FKKSA KUKTEM BKC 3413 Chapter 7 Chapter 2

FKKSA, KUKTEM BKC 3413: Chapter 7 Chapter 2 EVAPORATION 1

FKKSA, KUKTEM BKC 3413: Chapter 7 Content • Type of Evaporation equipment and Methods • Overall Heat Transfer Coefficient in Evaporators • Calculation Methods for Single Effect Evaporators • Calculation Methods for Multiple Effects Evaporators • Condenser for Evaporator • Evaporation using Vapor Recompression 2

BKC 3413: Chapter 7 FKKSA, KUKTEM Evaporation • Heat is added to a solution to vaporize the solvent, which is usually water. • Case of heat transfer to a boiling liquid. • Vapor from a boiling liquid solution is removed and a more concentrated solution remains. • Refers to the removal of water from an aqueous solution. • Example: concentration of aqueous solutions of sugar. In these cases the crystal is the desired product and the evaporated water is discarded. 3

FKKSA, KUKTEM BKC 3413: Chapter 7 Materials of construction Pressure and temperature Foaming or frothing Scale deposition Processing Factors solubility Temperature sensitivity of materials Concentration in the liquid 4

BKC 3413: Chapter 7 FKKSA, KUKTEM Processing Factors • Concentration dilute feed, viscosity , heat transfer coefficient, h concentrated solution/products, , and h . • Solubility concentration , solubility , crystal formed. solubility with temperature . • Temperature. heat sensitive material degrade at higher temperature & prolonged heating. 5

BKC 3413: Chapter 7 FKKSA, KUKTEM • Foaming/frothing. caustic solutions, food solutions, fatty acid solutions form foam/froth during boiling. entrainment loss as foam accompany vapor. • Pressure and Temperature pressure , boiling point . concentration , boiling point. heat-sensitive material operate under vacuum. • Material of construction minimize corrosion. 6

FKKSA, KUKTEM BKC 3413: Chapter 7 Effect of Processing Variables on Evaporator Operation. • TF TF < Tbp, some of latent heat of steam will be used to heat up the cold feed, only the rest of the latent heat of steam will be used to vaporize the feed. Is the feed is under pressure & TF > Tbp, additional vaporization obtained by flashing of feed. • P 1 desirable T [Q = UA(TS – T 1)], A & cost . T 1 depends on P 1 will T 1. 7

BKC 3413: Chapter 7 FKKSA, KUKTEM • PS will TS but high-pressure is costly. optimum TS by overall economic balances. • BPR The concentration of the solution are high enough so that the c. P and Tbp are quite different from water. BPR can be predict from Duhring chart for each solution such as Na. OH and sugar solution. • Enthalpy–concentration of solution. for large heat of solution of the aqueous solution. to get values for h. F and h. L. 8

BKC 3413: Chapter 7 FKKSA, KUKTEM 9

BKC 3413: Chapter 7 FKKSA, KUKTEM 10

FKKSA, KUKTEM BKC 3413: Chapter 7 vapor, V to condenser T 1 , y. V , HV feed, F TF , x. F , h. F. steam, S TS , HS P 1 T 1 heat-exchanger tubes condensate, S T S , h. S concentrated liquid, L T 1 , x. L , h. L Simplified Diagram of single-effect evaporator 11

BKC 3413: Chapter 7 FKKSA, KUKTEM • Single-effect evaporators; • the feed (usually dilute) enters at TF and saturated steam at TS enters the heat-exchange section. • condensed leaves as condensate or drips. • the solution in the evaporator is assumed to be completely mixed and have the same composition at T 1. • the pressure is P 1, which is the vapor pressure of the solution at T 1. • wasteful of energy since the latent heat of the vapor leaving is not used but is discarded. • are often used when the required capacity of operation is relatively small, but it will wasteful of steam cost. 12

BKC 3413: Chapter 7 FKKSA, KUKTEM Calculation Methods for Single-effect Evaporator. • Objectives: to calculate vapor, V and liquid, L flowrates. heat transfer area, A overall heat-transfer coefficient, U. Fraction of solid content, x. L. • To calculate V & L and x. L, solve simultaneously total material balance & solute/solid balance. F=L+V total material balance F (x. F) = L (x. L) solute/solid balance 13

BKC 3413: Chapter 7 FKKSA, KUKTEM • To calculate A or U, no boiling point rise and negligible heat of solution: calculate h. F, h. L, Hv and . where, = (HS – Hs) h = c. P(T – Tref) where, Tref = T 1 = (as datum) c. PF = heat capacity (dilute as water) HV = latent heat at T 1 solve for S: F h. F + S = L h L + V H V solve for A and U: q = S = U A T = UA (TS – T 1) 14

BKC 3413: Chapter 7 FKKSA, KUKTEM • To get BPR and the heat of solution: calculate T 1 = Tsat + BPR get h. F and h. L from Figure 8. 4 -3. get S & HV from steam tables for superheated vapor or HV = Hsat + 1. 884 (BPR) solve for S: F h. F + S = L h L + V H V solve for A and U: q = S = U A T = UA (TS – T 1) 15

BKC 3413: Chapter 7 FKKSA, KUKTEM Example 8. 4 -1: Heat-Transfer Area in Single-Effect Evaporator. A continuous single-effect evaporator concentrates 9072 kg/h of a 1. 0 wt % salt solution entering at 311. 0 K (37. 8 ºC) to a final concentration of 1. 5 wt %. The vapor space of the evaporator is at 101. 325 k. Pa (1. 0 atm abs) and the steam supplied is saturated at 143. 3 k. Pa. The overall coefficient U = 1704 W/m 2. K. calculate the amounts of vapor and liquid product and the heattransfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water. 16

BKC 3413: Chapter 7 F = 9072 kg/h TF = 311 K x. F = 0. 01 h. F. P 1 = 101. 325 k. Pa U = 1704 W/m 2 T 1 A=? S , TS , HS PS = 143. 3 k. Pa L=? T 1 , h. L x. L = 0. 015 FKKSA, KUKTEM V=? T 1 , y. V , HV S, TS , h. S Figure 8. 4 -1: Flow Diagram for Example 8. 4 -1 17

BKC 3413: Chapter 7 FKKSA, KUKTEM Solution; Refer to Fig. 8. 4 -1 for flow diagram for this solution. For the total balance, F=L+V 9072 = L + V For the balance on the solute alone, F x F = L x. L 9072 (0. 01) = L (0. 015) L = 6048 kg/h of liquid Substituting into total balance and solving, V = 3024 kg/h of vapor 18

BKC 3413: Chapter 7 FKKSA, KUKTEM Since we assumed the solution is dilute as water; cp. F = 4. 14 k. J/kg. K (Table A. 2 -5) From steam table, (A. 2 -9) At P 1 = 101. 325 k. Pa, T 1 = 373. 2 K (100 ºC). HV = 2257 k. J/kg. At PS = 143. 3 k. Pa, TS = 383. 2 K (110 ºC). = 2230 k. J/kg. The enthalpy of the feed can be calculated from, h. F = cp. F (TF – T 1) h. F = 4. 14 (311. 0 – 372. 2) = -257. 508 k. J/kg. 19

BKC 3413: Chapter 7 FKKSA, KUKTEM Substituting into heat balance equation; F h. F + S = L h L + V H V with h. L = 0, since it is at datum of 373. 2 K. 9072 (-257. 508) + S (2230) = 6048 (0) + 3024 (2257) S = 4108 kg steam /h The heat q transferred through the heating surface area, A is q = S ( ) q = 4108 (2230) (1000 / 3600) = 2 544 000 W Solving for capacity single-effect evaporator equation; q = U A T = U A (TS – T 1) 2 544 000 = 1704 A (383. 2 – 373. 2) Solving, A = 149. 3 m 2. 20

BKC 3413: Chapter 7 FKKSA, KUKTEM Example 8. 4 -3: Evaporation of an Na. OH Solution. An evaporator is used to concentrate 4536 kg/h of a 20 % solution of Na. OH in water entering at 60 ºC to a product of 50 % solid. The pressure of the saturated steam used is 172. 4 k. Pa and the pressure in the vapor space of the evaporator is 11. 7 k. Pa. The overall heat-transfer coefficient is 1560 W/m 2. K. calculate the steam used, the steam economy in kg vaporized/kg steam used, and the heating surface area in m 2. 21

FKKSA, KUKTEM BKC 3413: Chapter 7 F = 4536 kg/h TF = 60 ºC x. F = 0. 2 P 1 = 11. 7 k. Pa h. F. V, T 1 , y. V , HV U = 1560 W/m 2 T 1 A=? S=? TS , HS PS = 172. 4 k. Pa L, T 1 , h. L x. L = 0. 5 S, TS , h. S Figure 8. 4 -4: Flow Diagram for Example 8. 4 -3 22

BKC 3413: Chapter 7 FKKSA, KUKTEM Solution, Refer to Fig. 8. 4 -4, for flow diagram for this solution. For the total balance, F = 4536 = L + V For the balance on the solute alone, F x F = L x. L 4536 (0. 2) = L (0. 5) L = 1814 kg/h of liquid Substituting into total balance and solving, V = 2722 kg/h of vapor 23

BKC 3413: Chapter 7 FKKSA, KUKTEM To determine T 1 = Tsat + BPR of the 50 % concentrate product, first we obtain Tsat of pure water from steam table. At 11. 7 k. Pa, Tsat = 48. 9 ºC. From Duhring chart (Fig. 8. 4 -2), for a Tsat = 48. 9 ºC and 50 % Na. OH , the boiling point of the solution is T 1 = 89. 5 ºC. hence, BPR = T 1 - Tsat = 89. 5 -48. 9 = 40. 6 ºC From the enthalpy-concentration chart (Fig. 8. 4 -3), for TF = 60 ºC and x. F = 0. 2 get h. F = 214 k. J/kg. T 1 = 89. 5 ºC and x. L = 0. 5 get h. L = 505 k. J/kg. 24

BKC 3413: Chapter 7 FKKSA, KUKTEM For saturated steam at 172. 4 k. Pa, from steam table, we get TS = 115. 6 ºC and = 2214 k. J/kg. To get HV for superheated vapor, first we obtain the enthalpy at Tsat = 48. 9 ºC and P 1 = 11. 7 k. Pa, get Hsat = 2590 k. J/kg. Then using heat capacity of 1. 884 k. J/kg. K for superheated steam. So HV = Hsat + c. P BPR = 2590 + 1. 884 (40. 6) = 2667 k. J/kg. Substituting into heat balance equation and solving for S, F h. F + S = L h L + V H V 4535 (214) + S (2214) = 1814 (505) + 2722 (2667) S = 3255 kg steam /h. 25

BKC 3413: Chapter 7 FKKSA, KUKTEM The heat q transferred through the heating surface area, A is q = S ( ) q = 3255 (2214) (1000 / 3600) = 2 000 W Solving for capacity single-effect evaporator equation; q = U A T = U A (TS – T 1) 2 000 = 1560 A (115. 6 – 89. 5) Solving, A = 49. 2 m 2. Steam economy = 2722/3255 = 0. 836 26

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION vapor T 1 feed, TF (1) T 1 vapor T 2 (2) T 2 steam, TS vapor T 3 (3) to vacuum condenser T 3 condensate concentrate from first effect. concentrate from second effect. concentrated product Simplified diagram of forward-feed triple-effect evaporator 27

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION • Forward-feed multiple/triple-effect evaporators; - the fresh feed is added to the first effect and flows to the next in the same direction as the vapor flow. operated when the feed hot or when the final concentrated product might be damaged at high temperature. at steady-state operation, the flowrates and the rate of evaporation in each effect are constant. the latent heat from first effect can be recovered and reuse. The steam economy , and reduce steam cost. the Tbp from effect to effect, cause P 1 . 28

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Calculation Methods for Multiple-effect Evaporators. • Objective to calculate; temperature drops and the heat capacity of evaporator. the area of heating surface and amount of vapor leaving the last effect. • Assumption made in operation; no boiling point rise. no heat of solution. neglecting the sensible heat necessary to heat the feed to the boiling point. 29

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION • Heat balances for multiple/triple-effect evaporator. the amount of heat transferred in the first effect is approximately same with amount of heat in the second effect, q = U 1 A 1 T 1 = U 2 A 2 T 2 = U 3 A 3 T 3 - usually in commercial practice the areas in all effects are equal, q/A = U 1 T 1 = U 2 T 2 = U 3 T 3 - to calculate the temperature drops in evaporator, T = T 1 + T 2 + T 3 = T S – T 3 30

BKC 3413: Chapter 7 FKKSA, KUKTEM - hence we know that T are approximately inversely proportional to the values of U, - similar eq. can be written for T 2 & T 3 - if we assumed that the value of U is the same in each effect, the capacity equation, q = U A ( T 1 + T 2 + T 3 ) = UA T 31

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION vapor T 1 (1) steam, TS T 1 vapor T 2 (2) T 2 vapor T 3 (3) T 3 to vacuum condenser feed, TF condensate concentrated product Simplified diagram of backward-feed triple-effect evaporator 32

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION • Backward-feed multiple/triple-effect evaporators; fresh feed enters the last and coldest effect and continues on until the concentrated product leaves the first effect. advantageous when the fresh feed is cold or when concentrated product is highly viscous. working a liquid pump since the flow is from low to high pressure. the high temperature in the first effect reduce the viscosity and give reasonable heat-transfer coefficient. 33

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Step-by-step Calculation Method for Triple-effect Evaporator (Forward Feed) For the given x 3 and P 3 and BPR 3 From an overall material balance, determine VT = V 1 + V 2 + V 3 (1 st trial – assumption) Calculate the amount of concentrated solutions & their concentrations in each effect using material balances. Find BPR & T in each effect & T. If the feed is very cold, the portions may be modified appropriately, calculate the boiling point in each effect. Calculate the amount vaporized and concentrated liquid in each effect through energy & material balances. If the amounts differ significantly from the assumed values in step 2, and 4 must be repeated with the amounts just calculated. Using heat transfer equations for each effect, calculate the surface required for each effect If the surfaces calculated are not equal, revise the TS. Repeat step 4 onward until the areas are distributed satisfactorily. 34

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Ex. 8. 5 -1 : Evaporation of Sugar Solution in a Triple-Effect Evaporator. A triple-effect forward-feed evaporator is being used to evaporate a sugar solution containing 10 wt% solids to a concentrated solution of 50 %. The boiling-point rise of the solutions (independent of pressure) can be estimated from (BPR ºC = 1. 78 x + 6. 22 x 2 ), where x is wt fraction of sugar in solution. Saturated steam at 205. 5 k. Pa and 121. 1ºC saturation temperature is being used. The pressure in the vapor space of the third effect is 13. 4 k. Pa. The feed rate is 22 680 kg/h at 26. 7 ºC. the heat capacity of the liquid solutions is c. P = 4. 19 – 2. 35 x k. J/kg. K. The heat of solution is considered to be negligible. The coefficients of heat transfer have been estimated as U 1 = 3123, U 2 = 1987, and U 3 = 1136 W/m 2. K. If each effect has the same surface area, calculate the area, the steam rate used, and the steam economy. 35

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION V 1 = 22, 680 – L 1 F = 22680 T 1 x. F = 0. 1 TF = 26. 7 ºC T 3 V 3 = L 2 - 4536 T 2 P 3 = 13. 7 k. Pa (1) S =? TS 1 = 121. 1 ºC PS 1 = 205. 5 k. Pa V 2 = L 1 – L 2 (3) (2) TS 1 T 1 , L 1 , x 1 TS 2 TS 3 T 2 , L 2 , x 2 T 3 L 3 = 4536 x 3 = 0. 5 Fig. 8. 5 -1: Flow diagram for example 8. 5 -1 36

BKC 3413: Chapter 7 FKKSA, KUKTEM Solution, The process flow diagram is given in Fig. 8. 5 -1. . Step 1, From steam table, at P 3 = 13. 4 k. Pa, get Tsat = 51. 67 ºC. Using the BPR equation for third effect with x. L = 0. 5, BPR 3 = 1. 78 (0. 5) + 6. 22 (0. 52) =2. 45 ºC. T 3 = 51. 67 + 2. 45 = 54. 12 ºC. (BPR = T – Ts) Step 2, Making an overall and a solids balance. F = 22 680 = L 3 + (V 1 + V 2 + V 3) Fx. F = 22 680 (0. 1) = L 3 (0. 5) + (V 1 + V 2 + V 3) (0) L 3 = 4536 kg/h Total vaporized = (V 1 + V 2 + V 3) = 18 144 kg/h 37

BKC 3413: Chapter 7 FKKSA, KUKTEM Assuming equal amount vaporized in each effect, V 1 = V 2 = V 3 = 18 144 / 3 = 6048 kg/h Making a total material balance on effects 1, 2, and 3, solving F = 22 680 = V 1 + L 1 = 6048 + L 1, L 1 = 16 632 kg/h. L 1 = 16 632 = V 2 + L 2 = 6048 + L 2, L 2 = 10 584 kg/h. L 2 = 10 584 = V 3 + L 3 = 6048 + L 3, L 3 = 4536 kg/h. Making a solids balance on each effect, and solving for x, 22 680 (0. 1) = L 1 x 1 = 16 632 (x 1), x 1 = 0. 136 16 632 (0. 136) = L 2 x 2 = 10 584 (x 2), x 2 = 0. 214 10 584 (0. 214) = L 3 x 3 = 4536 (x 3), x 3 = 0. 5 (check) 38

FKKSA, KUKTEM BKC 3413: Chapter 7 Step 3, The BPR in each effect is calculated as follows: BPR 1 = 1. 78 x 1 + 6. 22 x 12 = 1. 78(0. 136) + 6. 22(0. 136)2 = 0. 36ºC. BPR 2 = 1. 78(0. 214) + 6. 22(0. 214)2 =0. 65ºC. BPR 3 = 1. 78(0. 5) + 6. 22(0. 5)2 =2. 45ºC. then, T available = TS 1 – T 3 (sat) – (BPR 1 + BPR 2 + BPR 3 ) = 121. 1 – 51. 67 – (0. 36+0. 65+2. 45) = 65. 97ºC. Using Eq. (8. 5 -6) for T 1 , T 2 , and T 3 T 1 = 12. 40 ºC T 2 = 19. 50 ºC T 3 = 34. 07 ºC 39

BKC 3413: Chapter 7 FKKSA, KUKTEM However, since a cold feed enters effect number 1, this effect requires more heat. Increasing T 1 and lowering T 2 and T 3 proportionately as a first estimate, so T 1 = 15. 56ºC T 2 = 18. 34 ºC T 3 = 32. 07 ºC The above data T 1, T 2 and T 3 are getting from iteration-s To calculate the actual boiling point of the solution in each effect, T 1 = TS 1 - T 1 = 121. 1 – 15. 56 = 105. 54 ºC. T 2 = T 1 - BPR 1 - T 2 = 105. 54 – 0. 36 – 18. 34 = 86. 84 ºC. TS 2 = T 1 –BPR 1 = 105. 54 – 0. 36 = 105. 18 ºC. T 3 = T 2 - BPR 2 - T 3= 86. 84 – 0. 65 – 32. 07 = 54. 12 ºC. TS 3 = T 2 –BPR 2 = 86. 84 – 0. 65 = 86. 19 ºC. 40

FKKSA, KUKTEM BKC 3413: Chapter 7 The temperatures in the three effects are as follows: Effect 1 Effect 2 Effect 3 Condenser TS 1 = 121. 1ºC TS 2 = 105. 18 TS 3 = 86. 19 TS 4 = 51. 67 T 1 = 105. 54 T 2 = 86. 84 T 3 = 54. 12 Step 4, The heat capacity of the liquid in each effect is calculated from the equation c. P = 4. 19 – 2. 35 x. F: L 1: L 2: L 3: c. PF c. P 1 c. P 2 c. P 3 = 4. 19 – 2. 35 (0. 1) = 4. 19 – 2. 35 (0. 136) = 4. 19 – 2. 35 (0. 214) = 4. 19 – 2. 35 (0. 5) = 3. 955 k. J/kg. K = 3. 869 k. J/kg. K = 3. 684 k. J/kg. K = 3. 015 k. J/kg. K 41

BKC 3413: Chapter 7 FKKSA, KUKTEM The values of the enthalpy H of the various vapor streams relative to water at 0 ºC as a datum are obtained from the steam table as follows: Effect 1: H 1 = HS 2 + 1. 884 BPR 1 = 2684 + 1. 884(0. 36) 2685 k. J/kg. S 1 = HS 1 – h. S 1 = 2708 – 508 = 2200 k. J/kg. Effect 2: H 2 = HS 3 + 1. 884 BPR 2= 2654 + 1. 884(0. 65) = 2655 k. J/kg. S 2 = H 1 – h. S 2 = 2685 – 441 = 2244 k. J/kg. Effect 3: H 3 = HS 4 + 1. 884 BPR 3 = 2595 + 1. 884(2. 45) = 2600 k. J/kg. S 3 = H 2 – h. S 3 = 2655– 361 = 2294 k. J/kg. 42

FKKSA, KUKTEM BKC 3413: Chapter 7 Write the heat balance on each effect. Use 0ºC as a datum. Fc. PF (TF – 0) + S S 1 = L 1 c. P 1 (T 1 – 0) + V 1 H 1 , , ………(1) 22680(3. 955)(26. 7 -0)+2200 S = 3. 869 L 1(105. 54 -0)+(22680 -L 1)2685 L 1 c. P 1 (T 1 – 0) + V 1 S 2 = L 2 c. P 2 (T 2 – 0) + V 2 H 2 ………(2) L 2 c. P 2 (T 2 – 0) + V 2 S 3 = L 3 c. P 3 (T 3 – 0) + V 3 H 3 ………(3) 3. 869 L 1(105. 54 -0)+(22680 -L 1)2244=3. 684 L 2(86. 84 -0)+(L 1 -L 2)2655 3. 68 L 2(86. 84 -0)+(L 1 -L 2)2294=4536(3. 015)(54. 1 -0)+(L 2 -4536)2600 Solving (2) and (3) simultaneously for L 1&L 2 and substituting into(1) L 1 = 17078 kg/h S = 8936 kg/h V 3 = 6532 kg/h L 2 = 11068 kg/h L 3 = 4536 kg/h V 1 = 5602 kg/h V 2 = 6010 kg/h 43

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Step 5, Solving for the values of q in each effect and area, 44

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Am = 104. 4 m 2, the areas differ from the average value by less than 10 % and a second trial is really not necessary. However, a second trial will be made starting with step 6 to indicate the calculation methods used. Step 6, Making a new solids balance by using the new L 1 = 17078, L 2 = 11068, and L 3 = 4536, and solving for x, 22 680 (0. 1) = L 1 x 1 = 17 078 (x 1), x 1 = 0. 133 17 078 (0. 130) = L 2 x 2 = 11 068 (x 2), x 2 = 0. 205 11 068 (0. 205) = L 3 x 3 = 4536 (x 3), x 3 = 0. 5 (check) 45

FKKSA, KUKTEM BKC 3413: Chapter 7 EVAPORATION Step 7. The new BPR in each effect is then, BPR 1 = 1. 78(0. 133) + 6. 22(0. 13)2 =0. 35ºC. BPR 2 = 1. 78(0. 205) + 6. 22(0. 205)2 =0. 63ºC. BPR 3 = 1. 78(0. 5) + 6. 22(0. 5)2 =2. 45ºC. then, T available = 121. 1 – 51. 67 – (0. 35+0. 63+2. 45) = 66. 0 ºC. The new T are obtained using Eq. (8. 5 -11), 46

BKC 3413: Chapter 7 FKKSA, KUKTEM These T’ values are readjusted so that T 1`= 16. 77, T 2`= 16. 87, T 3` = 32. 36, and T = 66. 0 ºC. To calculate the actual boiling point of the solution in each effect, (1) T 1 = TS 1 + T 1` = 121. 1 – 16. 77 = 104. 33ºC (2) T 2 = T 1 – BPR 1 - T 2` = 104. 33 – 0. 35 – 16. 87 = 87. 11 ºC TS 2 = T 1 – BPR 1 = 104. 33 – 0. 35 = 103. 98ºC (3) T 3 = T 2 – BPR 2 - T 3` = 87. 11 – 0. 63 – 32. 36 = 54. 12 ºC TS 3 = T 2 – BPR 2 = 87. 11 – 0. 63 = 86. 48 ºC. Step 8; Following step 4 to get c. P = 4. 19 – 2. 35 x, F: c. PF = 4. 19 – 2. 35 (0. 1) = 3. 955 k. J/kg. K L 1 : c. P 1 = 4. 19 – 2. 35 (0. 133) = 3. 877 k. J/kg. K L 2 : c. P 2 = 4. 19 – 2. 35 (0. 205) = 3. 705 k. J/kg. K L 3 : c. P 3 = 4. 19 – 2. 35 (0. 5) = 3. 015 k. J/kg. K 47

FKKSA, KUKTEM BKC 3413: Chapter 7 Then the new values of the enthalpy are, (1) H 1 = HS 2 + 1. 884 BPR 1 = 2682 + 1. 884(0. 35) = 2683 k. J/kg. S 1 = HS 1 – h. S 1 = 2708 – 508 = 2200 k. J/kg. (2) H 2 = HS 3 + 1. 884 BPR 2 = 2654 + 1. 884(0. 63) = 2655 k. J/kg. S 2 = H 1 – h. S 2 = 2683 – 440 = 2243 k. J/kg. (3) H 3 = HS 4 + 1. 884 BPR 3 = 2595 + 1. 884(2. 45) = 2600 k. J/kg. S 3 = H 2 – h. S 3 = 2655– 362 = 2293 k. J/kg. Writing a heat balance on each effect, and solving, (1) 22680(3. 955)(26. 7 -0)+2200 S = 3. 877 L 1(104. 33 -0)+(22680 -L 1)2683 (2) 3. 877 L 1(104. 33 -0)+(22680 -L 1)2243=3. 708 L 2(87. 11 -0)+(L 1 -L 2)2655 (3) 3. 708 L 2(87. 11 -0)+(L 1 -L 2)2293=4536(3. 015)(54. 1 -0)+(L 2 -4536)2600 L 1 = 17005 kg/h L 2 = 10952 L 3 = 4536 S = 8960 V 1 = 5675 V 2 = 6053 V 3 = 6416 48