Thermodynamics of separation processes

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1 Process EngineeringThermodynamics course # v Thermodynamics of separation processes Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik tel ; ron.zevenhoven@abo.fi Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 1/82 ÅA Separation and entropy Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 2/82

2 Separation process and entropy /1 Separation processes involve de-mixing of species or phases, in thermodynamic sense resulting in a decrease of the total entropy of these Entropy of components or phases < Entropy of mix One way of acccomplishing a separation is to create more phases; Gibbs phase rule gives for the degrees of freedom, f, for a system of p phases and n components: f = n + 2 p. f gives ΔS ; f gives ΔS. A new phase can be created by 1) adding a new component (absorption/desorption, extraction), or 2) by adding or removing energy (e.g. heat) (distillation, crystallisation) Creating an un-equilibrium gives rise to a separation (or a chemical reaction) if ΔG = ΔH-T ΔS < 0, or ΔS >ΔH/T Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 3/82 Separation process and entropy /2 m mix s mix Q out T out m A s A Steady - state mass balance,1st Law,2nd Law : m m Q T mix mix out out Q T m h out out mix m A A Q T m Q s in in A in B m m m B mix A h s B s A Q T mix m in in B m h A B m s Q mix A out s m mix B s B m B s B m mix s mix m liquid s liquid Q in T in m crystals s crystals Distillation Q out T out Crystallisation 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 4/82

Separation process and entropy /3 Clearly, some entropy must be produced to 1) compensate for ΔS < 0 resulting from the separation, and 2) create some driving force ΔG < 0, ΔS >

uniform production of entropy is preferable: the equipartitioning principle (TK87, B97, SAKS04) see also macroscopic organisation in natural processes!

It can be extended to considering stable, non-equilibrium states S > S min, with stability criterion dṡ gen /dt < 0.

3 Separation process and entropy /3 Clearly, some entropy must be produced to 1) compensate for ΔS < 0 resulting from the separation, and 2) create some driving force ΔG < 0, ΔS > ΔH/T. Proper engineering requires minimisation of this entropy production ΔS and the exergy losses T ΔS, and balancing this against economical considerations It was found that a uniform production of entropy is preferable: the equipartitioning principle (TK87, B97, SAKS04) see also macroscopic organisation in natural processes! For example, for a heat exchanger this implies that the driving force Δ(1/T) = constant. It can be extended to considering stable, non-equilibrium states S > S min, with stability criterion dṡ gen /dt < 0. Stable, unstable, neutral equilibrium Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 5/82 Pic: Separation process and entropy /4 Example taken from TK87 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 6/82

4 Separation process and entropy /5 µ = chemical potential Examples taken from TK87 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 7/82 ÅA Binary Distillation I (the McCabe-Thiele procedure) Assumed pre-knowledge: Continuous distillation see course Massöverföring & separationsteknik, #12 or re-wrap course Principles of process engineering (which summarises courses , and ) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 8/82

5 Continuous distillation (binary) /1 From MÖF-ST ÅA Separating a mixture of 2 components A and B, with A more volatile Liquid for absorption section produced by condensing some top product Gas for stripping section produced by boiling some bottom product Picture: WK92 Top product: more volatile component Top section: rectifying section absorbing the less volatile component Bottom section: stripping section desorbing the more volatile component Bottom product: less volatile component Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 9/82 Mass balance and equilibrium: absorption y in V y out L x in slope L/V y i = Kx i From MÖF-ST ÅA y out y i y i+1 V i x i-1 x i L x in x out V y in L x out mass balance: V y i +L x i = V y i+1 + L x i-1 working line y i+1 y i = (L/V) (x i-1 x i ) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 10/82

6 Mass balance and equilibrium: desorption / stripping V y out L x in y i = Kx i y out slope L/V From MÖF-ST ÅA y i y i+1 V i x i-1 x i L y in x out x in V y in L x out mass balance: V y i +L x i = V y i+1 + L x i-1 working line y i+1 y i = (L/V) (x i-1 x i ) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 11/82 Continuous distillation (binary) /2 From MÖF-ST ÅA Roughly equimolar exchange: 1 mol A liquid gas «-» 1mol B gas liquid As a result: Gas and liquid streams ~ constant in each section: L and V in top section, L and V in bottom section Feed enters there where it is similar to the mixture inside Picture: WK92 Top product: more volatile component Top section: rectifying section absorbing the less volatile component Bottom section: stripping section desorbing the more volatile component Bottom product: less volatile component Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 12/82

7 Continuous distillation /3 From MÖF-ST ÅA Some general considerations and assumptions: The equimolar exchange A B requires balanced energy exchange as well: vaporisation heats of A and B should be roughly the same Pressure and the temperature ranges must be chosen: Below the critical pressures and temperatures of the components The components must be thermally stable (for hydrocarbon fractions! of crude oil: below 300 ~ 350 C) The temperature at the top of the column should preferably be > 40 C, allowing for heat transfer with surrounding air or water Equilibrium data is needed, + data for feed and products specifications The necessary gas and liquid streams, the number of stages and heat consumption can be calculated A simple procedure based on local equilibrium and mass balances was suggested by McCabe and Thiele Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 13/82 Equilbrium data Example: C 3 ic 4 /1 From MÖF-ST ÅA pressures < critical pressures allowable for C 3 ic 4 : p < 37 bar 37 bar boiling points C: stability OK T boil ic 4 at 37 bar T boil C 3 at 37 bar Raoults law: y p total = x p K = y/x = p /p total Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 14/82

8 Equilbrium data Example: C 3 ic 4 /2 From MÖF-ST ÅA Chose 40 C as the temperatur at the condensor. At T min = 40 C, the vapour pressure of the more volatile component which is C 3 = 14 bar p 14 bar p C 3 at 40 C T boil ic 4 at 14 bar T boil C 3 at 14 bar Then T boil ~ 81 C for ic 4 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 15/82 Binary distillation, McCabe-Thiele /1 The McCabe-Thiele x,y diagram procedure for continuous binary distillation is limited to cases where the vapour and liquid mass flows are constant in both the top and bottom section of the column. This requires that the vaporisation / condensation heats for the two components are the same, and that mixing heat effects are negligible. For example EtOH/H 2 O: Δ vap h(etoh) Δ vap h(h 2 O) 40 kj/mol elementary courses (e.g. ÅA424302, Z12) Pic: WK92 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 16/82

$Binary distillation, McCabe-Thiele /2 q = fraction of feed that gives liquid on the feeding tray: L = L + q F, F q = ΔL and F (1-q) = ΔV From MÖF-ST ÅA 424302 q is related to the energy needed to$

9 Binary distillation, McCabe-Thiele /2 q = fraction of feed that gives liquid on the feeding tray: L = L + q F, F q = ΔL and F (1-q) = ΔV From MÖF-ST ÅA q is related to the energy needed to convert the feed completely into vapour. With enthalpies H for saturated liquid and saturated gas it is found that HG,sat HF q H H G,sat L,sat q >1 q = 1 0 < q < 1 q = 0 q < 0 Picture: after T68 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 17/82 Binary distillation, McCabe-Thiele /3 The McCabe-Thiele x,y diagram procedure is thus based on mass balances (giving straight operating lines) and chemical equilibrium data If this method cannot be used, a graphical method based on enthalpy composition (h,x or h,ξ) diagrams as given by Ponchon de Savarit can be used, based on local equilibrium, mass balances and energy balances elementary courses (e.g. ÅA424302, Z12) Pic: Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 18/82

From MÖF-ST ÅA 424302 Restrictions McCabe-Thiele method Molar heats of vaporisation should not differ more than ~10% Heats of dissolution or mixing are negligible Relative volatilities should be 1.

10 From MÖF-ST ÅA Restrictions McCabe-Thiele method Molar heats of vaporisation should not differ more than ~10% Heats of dissolution or mixing are negligible Relative volatilities should be 1.3 < α < 5 Reflux ratio s (L/V) should be R > 1.1 R min Number of trays N << 50 preferably Otherwise: operating lines are presumably not straight use a more exact method that includes an energy balance Source: CR83 Pic: Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 19/82 ÅA h,x / h,ξ, and s,x / s,ξ diagrams Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 20/82

11 Mixing heat for binary mixtures ξkg B 1-ξ kg A For this mixing process: h = ξ h a + (1-ξ) h b + q mix 1 kg mix with mixing heat q mix to keep temperature constant Mixing heat data can be used to produce h,x or h,ξ diagrams ξ= mass fraction for the most volatile compound x = molar fraction for the most volatile compound With molar masses M a and M b : x ; ξ Mb ξ M ξ x M x M a a b Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 21/82 Mixing heat, h,ξ diagram /1 NH 3 -H 2 O EtOH- H 2 O Pics: B01 Mixing heat data for NH 3 -H 2 O and for EtOH-H 2 O (liq) Schematic h,x or h,ξ diagram for three temperatures (t 1, t 2, t 3 ) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 22/82

12 Mixing heat, h,ξ diagram /2 Determination of heat of mixing, q mix, versus composition, ξ: a boiling liquid mixture (mass m, composition ξ f, enthalpy h f ) is fed into a calorimeter; a small amount of heat dq gives a small amount of vapour, dm (composition ξ d, enthalpy h d ) Mass balance, heat balance: Pic: B01 gives (eliminate m) q mix = dq/dm = Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 23/82 Mixing heat, h,ξ diagram /3 liquid gas i q mix c i b Construction of condensation line from boiling line, heat of vaporisation and heat of mixing For point P, which demixes into liquid F and gas D, the relative amounts are G/L = FP / PD (lever rule) Schematic h,ξ diagram: - Condensation line c - Boiling line b - Isotherms i Pics: B01 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 24/82

13 s,ξ diagram For 2nd Law analysis, exergy analysis or for tracking down irreversibilities in mixture applications (for example mixtures of refrigerants) s,ξ or s,x diagrams may be used. Note that during adiabatic mixing h mix = h a + h b, but s mix = s a + s b + Δs mix s,x and Δs mix for two ideal gases x Pics: B01 x Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 25/82 ÅA Flash vaporisation Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 26/82

Flash vaporisation, h,x diagram /1 Consider a flash vaporisation process, in which a binary liquid at temperature T 0, pressure p 0 is brought to a lower pressure p 1 using a throttling valve.

14 Flash vaporisation, h,x diagram /1 Consider a flash vaporisation process, in which a binary liquid at temperature T 0, pressure p 0 is brought to a lower pressure p 1 using a throttling valve. Temperature drops to T 1. Part of the liquid vaporises, giving a vapour that is more rich in the more volatile component. Mass balance: F = G + L Mass balance volatile component: x F F = y G + x L Heat balance: h F F = h G G + h L L, with heat of vaporisation Δ vap h= h G -h L at T 0. Pic: Z87 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 27/82 Flash vaporisation, h,x diagram /2 Initially, at p 0 the feed point A with x = x F is in the liquid region of the h,x diagram After the pressure reduction to p 1, point A lies in the 2-phase region, giving liquid (L) C + gas (G) B Amount ratio G/L equals = CA /AB, or G L x x F y x F h h F G hl h F G L Pic: Z87 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28/82

Flash vaporisation, example /1 220 F = 104 C 170 F = 77 C K71- Fig 2.11 K71- Fig 2.

15 Flash vaporisation, example /1 220 F = 104 C 170 F = 77 C K71- Fig 2.11 K71- Fig BTU = 1055,06 J 100 BTU/lb = 2,326 kj/kg Pics: K71 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 29/82 Flash vaporisation, example /2 w K71- Fig 2.11 v K71- Fig 2.12 v w Lever rule see for example: Pics: K71 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 30/82

16 ÅA Binary Distillation II (the Ponchon-Savarit method) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 31/82 Ponchon-Savarit method /1 Continuous distillation Feed F(kg/s), x f (kg/kg) Top product D, x d Bottom product W, x w Upward vapour flows V (kg/s), downward liquid flows L (kg/s) Liquid mass fractions x, vapour mass fractions y Reboiler heat input Q B (kw), condenser heat output Q C (kw) Pic: CR83 Top section I, index n (downwards) Bottom section II, index m (downwards) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 32/82

17 Ponchon-Savarit method /2 H L and H V are enthalpies of liquid and vapour streams (kj/kg) Mass balances top section stages (total, and for the most volatile species of the two) *) Pic: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 33/82 Ponchon-Savarit method /3 Heat balances (top section) H L d = H of liquid D (x = x d ) Constant H d = H L d + Q C /D: **) Pic: CR83 leaving system via top: D H L d + Q C Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 34/82

Ponchon-Savarit method /4 From the mass balances *) From mass balance *) and heat balances **) x Pic: CR83 Functions y n (x n-1 ) are all lines that go through point (pole) N = (x d, H d ) in the

18 Ponchon-Savarit method /4 From the mass balances *) From mass balance *) and heat balances **) x Pic: CR83 Functions y n (x n-1 ) are all lines that go through point (pole) N = (x d, H d ) in the (x,h) plot Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 35/82 Ponchon-Savarit method /5 The lines y n (x n-1 ) can be put in the h,x or h,ξ diagram (see next slide) Similarly, mass and heat balances for the bottom section: Pic: CR83 Top section I, index n (downwards) Bottom section II, index m (downwards) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 36/82

Ponchon-Savarit method /6 H L w = H of liquid W (x = x w ) Constant H w = H L w -Q B /W This then gives Note: this implies q = 1 leaving system via bottom: W H L w -Q B Pic: CR83 Functions y m (x m-1

19 Ponchon-Savarit method /6 H L w = H of liquid W (x = x w ) Constant H w = H L w -Q B /W This then gives Note: this implies q = 1 leaving system via bottom: W H L w -Q B Pic: CR83 Functions y m (x m-1 ), all lines go through point (pole) M = (x w, H w ) in (x,h) plot Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 37/82 Ponchon-Savarit method /7 A stream N can be visualised with composition = x d, enthalpy = H d, mass = V n L n-1 Similarly, M can be visualised as a stream with composition = x w, enthalpy = H w, mass = L m-1 V m Then also: F = M + N, and F x F = M x w + N x d Pic: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 38/82

20 Ponchon-Savarit method /8 The necessary number of theoretical stages can be determined graphically as shown Feed point F is shown on the boiling line (here, q =1) V 1 is vapour from first plate, L 1 is liquid from first plate to second plate, et c. Here, 7 theoretical stages needed. Pic: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 39/82 Ponchon-Savarit method /9 The heat removed in the condenser per unit mass D is q c = Q c /D. Lowering N N means that more stages are needed At N N m the number of stages Here, 7 theoretical stages needed. Pic: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 40/82

Ponchon-Savarit method /10 Using q c = Q c /D and the minimum heat requirements can be found: Increasing the reflux ratio requires more heat to be removed, point N upwards, reducing the number of

21 Ponchon-Savarit method /10 Using q c = Q c /D and the minimum heat requirements can be found: Increasing the reflux ratio requires more heat to be removed, point N upwards, reducing the number of stages Pic: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 41/82 Ponchon-Savarit method Example /1 1 kg/s x f = 0.3 x d = R = 1.08 R min Mass balances x w = Source: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 42/82

22 Ponchon- Savarit method Example /2 h, ξ diagram NH 3 -H 2 O MPa = 10 atm Source: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 43/82 Ponchon-Savarit method Example /3 Source: CR83 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 44/82

23 ÅA Ideal distillation column analysis Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 45/82 Analysis of ideal column /1 Exergy analysis shows that the minimum work for separation of 1 mol of mixture equals w min = Δex = T Δs For distillation with reboiler heat Q in and condenser heat Q out, the exergy loss ΔEx can be approximated as o 1 Ex Qin T Ttop T (if H F H B + H D!) bottom 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 1 Thus, the separation is accomplished by degrading Q in at T in to Q out (= Q in ) at T out < T in Pic: SAKS04 46/82

24 Analysis of ideal column /2 The Clausius Clapeyron expression for phase transitions based on the slope of the two-phase co-existence (boiling) curve in a p - T diagram, dp/dt = ΔS/ΔV ~ Δ vap H/(TΔV) gives for ideal gases: dp p vaph dt 2 R T p ln p sat, 1 sat sat, 2 H 1 T Also, for an ideal mixture the relative volatility for two similar* components is * Chemically similar, boiling points not too far apart, α ~ O(1) 1 T 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku vap R p p sat 1 sat 2 Pic: SAKS04 47/82 Analysis of ideal column /3 This then gives, for the distillation temperature range (T top, T bottom ): ln 12 vap R H 1 1 Tbottom T top which if α ~ 1, and using ln(1+z) z if z «1, it simplifies to: 12 1 vap R H 1 1 Tbottom T top Pic: SAKS04 This requires that - heat Q in is supplied using a Carnot heat pump -heat Q out is converted into work in a Carnot process, - minimum reflux is used 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 48/82

25 Analysis of ideal column /4 Assume a propane C 3, propene C = 3 distillation with x F = ½, q = 1, B/F = D/F = ½ (mol/s)/(mol/s) Note: for this equimolar mix, Δs mix is at its maximum (if ideal) Then per mole of feed, L/F = r D/F = ½ r, (r = reflux ratio* = L/D) above the feed, L/F = (½ r + 1) below the feed, and V/F = ½ (r + 1) in the whole column. Then the heat input at the reboiler is equal to Q in /F = V/F Δ vap H= ½ (r+1) Δ vap H Q out /F Pic: SAKS04 * Using r to distinguish it from gas constant R. 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 49/82 Analysis of ideal column /5 The minimum work required to separate the mixture can be found by considering a heat pump between T out and T in see Figure and exergy analysis gives (per mole feed): w R T ln2 for an equimolar mixture This can be used to define the minimum reflux ratio r min : W sep sep F min in Q F R T o R T ½ o o ln x 2ln2 ln 1 T r 1 H r 1 min i i x x i i ln x i vap i W F min min min in Q F T 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 12 o top T 1 bottom (α < 4!) Pic: SAKS04 100% efficient! (reversible) 50/82

26 ÅA Real distillation column analysis I: efficiency Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 51/82 Analysis of real column /1 What will be the efficiency of a real column for this case? Heat transfer requires temperature differences! Q in is supplied at 377K, Q out goes out at 298K The minimum heat input for the separation is Q min in / F = ½ (r min +1) Δ vap H with r min = 9.64 (α ), per mole feed Deviations from ideal thermodynamics r real = 15.9 (α ) Q real in / F = ½ (r real +1) Δ vap H r real /r min = 1.65 Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 52/82

27 Analysis of real column /2 Bottom section ΔT = 46 K, top section ΔT = 22 K The total thermodyn. eff. of the column is calculated as: Pic: SAKS04 r r min real 1 1 which gives η overall = = 9.3% The main sources of the inefficiency are the driving forces Δ(1/T) in reboiler and condenser The ratio Q min in /Q real in = 0.63 reflects exergy losses inside the column Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 53/82 Analysis of real column /3 The amount of exergy (work) lost in the column + the exergy used for the separation = the difference between the exergies of the reboiler heat and the condenser heat For the minimum work (exergy) needed it was shown that Pic: SAKS04 so that the work (exergy) lost in the column is equal to see Table on next slide Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 54/82

Analysis of real column /4 Efficiency of the distillation column: summary Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, 20500 Turku 28 jan 15 55/82 Analysis of real

28 Analysis of real column /4 Efficiency of the distillation column: summary Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 55/82 Analysis of real column /5 Efficiency of the distillation column: summary Reboiler Column Condenser Previous table,rated by normalising with Q real in T 1 T 0 R Pic: SAKS04 28 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 56/82

29 Binary distillation efficiency: overview Maximum efficiency: where x = mole fraction of the more volatile component Reboiler heat duty: where r = reflux ratio, = distillate product flow For Q REB Q COND : D Source: OFRGJ03 Reboiler duty exergy: Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 57/82 ÅA Real distillation column analysis II: exergy analysis Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 58/82

30 Exergy analysis /1 An analysis can also be made based on exergy flows: Note that still no enthalpy of mixing is considered, and ideal mixing entropy -R x lnx. Flowsheeting programs (should) give values for H s, S s, and Q s, making it easy to calculate exergy streams No chemistry physical exergy is sufficient! Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 59/82 Exergy analysis /2 Exergy balance for the equimolar C 3 /C 3= distillation * reversible case: * real case Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 60/82

31 Exergy analysis /3 Exergy streams for ideal and real case: summary = 0 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 61/82 Exergy analysis /4 The main difference between ideal and real performances are Ex(Q 1 ) and Ex(Q 2 ); note that Ex(Q 2 ) = 0 in the real case. The difference between the two cases as given on the previous two slides is + or, normalised per unit input exergy: (gives again value efficiency 0.093) Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 62/82

32 Exergy analysis /5 The analysis showed that the main losses were at the reboiler and the condenser, due to heat transfer over large temperature differences ΔT Instead of trying to reduce these ΔT s, waste heat from another process unit may be available Heat integration, which at the same time saves steam (for example) Pic: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 63/82 Example: H 2 O/EtOH distillation /1 A 5%/95% (w/w) EtOH/H 2 O mix is to be distilled into a 90% EtOH top product and pure water bottom product Feed (18 kg/s) and bottom product (17 kg/s) exchange heat: feed is heated C; bottom product cools C; the heat exchanged is 3690 kj/kg top product (1 kg/s) The reflux cooling heat q r = 2640 kj/kg top product; the heat reboiler q b = 5620 kj/kg top product Tempertures q r = 79 C; q b = 104 C m c p (top) = kj/k, m Δ vap H (top) = 980 kj m c p (bottom) = kj/k Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 64/82

33 Example: H 2 O/EtOH distillation /2 The minimum work needed for making a 90% EtOH mix from a 5% EtOH mix: see the mass balance in moles. The mole fractions for water are for the feed, for the distillate and 1.00 for the bottom product; this then gives with the molar amounts as indicated: for the minimum work per kg top product. See section 1.6 of Exergy analysis Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 65/82 Example: H 2 O/EtOH distillation /3 Assuming surroundings temperature T = 293K an exergy analysis can be made (exergy flows per kg top product): Reflux cooler Boiling top product Top product heat-up. 980 kj/kg for 10% H 2 O + 90% EtOH Bottom product. Bottom product after heat exchange Source: B01, H84. Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 66/82

34 Example: H 2 O/EtOH distillation /4 Assume that the bottom heat q b is delivered by an ideal (Carnot) heat pump: 5620 (1-293/377) = 1252 kj/kg top product exergy (Note the difference !) This goes to the cold bottom product (7%), top product (14%), reflux cooler (35%) and separation work (4%), total 60%. Thus, = 356 kj is lost as irreversibilities. For the heat exchanger, the losses per kg top product are which means that 70 kj/kg top product is lost in the column itself; which is ~ 40% of the minimum separation work (186 kj/kg top product.) Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 67/82 ÅA Distillation + heat pump and other improvements Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 68/82

35 One option: feed stream splitting Also known as hybrid distillation A membrane is used to split the feed into two streams, fed into the column at different locations Part of the separation is done (less irreversible) by the membrane lower reflux ratio and/or less stages Pics: SAKS04 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 69/82 A few other options Some GENERAL considerations for improving distillation column efficiency: Process control Feed pre-heat Reduce pressure drop by optimising the trays (or packing) Separate to lower purity and combine Pic, Source: SAKS04 with another separation process Use intermediate reboilers and condensers, bringing operating lines closer to the equilibrium line (but this may require more stages or trays) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 70/82

36 Double columns Pic, Source: SAKS0 Paraffin /olefin separations are very energy intensive. One example is propane / propylene, or C 3 (atm. b.p C) / C = 3 (atm. b.p C). For a single-column tray column distillation process trays would be needed, giving a very high (> 100 m) column, with reflux condensing using air or water A double-column process is preferred, with smaller (shorter) columns, using hot and cold water for condensing (column 2) and reboiling (column 1). Note that reboiler for column 2 = condenser for column 1. Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 71/82 Distillation column + heat pump /1 Improved efficiency can be obtained by using a heat pump: the condenser is removed and the reflux is obtained after heat exchange in the reboiler and with the feed, and throttling Drawback: complexity costs... Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 72/82

37 Distillation column + heat pump /2 Returning to the EtOH/water distillation (see the end of the previous section): the top product is compressed so that its temperature rises from 79 to 104 C (which implies p 2 = 2 bar) It can then give off heat to the water in the bottom of the column at 100 C. See points 1,2,3,4 in the h,ξ diagram below. The h, ξ diagram gives masses δ:φ:1 = (*-4):(*-1):(1-4) = 3.7:2.7:1 a.k.a. Mechanical Vapour Recompression (MVR) * pole point Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 73/82 Distillation column + heat pump /3 For the compressor work w c = δ (h 2 -h 1 ), and for the bottom heat exchanger Δq b = δ (h 3 -h 2 ) which is then found to be 3580 kj per kg top product. This is 64% of the conventional process value as in the previous section. Note compressor losses; and that stream 3 may be used to preheat the top product before the compressor. * pole point Source: B01, H84 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 74/82

38 Two-feed distillation Two-feed, or two-enthalpy feed can Allow for lowering the number of stages, N (same L/D, x D, x B ) Improve the separation, better X D, X B (same L/D, N) Allow for lowering reflux ratio r = L/D (same N, x D, x B ) 50% energy savings possible, but modifications are needed Source: D04, WP93 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 75/82 Adiabatic versus diabatic distillation In conventional distillation, heat is only added/removed at reboiler/condensor; no heat is added/removed on the trays: adiabatic operation In diabatic operation, heat can be added/removed by heat exchange, in principle on each tray Heat is removed from each tray above feed, added to each tray below feed. Reboiler much smaller or not needed. More complicated ( - process control), but more efficient: lower entropy production (decreases with increasing heat exchange area) Source: KBJG10, J10 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 76/82

HIDiC Heat integrated distillation column Rectification ( top ) section is operated at a higher pressure (= pressure of conventional column) than the stripping ( bottom ) section Heat is transferred

39 HIDiC Heat integrated distillation column Rectification ( top ) section is operated at a higher pressure (= pressure of conventional column) than the stripping ( bottom ) section Heat is transferred from rectification to stripping section Comparison for C 3 / C 3= splitter Source: J10, OFRGJ03 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 77/82 Final remarks Distillation accounts for ~3% of world energy consumption Heat integration can give significant energy efficiency improvements (but increased costs, more process control) Fully diabatic operation of a distillation column won t be practically, not all temperature levels will be available Most important are the heat exchanger at the lowest and highest trays, these can be integrated using for example a heat pump. Maybe integration with other columns is possible Further improvement can be obtained by varying column cross sectional area: in a diabatic column rectifier section the vapour flow decreases upwards, in the stripper section it is the opposite Source: KBJG10, J10 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 78/82

40 Sources 3 B01: Bart, G.C.J. Advanced thermodynamics (in Dutch) course compendium Delft Univ. of Technol., Delft (2001) B97: Bejan, A. Advanced engineering thermodynamics John Wiley & Sons (1997) Chapter 3 CRBH83 J.M. Coulson, J.F. Richardson, J.R. Blackhurst, J.H. Harker Chemical engineering, vol. 2, 3rd ed. Pergamon Press (1983) Chapter 11 D04: Demirel, D. Thermodynamic analysis of separation systems Separation Sci. & Technol. 39 (16) (2004) H84: Hoogendoorn, C.J. Advanced thermodynamics (in Dutch) course compendium Delft Univ. of Technol., Delft (1984) HS92: Humphrey, J.L, Siebert, A.F. Separation technologies: an opportunity for energy savings Chem Eng Prog 88 (March 1992) (1992) J10: Jana, A.K., Heat integrated distillation operation Appl. Energy 87, (2010) K71 King, C.J. Separation processes McGraw-Hill (1971) KBJG10: Kjelstrup, S., Bedeaux, D., Johanessen, E., Gross, J. Non-equilibrium thermodynamics for engineers, World Scientific (2010) OFRGJ03: Olujic, Z, Fakhri, F., de Graauw, J., Jansens, P.J. Internal heat integration the key to an energy conserving distillation column J. Chem Technol Biotechnol 78, (2004) Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 79/82 Sources 3 /cont d SAKS04: de Swaan Arons, J., van der Kooi, H., Sankaranarayanan, K. Efficiency and sustainability in the efficiency and chemical industries. Marcel Dekker, New York (NY) 2004 SH06 J.D. Seader, E.J Henley Separation process principles John Wiley, 2nd edition (2006) chapter 17 TK87: Tondeur, D., Kvaalen, E. Equipartition of entropy production. An optimality criterion for transfer and separation processes Ind. Eng. Chem. Res. 87 (26) (1987) T68 R.E. Treybal Mass transfer operations McGraw-Hill 2nd edition (1968) WK92 J.A. Wesselingh, H.H. Kleizen Separation processes (in Dutch: Scheidingsprocessen) Delft University Press (1992) WP93: Wankat, P.C., Kessler, D.P. Two-feed distillation: same composition feeds with different enthalpies Ind. Eng. Chem. Res. 32, (1993) Z87 F. Zuiderweg Physical separation methods (in Dutch: Fysische Scheidingsmethoden) TU Delft (1987) (vol. 1) Z12: Zevenhoven, R. Massöverföring & separationsteknik / Mass transfer & separation technology course material (in English!) Åbo Akademi Univ. (2012) Chapter 12 ÖS97 G. Öhman, H. Saxén Transportprocesser (in Swedish) Åbo Akademi Värmeteknik (1997) 3rd ed., Chapter 3 Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku 28 jan 15 80/82

41 ÅA PSIA = 6.8 atm 1 BTU/lb mole = J/mol Appendix: h,ξ & x,y diagram NH 3 H 2 O 6.8 atm Source: Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku Appendix: h,ξ diagram n-c 6 / n-c 8 ÅA HL, HG kj/mol n-c 6 - n-c kpa x,y, - HL, HG kj/mol n-c 6 - n-c kpa x,y, - T C x - HL kj/mol y - HG kj/mol Åbo Akademi Univ - Thermal and Flow Engineering Piispankatu 8, Turku

Separationsteknik / Separation technology

Separationsteknik / Separation technology 424105 1. Introduktion / Introduction Page 47 was added Nov. 2017 Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik