CHAPTERFIVE EVAPORATION 5.1 Climbing Film Evaporator Cold hearted orb that rules the night, And steals the colours from our sight. Red is gray and yellow white, But we decide which is right, And which is an illusion. MOODY BLUES and even when the vapor is a mixture.sometimes a highly viscous one .Downward-flow (falling-film) 3. it may have a very high boiling point elevation. upward-flow (climbing-film). and to measure the thermal efficiency of the climbing film evaporator. it differs from distillation in that the vapor is usually a single component. Short-tube evaporators Long-tube vertical evaporators .1 Object The object of this experiment is to investigate the effect of the variations in the feed rate on the concentration of the product. The liquid to be evaporated may be less viscous than water. or it may be so viscous that it will hardly flow. it differs from crystallization in that emphasis is placed on concentrating a solution rather than forming and building crystals. the effect of the operating steam pressure on the rate of evaporation achieved.1. thermal efficiency.Upward-flow (climbing-film) . it may tend to foam. or it may be damaged by the application of too high temperatures. 5. It may deposit scale on the heating surface.1 CLIMBING FILM EVAPORATOR Keywords: Evaporation. Evaporation differs from drying in that the residue is a liquid . 5.rather than a solid.2 Theory The vaporization of a liquid for the purpose of concentrating a solution consisting of a nonvolatile solute and volatile solvent is a common unit operation in chemical processing and is performed in many ways.1. The conditions under which evaporation is carried out in practice vary widely.Chapter 5: EVAPORATION 5. it may precipitate in crystals.Forced-circulation . This wide variety of problems has led to considerable variation in the types of mechanical construction used. Evaporation is conducted by vaporizing a portion of the solvent to produce a concentrated solution or a thick liquor. Evaporator types may be classified as follows: 1. no attempt is made in the evaporation step to separate the vapor into fractions. 2. Agitated-film evaporators . 1. Let the enthalpy of the feed per kilogram be hf.1. J/kg.Energy balance for single-effect evaporator.1. and therefore y is zero. There is taken out of the evaporator L kg of thick liquor. There is also V kg of vapor having a solute concentration of y and an enthalpy of H. Let F be the kilogram of the feed to the evaporator per hour. in which the heating surface is represented by a simple coil. In most evaporators. the vapor is pure water.Material and Heat Balances: Figure 5. The material balance equations for this case become: F = L+ V Total Material Balance: (5.1 . S kg of steam is supplied to the heating surface with an enthalpy of Hs (J/kg) and there is taken out S kg of condensate with an enthalpy of hc (J/kg). whose solute content is xf (x is the weight fraction). One simplifying assumption usually made is that in an evaporator there is very little .2) In order to furnish the heat necessary for evaporation. Figure 5.1.1) Solute Balance: Fx F = Lx L + Vy (5. whose composition in weight fraction of solute is xL and whose enthalpy is hL in joule per kilogram.1 is a highly simplified diagram of an evaporator. Q = ( MCp∆T) + λm (5. the overall heat transfer coefficient and the percentage evaporation rate of which affect the product concentration. J/m2•s•K total fixed heat transfer area.1. Since the available heat transfer area is fixed then for a given steam pressure and feed rate.6) (5. kg/s specific heat of feed.1. K log mean temperature difference.1. K heat transfer coefficient. kg/s evaporation rate. The heat balance equation is: Heat in feed + Heat in steam = Heat in thick liquor + Heat in vapor + Heat in condensate + Heat lost by radiation (5. This leads the assumption that the condensate will leave at the condensing temperature of the steam. the overall heat transfer coefficient is steady and the rate of evaporation is constant.1. J mass flow rate of feed.Chapter 5: EVAPORATION cooling of the condensate.4) which represents the heat balance of the evaporator.3) Neglecting losses by radiation and using the relevant symbols we get the following equation: Fh F + SH S = VH + Lh L + Sh C (5. J/kg•K temperature increase. m2 latent heat of evaporation.5) Q = U heatýng A h ∆Tlmheatýng + U evap A e ∆Tlmevap A = Ae + Ah ( ) ( ) (5.7) where Q M m cp : : : : total heat load. Variation in the feed rate will change the sensible heat loading. J/kg ∆T : ∆Tlm: U A λ : : : .1. 1.For an increase in the feed rate then the sensible heat load is larger. Water carried over by the steam is collected in the purge receiver and effectively lost from all dryness factor value (DF) is required for the steam supplied. The rate of heat transferred Q is dependent upon the overall heat transfer coefficient U. the heat transfer surface area A. that is: Q = UA∆Tlm (5.8) For a given system.85 to 1. Since varying the steam pressure varies the operating steam temperature. U is virtually constant so that the rate of heat transfer depends entirely upon the log mean temperature difference. and the log mean temperature difference ∆Tlm. the percentage evaporation will be lower so that the product is less concentrated. the rate of heat transfer and hence the rate of evaporation is changed. This may in practice vary from 0. Q = (mass) × (specific heat) × (temperature rise) + (evaporation) × (latent heat) (5. Heat taken up by the feed is the sensible heat and latent heat. The latent heat of the steam is used to preheat the feed into the evaporator and allowance must be made in the calculation for this. mass of water in the feed×Sp heat of feed× temperatur rise of water e steam used to preheat feed = Latent heat of steam (5.1. In practice heat is lost to the atmosphere and to the side streams purged from the system.1. Since there is little change in the overall heat transfer coefficient for heating and the ∆T will be constant then Ah must be larger.11) . A will be constant.1.10) Evaporative efficiency = (mass of steam condensed − mass of mass of water evaporated×100 preheat steam × DF ) - (5. The heat transfer area Ae must be correspondingly smaller and the evaporation rate will be less. Since the volume of the feed is higher.00.9) This experiment assumes as a basis that 1 kg of dry saturated steam condensing will evaporate 1 kg of boiling water. Figure 5.) Bottom receiving vessel 11. 1.) Water outlet 17.) Condensate drain 10.1.) Thermometer pocket .) Condensate receiver 8.2 .) Vacuum cock 14.) Feed inlet and drain valve 2.) Condenser 16.1.) Three-way cock 9.) Water inlet 15.) Calandria vent 6.2.) Cyclone 7.3 Apparatus The apparatus used in this experiment is shown in Figure 5.QVF climbing film evaporator.) Condensate outlet 3.) Top receiving vessel 13.1.) Stream inlet 4.) Vacuum and vent cock 12.) Calandria 5.Chapter 5: EVAPORATION 5. Open the feed inlet cock. i. Allow the liquid level in the calandria to reach the steam inlet neck before finally adjusting the feed rate. and control the feed rate by the rotameter. 10. For a minimum of 15 minutes operation ensure that the steam pressure.1). Measure the volume of the condensate collected.15). 8. Adjust the feed rate to a value and allow a minimum of 5 minutes of operation under these conditions so that the unit can settle down. feed rate. (No. 15.e. 12. Close the steam generator. 4. Close all drain cocks on the plant. 13. Turn on the cooling water to the condenser. Close the steam control valve and all other steam valves on the path between the steam generator and control table. Adjust steam pressure to a specified value. and vacuum are constant.1. 2. . Shut the unit down following the procedure described below: IMPORTANT: • • • Close the feedcock. In one of these five data calculate (find) the amount of steam used to evaporate the feed by placing the steam pipe into a specified amount of cold water.5. Once the unit is operating smoothly close the valve between the condensate receivers and note the liquid level in the graduated concentrate receiver. 11. Check that the steam trap is functioning correctly. Turn the steam generator on. 5. 6. 14. Prepare 30 dm3 of a 10 % w/w solution of glycerol in water. (No. At the end of the test period close the valve between the twin condensate receivers and note the level in the graduated concentrate receiver. come to steady state. Repeat the experiment keeping the flow rate constant and varying the pressure using two other different pressures. calculate the volume of the feed to the unit and the volume of the concentrate produced. 3. 9. The liquid will begin to boil and expanding bubbles will rise rapidly in the tube giving climbing film operation. 16. 1..4 Experimental Procedure Attention: During the experiment the floor gets wet. so wear suitable clothes. 7. Repeat the experiment with two other feed rates. 5. 11. Green. 1955...Chapter 5: EVAPORATION • • • • Stop the pump.5 Report Objectives 1. 7. Perry. 6. Calculate the percent of feed evaporated. Pergamon Press. New York. and J. Introduction to Chemical Engineering. Close the condensate cooling water valve. Leave all the drain cocks open. Badger. 9. L.85? 10. 2. Calculate the total heat required to heat the concentrate water and heat required to evaporate the condensate. 8. Banchero.1. Chemical Engineering Unit Operations. 3. 5. J. F. Open each of the drain cocks in turn and drain contents to a suitable receiver. State the factors that affect the evaporative efficiency. Comment on the effects of vacuum operation on evaporative efficiency. What is the enthalpy difference between 1 kg of steam having a DF of 0. H. Compare this value with the heat available from the steam. McGrawHill. and D. 3.95 and 1 kg of steam having a DF of 0. T. Why is glycerol used? What are the properties of a climbing film evaporator? Calculate the percent concentration of the concentrate. . and J.1. Richardson. W.6 References 1. 6th edition. Coulson. M. Calculate the heat transferred in producing the product concentrate and condensate. 1988. McGraw-Hill. R.. 5. 4. Perry’s Chemical Engineers’ Handbook. State the assumptions used in your calculations. 1962. which was condensed and collected from the steam trap. Calculate the evaporative efficiency. 2.