BKF3463: EVAPORATIONFKKSA, UMP1BKF3463: UNIT OPERATION 1Chapter 8:EVAPORATION2CONTENT1. Type oI Evaporation equipment and Methods2. Overall Heat TransIer CoeIIicient in Evaporators3. Calculation Methods Ior Single EIIect Evaporators4. Calculation Methods Ior Multiple EIIects Evaporators5. Condenser Ior Evaporator6. Evaporation using Vapor RecompressionTopic Outcomes Define Evaporation process Describe and determine the processing factors that involved in evaporation process identify the types of evaporators and distinguish the suitable evaporator for certain operations Describe other specific types of evaporators that are used in certain industries 3

EVAPORATONHeat is added to a solution to vaporize the solvent, which is usually water.Case oI heat transIer to a boiling liquid.Vapor Irom a boiling liquid solution is removed and a more concentrated solution remains.ReIers to the removal oI water Irom an aqueous solution.Example: concentration oI aqueous solutions oI sugar. In these cases the crystal is the desired product and the evaporated water is discarded.

Processing FactorsConcentration in the liquidsolubilitv1emperature sensitivitvof materialsFoaming or frothingPressure and temperatureScale depositionMaterials of construction

PROCESSNG FACTORS(1) Concentrationdilute Ieed, viscosity 3G, heat transIer coeIIicient, Tconcentrated solution/products, 3T, andG.(2) Solubilitvconcentration T, solubility G , crystal Iormed.solubility T with temperature T. (3) Temperature.heat sensitive material degrade at higher temperature & prolonged heating.

(4) Foaming/froting.caustic solutions, Iood solutions, Iatty acid solutions Iorm Ioam/Iroth during boiling.entrainment loss as Ioam accompany vapor.(5) Pressure and Temperaturepressure T, boiling point T.concentration T, boiling pointT.heat-sensitive material operate under vacuum.(6) Material of constructionminimize corrosion.BKF3463: EVAPORATIONFKKSA, UMP

ITRAYGrowing foam slows/ stops liquid from flowing downIPACKINGFoam effects less severe. but can cause some channelingGrowing foam leaves towerSour gasSour gasRich soln Rich solnLean solnLean solnSweet gas Sweet gasForming Effects

Type of Evaporation Equipment Horizontal tube type Vertical tube type Long tube vertical type Forced-circulation type Open kettle or pan Open-pan solar evaporator Falling-film-type evaporator Agitated-film evaporatorTo help protect your privacy, PowerPoint prevented this external picture from being automatically downloaded. To download and display this picture, click Options in the Nessage Bar, and then click Enable external content.10Wiped fiIm evaporatorOperating parameters Operating windowFeed rate kg/h 20 - 100.000Evaporation rate kg/h up to 40.000Heatingtemperature H up to 380 1)Pressure on process side barg -1 to 30Product viscosity at operating temperature mPas up to 70.000Residence time min < 1 2)Evaporation ratio (concentrate/feed) up to 1:50 3)Through put % 20 - 100 3)1)evaporators with electrical inductive heating up to 600 H2)in horizontal evaporators longer residence times possible3)depending on evaporator type11Effect of Processing Variables on Evaporator Operation(1) TFTF < 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. s the feed is under pressure & TF > Tbp, additional vaporization obtained by flashing of feed.(2) P1desirable AT T [Q = UA(TS T1)]A G & cost G.T1 depends on P1 will G T1.12(3) PST PS will T TS but high-pressure is costly.optimum TS by overall economic balances.(4) BPRThe concentration of the solution are high enough so thatthe cP and Tbp are quite different from water.BPR can be predict from Duhring chart for each solution such as NaOH and sugar solution.(5) Enthalpyconcentration of solution.for large heat of solution of the aqueous solution.to get values for hF and hL. 1311feed, F1F , xF , hF.steam, S1S , HSconcentrated liquid, L11 , xL , hLcondensate, S1S , hSvapor,Jto condenser11 , vJ , HJP1 11heat-exchangertubesSimplified Diagram of single-effect evaporator1steam, 1Sfeed, 1Fconcentratefrom firsteffect.vapor 11(1) 11(2) 12(3) 13concentratefrom secondeffect.concentratedproductcondensatevapor 12vapor 13to vacuum condenserSimplified diagram of forward-feed triple-effect evaporator1Simplified diagram of backward-feed triple-effect evaporatorsteam, 1Sfeed, 1Fvapor 11(1) 11(2) 12(3) 13concentratedproductcondensatevapor 12vapor 13to vacuum condenser1 The feed (usually dilute) enters at TFand saturated steam at TSenters 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 T1. The pressure is P1, which is the vapor pressure of the solution at T1. asteful 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.SNGLE EFFECT EVAPORATORS1CALCULATIONS a) vapor, V and liquid, L flowrates.b) heat transfer area, Ac) overall heat-transfer coefficient, U.d) Fraction of solid content, xL. To caIcuIate V & L and xL,- solve simultaneously total material balance & solute/solid balance.F = L + V total material balanceF (xF) = L (xL) solute/solid balanceCALCULATON METHODS FOR SNGLE-EFFECT EVAPORAT20 To caIcuIate A or U,- No boiling point rise and negligible heat of solution:calculate hF, hL, Hv and .where, = (HS hs)h = cP(T Tref)where, Tref = T1 = (as datum)cPF = heat capacity (dilute as water)HV= latent heat at T1solve for S:F hF + S = L hL + V HVsolve for A and U:q = S = U A AT = UA (TS T1)21 To get BPR and the heat of soIution:1) Calculate T1 = Tsat BPR2) Get hF and hL from Figure .-3.3) Get S & HV from steam tables for superheated vapor orHV = Hsat 1. (BPR)) Solve for S:F hF + S = L hL + V HV) Solve for A and U:q = S = U A AT = UA (TS T1)22A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt solution entering at 311.0 K (37.8 C) to a final concentration of 1.5 wt %. The vapor space of the evaporator is at 101.325 kPa (1.0 atm abs) and the steam supplied is saturated at 143.3 kPa. The overall coefficientU = 1704 W/m2.K. Calculate the amounts of vapor and liquid product and the heat-transfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water.xample 8.4-1: Heat-1ransfer Area in Single-ffect vaporator23& 174 W/m211A ?P1 11.325 kPaF 972 kg/h1F 311 KxF .1hF.S , 1S , HS PS 143.3 kPaL ?11 , hL xL .15 S, 1S , hSJ ?11 , vJ , HJFigure 8.4-1: Flow Diagram for xample 8.4-12$4luti43Refer to Fig. 8.4-1 for flow diagram for this solution.For the total balance,F = L + V9072 = L + VFor the balance on the solute alone,F xF = L xL9072 (0.01) = L (0.015)L = 6048 kg/h 41 liquidSubstituting into total balance and solving,V = 3024 kg/h 41 vap472Since we assumed the solution is dilute as water;cpF = 4.14 kJ/kg. K From steam table, (A.2-9)At P1 = 101.325 kPa, T1 = 373.2 K (100 C). HV = 2257 kJ/kg.At PS = 143.3 kPa, TS = 383.2 K (110 C). = 2230 kJ/kg.The enthalpy of the feed can be calculated from,hF = cpF (TF T1)hF = 4.14 (311.0 373.2)= -257.508 kJ/kg.2Substituting into heat balance equation;F hF + S = L hL + V HVwith hL = 0, since it is at datum of 373.2 K.9072 (-257.508) + S (2230) = 6048 (0) + 3024 (2257)S = 4108 kg steam /hThe heat q transferred through the heating surface area, A isq = S ()q = 4108 (2230) (1000 / 3600) = 2 544 000 WSolving for capacity single-effect evaporator equation;q = U A AT = U A (TS T1)2 544 000 = 1704 A (383.2 373.2)Solving, A = 149.3 m2.2xample 8.4-3: vaporation of an AaOH Solution.An evaporator is used to concentrate 4536 kg/h of a 20 % solution of NaOH in water entering at 60 C to a product of 50 % solid. The pressure of the saturated steam used is 172.4 kPa and the pressure in the vapor space of the evaporator is 11.7 kPa. The overall heat-transfer coefficient is 1560 W/m2.K.Calculate the steam used, the steam economy in (kg vaporized/kg steam)used, and the heating surface area in m2.2& 15 W/m211A ?P1 11.7 kPaF 453 kg/h1F CxF .2hF.S ?1S , HS PS 172.4 kPaL, 11 , hL xL .5 S, 1S , hSJ, 11 , vJ , HJFigure 8.4-4: Flow Diagram for xample 8.4-330$4luti43Refer to Fig. 8.4-4, for flow diagram for this solution.For the total balance,F = 4536=L + VFor the balance on the solute alone,F xF = L xL4536 (0.2) = L (0.5)L = 1814 kg/h 41 liquidSubstituting into total balance and solving,V = 2722 kg/h 41 vap4731To determine T1 = Tsat + BPRof the 50 % concentrate product, first we obtainTsat of pure water from steam table. At 11.7 kPa, Tsat = 48.9 C. From Duhring chart (Fig. 8.4-2), for a Tsat = 48.9 C and 50 % NaOH , the boiling point of the solution is T1 = 89.5 C. hence, BPR = T1 - Tsat = 89.5-48.9= 40.6 C From the enthalpy-concentration chart (Fig.8.4-3), forTF = 60 C and xF= 0.2 get hF= 214 kJ/kg.T1 = 89.5 Cand xL= 0.5 get hL= 505 kJ/kg.32For saturated steam at 172.4 kPa, from steam table, we getTS = 115.6 C and = 2214 kJ/kg.To get HV for superheated vapor, first we obtain the enthalpy at Tsat = 48.9 C and P1 = 11.7 kPa, get Hsat = 2590 kJ/kg.Then using heat capacity of 1.884 kJ/kg.K for superheated steam. So HV = Hsat + cP BPR= 2590 + 1.884 (40.6)=2667 kJ/kg.Substituting into heat balance equation and solving for S,F hF + S = L hL + V HV4535 (214) + S (2214) = 1814 (505) + 2722 (2667)$ = 3255 kg steam /h.33The heat q transferred through the heating surface area, A isq = S ()q = 3255 (2214) (1000 / 3600) = 2 002 000 WSolving for capacity single-effect evaporator equation;q = U A AT = U A (TS T1)2 002 000 = 1560 A (115.6 89.5)Solving, A = 49.2 m2.$team ec434my= 2722/3255 = 0.83631(/)O)/I1O^steam, 1Sfeed, 1Fconcentratefrom firsteffect.vapor 11(1) 11(2) 12(3) 13concentratefrom secondeffect.concentratedproductcondensatevapor 12vapor 13to vacuum condenserSimplified diagram of forward-feed triple-effect evaporator31(/)O)/I1O^ Forward-feed multiple/triple-effect evaporators;- the fresh feed is added to the first effect and flows tothe next in the same direction as the vapor flow.- operated when the feed hot or when the finalconcentrated product might be damaged at hightemperature.- 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 andreuse. The steam economy T, and reduce steam cost.- the Tbp G from effect to effect, cause P1 G.31(/)O)/I1O^Calculati43 Meth4ds 147 Multiple-e11ect Evap47at47s. 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 heatthe feed to the boiling point.31(/)O)/I1O^ Heat balances for multiple/triple-effect evaporator.- the amount of heat transferred in the first effect is approximately same with amount of heat in thesecond effect,q = U1 A1 AT1 = U2 A2 AT2 = U3 A3 AT3- usually in commercial practice the areas in alleffects are equal,q/A = U1 AT1 = U2AT2 = U3AT3- to calculate the temperature drops in evaporator,L AT = AT1 +AT2 +AT3 = TS T33- hence we know that L AT are approximatelyinversely proportional to the values of U,- similar eq. can be written for AT2 & AT3- if we assumed that the value of U is thesame in each effect, the capacity equation,q = U A (AT1 +AT2 +AT3 ) = UA L AT3 2 1111 1 11& & &&T T+ +LA = A31(/)O)/I1O^Simplified diagram of backward-feed triple-effect evaporatorsteam, 1Sfeed, 1Fvapor 11(1) 11(2) 12(3) 13concentratedproductcondensatevapor 12vapor 13to vacuum condenser01(/)O)/I1O^ Backward-feed multiple/triple-effect evaporators;- fresh feed enters the last and coldest effect andcontinues on until the concentrated productleaves the first effect.- advantageous when the fresh feed is cold orwhen concentrated product is highly viscous.- working a liquid pump since the flow is from lowto high pressure.- the high temperature in the first effect reduce the viscosity and give reasonable heat-transfer coefficient.1$tep-by-step Calculation Method for Triple-effect Evaporator (Forward Feed)or the given x3 and P3 and BPR3rom an overall material balance, determine VT V1 V2 V3(1sttrialassumption)Calculate the amount oI concentrated solutions & their concentrations in each eIIect using material balances.ind BPR & AT in each eIIect & LAT.II the Ieed is very cold, the portions may be modiIied appropriately, calculate the boiling point in each eIIect.Calculate the amount vaporized and concentrated liquid in each eIIect through energy & material balances.II the amounts diIIer signiIicantly Irom the assumed values in step 2, step 2, and 4 must be repeated with the amounts iust calculated.Using heat transIer equations Ior each eIIect, calculate the surIace required Ior each eIIectII the surIaces calculated are not equal, revise the ATS . Repeat step 4 onward until the areas are distributed satisIactorily. 2A 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 C = 1.78x + 6.22 x2 ), where x is wt fraction of sugar in solution. Saturated steam at 205.5 kPa and 121.1C saturation temperature is being used. The pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 22 680 kg/h at 26.7 C. the heat capacity of the liquid solutions is cP = 4.19 2.35x kJ/kg.K. The heat of solution is considered to be negligible. The coefficients of heat transfer have been estimated as U1 = 3123, U2 = 1987, and U3 = 1136 W/m2.K. If each effect has the same surface area, calculate the area, the steam rate used, and the steam economy.AMPL. 8.5-1 : vaporation of Sugar Solution in a 1riple- ffect vaporator31(/)O)/I1O^S ?1S1 121.1 CPS1 25.5 kPa(2) 131112F 228 xF .11F 2.7 C11 , L1 , x1J1 22,8 - L1(1)(3) J2 L1 - L2J3 L2 - 4531S1 1S3 1S2 12 , L2 , x213 L3 453x3 .5 P3 13.7 kPaFig. 8.5-1: Flow diagram for example 8.5-1

Solution,The process flow diagram is given in Fig. 8.5-1..$tep 1,From steam table, at P3 = 13.4 kPa, get Tsat = 51.67 C. Using the BPR equation for third effect with xL = 0.5,BPR3 = 1.78 (0.5) + 6.22 (0.52) =2.45 C.T3= 51.67 + 2.45= 54.12 C. (BPR = T Ts)$tep 2, Making an overall and a solids balance.F = 22 680 = L3 + (V1 + V2 + V3)FxF= 22 680 (0.1) = L3(0.5) + (V1 + V2 + V3) (0)L3 = 4536 kg/hTotal vaporized = (V1 + V2 + V3) = 18 144 kg/h

Assuming equal amount vaporized in each effect,V1 = V2 = V3 = 18 144 / 3 = 6048 kg/h Making a total material balance on effects 1, 2, and 3, solvingF = 22 680 = V1 + L1 = 6048 + L1, L1 = 16 632 kg/h.L1 = 16 632 = V2 + L2 = 6048 + L2, L2 = 10 584 kg/h.L2 = 10 584 = V3 + L3 = 6048+ L3, L3 = 4536 kg/h. Making a solids balance on each effect, and solving for x,22 680 (0.1) = L1 x1 = 16 632 (x1), x1 = 0.13616 632 (0.136) = L2 x2 = 10 584 (x2), x2 = 0.21410 584 (0.214) = L3 x3 = 4536 (x3), x3 = 0.5 (check)

$tep 3, The BPR in each effect is calculated as follows:BPR1 = 1.78x1 + 6.22x12= 1.78(0.136) + 6.22(0.136)2= 0.36C.BPR2 = 1.78(0.214) + 6.22(0.214)2=0.65C.BPR3 = 1.78(0.5) + 6.22(0.5)2=2.45C. then,LATavailable = TS1 T3 (sat) (BPR1 + BPR2 + BPR3 )= 121.1 51.67 (0.36+0.65+2.45) = 65.97C.Using Eq.(8.5-6) for AT1 , AT2 , and AT3AT1 = 12.40 C AT2 = 19.50 C AT3 = 34.07 C3 2 1111 1 11& & &&T T+ +LA = A )) 1136 1 ( ) 1987 1 ( ) 3123 1 () 3123 1 ( 97 . 65+ +=

However, since a cold feed enters effect number 1, this effect requires more heat. Increasing AT1 and lowering AT2 and AT3 proportionately as a first estimate, soAT1 = 15.56C AT2 = 18.34 C AT3 = 32.07 CTo calculate the actual boiling point of the solution in each effect,T1 = TS1 - AT1= 121.1 15.56 = 105.54 C. T2 = T1 - BPR1 - AT2= 105.54 0.36 18.34 = 86.84 C. TS2= T1 BPR1= 105.54 0.36= 105.18 C. T3= T2 - BPR2 - AT3= 86.84 0.65 32.07 = 54.12 C.TS3= T2 BPR2= 86.84 0.65= 86.19 C. The above data AT1, AT2 and AT3 are getting from iteration-s

The temperatures in the three effects are as follows:Effect 1 Effect 2 Effect 3CondenserTS1 = 121.1C TS2 = 105.18 TS3 = 86.19 TS4 = 51.67T1= 105.54 T2 = 86.84T3 = 54.12$tep 4,The heat capacity of the liquid in each effect is calculated from the equation cP= 4.19 2.35x.F: cPF= 4.19 2.35 (0.1)= 3.955 kJ/kg.KL1: cP1= 4.19 2.35 (0.136) = 3.869 kJ/kg.KL2: cP2= 4.19 2.35 (0.214) = 3.684 kJ/kg.KL3: cP3= 4.19 2.35 (0.5) = 3.015 kJ/kg.K

The values of the enthalpy H of the various vapor streams relative to water at 0 C as a datum are obtained from the steam table as follows:Effect 1:H1 = HS2 + 1.884 BPR1 = 2684 + 1.884(0.36)2685 kJ/kg.S1 = HS1 hS1= 2708 508= 2200 kJ/kg.Effect 2:H2 = HS3 + 1.884 BPR2= 2654 + 1.884(0.65) = 2655 kJ/kg.S2 = H1 hS2= 2685 441 = 2244 kJ/kg.Effect 3:H3= HS4 + 1.884 BPR3 = 2595 + 1.884(2.45)= 2600 kJ/kg.S3 = H2 hS3= 2655 361= 2294 kJ/kg.0Write the heat balance on each effect. Use 0C as a datum.FcPF (TF 0) + SS1 = L1cP1 (T1 0) + V1H1,,...(1)22680(3.955)(26.7-0)+2200S = 3.869L1(105.54-0)+(22680-L1)2685L1cP1 (T1 0) + V1S2 = L2cP2 (T2 0) + V2H2 ...(2)3.869L1(105.54-0)+(22680-L1)2244=3.684L2(86.84-0)+(L1-L2)2655L2cP2 (T2 0) + V2S3 = L3cP3 (T3 0) + V3H3 ...(3) 3.68L2(86.84-0)+(L1-L2)2294=4536(3.015)(54.1-0)+(L2-4536)2600Solving (2) and (3) simultaneously for L1&L2 and substituting into(1)L1 = 17078 kg/hL2 = 11068 kg/h L3 = 4536 kg/hS = 8936kg/h V1 = 5602kg/hV2 = 6010kg/hV3 = 6532kg/h11(/)O)/I1O^$tep 5, Solving for the values of q in each effect and area, )S qS61 110 460 . 5 1000 220036008936='+

'

= = )' qS62 1 210 492 . 3 1000 224436005602='+

'

= = )' qS63 2 310 830 . 3 1000 229436006010='+

'

= = )261 1114 . 11265 . 15 312310 460 . 5m

T &qA = =A= )262 2228 . 9534 . 18 198710 492 . 3m

T &qA = =A= )263 3331 . 10507 . 32 113610 830 . 3m

T &qA = =A=23 2 14 . 1043) (mA A AAm=+ +=21(/)O)/I1O^Am = 104.4 m2, 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.$tep 6,Making a new solids balance by using the new L1 = 17078,L2 = 11068, and L3 = 4536, and solving for x,22 680 (0.1) = L1 x1 = 17 078 (x1), x1 = 0.13317 078 (0.130) = L2 x2 = 11 068 (x2), x2 = 0.20511 068 (0.205) = L3 x3 = 4536 (x3), x3= 0.5 (check)31(/)O)/I1O^$tep 7. The new BPR in each effect is then,BPR1 = 1.78(0.133) + 6.22(0.13)2=0.35C.BPR2 = 1.78(0.205) + 6.22(0.205)2=0.63C.BPR3 = 1.78(0.5) + 6.22(0.5)2=2.45C. then,LATavailable = 121.1 51.67 (0.35+0.63+2.45) = 66.0 C.The new ATare obtained using Eq.(8.5-11), )CAA TTmH = =A= A 77 . 164 . 1044 . 112 56 . 151 1'1 )CAA TTmH = =A= A 86 . 164 . 1048 . 95 34 . 182 2'2 )CAA TTmH = =A= A 34 . 324 . 1041 . 105 07 . 323 3'3C T H = + + = LA 97 . 65 34 . 32 86 . 16 77 . 16

These AT' values are readjusted so that AT 1`= 16.77,AT 2`= 16.87,AT 3` = 32.36, and LAT= 66.0 C. Tocalculate the actual boiling point of the solution in eacheffect,(1) T1 = TS1 + AT 1` = 121.1 16.77 = 104.33C(2) T2 = T1 BPR1 - AT 2` = 104.33 0.35 16.87 = 87.11 CTS2 = T1 BPR1= 104.33 0.35 = 103.98C(3) T3 = T2 BPR2 - AT 3` = 87.11 0.63 32.36 = 54.12 CTS3 = T2 BPR2= 87.11 0.63 = 86.48 C.$tep 8;Following step 4 to get cP = 4.19 2.35x,F: cPF= 4.19 2.35 (0.1) = 3.955 kJ/kg.KL1: cP1= 4.19 2.35 (0.133) = 3.877 kJ/kg.KL2: cP2= 4.19 2.35 (0.205) = 3.705 kJ/kg.KL3: cP3= 4.19 2.35 (0.5) = 3.015 kJ/kg.K

Then the new values of the enthalpy are,(1) H1 = HS2 + 1.884 BPR1= 2682 + 1.884(0.35) = 2683 kJ/kg.S1 = HS1 hS1= 2708 508= 2200 kJ/kg.(2) H2 = HS3 + 1.884 BPR2= 2654 + 1.884(0.63) = 2655 kJ/kg.S2 = H1 hS2= 2683 440= 2243 kJ/kg.(3) H3 = HS4 + 1.884 BPR3= 2595 + 1.884(2.45) = 2600 kJ/kg.S3 = H2 hS3= 2655 362 = 2293 kJ/kg.Writing a heat balance on each effect,and solving,(1) 22680(3.955)(26.7-0)+2200S = 3.877L1(104.33-0)+(22680-L1)2683(2) 3.877L1(104.33-0)+(22680-L1)2243=3.708L2(87.11-0)+(L1-L2)2655(3) 3.708L2(87.11-0)+(L1-L2)2293=4536(3.015)(54.1-0)+(L2-4536)2600L1 = 17005 kg/hL2 = 10952L3 = 4536S = 8960V1 = 5675 V2 = 6053V3 = 6416

1(/)O)/I1O^Solving for q and A in each effect, )S qS61 110 476 . 5 1000 220036008960='+

'

= = )' qS62 1 210 539 . 3 1000 224336005675='+

'

= = )' qS63 2 310 855 . 3 1000 229336006053='+

'

= = )26'1 1116 . 10477 . 16 312310 476 . 5m

T &qA = =A= )26'2 2226 . 10587 . 16 198710 539 . 3m

T &qA = =A= )26'3 3339 . 10436 . 32 113610 855 . 3m

T &qA = =A=

1(/)O)/I1O^The average area Am = 105.0 m2to use in each effect.steam ec434my =025 . 289606416 6053 56753 2 1=+ +=+ +S' ' '

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