heaters1. OPEN FWH: the streams mix → high temperature steam with low temperature water atconstant pressure2. CLOSED FWH: a heat exchanger is used to transfer heat between the two streams but thestreams do not mix. The two streams can be maintained at different pressures.6

1. OPEN FWH:• working fluid passes isentropically through the turbine stages and pumps• steam enters the first stage turbine at state 1 and expands to state 2 - where a fractionof the total flow is bled off into an open feedwater heater at P 2• the rest of the steam expands into the second stage turbine at state point 3 - this portionof the fluid is condensed and pumped as a saturated liquid to the FWH at P 2• a single mixed stream exists the FWH at state point 6Analysis:• we must determine the mass flow rates through each of the components.By performing an mass balance over the turbineṁ 6 + ṁ 7 = ṁ 5 (1)If we normalize Eq. (1) with respect the total mass flow rate ṁ 1ṁ 6+ ṁ7= 1 (2)ṁ 5 ṁ 5Let the flow at state point 2 bey =ṁ6ṁ 5Thereforeṁ 7=1− y (3)ṁ 5Assuming no heat loss at the FWH, establish an energy balance across the FWHyh 6 +(1− y)h 2 =1· h 3andy = h 3 − h 2h 6 − h 2=ṁ6ṁ 51 − y = ṁ7ṁ 57

2. CLOSED FWH:• two variations exist(a) pump the condensate back to the high pressure line(b)– a steam trap is inserted in the condensed steam line that allows only liquid topass– liquid is passed to a low pressure region such as the condenser or a low pressureheater• the incoming feedwater does not mix with the extracted steam– both streams flow separately through the heater– the two streams can have different pressures8

Other Topics“IDEAL” RANKINE CYCLE:• too expensive to build• requires multiple reheat and regeneration cycles• approaches Carnot efficiencyTOPPING CYCLE (BINARY CYCLE):• involves two Rankine cycles running in tandem with different working fluids such asmercury and water• why:– typically a boiler will supply energy at 1300 − 1400 ◦ C– but T critical for water is 374.14 ◦ C∗ most energy is absorbed below this temperature∗ high ΔT between the boiler source and the water leads to a major source ofirreversibilities– T critical for mercury is about 1500 ◦ C∗ no need for superheating– combine the large enthalpy of evaporation of water at low temperatures with theadvantages of mercury at high temperatures– in addition, the mercury dome leads to a high quality at the exit of the turbine9

Refrigeration CycleReadingProblems11-1 → 11-7, 11-9 11-11, 11-44, 11-47, 11-104Definitions• a refrigeration system removes thermal energy from a low-temperature region and transfersheat to a high-temperature region.• the 1st law of thermodynamics tells us that heat flow occurs from a hot source to a coolersink, therefore, energy in the form of work must be added to the process to get heat to flowfrom a low temperature region to a hot temperature region.• refrigeration cycles may be classified as– vapour compression– gas compression• we will examine only the vapour compression systems• refrigerators and heat pumps have a great deal in common. The primary difference is in themanner in which heat is utilized.– Refrigerator ↓ C} {{ }takes heat from→H }{{}transf ers to– Heat Pump C }{{}takes heat from• this is simply a change in view point→H ↑} {{ }transf ers to• the Carnot cycle can serve as the initial model of the ideal refrigeration cycle.– operates as a reversed heat engine cycle - transfers a quantity of heat, Q L , from a coldsource at temperature, T LQ L = T L (s 3 − s 2 )Q H = T H (s 4 − s 1 )1

W in = Q net = Q H − Q L= (T H − T L )(s 3 − s 2 )The coefficient of performance (COP) is given byCOP = benefitcostwhere the benefit for a refrigeration process is the cooling load given as Q L . This is the net benefit,i.e. heat is removed from the cold space. For a heat pump, the benefit is the heat added to the hotspace, i.e. Q H .COP refrig = Q LW in=T LT H − T LCOP heat pump = Q HW in=T HT H − T LNote:COP heat pump =T HT H − T L= (T H − T L )+T LT H − T L=T LT H − T L+1= COP refrig +1The “1” accounts for the sensible heat addition in going from T L to T H .2

Vapour Compression Refrigeration CycleRoom Airsat. liquidQHCondensersuperheatedvapourExpansionValveh = h 3compressorgas4LEvaporator2 phase sat. vapourQFoodAssumptions for Ideal VCRC• irreversibilities within the evaporator, condenser and compressor are ignored• no frictional pressure drops• refrigerant flows at constant pressure through the two heat exchangers (evaporator and condenser)• stray heat losses to the surroundings are ignored• compression process is isentropic3

Refrigeration ProcessProcess1-2s:2s-3:DescriptionA reversible, adiabatic (isentropic) compression of the refrigerant.The saturated vapour at state 1 is superheated to state 2.⇒ w c = h 2s − h 1An internally, reversible, constant pressure heat rejectionin which the working substance is desuperheated and then condensedto a saturated liquid at 3. During his process, the working substancerejects most of its energy to the condenser cooling water.⇒ q H = h 2s − h 33-4 An irreversible throttling process in which the temperature andpressure decrease at constant enthalpy.⇒ h 3 = h 44-1 An internally, reversible, constant pressure heat interactionin which the working fluid is evaporated to a saturated vapourat state point 1. The latent enthalpy necessary for evaporationis supplied by the refrigerated space surrounding the evaporator.The amount of heat transferred to the working fluid in the evaporatoris called the refrigeration load.⇒ q L = h 1 − h 4The thermal efficiency of the cycle can be calculated asη = q evapw comp= h 1 − h 4h 2s − h 14

Common RefrigerantsThere are several fluorocarbon refrigerants that have been developed for use in VCRC.R11R12 CCl 2 F 2 dichlorofluoromethane- used for refrigeration systems at highertemperature levels- typically, water chillers and air conditioningR22 CHClF 2 has less chlorine, a little better for theenvironment than R12- used for lower temperature applicationsR134a CFH 2 CF3 tetrafluorethane - no chlorine- went into production in 1991- replacement for R12R141b C 2 H 3 FCl 2 dichlorofluoroethaneAmmonia NH 3 corrosive and toxic- used in absorption systemsR744 CO 2 behaves in the supercritical region-lowefficiencyR290 propane combustibleHow to Choose a RefrigerantMany factors need to be considered• ozone depletion potential• global warming potential• combustibility• thermal factorsOzone Depletion Potential• chlorinated and brominated refrigerants• acts as a catalyst to destroy ozone molecules5

• reduces the natural shielding effect from incoming ultra violet B radiationGlobal Warming Potential• gases that absorb infrared energy• gases with a high number of carbon-fluorine bonds• generally have a long atmospheric lifetimeCombustibility• all hydro-carbon fuels, such as propaneThermal Factors• the heat of vaporization of the refrigerant should be high. The higher h fg , the greater therefrigerating effect per kg of fluid circulated• the specific heat of the refrigerant should be low. The lower the specific heat, the less heatit will pick up for a given change in temperature during the throttling or in flow through thepiping, and consequently the greater the refrigerating effect per kg of refrigerant• the specific volume of the refrigerant should be low to minimize the work required per kgof refrigerant circulated• since evaporation and condenser temperatures are fixed by the temperatures of the surroundings- selection is based on operating pressures in the evaporator and the condenser• selection is based on the suitability of the pressure-temperature relationship of the refrigerant• other factors include:– chemical stability– toxicity– cost– environmental friendliness– does not result in very low pressures in the evaporator (air leakage)– does not result in very high pressures in the condenser (refrigerant leakage)6

Designation Chemical Ozone Depletion Global WarmingFormula Potential 1 Potential 2Ozone Depleting & Global Warming ChemicalsCFC-11 CCl 3 F 1 3,400CFC-12 CCl 2 F 2 0.89 7,100CFC-13 CClF 3 13,000CFC-113 C 2 F 3 Cl 3 0.81 4,500CFC-114 C 2 F 4 Cl 2 0.69 7,000CFC-115 C 2 F 5 Cl 1 0.32 7,000Halon-1211 CF 2 ClBr 2.2-3.5Halon-1301 CF 3 Br 8-16 4,900Halon-2402 C 2 F 4 Br 2 5-6.2carbon tetrachloride CCl 4 1.13 1,300methyl chloroform CH 3 Ccl 3 0.14nitrous oxide N 2 O 270Ozone Depleting & Global Warming Chemicals - Class 2HCFC-22 CHF 2 Cl 0.048 1,600HCFC-123 C 2 HF 3 Cl 2 0.017 90HCFC-124 C 2 HF 4 Cl 0.019 440HCFC-125 C 2 HF 5 0.000 3,400HCFC-141b C 2 H 3 FCl 2 0.090 580HCFC-142b C 2 H 3 F 2 Cl 0.054 1800Global Warming, non-Ozone Depleting Chemicalscarbon dioxide CO 2 0 1methane CH 4 0 11HFC-125 CHF 2 CF 3 0 90HFC-134a CFH 2 CF 3 0 1,000HFC-152a CH 3 CHF 2 0 2,400perfluorobutane C 4 F 10 0 5,500perfluoropentane C 5 F 12 0 5,500perfluorohexane C 6 F 14 0 5,100perfluorotributylamine N(C 4 F 9 ) 3 0 4,3001 - relative to R112 - relative to CO 27

Cascade Refrigeration System• combined cycle arrangements• two or more vapour compression refrigeration cycles are combined• used where a very wide range of temperature between T L and T H is required• the condenser for the low temperature refrigerator is used as the evaporator for the hightemperature refrigeratorAdvantages• the refrigerants can be selected to have reasonable evaporator and condenser pressures in thetwo or more temperature ranges8

Absorption Refrigeration SystemDifferences between an absorption refrigeration system and a VCRCVCRC• vapour is compressedbetween the evaporator andthe condenser• process is driven by workAbsorption RS• the refrigerant is absorbed byan absorbent material to form aliquid solution• heat is added to the processto retrieve the refrigerant vapourfrom the liquid solution• process is driven by heatAdvantages• since the working fluid is pumped as a liquid the specific volume is less than that of a gas (asin the VCRC compressor), hence the work input is much less.• there are considerable savings in power input because a pump is used instead of a compressor.• this is weighed off against the cost of extra hardware in an absorption systemCommon Refrigerant/Absorber CombinationsRefrigerantAbsorber1. ammonia water2. water lithium bromidelithium chloride9

ProcessliquidammoniaRoom AirQHCondenserammoniavapour onlySource ofHeatGeneratorQ*HExpansionValveweak ammoniawater solutioncoldregeneratorEvaporatorAbsorber2 phase dry vapourQLQ*LFoodstrong ammoniacold water solutionsinkpump• ammonia circulates through the condenser, expansion valve and evaporator (same as in theVCRC)• the compressor is replaced by an absorber, pump, generator, regenerator and a valve• in the absorber, ammonia vapour is absorbed by liquid water– the process is exothermic (gives off heat)– ammonia vapour is absorbed into the water at low T and P maintained by means ofQ ∗ L– absorption is proportional to 1/T ⇒ the cooler the better• the pump raises the solution to the pressure of the generator• in the generator, ammonia is driven out of the solution by the addition of Q ∗ H ,(endothermic reaction)• ammonia vapour is passed back to the condenser• a regenerator is used to recoup some of the energy from the weak ammonia water solutionpassed back to the absorber. This energy is transferred to the solution pumped to the generator.This reduces the Q ∗ Hrequired to vapourize the solution in the generator. It also reducesthe amount of Q ∗ Lthat needs to be removed from the solution in the absorber.10

Internal Combustion EnginesReading Problems9-3 → 9-7 9-37, 9-41, 9-47, 9-51, 9-56Definitions1. spark ignition:• a mixture of fuel and air is ignited by a spark plug• applications requiring power to about 225 kW (300 HP)• relatively light and low in cost2. compression ignition engine:• air is compressed to a high enough pressure and temperature that combustion occurswhen the fuel is injected• applications where fuel economy and relatively large amounts of power arerequiredThe Gasoline Engine1

• conversion of chemical energy to mechanical energy• can obtain very high temperatures due to the short duration of the power strokeAir Standard CycleASSUMPTIONS:• air is an ideal gas with constant c p and c v• no intake or exhaust processes• the cycle is completed by heat transfer to the surroundings• the internal combustion process is replaced by a heat transfer process from a TER• all internal processes are reversible• heat addition occurs instantaneously while the piston is at TDCDefinitionsMean Effective Pressure (MEP): The theoretical constant pressure that, if it acted on the pistonduring the power stroke would produce the same net work as actually developed in onecomplete cycle.MEP =net work for one cycledisplacement volume =W netV BDC − V TDCThe mean effective pressure is an index that relates the work output of the engine to it size(displacement volume).2

Otto Cycle• the theoretical model for the gasoline engine• consists of four internally reversible processes• heat is transferred to the working fluid at constant volumeThe Otto cycle consists of four internally reversible processes in series1 → 2 isentropic compression or air as the piston moves from BDC to TDC2 → 3 constant volume heat addition to the fuel/air mixture from an external source while thepiston is at TDC (represents the ignition process and the subsequent burning of fuel)3 → 4 isentropic expansion (power stroke)4 → 1 constant volume heat rejection at BDC3

The Otto cycle efficiency is given asη =1− T ( ) k−1 ( ) 1−k1 V2V1=1− =1−T 2 V 1 V 2If we letr = V 1V 2= V 4V 3= compression ratioThenη Otto =1− r 1−kWhy not go to higher compression ratios?• there is an increased tendency for the fuel to detonate as the compression ratio increases• the pressure wave gives rise to engine knock• can be reduced by adding tetraethyl lead to the fuel• not good for the environment4

Diesel Cycle• an ideal cycle for the compression ignition engine (diesel engine)• all steps in the cycle are reversible• heat is transferred to the working fluid at constant pressure• heat transfer must be just sufficient to maintain a constant pressureIf we letr = V 1V 2= compression ratio = V 4V 2r v = V 3V 2= cutoff ratio → injection periodthen the diesel cycle efficiency is given asη Diesel =1− 1 ( )( 1 rkv − 1 )r k−1 k r v − 15

Where we noteη Diesel =1− 1 ( )( 1 rkv − 1 )r k−1 k r v − 1} {{ }=1 in the Otto CycleComparison of the Otto and the Diesel Cycle• η Otto >η Diesel for the same compression ratio• but a diesel engine can tolerate a higher ratio since only air is compressed in a diesel cycleand spark knock is not an issue• direct comparisons are difficultDual Cycle(Limited Pressure Cycle)• this is a better representation of the combustion process in both the gasoline and the dieselengines• in a compression ignition engine, combustion occurs at TDC while the piston moves downto maintain a constant pressure6

Dual Cycle EfficiencyGivenr = V 1V 2= compression ratior v = V 4V 3= cutoff ratior p = P 3P 2= pressure ratioη Dual =1−r p r k v − 1[(r p − 1) + kr p (r v − 1)] r k−1Note: if r p =1we get the diesel efficiency.7

Stirling Cycle8

• reversible regenerator used as an energy storage device• possible to recover all heat given up by the working fluid in the constant volume coolingprocess• all the heat received by the cycle is at T H and all heat rejected at T L• η Stirling =1− T L /T H(Carnot efficiency)With perfect regenerationQ H = T H (s 2 − s 1 )Q L = T L (s 3 − s 4 )9

η = W netQ H= Q H − Q LQ H=1− Q LQ H=1− T L(s 3 − s 4 )T H (s 2 − s 1 )(1)From Gibb’s equationTds = du + Pdv = c v dT + Pdvif T = constant ⇒ Tds = Pdv ⇒ ds = PdvIntegrating givesT= Rdvvs 3 − s 4 = R ln( ) ( )v3v2= R ln = s 2 − s 1v 4 v 1Therefore s 3 − s 4 = s 2 − s 1 , and Eq. 1 givesη =1− T LT H⇒ Carnot efficiency10

Brayton CycleReading Problems9-8 → 9-10 9-78, 9-84, 9-108Open Cycle Gas Turbine Engines• after compression, air enters a combustion chamber into which fuel is injected• the resulting products of combustion expand and drive the turbine• combustion products are discharged to the atmosphere• compressor power requirements vary from 40-80% of the power output of the turbine (remainderis net power output), i.e. back work ratio = 0.4 → 0.8• high power requirement is typical when gas is compressed because of the large specificvolume of gases in comparison to that of liquidsIdealized Air Standard Brayton Cycle• closed loop• constant pressure heat addition and rejection• ideal gas with constant specific heats1

Brayton Cycle EfficiencyThe Brayton cycle efficiency can be written asη =1− (r p ) (1−k)/kwherewedefine the pressure ratio as:r p = P 2P 1= P 3P 42

Maximum Pressure RatioGiven that the maximum and minimum temperature can be prescribed for the Brayton cycle, achange in the pressure ratio can result in a change in the work output from the cycle.The maximum temperature in the cycle (T 3 ) is limited by metallurgical conditions because theturbine blades cannot sustain temperatures above 1300 K. Higher temperatures (up to 1600 K canbe obtained with ceramic turbine blades). The minimum temperature is set by the air temperatureat the inlet to the engine.3

Brayton Cycle with Reheat• T 3 is limited due to metallurgical constraints• excess air is extracted and fed into a second stage combustor and turbine• turbine outlet temperature is increased with reheat (T 6 >T ′ 4), therefore potential for regenerationis enhanced• when reheat and regeneration are used together the thermal efficiency can increase significantly4

Compression with Intercooling• the work required to compress in a steady flow device can be reduced by compressing instages• cooling the gas reduces the specific volume and in turn the work required for compression• by itself compression with intercooling does not provide a significant increase in the efficiencyof a gas turbine because the temperature at the combustor inlet would require additionalheat transfer to achieve the desired turbine inlet temperature• but the lower temperature at the compressor exit enhances the potential for regeneration i.e.a larger ΔT across the heat exchanger5

Brayton Cycle with Regeneration• a regenerator (heat exchanger) is used to reduce the fuel consumption to provide the required˙Q H• the efficiency with a regenerator can be determined as:η =Ẇnet˙Q H= 1 − ˙Q L˙Q H6

= 1 − c p(T 6 − T 1 )c p (T 3 − T 5 )= 1 − c p(T ′ 6 − T 1)c p (T 3 − T ′ 5 )⇒ (for a real regenerator)⇒ (for an ideal regenerator)= 1 − c p(T 2 − T 1 )c p (T 3 − T 4 )andη =1−( )Tmin(r p ) (k−1)/kT max• for a given T min /T max , the use of a regenerator above a certain r p will result in a reductionof ηwith an idealregeneratorwithout aregeneratorbetter with aregeneratorbetter withouta regeneratorT / T = 0.25T / T = 0.3131T / T = 0.2313rp, crrp7

Regenerator Effectivenessɛ =˙Q reg,actual˙Q reg,ideal= h 5 − h 2h ′ 5 − h 2= h 5 − h 2h 4 − h 2= T 5 − T 2T 4 − T 2Typical values of effectiveness are ≤ 0.7Repeated intercooling, reheating and regeneration ( will provide a system that approximates theEricsson Cycle which has Carnot efficiency η =1− T )L.T HBrayton Cycle With Intercooling, Reheating and Regeneration8

TTmaxQHQH, RQreg7sQ9sreg4s2sTminQL, IQLsCompressor and Turbine EfficienciesIsentropic Efficiencies(1) η comp = h 2,s − h 1= c p(T 2,s − T 1 )h 2 − h 1 c p (T 2 − T 1 )(2) η turb = h 3 − h 4h 3 − h 4,s= c p(T 3 − T 4 )c p (T 3 − T 4,s )(3) η cycle = W netQ H= Q H − Q LQ H=1− Q L=1− c p(T 4 − T 1 )Q H c p (T 3 − T 2 )Given the turbine and compressor efficiencies and the maximum (T 3 ) and the minimum (T 1 ) temperaturesin the process, find the cycle efficiency (η cycle ).(4) Calculate T 2s from the isentropic relationship,( )T (k−1)/k2,s P2=.T 1 P 1Get T 2 from (1).(5) Do the same for T 4 using (2) and the isentropic relationship.(6) substitute T 2 and T 4 in (3) to find the cycle efficiency.9

Jet PropulsionReading Problems9-11 9-117, 9-121Gas Turbines for Aircraft Propulsion• gas turbines are well suited to aircraft propulsion because of their favorable power-to-weightratio• the exhaust pressure of the turbine will be greater than that of the surroundings• gases are expanded in the turbine to a pressure where the turbine work is just equal to thecompressor work plus some auxiliary power for pumps and generators i.e. the net workoutput is zero• since the gases leave at a high velocity, the change in momentum that the gas undergoesprovides a thrust to the aircraft• typically operate at higher pressure ratios, often in the range of 10 to 25Conservation of Momentumwhere v ∗ iis the velocity of the aircraftd(Mom) x,cvdt=(Ṁom) x,in − (Ṁom) x,out + ∑ F xfor steady flow ⇒ d dt =0and 1

ṁ i v ∗ i − ṁ ev ∗ e + F T + P i A i − P e A e =0Since the air-fuel mass ratio is highṁ fuel

Turbojet EngineSections• a-1: diffuser– decelerates the incoming flow relative to the engine– a pressure rise known as a ram effect occurs, v ∗ (↓), P (↑)• 1-4: gas generator– compressor, combustor and turbine∗ 1-2: isentropic compression∗ 2-3: constant pressure heat addition∗ 3-4: isentropic expansion through the turbine during which work is developed– turbine power just enough to drive the compressor– air and fuel are mixed and burned in the combustion chamber at constant pressure– P T >> P atm• 4-5: nozzle– isentropic expansion through the nozzle, air accelerates and the pressure deceases– gases leave the turbine significantly higher in pressure than atmospheric pressure– gases are expanded to produce a high velocity, v ∗ e >> v∗ iresults in a thrust– v ∗ 1

Afterburner• similar to a reheat device• produces a higher temperature at the nozzle inlet, T 5 >T 4• results in an increase in velocity4

By performing a 1st law energy over the nozzle we can obtain an expression for the exit velocityin terms of the entrance temperature to the nozzle.dE↗ 0dt⎧ ⎡⎤⎫{]}⎪⎨= ˙Q↗ 0 +Ẇ↗ 0 + ṁ ⎢⎣ h 4 + (v∗ 4 )2⎪⎬⎥ − ṁ[h2 ⎦e + (v∗ e )2⎪⎩} {{ }2⎪⎭→ 0If we assume that the air velocity leaving the turbine is relatively small, the kinetic energy term at4 can be assumed to go to zero and we getv ∗ e==√2(h 4 − h e )√2c p (T 4 − T e )• exit velocity proportional to v ∗ e ∝ √ 2c p (T 4 − T e )• afterburner is used to increase T 4 to T 5• similar to a reheat device• produces a higher temperature at the nozzle inlet5

Other Types of Engines1. Turbo-Prop Engine• gas turbine drives the compressor and the propeller• most of the thrust is from the propeller• works by accelerating large volumes of air to moderate velocities• propellers are best suited for low speed (< 300 mph) flight• new turbo-props are suitable for velocities up to 500 mph• by-pass ratio of 100:1 or more• by-pass ratio defined as:bypass ratio =mass flow bypassing the combustion chambermass flow through the combustion chamber6

2. Turbo-Fan Engine (Ducted Turbo-Prop Engine)• best choice for fuel economy and speed• high speed exhaust gases are mixed with the lower speed air in the by-pass resulting ina considerable noise reduction• by-pass ratio can be adjusted• by-pass provides thrust for takeoff• the core provides thrust for cruising• typically used for speeds up to 600 mph• increasing the by-pass ratio results in increased thrust• typical by-pass ratios are 5-67

3. Ramjet• no moving parts• compression is achieved by decelerating the high-speed incoming air in the diffuser• aircraft must already be in flight at a high speed• used in aircraft flying above Mach 14. Pulse Jet Engine• similar to a ram jet but lets in a slug of air at a time and then closes a damper duringthe combustion stage• uses a shutter-type valve for damper control• can be used effectively at low velocities• used in German V1 missle• the combustion firing rate was approximately 40 cycles/sec with a maximum flightvelocity of 600 mph8

Non-Reacting Gas MixturesReading Problems13-1 → 13-3 13-52, 13-6014-1 → 14-7 14-32, 14-35, 14-68, 14-71, 14-7514-79, 14-103, 14-112Introduction• homogeneous gas mixtures are frequently treated as a single compound rather than manyindividual constituents• the individual properties of inert gases tend to be submerged, such that the gas behaves as asingle pure substance• eg. - air consists of oxygen, nitrogen, argon and water vapour. But dry air can be treated asa simple gas with a molar mass of 28.97 kg/kmole• equations can be derived to express the properties of mixtures in terms of the properties oftheir individual constituents• it is assumed that each constituent is unaffected by the other constituents in the mixtureP-V-T Relationships for Ideal Gas MixturesAmagat Model (law of additive volumes)• the volume of a mixture is the sum of the volumes that each constituent gas would occupy ifeach were at the pressure, P and temperature, T , of the mixtureT,P T,P T,Pm A m B remove m C = m A + m Bn A n B → n C = n A + n BV A V B partition V C = V A + V B• the volume that n i moles of a component i would occupy at P and T is called the partialvolume, V iV i = n iRTP1

The volume of the gas mixture is equal to the sum of the volumes each gas would occupy if it existedat the mixture temperature and pressure.j∑V = V ii=1Dalton Model (law of additive pressures)• the pressure of a mixture of gases is the sum of the pressures of its components when eachalone occupies the volume of the mixture, V , at the temperature, T , of the mixtureV, T V, T V, Tm A ,n A ,P A + m B ,n B ,P B = m C = m A + m Bn C = n A + n BP C = P A + P B• for a mixture of ideal gases the pressure is the sum of the partial pressures of the individualcomponentsThe pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existedalone at T and V .By combining the results of the Amagat and Dalton models i.e. (1) and (2), we obtain for ideal gas mixturesP iP = V iV = n inTherefore, Amagat’s law and Dalton’s law are equivalent to each other if the gases and the mixtureare ideal gases.2

Mixture PropertiesExtensive properties such as U, H, c p ,c v and S can be found by adding the contribution of eachcomponent at the condition at which the component exists in the mixture.U = ∑ U i = ∑ m i u i = m ∑ X i u i = mu= ∑ n i u i = n ∑ Y i u i = nuwhere u is the specific internal energy of the mixture per mole of the mixture.u = ∑ X i u ih = ∑ X i h ic v = ∑ X i c vic p = ∑ X i c pis = ∑ X i s iChanges in internal energy and enthalpy of mixturesu 2 − u 1 = ∑ X i (u 2 − u 1 ) i =h 2 − h 1 = ∑ X i (h 2 − h 1 ) i =∫ T2T 1c v dT = c v (T 2 − T 1 )∫ T2T 1c p dT = c p (T 2 − T 1 )s 2 − s 1 = ∑ X i (s 2 − s 1 ) i = c p ln T 2T 1− R ln P 2P 1These relationships can also be expressed on a per mole basis.Entropy Change Due to Mixing of Ideal Gases• when ideal gases are mixed, a change in entropy occurs as a result of the increase in disorderin the systems• if the initial temperature of all constituents are the same and the mixing process is adiabatic3

– temperature does not change– but entropy doesΔS = −(m A R A ln P AP+ m BR B ln P BP + ···)j∑= − m i R i ln P ii=1 P= −Rj∑i=1n i ln Y i4

Psychrometrics• studies involving mixtures of dry air and water vapour• used in the design of air-conditioning systems, cooling towers and most processes involvingthe control of vapour content in air• for T ≤ 50 ◦ C (P sat ≤ 13 kP a) ⇒ h ≈ h(T )– water vapour can be treated as an ideal gasDefinitionsMoist Air• a mixture of dry air and water vapour where dry air is treated as if it were a pure component• the overall mixture is given as ⇒ P = mRTVTotal PressureP = P a + P wP a = m aR a TVP w = m wR w TVwhere P a is the partial pressure of air and P w is the partial pressure of water vapour. Typicallym w

= ˜M w n w˜M a n a= ˜M w (P w V/RT )˜M a (P a V/RT )=⎛ ˜ ⎞ ( )M⎝ w Pw⎠˜M a P a= 0.622(PwP a)In addition ω can be written asω =0.622(PwP a)=0.622(PwP − P w)=0.622(φPsatP − φP sat)which can be rearranged in terms of relative humidityφ =PωP sat⎛⎝ω +˜M⎞w⎠˜M a=PωP sat (ω +0.622)Dry Bulb Temperature - the temperature measured by a thermometer placed in a mixture of airand water vapourWet Bulb Temperature6

• thermometer surrounded by a saturated wick• if air/water vapour mixture is not saturated, some water in the wick evaporates and diffusesinto the air → cooling the water in the wick• as the temperature of the water drops, heat is transferred to the water from both the air andthe thermometer• the steady state temperature is the wet-bulb temperatureSling Psychrometer - a rotating set of thermometers one of which measures wet bulb temperatureand the other dry bulb temperature. T DB and T WB are sufficient to fix the state of the mixture.The Psychrometric Chartwhere the dry bulb temperature is the temperature measured by a thermometer place in the mixtureand the wet bulb temperature is the adiabatic saturation temperature.7

An Adiabatic SaturatorHow can we measure humidity?• the adiabatic saturator is used to measure humidity• two inlets, single exit device through which moist air passes• air-water mixture of unknown humidity enters at a known pressure and temperature• if air/water mixture is not saturated, (φ

By definitionω 1 =(ṁwṁ a)1(4)ω 3 =(ṁwṁ a)3(5)From (2) and (1)) (ṁw,2=ṁ a,1⎛ ⎞ṁ w,3−⎜⎝ ṁ a,1⎟} {{ }⎠ṁ a,3) (ṁw,1= ω 3 − ω 1ṁ a,1Dividing (3) by ṁ a,1 and noting m a1 = m a3 and m w 2m a1= ω 3 − ω 1h a,1 + ω 1 h w,1 +(ω 3 − ω 1 ) h w,2 = h a,3 + ω 3 h w,3 (6)ω 1 = (h a,3 − h a,1 )+ω 3 (h w,3 − h w,2 )(h w,1 − h w,2 )9

If we assume:i) air is an ideal gas and (h a,3 − h a,1 )=c pa (T 3 − T 1 )ii) (h w,3 − h w,2 )=h g − h f = h fg (T 3 )iii) h w,1 ≈ h g (T 1 )iv) h w,2 = h f (T 2 )=h f (T 3 )then we can write ω 1 as a function of T 1 and T 3 onlyω 1 = c p a(T 3 − T 1 )+ω 3 h fg (T 3 )h g (T 1 ) − h f (T 3 )Aside: Since h f

Reacting Gas MixturesReading Problems15-1 → 15-7 15-21, 15-32, 15-51, 15-61, 15-7415-83, 15-91, 15-93, 15-98Introduction• thermodynamic analysis of reactive mixtures is primarily an extension of the principles wehave learned thus far• it is necessary to modify the methods used to calculate specific enthalpy, internal energy andentropyDefinitionsCombustion Process:• a fuel made up of hydrocarbons is said to have burned completely if:– all the carbon present in the fuel is burned to carbon dioxide– all the hydrogen is burned to water• if the conditions are not fulfilled the combustion process is incompleteCombustion Reactions:reactants → productsorfuel + oxidizer → products• in all cases the mass is conservedmass of products = mass of reactantsFuels:• fuel is simply a combustible substance• hydrocarbon fuels exist as liquids, gases and solids– liquids → gasoline - octane, C 8 H 181

– gases → methane, CH 4– solids → coalCombustion Air:• oxygen is required in every combustion reaction• in most combustion reactions air provides the needed oxygen• dry air is considered to be21% oxygen79% nitrogen}on a molar basismolar ratio = n N 2= 0.79n O2 0.21 =3.761 mole of air can then be written as [0.21 O 2 + 0.79 N 2 ]For convenience, we typically refer to air as [O 2 + 3.76 N 2 ] which is actually 4.76moles of air.Note: From the Amagat model, we know that a mixture at a known T and P (as is thecase with combustion reactions)n in = V iVTherefore, by expressing a mixture in terms of the number of moles we are also expressingit in terms of a volume fraction.• nitrogen does not undergo a chemical reaction in combustion since it is inertAir-Fuel Ratio:where:mass of airmass of fuel =AF =¯ AF⎛⎝˜M⎞air⎠˜M fuelmoles of air × ˜M airmoles of fuel × ˜M fuelAFAF- air fuel ratio on a mass basis- air fuel ratio on a molar basis˜M air= 28.97 kg/kmole2

Theoretical or Stoichiometric Air:• the minimum amount of air that supplies sufficient oxygen for complete combustion ofall carbon and hydrogen in the fuel - referred to as stoichiometric, 100% stoichiometricor theoretical• no free oxygen would appear in the products• greater than stoichiometric leads to free oxygen in the products• less than stoichiometric and C, CO, OH, H 2 will appear in the products since there isnot enough oxygen to form water or carbon dioxide (the actual proportions will dependon the temperature and the pressure)• normally the amount of air supplied is given as a percentage of the theoretical valuei.e. 150% = 1.5 × the theoretical air- referred to as 20% excess air, 120% stoichiometricEquivalence Ratio:• defined asequivalence ratio =AF actualAF theoretical• if the equivalence ratio is:– > 1 → lean mixture (excess air)– < 1 → rich mixture (not enough air)3

Conservation of Energy for Reacting SystemsEnthalpy of Formation• previous calculations involving enthalpy were all based on differences and the reference usedto determine enthalpy did not matter• when chemical reactions occur, reactants disappear and products are formed→ differences cannot be calculated for all substances involved• it is necessary to establish a common base to account for differences in composition• the enthalpy datum for reacting systems is set to zero at standard temperature and pressure– T ref =25 ◦ C → 298 K– P ref =1atm• h =0assigned to elements in their most stable form i.e. O 2 ,N 2 , C, etc.• Enthalpy of Formation: the energy released or absorbed when a compound is formed fromits stable elements at STPwhere h o fis the enthalpy of formation.Taking an energy balance over the combustion chamber shown above, we obtaina h o A + b ho B + c ho C} {{ }generally=0+ h o f −→ ho ABCThe left side of the equation is typically zero because h =0for elements in their stableform. The sign of h o findicates the direction of heat flow; +ve is endothermic and -ve isexothermic.4

Effects of Non-Standard Temperatureh(T,P)=h o f +(h T,P − h o T =25 ◦ C, P =1atm )} {{ }Δh at known temperatureswhereh o fis the heat resulting from a chemical change at T =25 ◦ C and P =1atmΔhis the heat resulting from a change in temperature (sensible heat) with respectto the reference temperature, T ref =25 ◦ C5

Enthalpy of Combustion• while the enthalpy of formation is related to elemental reactants −→ resulting in a singlecompound; the enthalpy of combustion is related to fuel + oxidizer as the reactants• Enthalpy of Combustion: the difference between the enthalpy of the products and the enthalpyof the reactants where complete combustion occurs at a given temperature and pressureQ = ∑ (mh) P − ∑ (mh) R = H P (T P ) − H R (T R )} {{ }H RPQ = ∑ (nh) P − ∑ (nh) R = H P (T P ) − H R (T R )} {{ }H RPwhereh c = H RP /kmole of fuelwith:+ve Q ⇒ endothermic−ve Q ⇒ exothermic• when enthalpy of formation data are available for all products and reactants the above equationcan be used• otherwise a calorimeter must be used to measure the enthalpy of combustion6

Heating Value• the heating value of a fuel is a positive value equal to the magnitude of the enthalpy ofcombustion when products are returned to the state of the reactants• two values are used– HHV: higher heating value - obtained when all the water formed by combustion is aliquid at the reference temperature–LHV: lower heating value - obtained when all the water formed by combustion is avapour as an ideal gas in the mixture of the products• the HHV exceeds the LHV by the energy required to vaporize the liquid formedHHV = LHV + (m · h fg) H2 Okmole of fuel= LHV +( ˜M · h fg ) H2 O · nH 2 On fuelwhereh fg (25 ◦ C) = 2, 442.3 kJ/kg˜M H2 O = 18.015 kg/kmoleAdiabatic Flame Temperature• if the system is perfectly insulated it cannot dispose of the LHV and the LHV goes intoheating the products above the reference temperature7

• under adiabatic conditions, the maximum temperature attained by the products when combustionis complete is called the adiabatic flame or adiabatic combustion temperatureH P (T ad )=H R (T R )∑n P (h o f + h − h0 )} {{ } P = ∑ n R (h o f + h − h0 )} {{ } RPRΔhΔhWe need to collect terms based on what we know or can readily calculate and what we donot know, i.e. terms that are a function of T ad .∑Pn P (h) P = ∑ } {{ }Rsensible heatfunction of T ad(n R (h − h o ) R − − ∑ )n P (h o ) PP} {{ }sensible heatfunction of T R or T ref+ ∑ Rn R (h o f ) R − ∑ n P (h o f ) PP} {{ }chemical heatfunction of T R or T refStep 1: Calculate the right hand side based on known values of T R and T ref .Step 2: Calculate the left hand side based on a guessed value of T ad .Step 3: Repeat Step 2, until LHS = RHS.8

Dew Point• since water is formed when hydrocarbon fuels are burned, the mole fraction of water vapourin the form of gaseous products can be significant• if the gaseous products of combustion are cooled at constant mixture pressure the dew pointtemperature is reached when water vapour begins to condense• since corrosion of duct work, mufflers etc. can occur, the knowledge of dew point temperatureis importantEvaluation of Entropy for Reacting SystemsThe 2nd law entropy equation can be written asS in − S} {{ out}+ S gen} {{ }= ΔS system} {{ }due to heat & mass transfer generation change in entropyFor a closed system, such as a combustion process, the entropy balance on the system can bewritten as∑ Q iT i+ S gen = S P − S R• a common datum must be used to assign entropy values for each substance involved in thereaction• an entropy of 0 for pure crystalline substances is obtained at absolute zero• the entropy relative to this datum is called absolute entropy• absolute entropy at 1 atm and temperature T is denoted as s o (T ) or s o (T ) for a per unitmass or per mole basis• while h was only a function of temperature for ideal gases, we must account for the effectsof both T and P in entropy9

• theentropyatanyvalueofT and P can be calculated aswheres(T,P)=s o (T ) −R ln} {{ }tablesP ref = 1atm(PiP ref)P i = partial pressure of i ′ th componentR = 8.31434 kJ/kmole · K• the partial pressure P i can also be written asP i = Y i Pands(T,P i )=s o i (T ) −Rln (Yi PP ref)where P is the mixture pressure and Y i is the mole fraction of the i ′ th component.10