1. RIVERA, ALYSSA A. 5ChE-C

2. Evaporation..  Evaporation is basically a separation step which uses heat transfer to separate products presenting differences at boiling point.  It differs from the other mass transfer operations such as distillation and drying.  The major requirement in the field of evaporation technology is to maintain the quality of the liquid during evaporation and to avoid damage to the product.

3. There are two main of ways of improving steam economy in evaporators. Mechanical vapor recompression. Multiple effect evaporator

5. Auxiliaries: 1.Heat exchanger 2.Separator 3.Condenser 4.Compressor

6. Feed stock  Sugar (C12H22O11) is an organic compound, colourless, sweet-tasting crystals that dissolve in water. In addition to providing a sweet taste and flavour, sugar performs a variety of functions in food products. Sugar beets accounted for 20% of the world's sugar production. Sugar beets, after washed and sliced go through a large tank called a diffuser where raw sugar juice is extracted. Sugar juice is now purified resulting to a solution called thin juice. This thin juice having a 10-14% solids is evaporated to 60% to be fed to a crystallizer to prudence sugar crystals.

7. Rationale for equipment selection: (Triple Effect, Backward Feeding) -Backward Feeding is preferred in this case because our feed has the possibility to be viscous and the feed is cold. Falling Film Evaporator -Short vertical evaporator are usually used for evaporating thin juice but because, we are requiring big capacity of feed rate. Falling film is used. This evaporator is also used for thin juice evaporation.

8. Falling Film Evaporator In a falling film evaporator, the liquid is fed at the top of the tubes in a vertical tube bundle. The liquid is allowed to flow down through the inner wall of the tubes as a film. As the liquid travels down the tubes the solvent vaporizes and the concentration gradually increases. Vapor and liquid are usually separated at the bottom of the tubes and the thick liquor is taken out. Evaporator liquid is recirculated through the tubes by a pump below the vapor-liquid separator.

9. Material and Energy Balance / Flow diagram

10. Enthalpy Balance: (Hs-hc)=Lo  1 st Effect : V o L o =(F-V 3 -V 2 )Cp(T 1 -T 2 )+V 1 L 1  2 nd Effect : V 1 L 1 =(F-V3)Cp(T2-T3)+V2L2  3 rd Effect: V 2 L 2 =FCp(T 3 -T f )+V 3 L 3 Heat Transfer Equations:  1 st Effect : V O L O =U 1 A 1 ∆T­ 1  2 nd Effect : V 1 L 1 =U 2 A 2 ∆T 2  3 rd Effect : V 2 L 2 =U 3 A 3 ∆T 3 Over-all Material Balance : Feed = Effluent + Liquid Solute Balance : Fx F = Lx L

11. Equipment Design Theoretical calculations Step 1: Over-all Material Balance : Feed = Effluent + Liquid 10000 kg/hr = E + L Solute Balance : Fx F = Lx L 10000(0.14) = (0.60)L Solving simultaneously:E = 7.667X10 3 kg/hr L= 2.333X10 3 kg/hr

12. Step 2: Assume pressure for 3 rd effect. (Setting it to 550mmHg Vacuum). Entering steam is at its saturation temperature. Assume over-all heat transfer coefficient (W/m 2 K): U1=2500, U2= 2000 and U3=1600 (Literature 200-600 Btu/ hr ft^2 F) P o =233.54KPa  T o ­ = 398.15K ∑∆T= 57.591K

13. Step 3: Temperature and enthalpy profile  T o =398.15KL 0 =2198.165 kJ/kg  T 1 =383.044KL 1 =2240.316 kJ/kg  T 2 =364.162KL 2 =2289.5605 kJ/kg  T3=340.559KL 3 = 2300.8843kJ/kg Step 4: Energy Balance in 3 Evaporators 1 st Effect : V o L o =(F-V 3 -V 2 )Cp(T 1 -T 2 )+V 1 L 1 2 nd Effect : V 1 L 1 =(F-V3)Cp(T2-T3)+V2L2 3 rd Effect: V 2 L 2 =FCp(T 3 -T f )+V 3 L 3

14. Step 5: Solve for stream rates V o,V 1,V 2, V 3 1 st eq: V o (2198.165)=(10000-V3-V2)(3.8029)(109.894-91.012)+ V1(2240.316) 2 nd eq: V 1 (2240.316)=V 2 (2289.5605) + (10000-V 3 )(3.8029)(91.012-67.409) 3 rd eq; V 2 (2289.5605)=(10000)(3.8029)(67.409-20)+(V3)(2300.8843) 4 th eq: 7.667X10 3 =V1+V2+V3 V o = 3.31283x10 3 V 1 = 3.07708x10 3 V 2 =2.6932x10 3 V3=1.89638x10 3

15. Step 6: Check if the three areas are equal 1 st Effect : V O L O =U 1 A 1 ∆T­ 1 2 nd Effect : V 1 L 1 =U 2 A 2 ∆T 2 3 rd Effect : V 2 L 2 =U 3 A 3 ∆T 3 A 1 = 53.564 m 2 A mean = 49.875m 2 A 2 = 50.706 m 2 A 3 = 45.356 m 2

16. * Area must be less than 3 % difference with the A mean

17.  Repeat Step 3: temperature and enthalpy profile  T o =398.15KL 0 =2198.165 kJ/kg  T 1 =381.725KL 1 =2243.8754 kJ/kg  T 2 =362.289KL 2 =2294.2492 kJ/kg  T3=340.558KL 3 = 2300.8421kJ/kg  Step 4: Energy Balance in 3 Evaporators 1 st Effect : V o L o =(F-V 3 -V 2 )Cp(T 1 -T 2 )+V 1 L 1 2 nd Effect : V 1 L 1 =(F-V3)Cp(T2-T3)+V2L2 3 rd Effect: V 2 L 2 =FCp(T 3 -T f )+V 3 L 3  Step 5: Solve for stream rates V o,V 1,V 2, V 3 V o = 3.3033x10 3 V 1 = 3.05841x10 3 V 2 =2.69978x10 3 V3=1.90847X10 3

19. A= *D*L  Diameter and length has a ratio of 1:4 L=4D 49.83= *D*(4D) D=1.9793m L= 4D= 7.9173m  Maximum Allowable Working Pressure: 233.54kPa*(14.7/101.325)= 33.88145 psia 33.88145+ 25= 58.88psia  Maximum Allowable Temperature: 125°C+ 30°C= 155°C= 311°F

21. Equipment Specification Material of Construction Stainless steel (304) will be used as the material of construction because it does not readily corrode, rust or stain with this kind of feed solution as ordinary steel does.

22. Shell Specifications a. Material: stainless steel: type (304) b. Maximum allowable stress of stainless steel = 14884.5psi c. Length: 7.916 meters d. Diameter: 1.979 meters e. Thickness: 6.9236 mm f. Evaporator heads(thickness): hemispherical(4.958mm) & flat head(71.174mm)

23. Pipe Inside the Evaporator Specifications (Assume: Based on Heuristics) a. Material: 3 1/3 Schedule 40 stainless steel b. Length: 5 meters c. Inside Pipe Diameter: 3.5 in d. Outside Pipe Diameter: 3.334 in e. Number of tubes: 20 tubes (maximum)

24. Heuristics Maximum allowable PRESSURE AND TEMPERATURE -Operating pressure inside is up to 233.54KPa. Design pressure for these evaporators are equal to operating pressure plus 25psig or 10%. -Design temperature is equal to Operating temperature plus 30°C or 50°F. Shell and head thickness is obtain with design pressure and design temperature. Computing for diameter and length: having a ratio of 1:4. Maximum allowable stress is dependent on the material of construction and the design temperature. Internal tubes for the evaporator can be range to 19–63mm (0.75–24.8 in.) in diameter and 3.66–9.14m (12–30 ft) long. For corrosion allowance, 3mm is consider because the feed is not corrosive. Also welded joint efficiency 1.00 because it is assume that the equipment is a single/ double-welded butt joint that is fully radiographed.

25. 3D DIAGRAM

26. Prototype

27. REFERENCES:  http://nptel.ac.in/courses/103107096/module4/lecture2/lecture2.pdf http://nptel.ac.in/courses/103107096/module4/lecture2/lecture2.pdf  http://www.spxflow.com/cn/assets/pdf/Evaporator_Handbook_10003_01_08_2008_US.pdf http://www.spxflow.com/cn/assets/pdf/Evaporator_Handbook_10003_01_08_2008_US.pdf  https://www.bma-worldwide.com/products/sugar-and-sweeteners/tubular-fallingfilm- evaporator.html  http://multiple-effect-evaporation.webs.com/ http://multiple-effect-evaporation.webs.com/  http://rajprocessequipment.com/download/EVAPORATOR.doc http://rajprocessequipment.com/download/EVAPORATOR.doc