PM3125: Lectures 13 to 15 Content of Lectures 13 to 17: Evaporation: Factors affecting evaporation Evaporators Film evaporators Single effect and multiple effect evaporators Mathematical problems on evaporation Principal reference: Chapter 8 in C.J. Geankoplis, Transport Processes and Unit Operations, 3rd Edition, Prentice-Hall of India Prof. R. Shanthini 28 May 2012

Evaporation Suppose that we have a dilute solution of a solute (say, sugar) dissolved in a solvent (say, water) We need to remove part of the solvent (water) to produce a concentrated solution This can be achieved by heating the solution so as to evaporate the solvent (water) This process is known as evaporation. The industrial equipment used for this purpose in known as an evaporator. Prof. R. Shanthini 28 May 2012

Types of evaporators Open kettle or pan evaporator: Prof. R. Shanthini 28 May 2012 3

Types of evaporators Open kettle or pan evaporator: Pan Pressure gauge Concentrate Condensate Pan Boiler Jacket Pressure gauge Steam Prof. R. Shanthini 28 May 2012 4

Types of evaporators Open kettle or pan evaporator: simplest form of evaporators inexpensive simple to operate very poor heat economy in some cases paddles and scrapers for agitation are used Prof. R. Shanthini 28 May 2012 5

Types of evaporators Horizontal-tube evaporator: Vapour Dilute feed Steam inlet Condensate Concentrated product Prof. R. Shanthini 28 May 2012 6

Types of evaporators Horizontal-tube evaporator: - relatively cheap used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales poor liquid circulation (and therefore unsuitable for viscous liquids) Prof. R. Shanthini 28 May 2012 7

Types of evaporators Vertical-type short-tube evaporator: - Liquid is inside the tubes - Steam condenses outside the tubes - used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales Prof. R. Shanthini 28 May 2012 8

Types of evaporators Vertical-type short-tube evaporator: Vapour Dilute feed Steam inlet Condensate Concentrated product Prof. R. Shanthini 28 May 2012 9

Types of evaporators Falling-film-type evaporator: Prof. R. Shanthini 28 May 2012 10

Types of evaporators More types are given in additional handouts uploaded at the course website. Prof. R. Shanthini 28 May 2012 11

Factors effecting evaporation: Concentration in the liquid: Liquid feed to an evaporator is relatively dilute. So its viscosity is low, and heat-transfer coefficient high. As evaporation proceeds, the solution becomes concentrated. So viscosity increases and heat-transfer coefficient drops. Density and the boiling point of solution also increase. Prof. R. Shanthini 28 May 2012

Factors effecting evaporation: Solubility: As solution is heated, concentration of the solute in the solution increases. In case the solubility limit of the solute in solution is exceeded, then crystals may form. Solubility of the solute therefore determines the maximum concentration of the solute in the product stream. In most cases, the solubility of the solute increases with temperature. This means when a hot concentrated solution from an evaporator is cooled to room temperature, crystallization may occur. Prof. R. Shanthini 28 May 2012 13

Factors effecting evaporation: Temperature sensitivity of materials: Pharmaceuticals products, fine chemicals and foods are damaged when heated to moderate temperatures for relatively short times. So special techniques are employed to reduce temperature of the liquid and time of heating during evaporation Prof. R. Shanthini 28 May 2012 14

Factors effecting evaporation: Foaming and frothing: Solutions like organic compounds tend to foam and froth during vaporization. The foam is carried away along with vapor leaving the evaporator. Entrainment losses occur. Prof. R. Shanthini 28 May 2012 15

Factors effecting evaporation: Pressure and temperature: The boiling point of the solution is related to the pressure of the system. The higher the operating pressure of the evaporator, the higher the temperature at boiling. Also, as the concentration of the dissolved material in solution increases by evaporation, the temperature of boiling may rise (a phenomenon known as boiling point rise/elevation). To keep the temperatures low in heat-sensitive materials, it is often necessary to operate under atmospheric pressure (that is, under vacuum). Prof. R. Shanthini 28 May 2012 16

Factors effecting evaporation: Scale deposition: Some solutions deposit solid materials (called scale) on the heating surfaces. The result is that the overall heat-transfer coefficient (U) may drastically decrease, leading to shut down of the evaporators for cleaning purposes. Prof. R. Shanthini 28 May 2012 17

Factors effecting evaporation: Materials of construction: Evaporators are made of some kind of steel. However many solutions attack ferrous metals and are contaminated by them. Copper, nickel, stainless steels can also be used. Prof. R. Shanthini 28 May 2012 18

Method of operation of evaporators Single-effect evaporation: When a single evaporator is used ,the vapor from the boiling liquid is condensed and discarded. This is called single effect evaporation. It is simple but utilizes steam ineffectively. To evaporate 1 kg of water from the solution we require 1-1.3 kg of steam. Increasing the evaporation per kg of steam by using a series of evaporators between the steam supply and condenser is called multiple effect evaporation Multiple-effect evaporation: Prof. R. Shanthini 28 May 2012 19

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Feed: F – mass flow rate xF – mass fraction of solute in feed TF – temperature of feed hF – enthalpy of feed Vapour leaving the evaporator: V – mass flow rate yV – mass fraction of solute in vapour T1 – temperature of vapour HV – enthalpy of vapour Concentrate leaving the evaporator: L – mass flow rate xL – mass fraction of solute in concentrate T1 – temperature of concentrate hL – enthalpy of concentrate Steam: S – mass flow rate PS – steam pressure TS – steam temperature HS – enthalpy of steam hS – enthalpy of condensate P – pressure in the evaporator T1 – temperature in the evaporator Prof. R. Shanthini 28 May 2012 20

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Overall material balance: F = L + V Solute balance: F xF = L xL + V yV If the vapour is free of solute: F xF = L xL Heat balance: F hF + S HS = L hL + V HV + S hS Rewriting: F hF + S (HS - hS) = L hL + V HV F hF + S λ = L hL + V HV where λ = HS - hS Prof. R. Shanthini 28 May 2012 21

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Energy lost by the steam q = S λ = S (HS – hS) In case of no energy loss to the environment, q amount of energy gets transferred from steam to the solution through the tube wall of area A and overall heat transfer coefficient U. Therefore, q = U A ΔT = U A (TS – T1) Prof. R. Shanthini 28 May 2012 22

Example 1: A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt solution entering at 38ºC 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 150 kPa. The overall coefficient U = 1704 W/m2.K. Calculate the amounts of vapor and liquid products and the heat-transfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water. Prof. R. Shanthini 28 May 2012

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Data provided: F = 9072 kg/h xF = 1 wt % = 0.01 kg solute / kg feed TF = 38ºC xL = 1.5 wt % = 0.015 kg solute / kg liquid product P = 101.325 kPa (1.0 atm abs) PS = 150 kPa U = 1704 W/m2.K T1 = saturated temperature at P (= 101.325 kPa) = 100ºC TS = saturated temperature at 150 kPa = 111.4ºC Prof. R. Shanthini 28 May 2012 24

Calculation methods for single-effect evaporators Data provided: F = 9072 kg/h xF = 0.01 kg solute / kg feed TF = 38ºC xL = 0.015 kg solute / kg liquid product P = 101.325 kPa; T1 = 100ºC PS = 150 kPa; TS = 111.4ºC U = 1704 W/m2.K Available equations: Overall material balance: F = L + V Solute balance: F xF = L xL (no solute in the vapour) Heat balance: F hF + S λ = L hL + V HV where λ = HS – hS q = S λ = U A ΔT = U A (TS – T1) Amounts of vapor and liquid products = ? F, xF and xL are known, and therefore L = 6048 kg/h and V = 3024 kg/h Prof. R. Shanthini 28 May 2012 25

Calculation methods for single-effect evaporators Data known: F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h TF = 38ºC P = 101.325 kPa; T1 = 100ºC PS = 150 kPa; TS = 111.4ºC U = 1704 W/m2.K Available equations: Heat balance: F hF + S λ = L hL + V HV where λ = HS – hS q = S λ = U A ΔT = U A (TS – T1) Heat transfer area A = S λ / U (TS – T1) = ? S λ = L hL + V HV – F hF = (F – V) hL + V HV – F hF = F (hL – hF) + V (HV – hL ) = F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa ) Prof. R. Shanthini 28 May 2012 26

Calculation methods for single-effect evaporators Data known: F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h TF = 38ºC P = 101.325 kPa; T1 = 100ºC PS = 150 kPa; TS = 111.4ºC U = 1704 W/m2.K Heat transfer area A = S λ / U (TS – T1) = ? S λ = F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa) F, T1 , TF and V are already known. Cp = 4.14 kJ/kg.K (assumed) Latent heat of vapourization at 101.325 kPa = 2256.7 kJ/kg Therefore, S λ = (9072) (4.14) (100 – 38) +(3024) (2256.7) kJ/h = 9152862 kJ/h Prof. R. Shanthini 28 May 2012 27

Calculation methods for single-effect evaporators Data known: F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h TF = 38ºC P = 101.325 kPa; T1 = 100ºC PS = 150 kPa; TS = 111.4ºC U = 1704 W/m2.K Heat transfer area A = S λ / U (TS – T1) = ? S λ = 9152862 kJ/h = 9152862 * 1000 / 3600 W T1 and TS are known U = 1704 W/m2.K Therefore, A = S λ / U (TS – T1) = [9152862 * 1000 / 3600] / [1704 * (111.4 – 100)] = 130. 9 m2 Prof. R. Shanthini 28 May 2012 28

Effect of feed temperature: Effects of processing variables on evaporator operation: Effect of feed temperature: - The inlet temperature of the feed has a large effect on the evaporator operation. When feed is not at its boiling point, steam is needed first to heat the feed to its boiling post and then to evaporate it. Preheating the feed can reduce the size of evaporator heat-transfer area. Prof. R. Shanthini 28 May 2012 29

Effects of processing variables on evaporator operation: Effect of pressure: Pressure in the evaporator sets the boiling point of the solution (T1). Steam pressure determines the steam temperature (Ts) Since q = U A (TS – T1), larger values of (TS – T1) will help reduce the heat-transfer area needed and hence the cost of evaporator. Vacuum can be maintained in the solution side using a vacuum pump. For example, if the pressure in Example 1 is reduced to 41.4 kPa, boiling point of water reduces to 349.9 K and that would increase the (TS – T1) from 10 K to 33.3 K. A large decrease in heat-transfer area would be obtained. Prof. R. Shanthini 28 May 2012 30

Effect of steam pressure: Effects of processing variables on evaporator operation: Effect of steam pressure: High pressure provides high Ts values, and hence TS – T1 will increase. High pressure steam is however more costly. Therefore, overall economic balances must be considered to determine the optimum steam pressure. Prof. R. Shanthini 28 May 2012 31

Boiling point rise of solutions: In example 1, the solution is assumed to be dilute enough to be considered to have the same thermal properties as water. It is not true always. For concentrated solutions, heat capacity and boiling point are quire different from that of water. Duhring’s rule is an empirical law that relates the boiling point of a solution to the boiling point of the solvent at different pressures for a solution of given concentration. Prof. R. Shanthini 28 May 2012 32

Boiling point rise of solutions (an example): Prof. R. Shanthini 28 May 2012 Duhring plot for boiling point of sodium chloride solutions 33

Enthalpy-concentration charts of solutions: See the handout. Prof. R. Shanthini 28 May 2012 34

Example 2: An evaporator is used to concentrate 4536 kg/h of a 20% NaOH solution entering at 60ºC to a product of 50% solids. The pressure of the saturated steam used is 170 kPa and the vapor space pressure of the evaporator is at 12 kPa. The overall coefficient U is 1560 W/m2.K. Calculate the steam used, the steam economy (in kg vapourized / kg steam used) and the heating surface area. Prof. R. Shanthini 28 May 2012

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Data provided: F = 4536 kg/h xF = 20 wt % = 0.2 kg solute / kg feed TF = 60ºC xL = 50 wt % = 0.5 kg solute / kg liquid product P = 12 kPa = 0.12 bar PS = 170 kPa = 1.7 bar U = 1560 W/m2.K T1 ≠ saturated temperature at P (= 0.12 bar) = 49.4oC TS = saturated steam temperature at 1.7 bar = 115.2oC Prof. R. Shanthini 28 May 2012 36

Calculation methods for single-effect evaporators Amounts of vapor and liquid products = ? F, xF and xL are known, and therefore L = 1814 kg/h and V = 2722 kg/h Steam used = ? S λ = L hL + V HV - F hF hF = enthalpy of 20% solution at 60oC = 214 kJ/kg (using the enthalpy–concentration chart) hL = enthalpy of 50% solution at T1 = ? (using the enthalpy–concentration and boiling-point rise charts) Saturated temperature at P (= 0.12 bar) = 49.4oC Using the boiling-point rise chart, we get 89.5oC ( read against 49.4oC and 50 wt%) as the boiling point of the solution. That is T1 = 89.5oC hL = enthalpy of 50% solution at 89.5oC = 505 kJ/kg (using the enthalpy–concentration chart) HV = enthalpy of superheated steam at 89.5oC and 0.12 bar = 2667 kJ/kg (using the superheated steam table) Prof. R. Shanthini 28 May 2012 37

Calculation methods for single-effect evaporators Amounts of vapor and liquid products = ? F, xF and xL are known, and therefore L = 1814 kg/h and V = 2722 kg/h Steam used = S = ? S λ = 1814 x 505 + 2722 x 2667 – 4536 x 214 = 7204940 kJ/h λ = latent heat of vapourization of water at 1.7 bar and 115.2oC = 2216 kJ/kg (using the steam table) Therefore S = 7204940 / 2216 kg/h = 3251 kg/h Steam economy = kg vapourized / kg steam used = V / S = ? Steam economy = 2722 / 3251 = 0.837 A = S λ / U (TS – T1) = [7204940 * 1000 / 3600] / [1560 * (115.2 – 89.5)] = 49. 9 m2 Prof. R. Shanthini 28 May 2012 38

Example 3 (Repeat Example 2 assuming that the thermal properties of the liquid in the evaporator can be approximated by those of water): An evaporator is used to concentrate 4536 kg/h of a 20% NaOH solution entering at 60ºC to a product of 50% solids. The pressure of the saturated steam used is 170 kPa and the vapor space pressure of the evaporator is at 12 kPa. The overall coefficient U is 1560 W/m2.K. Calculate the steam used, the steam economy (in kg vapourized / kg steam used) and the heating surface area. Prof. R. Shanthini 28 May 2012

Calculation methods for single-effect evaporators Steam, S PS, TS, HS Feed, F xF, TF, hF Condensate, S PS, TS, hS Vapour, V yV, T1, HV Concentrate, L xL, T1, hL P T1 Data provided: F = 4536 kg/h xF = 20 wt % = 0.2 kg solute / kg feed TF = 60ºC xL = 50 wt % = 0.5 kg solute / kg liquid product P = 12 kPa = 0.12 bar PS = 170 kPa = 1.7 bar U = 1560 W/m2.K T1 = saturated temperature at P (= 0.12 bar) = 49.4oC TS = saturated steam temperature at 1.7 bar = 115.2oC Prof. R. Shanthini 28 May 2012 40

Calculation methods for single-effect evaporators Amounts of vapor and liquid products = ? F, xF and xL are known, and therefore L = 1814 kg/h and V = 2722 kg/h Steam used = ? S λ = L hL + V HV - F hF = (F – V) hL + V HV – F hF = F (hL – hF) + V (HV – hL ) = F Cp (T1 - TF) + V (Latent heat of vapourization at 0.12 bar) = (4536) (4.14) (49.4 – 60) +(2722) (2383) kJ/h (Cp = 4.14 kJ/kg.K is assumed) = 6301078 kJ/h λ = latent heat of vapourization of water at 1.7 bar = 2216 kJ/kg Therefore S = 6301078 / 2216 kg/h = 2843.5 kg/h Steam economy = kg vapourized / kg steam used = V / S = ? Steam economy = 2722 / 2843.5 = 0.957 A = S λ / U (TS – T1) = [6301078 * 1000 / 3600] / [1560 * (115.2 – 49.4)] = 17 m2 Prof. R. Shanthini 28 May 2012 41

Calculation methods for single-effect evaporators Compare the solutions of Examples 2 and 3 and discuss the importance of considering the boiling pint rise and enthalpy change of concentrated solution. Prof. R. Shanthini 28 May 2012 42

Double-effect evaporators Prof. R. Shanthini 28 May 2012 43

Calculation methods for double-effect evaporators If liquid is to be evaporated in each effect, and if the boiling point of this liquid is unaffected by the solute concentration, then writing a heat balance for the first evaporator: q1 = U1 A1 ΔT1 = U1 A1 (TS – T1) Similarly, in the second evaporator, remembering that the "steam" in the second is the vapour from the first evaporator and that this will condense at approximately the same temperature as it boiled, since pressure changes are small, q2 = U2 A2 ΔT2 = U2 A2 (T1 – T2) If the evaporators are working in balance, then all of the vapours from the first effect are condensing and in their turn evaporating vapours in the second effect. Also assuming that heat losses can be neglected, there is no appreciable boiling-point elevation of the more concentrated solution, and the feed is supplied at its boiling point,                 q1 = q2 Prof. R. Shanthini 28 May 2012 44

Calculation methods for double-effect evaporators If the evaporators are working in balance, then all of the vapours from the first effect are condensing and in their turn evaporating vapours in the second effect. Also assuming that heat losses can be neglected, there is no appreciable boiling-point elevation of the more concentrated solution, and the feed is supplied at its boiling point,                 q1 = q2 That is, U1 A1 ΔT1 = U2 A2 ΔT2 Further, if the evaporators are so constructed that A1 = A2, the foregoing equations can be combined.   U2 / U1 = ΔT1 / ΔT2 That is, the temperature differences are inversely proportional to the overall heat transfer coefficients in the two effects. This analysis may be extended to any number of effects operated in series, in the same way. Prof. R. Shanthini 28 May 2012 45

Example 4: Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h-1 of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m-2 s-1 °C-1 in the first, second and third effects, respectively. Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect. Source: http://www.nzifst.org.nz/unitoperations/evaporation2.htm Prof. R. Shanthini 28 May 2012