L/O/G/O 單元操作 ( 三 ) Chapter 21 Distillation 化學工程系李玉郎

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1 /O/G/O 單元操作 ( 三 ) Chapter 21 istillation 化學工程系李玉郎

2 ifferent methods: Batch: Single stage without reflu -- Simple dist. Multi-stage with reflu Continuous: Single stage without reflu--flash dist. Multi-stage with reflu -- rectification

3 Flash istillation ( 急驟蒸餾或平衡蒸餾 ) Flash istillation ( 急驟蒸餾 or 平衡蒸餾 ) c : pressure-reduced value d : vapor-liquid separator separated streams are in equilibrium For binary mitures: Assume 1 mole feed seperated into: f mole vapor, y (1-f) mole liquid, B Material balance to the more volatile component : F = y f + B (1-f) (21.1) y y y B y f F 1 (21.2) f f To solve B and y, equilibrium data should be provided ( B,y ) : intersection of line (21.2) and equilibrium curve. 3

4 f = f (T,P), depends on the enthalpy balance Eq (21.2) y 1 f f f F (21.2) A straight line, slope : 1 f f etting = F in Eq (21.2) y = F, that is ( F, F ) is a point on line Eq (21.2) The material-balance line crosses the diagonal at = F for all values of f. 4

5 slope : 1 f f

6 Eample 21.1 A miture of 50 mole percent benzene and 50 mole percent toluene is subjected to flash distillation at a separator pressure of 1 atm. The vapor-liquid equilibrium curve and boiling-point diagram are shown in Figs and Plot the following quantities, all as functions of f, the fractional vaporization: the temperature in the separator (T), the composition of the liquid leaving the separator ( B ), and the composition of the vapor leaving the separator (y ).

7 Feed point ( F, F ) = (0.5,0.5), Operation line: f y 1 f A given f value ( F, F ) op line Intersection ( B,y ) with EQ line Slope (1-1/f) Solution f F 0.67 dew point line (y) bubble point line ()

8

9 Continuous istillation with Reflu : Rectification Flash distillation etensively used in petroleum refining, Not effective in separation components of comparable volatility, which requires the use of distillation with reflu. Rectification continuous distillation with reflu

10 liquid and vapor leaving the plate are brought into equilibrium n+1 n n-1 y n+1 y n yn-1

11 A : more volatile component B : less volatile component (in dew point) liquid (heat of vaporization) low boiler, A high boiler, B (heat of condensation) vapor (in bubble point) heat to vaporize A = heat released in condensation B

12 Combination rectification and stripping Reflu provides the downflowing liquid in the rectifying section that is needed to act with the up flowing vapor. Without the reflu, no rectification would occur in the rectifying section Feed pre-heater

13 Overall material balances for two-component systems Feed : F mole/h, F Overhead product: mole/h, Bottom product: B mole/h, B. Two independent overall material balances can be written. Total-material balance F = B + (21.3) Component A balance F F = + B B (21.4) Eliminating B from these equations gives Eliminating gives Material Balances in Plate Columns (21.5) Equations (21.5) and (21.6) are ture for all values of the flows of vapor and liquid within the column F B F F B B F B (21.6) 13

14 Net Flow rates a = a + = a - a a X a X n+1 = n + = n+1 - n > F X F m = m+1 + B B = m m+1 b = b + B B = b - b < B X B

15 Net flow rate A material balance around the condenser and accumulator in Fig gives = a - a 1st plate (21.7) Considering the part of the plant enclosed by control surface I in Fig A total material balance around this control surface gives = n+1 n plate n (21.8) is the net flow rate of material upward in the upper section of the column. Regardless of changes in and, their difference is constant and equal to. Similar material balances for component A give the equations = a y a a a = n+1 y n+1 n n (21.9) Quantity is the net flow rate of component A upward in the upper section of the column. It, too, is constant throughout this part of equipment.

16 For Stripping section: B = b - b = m - m+1 (21.10) B B = b b - b y b = m m - m+1 y m+1 (21.11) The net flow rate of total material is B; of component A is B B.

17 Because there are two sections in the column, there are also two operating lines. For the rectifying section Based on Eq.(20.7) in Chap. 20 the operating line for this section is y n1 n1 ya n1 Substitution for a y a a a from Eq. (21.9) gives y n1 n n n n1 n a n1 The slope of the line defined by Eq. (21.13) is, as usual, the ratio of the flow of the liquid stream to that of the vapor stream. n+1 = n + Operating lines a a (21.12) = a y a a a (21.13) n yn1 n (21.14) n n 17

18 n+1 y n+1 = n n + y n1 n n1 n n1 (21.13) m m = m+1 y m+1 + B B m+1 y m+1 = m m - B B 18

19 For striping section A material balance over control surface 2 in Fig gives m+1 y m+1 = m m -B B (21.15) m BB y m1 m (21.16) m1 m1 Eq(21.16) is the operating line in the stripping section. Net flow rate in striping section: B = b - b = m - m+1 y m m BB 1 m (21.17) B B m m 19

20 Slope of op line : for rectifying section, slope = for stripping section, slope = n n1 n n m m B <1 >1 The slope of the operating line in the rectifying section is always less than 1.0 In the stripping section the slope is always greater than 1.0. When n, m are constant, the op lines are straight, when n. m are not constant, the op lines are curved, and the position of a curved op line should be determined by mass balance and enthalpy balance. 20

21 Number of Ideal plates: McCabe-Thiele Method Require: (1) Equilibrium curve (2) Operation line straight line: constant molar flow curve line: enthalpy balance is required Number of ideal plate: (1) step by step computation (2) McCabe-Thiele construction on -y diagram. Constant molar overflow: straight operating lines results from nearly equal molar heats of vaporization between components A and B Subscripts: n-1, n, n+1; m-1, m, m+1 are dropped, upper column, lower section 21

22 Eq (21.14) (divided by ) Slope =, intercept = R R y n n n n n n n R R R y R R 1 1 R 22 Reflu ratio (21. 18) (Used in this tet book) (21.19): operation line for rectifying section condenser

23 In Eq (21.19), set n = (, ) is a point at the upper end of OP line or OP line intersects the diagonal at point (, ) n R R R y To determine the operation line, one point is required n n R R R y

24 According to the properties of OP line, the upper terminus is ( c, y 1 ) Condenser and top plate (1). Total condenser: condense all the vapor from plate 1 y 1 = = c, so, ( c,y 1 )= (, ) ( C,y 1 ) (, ) abc represents the top plate

25 (2) Partial and final condenser: y =, y in equilibrium with C. Condenser is equivalent to an additional theoretical stage. y 1 c = y' Tc, c T 1 c 1 condenser y 1, 1 Plate 1 a ( c, y 1 ) end point of column a'b c the top plate

26 If condensate is at its bubble point, T c = T bc, = c, = c ( 1 ) If the reflu (condensate) is cooled below the bubble point, T c < T bc, then 1 <, > c Δ = - c C ( T T C pc bc C (21.21) The actual reflu ratio: C ) Tc, c T 1 = c 1 y' 26 condenser y 1, 1 Plate 1 λ C : heat of vaporization of condensate) C C [ 1 CPC ( Tbc TC ) / C ] (21.22) Normally T bc = T 1 (bubble point of condensate) (p676 description)

27 Bottom plate and re-boiler Eq. (21.17) y m1 m B m m BB B m Constant. y m1 B m B B B (21.23) Find a point on the operation line: Set m = B y m+1 = B (conc. of bottom product) Eq. (21.23) OP line for stripping section, crosses the diagonal at ( B, B ) (21.23) OP line constructed using (1) End point ( B, B ), (2) slope B ( ) 27

28 y b =y N y r y r B abc: bottom plate B reboiler: (An ideal plate) Figure Graphical construction for bottom plate and boiler: Triangle cde, reboiler ; triangle cde, bottom plate. C( b,y r ): end point of the bottom plate

29 Molar flow rate:,, above the feed plate below and of the feed plate Feed plate relationship depends on thermal condition of the feed efine: q = moles of liquid to stripping section per mole of feed (a) Cold feed, q>1, F, (b) Saturated liquid, q=1, F, (c) Partially vapor, 0<q<1, F, (d) Saturated vapor, q=0,, F F (e) Superheated vapor, q<0,, F 29

30 (a) Cold feed, q>1 (b) Saturated liquid, q=1 (c) Partially vapor, 0<q<1 F, F, F, (d) Saturated vapor, q=0 (e) Superheated vapor, q<0, F, F 30

31 For cold-liquid feed: For superheated vapor q C C q P ( Tb T ( d TF T ) P F 1 (21.24) ) (21.25) C P, C P = specific heat of liquid and vapor, respectively T F = temp. of feed T b, T d = bubble point and dew point of feed, respectively λ = heat of vaporization 31

32 For 1 mole feed, to liquid: q, to vapor: (1-q) F mole feed, to liquid: Fq, to vapor: F(1-q) qf ( 1 q) F Material-balance for rectifying and stripping sections y n = n-1 + (21.28) or qf (21.26) or Feed line ( 1 q) F (21.27) y m m1 B B (21.29) et y n = y m, n-1 = m-1 Eq. (21.28) Eq (21.29) y( ) ( ) B B (1-q)F -qf F F y 1 q q F 1 q (21.30) (21.31) Feed line depends on F and q slope: -q/(1-q), crosses the diagonal at ( F, F ) The positions where rectifying line and striping line intersect 32

33 Method: (1) locate the feed line: point ( F, F ), slope: (2) locate the rectifying line: point (, ); intercept (3) draw the stripping line Construction of operation lines R 1 ; slope: q 1 q R R 1 point ( B, B ); intersection of feed and rectifying lines Satu. vapor. Satu. iq. Cold feed (, ) a: cold feed: q > 1, slope > 0 b: saturated liq.: q = 1, slope c: partially vapor: 0 < q < 1, slope < 0 d: saturated vapor: q = 0, slope = 0 e: superheated vapor: q < 0, slope >0 ( B, B ) 33

34 Transfer can be occurred between a - b Feed plate location optimum location Optimum feed plate No. of plate: 7+1 reboiler feed on plate 5 If feed on plate 7 (----) composition of liquid on feed plate Fig X F : Feed composition No. of plate: 8 + a reboiler Optimum location the steps transfer from rectifying line to stripping line, so that the plate number is as small as possible

35 Construction can start either at bottom or at top Optimum location the steps transfer from rectifying line to stripping line, so that the plate number is as small as possible Criterion transfer is made immediately after the value is less than the coordinate of the intersection of the two operating lines Feed plate the triangle that has one corner on the rectifying line, and one corner on the stripping line. Feed on the wrong plate may seriously affect the column performance. Feed close to a or b will lower the quality of both the top and bottom product

36 Heat effects of entire unit condenser and reboiler Heat added in the reboiler : q r small) Amount of stream required: λ s : latent heat of steam λ: molar latent heat of miture If q = 1, feed at saturated liquid, heat supplied in reboiler = heat removed in the condenser = ( ) ( ) Amount of cooling water required: m W T 2 -T 1 = temp. rise of cooling water Heating and cooling requirements T 2 T 1 m s s ( ) (21.33) (sensible heat change of liquid is (21.32) F, H F b X b H b 36 q c = λ up q r = λ low B, H B, H

37 , H q c = λ up F, H F b X b H b B, H B qr = λlow 37

38 Base: Benzene X F = 0.4 X = 0.97 X B = 0.02 R = / = 3.5/1 λ B = 7360 cal/gmole λ T = 7960 cal/gmole (a): =?, B=? (b): N =?, feed plate? (c): steam required? (kg/h) (d): cooling water required? (m 3 /h)

39 X F = 0.4 X = 0.97 X B = =85.84 F = + B, F XF = + B B F F B B

40 (b). Calculate number of ideal stage and feed position (i) Feed at T b (95 ) : q=1 (saturated liquid) feed line: ( F, F ) = (0.44,0.44) slope = -q/(1-q) = rectifying line: (, ) = (0.974,0.974) intercept = /(R +1) = 0.974/(3.5+1) = stripping line: ( B, B ) = (0.0235,0.0235) intersection of feed line and rectifying line Number of ideal stages: 11 + one reboiler Feed plate: The 7 th plate

41 b-(ii): feed at liquid of 20 (T b = 95,C P = 0.44 cal/g. ) q C ( Tb T P F (95 20) = 1.37 Slope of feed line: -q/(1-q) = 3.70, ( F = 0.44) ) C P = = cal/g-mol, Number of ideal stages: 10 + one reboiler Feed plate: The 6 th plate 41

42 (b)-(iii): feed is a miture of 2/3 vapor, 1/3 liquid. By definition: q = 1/3 Slope of q line = - q/(1-q) = -0.5, ( F = 0.44) No of ideal plates: 12 + one reboiler Feed plate: the 7th plate

43 (c) Calculate the steam required per hours. m s, calculate first s = +(1-q)F = F(1-q) = + = (/ + 1 ) = (R +1) = 153.4(3.5+1) = 690 (kg-mole/h) = (1-q) Because at the bottom of colum, it is more reasonable to use the λ value of higher boiler compound (toluene) λ = 7960 (cal/gmole) m s s = + Fq - F ms 7960[ (1 q)]

44 (d) Calculate the cooling water required (in: 25, out: 40 ) ρ= kg/m 3 ) m w C PW ( T T1) 2 (7360 is more reasonable) (40 25) olume flow rate = m3 /h Kg/h 44

45 cold q Steam req. No. ideal plate (N) Satu. iq. Cold feed Satu. vapor If the feed is more cold, N decreases, steam required increases. The total energy requirement including (1) reboiler, (2) preheater about the same Reasons for preheating the feed: (1). Keep up = down (2). Use the energy of the hot-liquid stream, such as the bottom product. 45

46 Slope of rectifying line: R /(R +1) = (/) Minimum number of plates (N min ) R increase slope increases N ideal decreases R infinite total reflu, (=), (/ = 1), slope = 1 (R = / = ), OP line = diagonal At total reflu, the number of plates is a minimum, but the rates of feed and of both the overhead and bottom products are zero one limiting case in the operation of fractionating columns. No feed, no discontinuity Between the upper and lower section 46

47 Slope of rectifying line: R /(R +1) = (/) R increase slope increases N ideal decreases 最小理論板數 R ( 全回流 ) 47

48 Simple method for calculating N min --- for ideal miture and constant α AB α AB : relative volatility y / Ae Ae AB (21.34) ybe / Be For 2 component system: y Be = 1 y Ae, Be = 1 Ae Eq (21.34) y e AB e 1 ( 1) AB e (21.35) Raoult s law: P A = P A A, y A = P A /P = P A A /P, P B = P B B, y B = P B /P = P B B /P y A / A = P A /P y B / B = P B /P AB y y A B / / A B P P A B '/ P '/ P P P A B ' ' P A /P B does not change much with temperatures column, so the relative volatility can be taken as constant. 48

49 For a binary system y A /y B and A / B may be replaced by y A /(1-y A ) and A /(1- A ), so Eq. (21.34) can be written for plate n+1 as AB y y Ae Be / / Ae Be y Ae Ae / y / Be Be y n1 n1 AB 1 yn 1 1 n 1 (21.37) ---- EQ line For the total reflu, = 0 and / = 1, slope=1, the operating line is the 45 line. OP line: ( n, y n+1 ), y n+1 = n --- OP line combine EQ and OP lines: n n1 AB 1 n 1 n 1 (21.38) At the top of the column, if a total condenser is used, y 1 =, so Eq.(21.37) becomes for n=0, Eq. (21.37) 1 AB (21.39) 49

50 1 AB 1 1 Writing Eq. (21.38) for a succession of n plates gives 1 2 n=1 AB n 1 n AB 1 1 n 1 If Eq. (21.39) and all the equations in the set of Eqs. (21.40) are multiplied together: ( n (21.39) (21.40) n n AB ) (21.41) n n n1 AB 1 n 1 n 1 For the entile column, n=n min +1 reboiler, n = B, (21.41) gives 1 ( AB ) N min 1 B 1 B 50

51 Solving the equation for N min by logarithms gives 1 ( AB ) N min 1 B 1 B N min ln[ (1 ln B ) / AB B (1 ) 1 (21.42) Equation (21.42) is the Fenske equation, which applied when α AB is constant. If the change in the value of α AB from the bottom of the column to the top is moderate, a geometric mean of the etreme values is recommended α AB. 51

52 Minimum reflu, R m minimum reflu ratio M a a min When R decreases R slope decreases ( line upward) R 1 OP lines moves upwards toward the EQ line, R no. of plates increases. ( N : ) a min --- No of plates is infinity. (, y ) At R m (minimum reflu) OP line, feed line intersect at EQ line (ad, bd, fd) Rm y' slope : Rm 1 ' At R m, the plates required to obtain is. If R < R m, even the N is, the separation y' Rm (21.44) B cannot be approached. y' ' B For EQ curve which is concave downward (Eq 24.11) 52

53 for system with EQ curve of concavity upward --- ethanol / water ine correspondings to the minimum reflu ac small R The OP line that is tangent to the EQ line. higher R feed line R, min 53

54 At minimum reflu ratio, the number of ideal plates is infinite. There are regions where liquid or vapor concentrations do not change from plate to plate ( n = n-1 ; y n+1 = y n ). These regions are called invariant zone, or pinch point ( 挾點 ) The d point in Fig (intersection of q line and EQ line) the concentrations at feed plate the concentrations of invarient zone above feed plate : / below feed plate : / Invarient zone, ( pinch point ) the only difference of the two zones feed plate rectifying section invariant zone stripping section 54

55 Optimum reflu ratio --- consider the total fied charges (1) Fied charges on the heat echange equipment, (the reboiler and condenser) increase steadily with the reflu ratio (2) Fied charges on equipment cost of unit: total plate area (No. of plates cross sectional area) N Area for a given production ( is constant ), when R is increased,, No. of plate decreases (N ),, increase, column diameter increases. (Area ) Total cost: (1) + (2) drop sharply at first with increasing R, then pass through a very shallow minimum optimum reflu ratio R m At optimum reflu ratio: Steam cost 2/3 total cost R (optimum) : not much greater than R m 55

56 Effect of reflu ratio when R,, q c q c = λ up, H, q r N, decreases F, H F Area, increases b X b H b B, H B qr = λlow 56

57 When energy costs are relatively high R (optimum) closer to R m Actually, most plants are operated at reflu ratios somewhat above the optimum, because the total cost is not sensitive to reflu ratio in this range. EXAMPE What are (a) the minimum reflu ratio and (b) the minimum number of plates for cases (b)(i), (b)(ii), and (b)(iii) of Eample 21.2? Solution: in eample = F = 0.44 (a) minimum reflu ratio: intersection of feed line EQ line, (, y ) (, y ) (0.972, 0.972) q = 1 = 1.37 = slope : R R m 1 m y y R m y' y' ' 57

58 (b) minimum number of plates the OP lines coincide with the diagonal, the results have no difference between the three cases. 8 ideal plates + one reboiler Fig eample 21.3 (b) 58

59 Nearly pure products For McCabe - Thiele Method, when either the bottom or overhead product is nearly pure, a single diagram covering the entire range of concentrations is impractical because the steps near = 0 and = 1 become small. This problem can be solved by : (1) Use auiliary diagram of large scale for the ends (2) Computer calculation (3) Eq (20.27) when EQ line and OP line are both straight Raoult s law applies to the major component Henry s law applies to the minor component 59

60 EXAMPE A miture of 2 mol percent ethanol and 98 mol percent water is to be stripped in a plate column to a bottom product containing not more than 0.01 mol percent ethanol. Steam, admitted through an open coil in the liquid on the bottom plate, is to be used as a source of vapor. The feed is at its boiling point. The steam flow is to be 0.2 mol per mole of feed. For dilute ethanol-water solutions, the equilibrium line is straight and is given by y e = 9.0 e. How many ideal plates are needed? Solution: dilute solution, constant, y e = 9.0 e EQ line is straight Eq (20.27) can be used * * ln[( ya ya) /( yb yb)] N * * ln[( yb ya) /( yb ya)] y b = 0 b = calculate y a by material balance steam ( a b ) ( ) ( ya yb) ( a b ) ya yb * y a 9.0 * y b 9.0 a b a = 0.02 n[( ) /( )] N 7.6 ideal plates n [( ) /( )] 60 y a

61 Enthalpy Balances ariations in and ( ideal solution is assumed ) T d enthalpy balances material balances, phase equilibria T b : benzene = 80 o C toluene = o C Reference Temp. = 80 o C (liquid) apor: dew point iquid: bubble point H C T P, T ref at T b H H,ByB H,TyT ( function of T d ) toluene at T b = o C (benzene) benzene 80 o C T d Tref C P, B T d Tref C P, T H, B B, ref, H, T T, ref, 61

62 _

63 The enthalpy balance for the entire system FH FH F F q H BH q (21.45) r H BH B B c ( feed at T b ) q c = λ up, H qr q c ( only one is independent ) F, H F control q r R (Eq 21.45) q c b X b H b B, H B qr = λlow 63

64 Enthalpy balance in rectifying and stripping sections Enthalpy balance within control surface I n H y 1, n1 n H, n H (21.46) Enthalpy balance to condenser q c a H H H a y, a H a H q c qc a H y, a a H H (21.47) (21.47) into (21.46) H H H n 1 y, n1 n, n a y, a a H (21.48) 64

65 (21.47) into (21.46) material balance y known : n, solve : n, n+1, y n+1 n1 n 1 n n n n1 n1 Find n+1 and n at specified n value Known : a H y,a, RH,, f (y n+1 ) f ( n ) a H n 1 y, n1 n, n a y, a RH (21.48) (21.49) (21.50) by trial and error, using (21.48), (21.49), (21.50) T Chosen : n b H,n (From Fig 21.24) (n) (n-1) Trial and error : (1) assume n+1 = a, n = a (constant molar flow rate) (2) calculate y n+1 (Eq 21.49) H y,n+1 (Fig 21.24) ; (3) n = n+1 (Eq 21.50) (4) calculate new n+1 by Eq (21.48), n by (21.50) ; (5) repeat steps (2) (4) 65 until (y n+1 ) (y n+1 ) step m step m+1 H H

66 H,n H y,n+1 T b try n = a Assume n y n+1 new new y n+1 f (y n+1 ) f ( n ) H OP line n+1 = a H H n 1 y, n1 n, n a y, a T d a RH n enthalpy OP net flow rate line n+1 compare or repeat y n1 n n n1 n1 1 a n n H H

67 m H,m + q r = m+1 H y,m+ +BH B enthalpy balance for stripping section ( control surface II ) enthalpy balance : H individual material balance : overall material balance : H m 1 y, m1 m, m m y m1 m1 m B m1 q m r BH B B m1 B (21.51) (21.52) (21.53) 67

68 Find m+1 and m at specified m value trial and error Known : BH B, B, B, q r ( from ) Choose : m H,m T b H,n T b Assume n y n+1 new new y n+1 Trial and error : try n = b H y,n+1 OP line n+1 = b T d (1) assume m+1, m ( first m+1 = b, m = b ) (2) calculate y m+1 ( from Eq ) H y,m+1 (3) calculate new m+1 ( using and ) (4) calculate new m ( Eq ) H m 1 y, m1 m, m y m1 m1 n H enthalpy OP net flow rate line n+1 compare or repeat m m m1 m B B B m1 q r BH B (5) repeat steps (2) (4) 68

69 Solution : Base component : Benzene a F = 0.5, = 0.98, B = 0.02 Assume F = 100 mole F F B B = 50 mole, B = 50 mole 69

70 Calculate R ( = 1.2R m ), q = 1.0 Eq (21.3) R M y y ( based on constant ) = 0.5 = F, y = 0.72 ( EQ curve) y RM 1.18 y a R R 1.2RM 1.42 ( ) a = R = 1.42 = = 71 mole a 1 = R + = = 121 mole 1 = R = 71 mole a 70

71 Enthalpy balance Eq (21.48) n1h y, n1 nh, n a H y, a RH a Reference Temp : 80 o C ( T b of Benzene) H 0 ( nearly pure benzene at T b ) HR H CP, T 80 H,a for benzene vapor at T o C, H y,b = ( T 80) --- (A) enthalpy of vaporization at T o C H Cp ( Tb T) Cpv ( T Tb ) H, b H H, b ( T T)( Cp calculate first for toluene vapor at T o C, H y,t = ( T 80) --- (B) at T = 80 o C ΔH = ( )(40 30) = 8174 (cal / mole) b at 80 o C Cp So, toluene vapor at T o C, H y,t = (T 80 ) v ) a a liq. Toluene T o C T b (110.6) ΔH Cp l Cp v vaporization ΔH b = 7960 vapor of toluene T o C T b (110.6)

72 Evaluate H y,1 ( H y,a ) y B Top plate temp. 80 o C y T H y,1 = 0.98 H y,b H y,t = = 7376 (cal mole) Eq (A) and Eq (B) H y,b = ( T 80) --- (A) H y,t = ( T 80) --- (B) Evaluate intermedia point y n+1, n Pick n = 0.5, T b = 92 o C ( from Fig 21.3 ) H, n C P, (92 80) ( )(92 80) 438 cal mole C P, : mean value 72

73 H,n H y,n+1 T b try n = a Assume n y n+1 new new y n+1 OP line n+1 = a T d n enthalpy OP net flow rate line n+1 compare or repeat

74 To estimate y n+1 : use : n = 1 = 71 mole n+1 = 1 = 121 mole / value to draw the OP ine, and get y n+1, or ( ) n yn 1 n y n n1 Fiqure McCabeThiele diagram for Eample 21.5 benzene-toluene distillation: dashed line y n dew point T d 93 o C ( Fig 21.3 ) Benzene : 0.7, Toluene : 0.3 H y,b = ( T 80) --- (A) H y,t = ( T 80) --- (B)

75 Calculate H y,n+1 ( at 93 o C ) H y,n+1 = 0.7 H yb,n H yt,n+1 = 0.7( ) + 0.3( ) = 7942 Eq (21.48) -- n+1 H y,n+1 = n H,n + 1 H y,1 R H, ( n = n+1 ) n+1 (7942) = ( n+1 50) (7376) H,n H y,b = ( T 80) --- (A) H y,t = ( T 80) --- (B) n+1 = mole, n = 66 mole new n, n+1 ( 1 = 121 ) ( 1 = 71 ) 0 Calculate new y n+1 by Eq ( ) mass balance Eq ( ) y n1 n n1 n n close enough to ( n, y n+1 ) = (0.5, 0.707) 75

76 Similar calculation 1 = 71 1 = 121 pick n = 0.7 y n H y,n+1 y n+1 T b H,n gives : n+1 = 118, n = 68, y n+1 = ( n, y n+1 ) = (0.7, 0.818) T d enthalpy balance n+1 n (0.818) OP line : solid line in Fig Fiqure McCabeThiele diagram for Eample 21.5 benzene-toluene distillation:, based on constant molar overflow;, based on enthalpy balance. 76

77 To estimate the OP line of stripping section, q r should be known first, use of overall enthalpy balance F H F + q r = H + B H B + q c H F : feed at boiling point ( F = 0.5, T F = 92 o C ) H H : = 0 F CP, ( T 80) ( )(92 80) 438 cal H B : T B 111 o C ( close to T b of toluene, o C ), benzene = 0.02,, toluene = (C) mole H B = ( )(111 80) = 1236 cal mole q c : 1 H y,1 = q c + ( a + )H 0 q c = = 892,496 cal From Eq(c) q r = , = 910,496 cal ( 2% error for q r q c ) 77 a H 1 H y,1 q c H

78 Estimate, b, b enthalpy balance around reboiler : q r + b H,b = b H y,b + B H B b + B H y,b : about 5% benzene at 111 o C ( Fig 21.3 ) H y,b = 0.05 H y,b (benzene) H y,b (toluene) = 0.05( ) ( ) = 9141 cal mole?? H,b : about 4% benzene at 110 o C ( Fig 21.3 ) b, H,b = 4% 110 o C H,b = ( )(110 80) = 1192 cal mole () 111 o C, y = 0.05 b, H y,b q r B, H B 111 o C, B =

79 b = b + 50, into Eq () q r + b H,b = b H y,b + B H B ( b + 50) Eq () b q r 50H H y,b,b H BH,b B (1192) 50(1236) b another method to estimate the approimate value b q H r

80 Calculate an intermediate value of m+1 Eq (21.51) m+1 H y,m+1 = m H,m + q r B H B Eq (21.52) y m1 m m1 m B B m1 Eq (21.53) m = m+1 + B pick m = 0.4 ( T b = 95 o C ) (1) first assume m = b = ; m+1 = b = ( constant molar flow ) get y m+1 = 0.55 ( from fig 21.25, dashed line ) ( T m+1 = 97 o C ) (2) calculate H,m ; H y,m+1 H,m = ( )(95 80) = 558 y m+1 = 0.55 T d = 97 o C H y,m = 0.55[ (97 80)] [ (97 80)] = 8194 H y,b H y,t 80

81 Eq (21.51) m+1 H y,m+1 = m H,m + q r B H B 8194 m+1 = ( m ) , m+1 = ; m = = ( b = 114.3) ( b = 164.3) Almost no change in and in the stripping section OP line can be drawn as straight line ( B, B ) = (0.02, 0.02), intersection of upper OP line q line No. of ideal stages : (i) 27 ideal stage, consider the variation of, (solid line in Fig 21.25) (ii) 21 ideal stage ; assume, are constant (dashed line) 81

82 No. of ideal stages : constant, (i) 27 ideal stage, consider the variation of, (solid line in Fig 21.25) (ii) 21 ideal stage ; assume, are constant (dashed line) ( n, y n+1 ) = (0.5, 0.707) ---- variation of, ( n, y n+1 ) = (0.5, 0.7) ---- constant,

83 Rectifying section : decreases 7% (mainly due to higher molar heat of vaporization of toluene) λ T > λ B Sensible heat: high small vapor : C P, liquid : C P, small high >, C P, > C P, effect canceled Stripping section :, almost no change F 50% T 50% B 1 = 71 n n-1 n = 0.5 n = 66 Be. To. 1 = 121 y n y n+1 = n+1 = 116 small small high high C P, C P, C P, < C P, Heat gain by liq. Heat released by vapor part of sensible heat of liquid should be supplied by condensation of vapor b = m = 0.4 m = y m+1 = 0.55 m+1 = b = 114.3

84 The slight upward shift of operation lines may be important when operation close to the R m ( minimum reflu ratio ). When R 2R m,the effects of operatingline curvature would be very small. for straight OP line, slope = for curved OP line, slope local value of n n 1 84

85 Reason : Eq (21.49) n+1 y n+1 = n n + n = n+1 or n+1 = n + Eq (21.49) becomes n+1 y n+1 = ( n+1 ) n + n+1 (y n+1 n ) = ( n ) --- (21.54) Eq(21.55) Eq(21.54) for stripping section n n1 or ( n + )y n+1 = n n + n (y n+1 n ) = ( y n+1 ) --- (21.55) n yn 1 (21.56), slope of line connecting (, ), ( n, y n+1 ) m m1 n 1 y m1 m B B n ( B, B ), ( m, y m+1 ) 85

86 File download (PF files): in FTP of ab. 主機位址 : 使用者名稱 : units 密碼 : units 連接埠 : 650

87 m m H,m m+1 y m+1 H y,m+1 b b B, B, q r H B