1. Design of agitated Heat transfer vessel

2. GROUP MEMBERS SABA 06-CHEM-02 FARIHA 06-CHEM-16 SHAZIA 06-CHEM-38

3. Dual Shaft Paste Mixer The Dual Shaft Mixer includes an Anchor agitator and a High Speed Disperser. The anchor feeds product into the high speed disperser blade and ensures that the mixture is constantly in motion. The anchor can be provided with scrapers to remove materials from the interior vessel walls to enhance the heat transfer capabilities of the mixer. Both agitators are available for variable speed operation The Dual Shaft Mixer includes an Anchor agitator and a High Speed Disperser. The anchor feeds product into the high speed disperser blade and ensures that the mixture is constantly in motion. The anchor can be provided with scrapers to remove materials from the interior vessel walls to enhance the heat transfer capabilities of the mixer. Both agitators are available for variable speed operation

4. Dual Shaft Paste Mixer

6. Jacked and coil heat transfer agitator vessel

7. Heat transfer coil in side vessel

8. Use of baffles

9. Applications Mixing tanks are typically used in the production of viscous fluids such as petroleum, plastics, paints, paper, cosmetics, and food. The mechanical agitation of fluid in these vessels can significantly increase the rate of heat transfer between the process and cooling fluids. Since the 1950s, a number of authors have explored heat transfer for Newtonian fluids in a variety of agitated vessel configurations. There has been a limited amount of research performed, however, for heat transfer to non-Newtonian (power law) fluids. Mixing tanks are typically used in the production of viscous fluids such as petroleum, plastics, paints, paper, cosmetics, and food. The mechanical agitation of fluid in these vessels can significantly increase the rate of heat transfer between the process and cooling fluids. Since the 1950s, a number of authors have explored heat transfer for Newtonian fluids in a variety of agitated vessel configurations. There has been a limited amount of research performed, however, for heat transfer to non-Newtonian (power law) fluids.

10. chemical, food, drugs dyes and other industries

11. Use for Pharmaceutical Application

12. used in chemical and pharmaceutical industries

13. Petro-Chemical, Polymers,Coatings & Adhesives,Agricultural & General Chemicals,Plastics&Rubber,Food & Beverage Industry. Petro-Chemical, Polymers,Coatings & Adhesives,Agricultural & General Chemicals,Plastics&Rubber,Food & Beverage Industry.

14. Homogeneous Batch/Semi-batch Reactions Homogeneous batch/semi-batch are the most common reaction type. They can be as simple as adding measured quantities of two or more reactants to a vessel and mixing. Heat is normally removed or added through a jacket, heating mantle or external heat exchanger Homogeneous batch/semi-batch are the most common reaction type. They can be as simple as adding measured quantities of two or more reactants to a vessel and mixing. Heat is normally removed or added through a jacket, heating mantle or external heat exchanger

16. STATEMENT STATEMENT Calculate the time required to heat 1268 gal. of liquid from 80 O F to 180 O F in a jacketed, agitated vessel conforming to standard configuration as shown in the figure. The vessel is assumed to be clean, free of fouling films and heated with 212 O F steam. Calculate the time required to heat 1268 gal. of liquid from 80 O F to 180 O F in a jacketed, agitated vessel conforming to standard configuration as shown in the figure. The vessel is assumed to be clean, free of fouling films and heated with 212 O F steam.

17. Data Given Liquid Properties: Cp = 0.6 Btu/(lb) ( O F) k= 0.1Btu/(hr)(sq.ft)( O F/ft) μ= 1000cp (at80F) ρ=60lb/ft3 = 8.02lb/gal Cp,k and ρ are assumed to be constant

18. Steam properties: h s =1000 Btu/(hr)(sq.ft)( O F) Vessel Properties: Wall thickness= 0.125 in K of vessel = 9.4 Btu/(hr)(sq.ft)( O F/ft) Eq used for calculating heat transfer coefficient N NU = 0.73(N RE ) 0.65 (N PR ) 0.33( μ w / μ b ) 0.14

20. STEP 1: Diameter of the vessel D T =6 ft. Diameter of the vessel D T =6 ft. 6-ft diameter agitated vessel conforming to standard configuration as shown in the figure. The vessel is equipped with the 2 ft diameter 6- blade, flat-blade turbine impeller running at 100 rpm. Diameter of impeller = 2ft Diameter of impeller = 2ft Impeller blade width=0.5 ft Impeller blade width=0.5 ft Impeller blade height=0.4 ft Impeller blade height=0.4 ft Baffle width=0.6 ft Baffle width=0.6 ft

21. STEP 2: Inside area of the vessel = πDL (L=D) = 3.14(6)(6) = 3.14(6)(6) = 113 ft2 = 113 ft2

22. STEP 3: Reynolds number evaluation at Reynolds number evaluation at 80F,130F,180F by using, N RE = ρN D i 2 /μ N RE = ρN D i 2 /μ Here, we assume density to be constant over the temperature range. Viscosity values at different temperatures is; μ(80F) = 1000cp (from table) μ(80F) = 1000cp (from table) μ(130F) = 270 cp μ(130F) = 270 cp μ(130F) = 84cp μ(130F) = 84cp

24. Diameter of impeller, D i = 2 ft From these values, the Reynolds number is, N RE (80F) = (60)(6000)(4)/(2420) N RE (80F) = (60)(6000)(4)/(2420) = 595 = 595 N RE (130F)= 2200 N RE (130F)= 2200 N RE (130F)=7080 N RE (130F)=7080

25. STEP 4: The Prandtl number calculations at 80F,130F,180F are, The Prandtl number calculations at 80F,130F,180F are, N PR = C P μ /k N PR = C P μ /k N PR (80F) = (0.6)(2420)/(0.1) N PR (80F) = (0.6)(2420)/(0.1) = 14500 = 14500 N PR (130F) = 3920 N PR (130F) = 3920 N PR (180F) = 1220 N PR (180F) = 1220

26. STEP 5: Approximate the value of inside heat transfer coefficient from given equation; Approximate the value of inside heat transfer coefficient from given equation; N NU = h i D T /k = 0.73(N RE ) 0.65 (N PR ) 0.33 substituting the appropriate values into this relationship gives: substituting the appropriate values into this relationship gives: h i (80F) =18.4 Btu/(F) (hr) (sq.ft) h i (130F) =27.6 Btu/(F) (hr) (sq.ft) h i (180F) =40.4 Btu/(F) (hr) (sq.ft)

27. STEP 6: The wall temperature from the above heat transfer coefficients are calculated and used to evaluate the viscosity of liquid at vessel wall The wall temperature from the above heat transfer coefficients are calculated and used to evaluate the viscosity of liquid at vessel wall The wall temperature is estimated from the approximate equation: The wall temperature is estimated from the approximate equation: T W = T S – [(T S -T B )/1+(hsA O /hiA i )] T W = T S – [(T S -T B )/1+(hsA O /hiA i )] here, A O = A i here, A O = A i

28. Solving wall temperature T W (at T B = 80F) = 209.6 F T W (at T B = 130F) = 209.8 F T W (at T B = 180F) = 210.7 F Viscosity values at different temperatures is; μ(209.6F) = 47cp (from table) μ(209.6F) = 47cp (from table) μ(209.8F) = 47cp μ(209.8F) = 47cp μ(210.7F) = 46 cp μ(210.7F) = 46 cp

30. Calculate the viscosities ratios equals to μ w / μ b  At T B =80F and T W = 209.6F V IS =47/1000=0.047  At T B =130F and T W = 209.8F V IS =47/270=0.174  At T B =80F and T W = 209.6F V IS =46/85=0.541

31. STEP 7: N NU = 0.73(N RE ) 0.65 (N PR ) 0.33( μ w / μ b ) 0.14 As N NU = h i D T /k h i (80F) =38.4 Btu/(F) (hr) (sq.ft) h i (130F) =42.2 Btu/(F) (hr) (sq.ft) h i (180F) = 46.7Btu/(F) (hr) (sq.ft)

32. STEP 8: 1/Ui = 1/hi+ x/k +1/hs hi= as calculated in step 7 x= 1/8 in =0.0104 ft K= 9.4 Btu/(hr)(sq.ft)(F/ft) for the vessel wall h s = 1000 Btu/(F) (hr) (sq.ft) U i (Tb=80 F) =0.0281 Btu/(F) (hr) (sq.ft) U i (Tb=130 F) =38.7 Btu/(F) (hr) (sq.ft) U i (Tb=180 F)= 42.5 Btu/(F) (hr) (sq.ft)

35. STEP 9: Time taken to heat the liq over each temp increment is calculated by this formula T hr =(MC P /U i A i ) ln [T S – T I /T S –T f ] Liquid mass= π(D t 2 /4)(H t )(ρ) = π(36/4)(6)(60) = π(36/4)(6)(60) =10180 lb =10180 lb

36. Time reqd to heat the nass from 80 to 100 F is calculated as t(80-100F) t(80-100F) = (10180)(0.6) ln (212-80) (36.15)(113) (212-100) (36.15)(113) (212-100) =0.242 hr =0.242 hr t(100 -120F) = 0.283 hr t(100 -120F) = 0.283 hr t(120 -140F) = 0.340 hr t(120 -140F) = 0.340 hr t(140-160F) = 0.439 hr t(140-160F) = 0.439 hr t(160-180F) =0.630 hr t(160-180F) =0.630 hr Total Time = 1.934 hr

37. Approximate method: T hr =(MC P /U i A i ) ln [T S – T I /T S –T f ] T hr =(10180*0.6/38.7*113) *ln [212-80/212 -180] *ln [212-80/212 -180] T hr =1.981 hr Value differ by only 2.4%