JSS MAHAVIDYAPEETHA JSS

1 JSS MAHAVIDYAPEETHA JSS Science and Technology University Sri Jayachamarajendra College of Engineering JSS Technical Institutions Campus, Mysuru 570006 Chemical Engineering Lab Instruction Manual For 4 th Semester B.E. degree Department of Polymer Science and Technology

2 Outcome Based Education Statements Sri Jayachamarajendra College of Engineering VISION: Be an international leader in engineering education, research and application of knowledge to benefit society globally. MISSION M1: To synergistically develop high-quality manpower and continue to stay competitive in tomorrow's world. M2: To foster and maintain mutually beneficial partnerships with alumni, industry and government through public services and collaborative research M3: To create empowered individuals with sense of identity. Department of Polymer Science and Technology VISION: To excel in polymer engineering education and research and to serve as valuable resource for multi-faceted industry and society. MISSION M1: To provide well balanced curriculum and conducive environment to excel in polymer and allied engineering disciplines. M2: Creating state-of-the-art facilities for polymer research and make the best use of the facilities M3: To undertake collaborative projects for long term interactions with academia and industries. Program Outcomes (POs)[Graduate Attributes] Engineering Graduates will be able to: PO1 Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals, and an engineering specialization to the solution of complex engineering problems. PO2 Problem analysis: Identify, formulate, review research literature, and analyze complex engineering problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and engineering sciences.

3 PO3 Design/development of solutions: Design solutions for complex engineering problems and design system components or processes that meet the specified needs with appropriate consideration for the public health and safety, and the cultural, societal, and environmental considerations. PO4 Conduct investigations of complex problems: Use research-based knowledge and research methods including design of experiments, analysis and interpretation of data, and synthesis of the information to provide valid conclusions. PO5 Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern engineering and IT tools including prediction and modeling to complex engineering activities with an understanding of the limitations. PO6 The engineer and society: Apply reasoning informed by the contextual knowledge to assess societal, health, safety, legal and cultural issues and the consequent responsibilities relevant to the professional engineering practice. PO7 Environment and sustainability: Understand the impact of the professional engineering solutions in societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable development. PO8 Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the engineering practice. PO9 Individual and team work: Function effectively as an individual, and as a member or leader in diverse teams, and in multidisciplinary settings. PO10 Communication: Communicate effectively on complex engineering activities with the engineering community and with society at large, such as, being able to comprehend and write effective reports and design documentation, make effective presentations, and give and receive clear instructions. PO11 Project management and finance: Demonstrate knowledge and understanding of the engineering and management principles and apply these to one s own work, as a member and leader in a team, to manage projects and in multidisciplinary environments. PO12 Life-long learning: Recognize the need for, and have the preparation and ability to engage in independent and life-long learning in the broadest context of technological change.

4 Program Specific Outcomes (PSOs) Graduates receiving the Bachelor of Engineering in Polymer Science and Technology will be- PSO1: able to address the issues related to polymeric material design with the knowledge of synthesis, structure property relationships and compounding. PSO2: able to utilize the knowledge of polymer processing, testing, characterization, product and mold design to produce acceptable engineering products. PSO3: capable of taking up research/ higher education in premier institutes and employment in polymer industries viz. rubbers, plastics, composites, adhesives, paints, etc. Program Educational Objectives (PEOs) PEO1: Provide graduates with strong fundamentals of science and engineering for a successful career in Polymer Science & Technology. PEO2: Enable graduates to pursue higher education and take up innovative research so as to solve multidisciplinary issues. PEO3: Equip graduates with technical skills and moral values for being responsible individuals. Program Specific Criteria (PSCs) PSC1: Apply chemistry, physics to understand structure, properties, rheology, processing, performance of polymeric materials systems, for selection and design of suitable components and process to address the concerned issues. PSC2: Career development of members through research, consultancy and professional activities. PSC3: Provide collaborative learning opportunities through professional societies, industries, institutes and alumni fraternity for promoting PST education along with professional growth of associated individuals.

5 CONTENTS ************** Expt. No. TITTLE OF THE EXPERIMENT 1. VAPOUR LIQUID EQUILIBRIUM 2. TRAY DRIER 3. SIMPLE DISTILLATION 4. STEAM DISTILLATION 5. PACKED BED DISTILLATION 6. FLOW THROUGH FLUIDIZED BED 7. HEAT TRANSFER IN A JACKETED VESSEL 8. HEAT TRANSFER IN A PACKED BED 9. DOUBLE PIPE HEAT EXCHANGER 10. HEAT TRANSFER THROUGH BARE AND FINNED TUBE HEAT EXCHANGER 11. SINGLE EFFECT EVAPORATOR 12. VERICAL CONDENSER 13. CALIBRATION OF THERMOCOUPLE

6 EXPERIMENT 1 VAPOR-LIQUID EQUILIBRIUM Aim: To obtain the vapor-liquid equilibrium data for the given binary system. Apparatus: 500 ml Round bottom flask with necessary joints, a condenser, condensate collector, specific gravity bottles. Theory: consider a binary mixture AB in which A is more volatile. Therefore vapor pressure of A at any temperature is greater than that of B. According to the phase rule F= C-P+2 there has to be F= 2-2+2=2. i.e., degree of freedom for such a system is two. Thus the vapor-liquid equilibrium between pure components A and B in a mixture is its vapor pressure-temperature relationship. For binary mixtures, an additional variable, concentration must be considered. In general VLE can be represented by P xy diagram for a fixed temperature or T xy diagram for a fixed pressure. It is customary to represent concentration in terms of mole fraction in VLE calculations. The mole fraction of the more volatile component (A) in the liquid is designated as x and that in the vapor phase as y*. Procedure: 1. Take 100 ml of the given unknown mixture in a 500 ml round bottomed (RB) flask and heat the flask over a mantle heater. 2. The vapor condenses and collects in the collecting chamber. Pour this condensate back to the flask and continue heating for about 20 minutes and note down the boiling point. Run off the condensate formed during this time and find its specific gravity. 3. Transfer this condensate along with the sample taken for finding specific gravity into the RB flask with 15 ml of pure methanol. 4. Heat the RB flask again for 15 minutes and note down the boiling point. The condensate which is collected at the end of 15 minutes is run-off to find the specific gravity. 5. Again transfer this condensate along with an additional 25 ml of pure methanol. Heat the RB flask for 15 minutes and record the boiling point. Collect the condensate and find its specific gravity.

7 6. Repeat this procedure for additional 40 and 60 ml of methanol. 7. Find out the specific gravity of the original mixture. 8. Prepare a specific gravity v/s mole fraction chart. 9. Prepare T xy plot (x, y * on X-axis and T on Y-axis) for experimental values and compare them by plotting the literature values. 10. Also report % deviation of each set of experimental x, y * with that of literature values for corresponding temperatures. Observations and Calculations: 1. Calibration chart: Let A = Methanol and B = water ρ A = 0.792 g/cm³; ρ B = 1.0 g/cm³ Sample No Vol.of A (ml) Vol. of B (ml) Wt. of A (g) Wt. of B (g) Moles of A Moles of B Mole fraction of A Sp.gr.

8 Calculation of volumetric composition of methanol and water in the original mixure: Sp.gravity of the original mixure = Mole fraction of methanol in the original mixure, X A = (from calibration chart) V A = [ X A (MW) A ]/ ρ A =.cm³ V B = [ X B (MW)B]/ ρ B =.cm³ V t = V A + V B V t cm³ mixture contains V A cm³ of methanol. 100 cm³ mixture contains [V A x 100/ V t ] =.cm³ (say V 1 *) Vol. of water in the original mixture = [100-V 1 *] = cm³ (say V 2 *) Vapor liquid equilibrium calculations: Trial No. Vol. of methanol added (ml) Vol. of mixture after addition (ml) B.P ( C) Sp. gr. of the condensate 1 2 3 4 5

9 VLE - DATA Trial No. Vol. of B, (V 2 ) (ml) Vol. of A, (V 1 ) (ml) Mole fraction of A in liquid phase (x) using equation given below Sp. gr. of the condensate Mole fraction of A in vapor phase, (y*) from calibration chart 1 V 2 * V 1 * 2 V 2 * V 1 * + 15 3 V 2 * V 1 * + 40 4 V 2 * V 1 * + 80 5 V 2 * V 1 * + 140 Results : [ V 1 x ρ A / (MW) A ] X = ------------------------------------------------------ [ V 1 x ρ A / (MW) A ] + [V 2 x ρ B / (MW) B ] (1) Composition of given mixture : 100 ml of given mixture contains Methanol =.ml Water =..ml (2) VLE data for methanol water system. T ( C) X (Exptl.) X (Lit.) % Deviation Y* (Exptl.) Y* (Lit.) % Deviation (3) Plot of T (on Y-axis) Vs. x, y*(x-axis) for experimental and Literature values.

10 EXPERIMENT 2 TRAY DRIER Aim: To conduct the phenomenon of drying and to find out (a) equilibrium moisture content (X * ) and (b) critical moisture content (X c ) Apparatus: Tray drier set up, stop-watch and thermometer. Theory: The phenomenon of removal of relatively small amount of water or other liquids from a solid material is called drying. This can be achieved by means of evaporationg the liquid by direct heating (i.e. adiabatic drying where the wet solid is directly exposed to hot air) or by indirect heating (i.e. nonadiabatic heating where the heat is transferred from an external medium such as condensing steam through a metal surface) Consider a bed of wet solids over which hot air is circulated. When the temperature, humidity, velocity and direction of flow of the air across the drying surface are constant, the drying process is said to be under constant drying conditions. The rate of water evaporated per unit area of drying is known as the rate of drying. If the water content is above a particular level, the water forms a film over the solid and until this film is depleted drying proceeds at a constant rate and this period of process is called constant rate period. The driving force of heat and mass transfer in this region is the difference in temperature. After the film disappears a concentration gradient between the hot upper layer and the cooler lower layer with more moisture results in a slower and a falling rate of drying. The moisture, before and after depletion of film is called bound and unbound moisture respectively. The moisture content at the time of complete depletion of film is called critical moisture content. The moisture content at the end of falling rate period is the equilibrium moisture content. This is the moisture content which is in equilibrium with the incoming air conditions and beyond this limit no more drying will occur.

11 Formulae used: Moisture content = [wt. of moisture present / wt. of dry solid] = x Rate of drying, N = [dx. S / A dt] = -------kg/m² s Where A = area f drying pan in m² S = wt. of dry sand in kg dt = time taken for the removal of moisture content equal to dx, in seconds Procedure: 1) Weigh out about 100 g. of given material into the pan. 2) Add 32-35 ml of water and make a uniform paste. The sand should be equally distributed in the pan. 3) Then take the weight of sand and water in the balance provided in the drier. 4) Remove the pan from the balance and cover it with a polythene sheet so that no water gets evaporated. 5) Switch on the blower and heater and wait till the air attains a temperature of 60 C. 6) Place the pan on the balance after removing polythene sheet. 7) Start the stop watch and note down the time required for removal of 1 g. of moisture. 8) At the same time remove 1g. weight from the balance and again note down the time for the removal of 1 g. of moisture cumulatively. 9) Repeat the procedure until 30-31 g. of moisture is removed. 10) Find out the rate of drying and plot the graph between moisture content vs. time and rate of drying vs. moisture content. Find out the values of critical moisture content and equilibrium moisture content from the graph. Observations and calculations: Tray diameter (d) = 162 mm = m Wt. of material taken (S) = 100 g =. kg Wt. of water added =..g =..kg Area of tray (A) = [πd²/4] =.m²

12 Sl. No 1 2 3 4...... Wt. of moisture removed (kg) Moisture content, x dx Time (sec) dt N = [dx. S/A dt] (kg/m² s) 30 Result: Critical moisture content (X c ) =. Equilibrium moisture content (X * ) =.

13 EXPERIMENT 3 SIMPLE DISTILLATION Aim: To verify Rayleigh s equation for the given acetone-water system. Apparaturs: 3-necked flask, specific gravity bottle, chemical balance, water cooled condenser, thermometer etc. Theory: Rayleigh s equation is given by: In (F/W) = [/( )] Distillation is the separation of constituents of a liquid mixture via partial vaporization of the mixture and separate recovery of vapor and residue. Distiallation can be carried out by simple methods such as flash distillation, simple or differential distillation and steam distillation or by rectification or fractionation where only a part of the condensate is taken as product and the rest is recycled as reflux. Flash distillation consists of vaporizing a definite fraction of the liquid in such a way that the evolved vapor is in equilibrium with the residual liquid, separating the vapor from the liquid and condensing the vapor. Differential or simple distillation is considered to be made up of many steps of flash distillation. The difference between these two is the composition in flash distillation is equilibrium composition whereas in simple distillation equilibrium composition is obtained only at the instant of vapor production. The process is governed by Rayleigh s equation which is drived based on the material balance for the system. Let n 0 moles of a binary mixture is charged to batch still. Assume that at any time during the course of distillation, there are n moles of liquid in the still with mole fraction of the more volatile component (A) as x and mole fraction of vapor as y* in equilibrium with the liquid. Then the total moles of A left in the still, n A will be n A = n x (1) If a small amount of liquid dn is vaporized, the change in moles of component A is y* dn = dn A (2)

14 Upon differentiating equation (1), we get dn A = n dx + x dn (3) From (2) and (3), we get y* dn = n dx + x dn (4) By rearranging eq. (4), we get dn/n = [ 1 / (y* - x ) ] dx (5) Integrating eqn (5) between limits x w and x f, we get = (/ )= [1/( ) ] dx The above equation is known as Rayleigh s equation and is used to find out the composition of distillate. Notations: F = no. of moles of feed. W = no. of moles of residue x f = mole fraction of more volatile component (acetone) in feed x w = mole fraction of more volatile component (acetone) in residue x, y* = equilibrium data for acetone-water mixture Procedure: (1) Take 100 ml of acetone (component A) and 100 ml of water (component B) in a distillation flask. (2) Set up the distillation apparatus as shown in the figure. (3) Carry out the distillation till 90-95 ml of distillate gets collected in the receiver. (4) See that the distillate does not get evaporated. (5) Collect the residue and distillate. (6) Find the volume and specific gravity of residue. (7) Prepare mixtures of various compositions of acetone and water and find out their specific gravities. Plot a graph of specific gravity vs.mole fraction of acetone.

15 (8) Using the above graph find out the mole fraction of acetone in residue. (9) Calculate the value of ln (F/W) and compare it with the area under the curve [1/(y*-x) vs. x within limits x w and x f. Observations and calculations Volume of acetone = 100 ml Volume of water = 100 ml Volume of the feed = 100 + 100 = 200 ml Volume of distillate =.ml Volume residue = ml Density of acetone = 0.792 g/cm³ Wt. of empty sp. gravity bottle = W 1 = g Wt of empty sp. gravity bottle + water = W 2 = g Wt. of empty sp. gravity bottle + residue = W 3 = g Sp. gravity of residue = [W 3 W 1 ] / [ W 2 W 1 ] Mole fraction of acetone in residue, X W can be found out from the standard graph of Sp. gravity vs. mole fraction, corresponding to this specific gravity. CALIBRATION CHART Sl. No 1 Vol. of water (ml) 10 Vol. of Acetone (ml) 0 Wt. of Mixture (g) Wt. of water (g) Wt. of acetone, (g) gmol of water gmol of acetone Mole fraction of acetone Sp. gravity of mixture 2 7 3 3 6 4 4 4 6 5 2 8 6 1.5 8.5 7 1 9 8 0.5 9.5 9 0 10

16 Calculation of x f : Moles of acetone in feed = [Volume x density / mol. Wt. of acetone] = [100 x 0.792 / 58 ] =. Moles of water in feed = [Volume x density / mol. Wt of water] = [100 x 1.0 / 18] =. Mole fraction of acetone in feed, x f = mole of acetone / [moles of acetone + moles of water] (1) F = no. of moles of feed = no. of moles of acetone + no. of moles of water =. (2) AMW of residue = [ x w. MW of acetone ] + [ (1- x w ) MW water ] (3) W = moles of residue = [ Volume of residue x density of residue ] / AMW (4) Obtain the equilibrium data x Vs. y* from perry s Hand book. Result: Rayleigh s equation for the given acetone-water system is verified. In (F/W) = [1/( )]! " # " $

17 EXPERIMENT 4 STEAM DISTILLATION Aim: To study the characteristics of steam distillation. To determine the values of vaporization efficiency (η v ) and thermal efficiency (η t ) for steam distillation of aniline Apparatus: Distillation flask, steam generator, water cooled condenser, thermometer, separating funnel, specific gravity bottle, measuring jar. Theory: Steam distillation is the term applied to a batch of continuous distillation process with open steam. The liquid is distilled by feeding open steam directly into the distillation still, so that the steam carries with it the vapors of volatile liquid component and is then condensed to separate the liquid from water. Steam distillation is possible only when, 1. The substance does not react with steam at the given conditions of temperature and pressure. 2. The substance is insoluble in water. Steam distillation method is used for the separation of high-boiling substances from non-volatile impurities or for the removal of very high-boiling volatile impurities from still higher-boiling substances. The process has special value where it is desired to separate substances at temperature lower than their normal boiling points because of heat sensitivity or other reasons. The condensed organic liquid and water do not miscible in each other. Therefore partial pressure of each component is equal to vapor by each liquid. If P is total pressure, P A and P w are vapor pressure of organic liquid and water respectively. Then P = P A + P W. For ambient distillation, P = 1 atm = 101.3 kpa. Therefore P A v/s T and (101.3 P w ) v/s T plots intersects and this point corresponds to distillation temperature. Steam Requirements: Let y A and y w represent mole fraction of organic liquid and H 2 O respectively in vapor phase. Then by Dalton s law P A = Py A = P [N A /(N A +N W )] and P W = Py w = P [ N W / N A + N W ) ] where N A and N W are moles of A and W and P A and P W are partial pressure of A and W.

18 Therefore P A /P W = N A /N W = (W A /M A ). We have P A = P A and Po W =P W for steam distillation. Therefore W w /W A = (P W /P A ). (M W /M A ) (1) Obtain P W and P A at distillation temperature from literature. The term (W w /W A ) is the amount of open steam condensed per unit weight of liquid condensed however the actual requirement is higher. In equation (1) vapor pressure of liquid is assumed that be equal to its partial pressure. FORMULAE TO BE USED: Vaporization efficiency, η v (wt. of aniline distilled/unit wt. of steam) actual [W s /W w ] actual η v = --------------------------------------------------------- = ---------------------- (wt. of aniline distilled / unit wt. of steam) ideal [W s /W w ] Ideal [Vol. of aniline in distillate x density. of aniline in distillate] [W s / W w ] actual = --------------------------------------------------------------------------- [Vol. of water in distillate x density. of water in distillate] [W s / W w ] ideal = [P s M s ] / [P w M w ] Where P s = Vapor pressure of aniline at distillation temperature P w = Vapor pressure of water at distillation temperature M s and M w = Molecular weights of aniline (93.12) and water (18) Thermal efficiency, η t [Distillation requirement of steam/unit wt. of sample distilled] η t = ------------------------------------------------------------------------------ [Actual requirement of steam / unit wt. of sample distilled] i.e. η t = S t / S a where,

19 (1+R) C p, s (T d T r ) + λ s [P w M w ] S t = -------------------------------------- + ------------------- λ w + (T s T d ) C p, w [ P s M s ] [Volume of H 2 O in residue x density of H 2 O in residue] + [volume of H 2 O in distillate x density of H 2 O in distillate] S a = --------------------------------------------------------------------------------------------------------- [volume of ANILINE in distillate x density of ANILINE in distillate ] Wt. of aniline in residue R = ----------------------------------- Wt. of aniline in distillate T r = room temperature T d = distillation temperature T s = Steam temperature C p,s = specific heat of aniline at T d C p,w = specific heat of water at T d λ s = latent heat of vaporization of aniline at T d λ w = latent heat of vaporization of water at T d Note that [λ s / λ w ] = [M w / M s ] [d(ln P w ) / d(ln P s )] Where, M w and M s are molecular weights of water and aniline respectively. λs = [ M w / M s ] x [ d(ln P w ) / d ( ln P s ) ] [λ w ] PROCEDURE: 1. Take 100 ml aniline in the distillation flask and set up the apparatus. 2. Pass the steam at a pressure of 0.25 kg/cm² 3. Note down the distillation temperature. 4. Pass the steam till about 60-70 % of the liquid distills. 5. Stop the steam and allow the residue and distillate to get cooled. 6. Using the separating funnel, separate aniline and water in residue and distillate. 7. Measure the volume of aniline and water in residue and in distillate and find out specific gravities.

20 8. Draw a graph of ln P w vs ln P s and calculate the slope. Then calculate λ s. 9. Draw Hans-brandt chart [plot of (760 P s ) vs. temperature and P w vs. temperature) to find out the theoritical distillation temperature and compare it with the actual value. OBSERVATIONS AND CALULATIONS: Volume of aniline taken = 100 ml Distillation temperature T d = C Room temperature T r = C Steam temperature T s = C (from steam tables) Steam pressure, P = 0.25 kg/cm² - 0.3 kg/ cm² C p, s =. (from perry s H.B) C p, w = (from Perry s H.B) λ w = (from Perry s H.B) To get λ s, obtain Vapor pressure data for aniline and water (from Perry s H.B) Then λ s = [M w / M s ] [d(ln P w ) / d (ln P s ) ] x [λ w ] [d(ln P w ) / d(ln P s )] is given by slope of ln P w Vs. ln P s plot Wt. of empty sp. gr. bottle = W 1 =..g Wt. of empty sp. gr. bottle + water = W 2 = g Description Water Distilate Aniline Volume (ml) Wt. of sp. gr Bottle + soln. (g) Specific gravity Resiude Water Aniline To plot ln P w Vs. P s graph and Hans-brandt graph, the data extracted from Perry s Hand book is given below.

21 Sl. No. 1 Temperature (ºC) 34.8 P s (mm Hg) 1 P w (mm Hg) 41.71 (P-P s ) P=760 mm Hg 759 ln P s ln P w 2 57.9 5 136.08 755 3 69.4 10 223.73 750 4 82.0 20 384.90 740 5 96.7 40 674.60 720 For calculation of η v and η t [ W s / W w ] actual η v = ---------------------- X 100 =..% [ W s / W w ] ideal η t = [ S t / S a ] x 100 =..% (Show one specimen calculation to find out specific gravity) Results: For steam distillation of aniline Vaporization efficiency, η v =.. % Thermal efficiency, η t =.%

22 EXPERIMENT 5 PACKED BED DISTILLATION Aim: To find out (a) number of transfer units (NTU) and (b) height of transfer unit (HTU) using Fenske s equation and Mc cabe-thiele method under total reflux condition. Apparatus: Packed bed column, 3- necked flask, specific gravity bottle, chemical balance, water cooled condenser, thermometer etc. Theory: Fractionation can be carried out in a plate column or a packed column. A packed column is usually specified when plate devices would not be feasible because of undesirable fluid characteristics or some special design requirement. Fractionation by means of a plate column is stage-wise process, a discontinuos change in the composition of the components being effected at each plate. In a packed column the liquid and vapour are in continuous countercurrent contact and fractionation is not stagewise. Hence, the difficulty of separation to be accomplished is best characterized in terms of transfer units rather than theoretical plates. The efficiency of packed column is most frequently expressed in terms of height of packing to give a separation of one transfer unit ( HTU). The height of packing to give a separation of one theoretical plate is HETP. Under total reflux conditions, where the ratio of slop of equilibrium curve to slope of operating line is unity, a transfer unit and a theoretical plate become identical. So in the case of an ideal binary mixture, the packed height can be determined by finding out the number of stages using the Mc Cabe Thiele diagram under total reflux conditions or by using the Fenske s equation which gives the minimum number of stages for a specified separation (NTU). Fenske s equation is given a ln x D ( 1 x w ) / X w ( 1 X D ) NTU = ------------------------------------------ - 1 ln ( α avg ) where, NTU minimum no. of stages for the specified separation under total reflux condition.

23 x D mole fraction of acetone in distillate x W mole fraction of acetone in residue α avg relative volatility Plot the Mc cabe-thiele diagram and find out the number of stages for the specific separation. Get the equilibrium data, x vs y for acetone-water mixture from perry s Hand book. Procedure: 1. Take 100 ml of water and 100 ml acetone in a round bottom flask. 2. Set up the apparatus. 3. Maintain the temperature of distillation around 55-60 C. 4. Allow the distillate to overflow for half an hour so that the vapors produced during the distillation are completely condensed and the condensate is refluxed. This provides the total reflux condition. 5. Collect the residue and distillate and find out their specific gravities. 6. Prepare mixtures of various compositions of acetone and water and find out their specific gravities. Plot a graph of sp.gravity vs. mole fraction of acetone. 7. Using the above graph, find out the values of mole fractions of acetone in the residue and distillate. 8. Calculate the values of NTU and HTU using Fenske s equation and Mc Cabe-Thiele method. Observations and Calculations: Volume of acetone = 100 ml Volume of water = 100 ml Room temperature =. C Distillation temperature = C Molecular weight of acetone = 58 Molecular weight of water = 18 Height of packing, Z = 43 cm Wt. of empty sp. Gravity bottle = W 1 =..g Wt. of empty sp. Gravity bottle + water = W 2 = g

24 Wt of empty sp. Gravity bottle + mixture = W 3 = g Sp. gr. of mixture = [W 3 W 1 ] / [W 2 W 1 ] g moles of acetone = [Volume of acetone x density ] / Molecular Weight. Mole fraction of acetone = [ Moles of acetone ]/{[ moles of acetone] + [ moles of water]} Calibration chart readings: Sl. No. 1 Vol. of Water (ml) 10 Vol. of Acetone (ml) 0 Wt. of mixture (g.) Wt. of water, (g.) Vol. of acetone (g.) gmol of water gmol of acetone Mole fraction of acetone Sp. gr. of mixture 2 7 3 3 6 4 4 4 6 5 2 8 6 1.5 8.5 7 1 9 8 0.5 9.5 9 0 10 Specific gravity of residue = Specific gravity of distillate = Vapor pressure of acetone Relative Volatility, α = ------------------------------------ Vapor pressure of water α 1 = Relative volatility at room temperature = P A1 / P B1 = α 2 = Relative volatility at distillation temperature = P A2 /P B2 =. Where P A2, P A2 are vapour pressure of acetone at room temperature and at distillation temperature. P B1, P B2 are vapour pressure of water at room temperature and at distillation temperature. α avg = [ α1 + α2 ] / 2

25 By Fenske s equation ln [ x D ( 1 x W ) / x W ( 1 x D ) ] NTU = ------------------------------------------ - 1 ln ( α avg ) HTU = Z/NTU Result: By Fenkse s equation By Mc Cabe -Thiele method NTU HTU

26 EXPERIMENT 6 FLOW THROUGH FLUIDIZED BED Aim : (a) To observe and study the behaviour of bed during fluidization (b) To determine the effect of velocity of fluid on pressure drop (c) To find effect of porosity on fluid velocity (d) To determine the minimum fluidization velocity from the experimental data and to compare it with the value obtained from theoretical relations. Apparatus: Fluidized bed experimental set-up, meter scale, 250 ml measuring jar, 100 ml beaker. Theory: Fluidization is one of the methods available for contacting granular solids with fluids. The main advantages of fluidized bed are greater interfacial surface area of contact and high rates of heat transfer. Fluidized beds find application in catalytic cracking, oxidation of naphthalene to phthalic anhydride and many other chemical reactions which involves fluids and solids. When a fluid passes upward through a bed of solids, there will be certain pressure drop across the bed required to maintain the fluid flow. Depending upon the bed geometry, fluid velocity and particle characteristics, the following phenomena occur with gradual increase in fluid velocity. At low fluid velocity, there is a pressure drop across the bed but the particles are stationary and flow in such cases is just through a fixed bed. As the velocity is gradually increased, a certain velocity is reached when the bed starts expanding. At this point the pressure drop across the bed equals the mass per unit area of the bed. This point is known as FLUIDZATION point or POINT OF ONSET OF FLUIDIZTION or INCIPIENT FLUIDIZATION. The movement of solids at superficial velocities far below the terminal settling velocities of the solid particles and the process correspond to a situation which is approximately equivalent to hindered settling. The pressure drop is maximum at some point and at this point the force exerted by fluid must not only act against the force of gravity on the paticles but it must also overcome friction forces locking the particles together. Once the

27 particles are separated, these forces drop off and the pressure required to maintain fluidization becomes less. As the velocity is still further increased, the pressure drop continues to remain constant until the bed has assumed the loosest stable form of packing. With the increase in fluid velocity the particles separate still further from one another, the bed continues to expand, the porosity increases but the pressure difference does not change. Particulate fluidization occurs when the difference in density between the particles and fluid is small. Procedure 1) Determination of initial porosity of the bed (ε o ) Take the given packing material in a 100 ml beaker and note down height of bed in the beaker. Calculate the bed volume (V 1 ) using, V 1 = C/S area of beaker X bed height in beaker Add water into the beaker just to fill up to bed surface. Drain out this water in to measuring jar and note down the volume of water (V 2 ) Calculate initial porosity from eqn. (1) (given in next page) 2) Open the gate valve so that water enters into the column and close the valve when the water is filled upto the full length of the bed. Note down the bed length (L o ). 3) To start the experiment, set all the manometer tappings and slowly open the gate valve. Wait until there are no air bubbles in flow line and in manometer line. 4) Note down the manometer reading, rotameter reading and bed height. 5) Gradually increase the flow rate of water and follow step 4. Repeat the experiment for different flow rates. 6) Continue till the bed is fluidized and finally becomes turbulent (i.e. till there is no appreciable change in pressure drop indicated by manometer). 7) Plot P/L as function of superficial velocity, and determine the minimum fluidization velocity from the plot. 8) Plot P vs. superficial velocity and porosity vs. superficial velocity. 9) Calculate minimum fluidization velocity from equation (7). 10) Report minimum fluidization velocity obtained by experiment and form theory.

28 Equations used: To find initial porosity, ε o : Volume of packing material = V 1 Volume of water present in voids = V 2 Initial porosity, ε o = V 2 / V 1 (1) To find porosity at any time, ε: Initial bed height = L o Bed height at any time = L L o (1 ε o ) = L (1 ε) OR ε= 1 [L 0 (1 ε 0 ) / L] To find superficial velocity, u: (2) Volumetric flow rate as indicated by rotameter = Q lit./min. = Q x ( 1 x 10-3 ) / 60 m³/s Superficial velocity, u = volumetric flow rate / Area of cross section of column i.e u = Q / (πd i ² / 4 ) (3) where D i is the internal diameter of the column in meter To calculate pressure drop, P across the bed: Manometer reading = R m =. cm =..10-2 m P = (ρ CCl4 ρ H2O ) x g x R m.n/m² (4) Where ρ CCl4 = 1594 kg/m³; ρ H2O = 1000 kg/m³ and g = 9.81 m/s² To find minimum fluidization velocity, u mf From Kozney Carman equation (a special case of Ergun s equation) we have [ P D p ² ε³] ------------------ = 150 L u µ (1 ε)²

29 At u = u mf i.e., at minimum fluidization velocity ( P / L) D P ² ε³ u mf = ------------------------- (6) 150 µ (1 ε)² At the onset of fluidization the pressure drop across the bed equals the weight of bed per unit area of cross section. Let ε = ε mf = Porosity at minimum fluidization condition, then from eqn. (6) we have D p ² g ( ρ P ρ H2O ) ε³ mf u mf = ----------------------------------------- 150 µ (1 ε mf ) Where D p is the diameter of the packing material (4.6 mm) ρ p is the density of glass beads (2220 kg/m 3 ) Observations and Calculations: Volume of packing material taken for initial porosity measurement = V 1 Volume of packing material taken for initial porosity measurement = V 2 Volume of water filled into the voids = V2 Initial porosity, ε o = V 2 /V 1 Initial bed height, L o =.m. Internal diameter of the column, Di = 47 x 10-3 m Viscosity of water at room temperature = = 1cP = 1 x 10-3 N s/m²

30 Sl. No. Q Manometer reading Bed LPM m³/s LLR (cm) RLR (cm) R m (cm) R m (m) height L (cm) U = Q/A, (m/s) Porosity (ε) P (N/m²) P/L (N/m² /m) Result: Minimum fluidization velocity, u mf (Experimental) = m/s Minimum fluidization velocity, u mf (Calculated) =. m/s

31 EXPERIMENT 7 HEAT TRANSFER IN A JACKETED VESSEL Aim: To find out the overall heat transfer coefficient. Apparatus: Jacketed vessel, steam source, thermometers, stop watch Theory: Heat transfer surfaces which is in the form a jacket as in the present case are often used in the agitated vessels. When a liquid is heated in stirred tank by condensing vapor in the jacket, the controlling resistance is usually that of the liquid in the tank. Consider a well agitated vessel containing m kg of liquid of specific heat C p. It contains a heat transfer surface of area A heated by a constant temperature medium such as condensing steam at temperature T s. If the initial temperature of the liquid is T a, then its temperature T b at any time t T can be found as follows. Rate of accumulation of energy = [energy input energy output (=0)] (1) For a batch of liquid with no flow in or out and no chemical reaction, eqn (1) becomes m C p (dt/dt) = UA (T s T ) + 0 (2) Assuming U to be constant, eqn (2) can be integrated between the limits t = 0 T = T a t = t T T = T b Equation (2) can be rearranged and integrated between the limits T a and T b for temperature and 0 and t T for time to get the following equation i.e., ln (T s T a ) / (T s T b ) = [U A t T / m C p ] (3)

32 In equation (3), A, m and C p are constants. Therefore a plot of LHS vs t T should result a straight line with slope = (U A / m Cp). Since the values of A, m and C p are known for the given system, the overall heat transfer coefficient U can be evaluated. Experimental Set-up: The jacketed vessel essentially consists of an impeller driven by a fractional HP motor. A drain valve is provided at the bottom of the jacketed vessel to remove the condensed steam. One more valve is provided to drain out water from the vessel. Procedure: 1. Fill the vessel with water and note down the height of water. Also note down the initial temperature of the water. 2. Agitate the water at a constant speed. 3. Open the steam valve and set steam pressure at 0.5 kgf / cm². 4. Starting with a temperature (say 25 C, note down the time required for every 5 degree rise in temperature and record times till the water temperature is about 85 C. 5. Repeat steps 1 4 for different steam pressure (1 kgf / cm²). 6. Tabulate the results and plot ln (T s T a ) / (T s T b ) against t T and calculate U. Observations and Calculations: 1. Diameter of the vessel = D = 30 cm = 0.30 m 2. Height to which water is filled = h = 0.30 m 3. Mass of water taken, m = (πd² / 4) h ρ H2O =..kg. 4. Specific heat of water at constant pressure, C p = 4.183 x 10³ J/Kg K Trial - 1 Steam pressure = 0.5 kgf/cm² (i.e. guage pressure) Steam temperature = T s = K [at Absolute pr. (= gauge pr. + Atm. Pr.)] from steam tables Water temperature in the beginning (i.e. At t = 0), T a =.. C (.K).

33 Trial -2 Steam pressure = 1 kgf / cm² (i.e. gauge pressure) Steam temperature = T s =.K (at Absolute pr. (= gauge Pr. + Atm. Pr.)] from steam tables Water temperature in the beginning (i.e. At t = 0), T a =.. C (.K) Water temperature in the beginning, T a =.K Steam pressure = 0.5 (or 1.0) kgf/cm² Sl. No. Temp. T b ( C) Temp. T b (K) Time, t T (seconds) ln [(T s T a )/ (T s T b )] Plot ln (Ts Ta) / (Ts Tb) coefficient, U using the equation. vs. t T and find out the slope. Calculate overall heat transfer Slope. m C p U = ----------------------- = W/m² K A Where A is the total heat transfer area given by A = (πd²/4) + (πdh) =.m² Result : (1) U at steam pressure.kgf/cm² = W/m² K. (2) U at steam pressure kgf/cm² =..W/m²K.

34 EXPERIMENT 8 HEAT TRANSFER IN A PACKED BED Aim: To determine the overall heat transfer coefficient and the inside heat transfer coefficients in packed bed. Equipment used: Packed bed heat exchanger, pressure gauge, steam source, flow meters, thermometers. Theory: Solids in divided form such as powders, pellets and lumps are heated or cooled in chemical processing for a variety of objectives such as drying, chemical reactions oxidation etc. For such operations, solids are packed in a column and heated by a hot fluid either directly or indirectly. If the solids are separated from the heating medium by wall as in the present case heat is transferred by conduction. Formulae to be used: The rate of heat transfer between a hot and cold fluids is given by, Q = m c C p,c T c = U A T ln (1) Where T c = T c2 T c1 (2) And (T h1 T c2 ) ( T h2 T c1 ) T ln = ----------------------------------------- (3) ln [ (T h1 T c2 ) / (T h2 T c1 ) ] The area of heat transfer is, A = (πd² i L) (4) For a packed tower, the individual heat transfer coefficient, h i is given by, (Reference, Perry 6 th edn. P.10-46)

35 h i = 0.813 (k/d i ) (N re. P ) 0.9 exp (-6 D p /D i ); for (D p / D i ) < 0.35 (5) and h i = 0.125 (k/d i ) (N re. P ) 0.75 for 0.35 < (D p /D i ) < 0.60 (6) where N Re.P = D p G/µ c G = Mass flux = m c / (πd² i / 4 ) m c = Mass flow rate of cold fluid, = V c / (60 x 10³ ) x 1000 = kg/s V c = Volumetric flow rate of cold fluid, in lit/min Vol. flow rate in m³/s V c (60 x 10³) U = ---------------------------------- = ---------------------- =..m/s Area of c/s in m² (πd² i / 4 ) Procedure: 1. Open the inlet and outlet valves of cold water and hot water lines. 2. Maintain a constant flow rate in the inlet of cold water and hot water 3. Open the valves for steam entering the system and maintain a constant steam pressure. 4. Note down the inlet and outlet temperatures of hot and cold water. 5. Change cold water flow rate and note down the temperature readings. 6. Repeat the procedure for a different hot water flow rate. Observations and calculations: Length of bed = L = 1000 mm = m Packed bed inner diameter = D i = 41 mm = m Packed bed saddle diameter = D p = 12 mm =..m Steam pressure maintained =..kg/cm²

36 Expt. T c1, C T c2, C T h1, C T h2, C V c V h No (K) (K) (K) (K) LPM LPM 5 10 10 10 15 10 20 10 Properties Table: Take all the properties at average temperature of cold fluid. T c (avg) ρ c C p,c µ k (K) (kg/m³) (J/kg K) (N s/m²) (w/m K) m c = Mass flow rate of cold fluid = [ V c / (60 x 10³ ) ] x 1000 =.kg/s Q c = m c C p,c T c in Watts T c = T c2 T c1 U = Q c / A T ln Where A = πd i L = m² (T h1 T c2 ) ( T h2 T c1 ) T ln = ----------------------------------------- ln [ (T h1 T c2 ) / (T h2 T c1 ) ] Notations: Q (and Q c ) - amount of heat transferred in watts m c C p,c - mass flow rate of cold water in kg/s - Specific heat of cold water in J/Kg. K

37 T c1, T c2 Inlet and Outlet temperatures of cold water in K U Overall heat transfer coefficient in W/m² K u Velocity of fluid, m/s A area of heat transfer in m² T ln log mean temperature difference between the two flow streams L length of packed bed,m k Thermal conductivity of water in W/m K N Re Reynolds number D p particle diameter of packing material, m D i packed bed inner diameter in meter V c,v h Volumetric flow rate of cold and hot water lit per min. (LPM) h I Inside heat transfer coefficient in W/m² K G mass flux in kg/m²s µ - viscosity of water in N s/m² Results: Overall heat transfer Coefficient,U (w/m² K) Inside heat transfer coefficient, h i (w/m² K) Expt. 1 Expt. 2 Expt. 3 Expt. 4

38 EXPERIMENT 9 DOUBLE PIPE HEAT EXCHANGER Aim: 1. To find out the individual heat transfer coefficients on the hot fluid side and cold fluid side 2. To determine the overall heat transfer coefficient. Equipment s required: Double pipe heat exchanger, Steam source, Pressure gauge, Flowmeters, Thermometers, Control valves to obtain co-current and counter current flows. Note : In the present set-up hot fluid flows in inner tube and cold fluid in outer tube. Theory: In many heat transfer processes, heat is transferred from one fluid to another through a solid wall. For e.g. tubular heat exchangers are frequently used in industry where heat transfer between two fluid streams is desired. The exchanger symmetry, fluid property and flow rates are the parameters forms the basis for design of heat exchangers. For a double pipe heat exchanger, the rate of heat transfer to the cold fluid from the hot fluid is given by, Q = W h C p,h T h (Watts) (1) Where Tc = T c2 T c1 T ln = log mean temperature difference given by: (T h1 T c2 ) (T h2 T c1 ) T ln = ------------------------------------- (2) ln [(T h1 T c2 )/ (T h2 T c1 )] for counter-current flow (T h1 T c2 ) (T h2 T c2 ) T ln = ------------------------------------- (3) ln [(T h1 T c1 )/ (T h2 T c2 )] for co-current flow

39 Individual heat transfer coefficients: (1) Inside heat transfer coefficient (h i ) For turbulent flow: N Re = 10000 4,00,000, N pr = 0.7 120, the inside heat transfer Coefficient is given by Dittus Boelter equation h i = 0.023 (k h /d i )(N Re,h ) 0.8 (N pr,h ) 0.4 (4) For laminar flow: h i = 1.75 (k h /d i )( w h C. p.h / k h L ) 1/3 (5) Reynolds number and Prandtl Number for hot fluid are defined as: N Re,h = (d i u h ρ h )/ µ h (6) N pr,h = C p,h µ h / k h (7) Velocity of the hot fluid, u h flowing through the pipe is given by Vol. flow rate in m³/s [(V h / (60 x 10³)] u h = -------------------------------- = ----------------------------- =...m/s (8) Area of c/s in m² (πd² i /4) (2) Outside heat transfer coefficient (h 0 ) The outside heat transfer coefficient, h 0 is given by h 0 = 0.023 (k c / D e ) ( N Re,c ) 0.81 ( N Pr,c ) 0.33 (9) where D e is the equivalent diameter given by D e = D i d 0. N Re,c and N Pr,c are given by equations N Re,c = (D e u c ρ c )/ µ c (10) N Pr,c = C p,c µ c / k c The velocity of the cold fluid is V c / (60x10³) u c = ------------------------ =-----------------=..m/s (12) (πde² / 4) (11) (3) The overall heat transfer co-efficient, U 0 For clean tubes, the overall heat transfer co-efficient, U 0 is given by the equation 1/U 0 = (1/h 0 ) + (d 0 /h i d i ) + (x w d 0 / k m d ln ) (13) Where d ln is the log-mean diameter defined as: d ln = (d 0 - d i ) / [ln (d 0 / d i )] (14)

40 Procedure: 1. Open the inlet outlet valves of cold water and hot water lines. 2. Maintain constant flow rates of cold and hot water by approximately opening the corresponding valves. 3. For parallel flow (co-current flow), open valves (1) and (3) and close the valves(2) and (4) 4. Now open the steam valve and adjust the steam pressure for a constant value (>0.4 kgf/cm²) 5. Record the inlet and outlet temperatures of hot cold fluids. 6. Keeping hot water flow rate constant, change the cold water flow rate. 7. Repeat step (5) and (6) for different flow rates of cold fluid. 8. For counter current flow, open valves (2) and (4) and close the valves (1) and (3) 9. Repeat steps (2) - (7) Observations: Inner pipe dimension: Outside diameter, d 0 = 34 mm =....m Inside diameter, d i = 29 mm = m Wall thickness, x w = (d 0 - d i ) / 2 = 2.5 mm = m Outer pipe dimensions: Inside diameter, D i = 54 mm =.m Thermally conductivity of the wall, k m = 62.8 W/m K. Effective length of each tube, L = 1.5 m Number of tubes, N = 4 Steam pressure = kgf / cm² Co-current flow: Expt. T c1, C T c2, C T h1, C T h2, C No. (K) (K) (K) (K) V c V h LPM LPM 1 10 10 2 12.5 10 3 15 10

41 Counter-current flow: Expt. T c1, C T c2, C T h1, C T h2, C No. (K) (K) (K) (K) V c V h LPM LPM 1 10 10 2 12.5 10 3 15 10 Specimen calculations: Note: All the properties like ρ h, µ h, k h and C p,h are to be taken at the average fluid temperature. 1. Calculate mass flow rate of hot fluid in kg/s = W h = [V h / (60x10³ )] x ρ h 2. Calculate mass flow rate of cold fluid in kg/s = W c = [V c / (60x10³ )] x ρ c 3. Calculate T ln from eqn. (2) or eqn. (3) 4. Calculate Q from eqn. (1), i.e. Q = W h C p,h T h (in Watts) 5. Area of heat transfer, A = πd i L N = π x ( 29 x 10-3 ) x 1.5 x 4 = 0.5466 m² 6. Overall heat transfer coefficient, U = Q / (A T ln ) 7. Calculate u h from eqn. (8). 8. Compute N Re,h and N Pr,h from eqns (6) and (7) 9. Compute h i from eqn. (4) or from eqn. (5) based on N Re range 10. Calculate N Re,c and N Pr,c from eqns. (10) and (11) 11. Calculate velocity of cold fluid using eqn. (12) 12. Calculate h 0 from eqn. (9) and d ln from equation (14). 13. Calculate overall heat transfer coefficient for a clean tube, U 0 from eqn. (13)

42 Properties Table: Take all the properties at average fluid temperature. T c (avg) = [ T c1 + T c2 ] / 2 (k) ρ c (kg/m³) COLD FLUID C p,c (J/kg k) Co - current µ c (N s/m²) k c (w/m k) Counter- current HOT FLUID T c (avg) = [ T h1 + T h2 ]/2 K ρ h, (kg/m³) C p,h (J/kg k) Co - current µ h (N s/m²) k h (w/m k) Counter- current

43 Notations: A - Area of heat transfer C p,h - Specific heat of hot water in J/kg K C p,c - Specific heat of cold water in J/Kg K d i - Internal diameter of inside tube, m D i - Internal diameter of outside tube, m d 0 - Outer diameter of inside tube, m h i - inside heat transfer coefficient in W/m² K k h - Thermal conductivity of hot fluid in W/m K k c - Thermal conductivity of cold fluid in W/m K k m - Thermal conductivity of wall in W/m K L - Effective length of each tube (=1.5m) N - Number of tubes (=4) Q Rate of Heat Transfer T h1, T h2 - Inlet and outlet temperatures of hot fluid in K Tc 1, Tc 2 - Inlet and outlet temperatures of cold fluid in K U 0 - Overall heat transfer coefficient in W/m² K u c, u h - Velocity of cold and hot fluid, m/s V c, V h - Volumetric flow rate of cold and hot fluid, lpm W c, W h - Mass flow rate of cold and hot fluid, kg/s X w - Thickness of wall, m ρ c, ρ h - Densities of cold and hot fluid, kg/m 3 µ c, µ h - Viscosities of cold and hot fluid, Ns/m 2 RESULT: Co-current flow Expt. No. Q (m 3 /s) U 0 (w/m² k) h i (w/m² k) h 0 (w/m² k) Counter-current flow Expt. No. Q (m 3 /s) U 0 (w/m² k) h i (w/m² k) h 0 (w/m² k)

44 EXPERIMENT 10 HEAT TRANSFER THROUGH BARE AND FINNED TUBE HEAT EXCHANGERS Aim: To determine: (i) Overall heat transfer coefficient, U (ii) Air side heat transfer coefficient, h a (iii) Fin efficiency Apparatus: Bare and Finned tube heat exchanger set up, steam source, pressure gauges, beakers. Theory: Difficult heat exchange problems occur when of two fluid streams has a much lower heat-transfer coefficient than the other. A typical case is heating a fixed gas, such as air by means of condensing steam. The individual coefficient for the steam is typically 100-200 times that for the air stream, consequently, the overall heat transfer coefficient is essentially equal to the individual coefficient for the air, resulting a low capacity of unit area of heating surface. As a result tube length required is increased considerably to provide reasonable capacity. To conserve space and to reduce the cost of equipment in these cases, extended surfaces may be used in the form of fins, pegs, disks and other appendages thereby the outside area is multiplied several times. The fluid stream having low coefficient (eg. air) is brought into contact with the extended surface and flows outside the tubes, while the other fluid having the high coefficient flows through the tubes. Fin efficiency is defined as the ratio of mean temperatures difference from surface to fluid divided by the temperatures difference from fin to fluid at the base or root of the fin. Procedure: 1. Fill a bucket with cold water (to approx. half the volume) and note down the weight. 2. Drain out the residual water in bare and finned tube by opening taps at the bottom. 3. Close the taps. 4. Close the top valve of bare and finned tube heat exchanger and open the main line steam valve.