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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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