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Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
Balance evaporador 2012
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Balance evaporador 2012

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  • 1. NAMP PIECE First Example: A Single Effect Evaporator (to be done in Excel)Module 9 – Steady state simulation 1
  • 2. NAMP PIECE Evaporation Function is to concentrate solution What affects evaporation? • Rate at which heat is transferred to the liquid • Quantity of heat required to evaporate mass of water • Maximum allowable temperature of liquid • Pressure which evaporation takes placeModule 9 – Steady state simulation 2
  • 3. NAMP PIECE Single Effect Vertical Evaporator Three functional sections • Heat exchanger • Evaporation section • liquid boils and evaporates • Separator • vapor leaves liquid and passes off to other equipment Three sections contained in a vertical cylinderModule 9 – Steady state simulation 3
  • 4. NAMP PIECE • In the heat exchanger section (calandria), steam condenses in the outer jacket • Liquid being evaporated boils on inside of the tubes and in the space above the upper tube stack • As evaporation proceeds, the remaining liquors become more concentratedModule 9 – Steady state simulation 4
  • 5. NAMP PIECE Diagram of Single Effect Evaporator Vapor V Tf, xf, hf, ṁf Tv, yv, Hv, ṁV U = J/m2 s oC Feed F P = kPa Ts, Hs, ṁs A = ? m2 Condensate S Ts, hs, ṁs Steam S Concentrated liquid L TL, xL, hL, ṁLModule 9 – Steady state simulation 5
  • 6. NAMP PIECE Material and Heat Balances q = UAΔT ΔT = Ts – TL Heat given off by vapor ṁF = ṁL + ṁV λ = H s – hs ṁFxF = ṁLxL + ṁVyV ṁFhF + ṁsHs = ṁLhL + ṁVHV+ ṁshs ṁFhF + ṁsλ = ṁLhL + ṁVHV q = ṁs(Hs-hs) = ṁsλ ṁsλ – ideal heat transferred in evaporatorModule 9 – Steady state simulation 6
  • 7. NAMP PIECE Finding the Latent Heat of Evaporation of Solution and the Enthalpies • Using the temperature of the boiling solution TL, the latent heat of evaporation can be found; • The heat capacities of the liquid feed (CpF) and product (CpL) are used to calculate the enthalpies of the solution.Module 9 – Steady state simulation 7
  • 8. NAMP PIECE Property Effects on the Evaporator • Feed Temperature – Large effect – Preheating can reduce heat transfer area requirements • Pressure – Reduction • Reduction in boiling point of solution • Increased temperature gradient • Lower heating surface area requirements • Effect of Steam Pressure – Increased temperature gradient when higher pressure steam is used.Module 9 – Steady state simulation 8
  • 9. NAMP PIECE Boiling-Point Rise of Solutions • Increase in boiling point over that of water is known as the boiling point elevation (BPE) of solution • BPE is found using Duhring’s Rule – Boiling point of a given solution is a linear function of the boiling point of pure water at the same pressureModule 9 – Steady state simulation 9
  • 10. NAMP PIECE Duhring lines (sodium chloride) http://www.nzifst.org.nz/unitoperations/evaporation4.htmModule 9 – Steady state simulation 10
  • 11. NAMP PIECE Problem Statement (McCabe 16.1 modified) A single-effect evaporator is used to concentrate 9070 kg/h of a 5% solution of sodium chloride to 20% solids. The gauge pressure of the steam is 1.37 atm; the absolute pressure in the vapor space is 100 mm Hg. The overall heat transfer coefficient is estimated to be 1400 W/m2 oC. The feed temperature is 0oC. Calculate the amount of steam consumed, the economy, and required heating surface. First Example Excel SpreadsheetModule 9 – Steady state simulation 11
  • 12. NAMP PIECE 1. Draw Diagram and Label Streams 9070 kg/h feed, Vapor V Tv, 0% solids, 0oC, 5% solids, Hv, ṁv hF U = 1400 Feed F W/m2 oC P= 100 mm Hg Ts, Hs, 1.37 atm gauge, ṁs q=? Condensate S Ts, hs, ṁs Steam S A=? Liquor L TL, 20% solids, hL, ṁLModule 9 – Steady state simulation 12
  • 13. NAMP PIECE 2. Perform Mass Balances ṁF = ṁL + ṁV [9070 kg/h = ṁL kg/h+ ṁV kg/h] ṁFxF = ṁLxL + ṁVyV (note that yv is zero because only vapor is present, no solids) [0.05 * 9070 kg/h = 0.2 * ṁL kg/h + 0] • Can solve for ṁv and ṁL ṁV = 6802.5 kg/h, ṁL = 2267.5 kg/hModule 9 – Steady state simulation 13
  • 14. NAMP PIECE 3. Perform Heat Balances to find the Economy The economy is defined as the mass of water evaporated per mass of steam supplied. ṁFhF + ṁSHS = ṁLhL + ṁVHV+ ṁShS ṁFhF + ṁSλ = ṁLhL + ṁVHV q = ṁS(HS- hS) = ṁSλModule 9 – Steady state simulation 14
  • 15. NAMP PIECE Needed Data • Boiling point of water at 100 mm Hg = 51oC (from steam tables) www.nzifst.org.nz/unitoperations/appendix8.htm • Boiling point of solution = 88oC (from Duhring lines) http://www.nzifst.org.nz/unitoperations/evaporation4.htm • Boiling point elevation = 88 – 51 = 37oC • Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88oC and 100 mm Hg [.133 bar]) = 2664 kJ/kg (F&R, p.650) – also called the latent heat of evaporation • Heat of vaporization of steam (Hs-hs = λ ) at 1.37 atm gauge [20 lbf/in2] = 939 Btu/lb = 2182 kJ/kg (McCabe, App.7, p.1073)Module 9 – Steady state simulation 15
  • 16. NAMP PIECE Finding the enthalpy of the feed 1. Find the heat capacity of the liquid feed yNaCl=0.05 feed is 5% sodium chloride, 95% water ywater=0.95 Cp,water=4.18 kJ/kgoC Cp,mix = x iCpi all mixture Cp,NaCl=0.85 kJ/kgoC components (Cp)F = 0.05*0.85 + 0.95*4.18 = 4.01 kJ/kgoC 2. Calculate Enthalpy (neglecting heats of dilution) hF = Cp,F (TF - Tref ) hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kgModule 9 – Steady state simulation 16
  • 17. NAMP PIECE Finding the enthalpy of the liquor yNaCl=0.20 1. Find the heat capacity of the liquor ywater=0.80 feed is 20% sodium chloride, 80% water Cp,water=4.18 kJ/kgoC Cp,mix = x iCpi all mixture Cp,NaCl=0.85 kJ/kgoC components Cp,L = 0.20*0.85 + 0.80*4.18 = 3.51 kJ/kgoC 2. Calculate Enthalpy (neglecting heats of dilution) hL = Cp,L (TL - Tref ) hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kgModule 9 – Steady state simulation 17
  • 18. NAMP PIECE Heat Balances ṁLhL + ṁVHV - ṁFhF = ṁSHS - ṁShS = ṁS(HS- hS) = ṁSλ λ = (HS-hS) = 2182 kJ/kg (2267.5 kg/h *309.23 kJ/kg) + (6802.5 kg/h * 2664 kJ/kg) – (0) = ṁS (HS-hS) q = ṁS (2182 kJ/kg) ṁs=8626.5 kg/h q = 8626.5 kg/h*2182 kJ/kg = 1.88x107 kJ/h = 5228621 W = 5.23 MWModule 9 – Steady state simulation 18
  • 19. NAMP PIECE Find the Economy = ṁV/ṁS 6802.5 kg/h Economy = = 0.788 8626.5 kg/hModule 9 – Steady state simulation 19
  • 20. NAMP PIECE 4. Calculate Required Heating Surface Condensing temperature of steam (1.37 atm gauge = 126.1oC q = UAΔT A = q/UΔT 5228621 W A= = 98.02 m2 W 1400 2 o (126.1 - 88) o C m CModule 9 – Steady state simulation 20

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