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Clalandria Evaporator

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Che492-Unit Operations Laboratory II Experiment no. 5 Calandria Evaporator
Dr. Siddhartha Panda Dr. Deepak Kunzru

Name of TA:
Abir Ghosh Date of experiment: 23/9/2013 Date of submission: 30/9/2013

Group No: 2
Roll no. 10059 10062 10067 Name Akshay Bansal Akshit Gupta Aman Jain


Introduction Objective Theory and formulae Apparatus required Details required Observations and Calculations Sample calculation Results and Discussion Conclusion Precautions Sources of error Nomenclature Reference 3 4 4 6 6 6 8 8 8 9 9 9 9


The functional requirement of an evaporator is to vaporize a chemical liquid in order to achieve certain industrial benchmarks such as concentrating a solution, purification or retaining a number of useful solvents or to carry out crystallization on an industrial scale by achieving the limiting concentration. The fundamental working principle of an evaporator is characterized by heating the solution up to the boiling point of the solvent (in case of solid solute) or to the dew point of the solution (in case of liquid solute). The various types of industrially applicable evaporators are described as under: 1.) Direct heating evaporators-In these units, direct heating of the solution is carried out using solar energy or the heat energy of the gases which are brought to very high temperature using natural gas combustion. These are mainly used to extract salt from sea-water on an industrial scale. 2.) Natural Circulation evaporators-It is the steam-heated evaporator that is the most widely used unit in the process industries. In this unit, feed is passed over a horizontal/vertically aligned network of heating tubes through which high temperature steam flows and heat transfer takes place from steam to liquid. 3.) Multiple effect evaporators-This type of evaporator utilizes multiple heating of the solution instead of just a single stage process to increase the efficiency of evaporation which comes at the cost of proportionally increasing energy consumption Calandria evaporator-It is a vertical type of evaporator with tube sheets extending across the body and central downtake. The feed is introduced in the tubes of cylindrical equipment known as Calandria with steam flowing inside the shell. Here circulation is due to the difference in specific gravity between bulk liquid and heated liquid.

Figure 1: Calandria evaporator


   To concentrate a solution of sodium carbonates using Calandria Evaporator To perform mass and energy balance over the evaporator To determine the overall heat transfer coefficient of the evaporator

Theory and formulae
In the setup of Calandria evaporator, the feed flows through the tubes while the superheated steam flows across the shell space. For the calculation of overall heat transfer rate, we can write Q=U*A*ΔT Q=heat transfer per unit time A=effective area of heat transfer ΔT= Is the difference between temperature at which the steam condenses at the shell side pressure and the temperature at which the solution boils at thetube side (atm) pressure. U=Overall heat transfer coefficient For a Calandria evaporator effective heat transfer area can be computed by considering the total cylindrical tube side area and the central downcomer area A= n*(π *d*l) tube + (π*D*L) downcomer n=total no. of tubes in Calandria From mass balance (at steady state) a.) Overall mass balance Wf=Wp+Wv Wv = Rate of condensation of vapor from condenser, kg/sec. Wp = Flow rate of product, kg/sec. Wf = Flow rate of feed, kg/sec. Converting this equation into volumetric flow rate & density terms Vf*ρf=Vp*ρp + Vv*ρv ρf = density of feed stream (g/ml) ρp = density of product stream(g/ml) (iv) (iii) (ii) (i)


ρv = density of vapor stream (g/ml) From species balance on Na2CO3 xf*Wf = xp*Wp (since no solute in vapor) (v)

From the density-concentration calibration chart we have Density = 0.011*x + 0.983 From Energy balance Q=Wp*hp+Wv*hv-Wf*hf hi = enthalpy of i-th stream(J/kg) Q is as defined in (i) (vi) (density = [g/ml]; x= [wt. %])

Figure 2 water v/s concentration: Heat of solution for Na2CO3 solution in

Figure 3: Density v/s concentration of Na2CO3


Apparatus required
     Electricity Supply: Single Phase, 220 V AC, 50 Hz, 32 Amps MCB with earth connection Water supply: Initial fill Drain Required Floor Area Required: 1.5 m x 1 m Sodium carbonate (Na2CO3): 2 kg

Details required
• • Downcomer: Material Stainless Steel, Inner diameter 77 mm, length 200 mm Inner tubes: 10 nos., Material Stainless steel, Inner diameter 22 mm, outer diameter 25.5 mm, length 200 mm

1.) 2.) 3.) 4.) 5.) 6.) A 5% solution of sodium carbonate is prepared All the valves are closed and cooling water tank is filled with water The steam generator is filled with water and the electricity supply is made to the setup The set point of steam is set between 110-120 deg C by using the set button on DTC The feed tank is filled with the prepared solution The rotameter is fully opened and the vacuum pump is switched on to allow the feed to enter into the evaporator till the marked level on the evaporator is reached 7.) Once the marked level has been reached, the rotameter and the vacuum pump are closed to ensure no more feed is entering into the evaporator 8.) Now both the heaters are switched on to allow the evaporator temperature reach 100 deg C 9.) Once the desired temperature of the evaporator is reached, the rotameter is set to the required feed rate 10.) The steam control valve and the bottom product flow control valve are handled simultaneously to achieve steady state level of the solution in the evaporator 11.) Keep observing the liquid level in the evaporator by constantly looking through the front glass 12.) Once it has reached the steady state, note down the temperature readings and the volume of various streams collected in their respective measuring cylinders and record the time taken to collect the samples using stop watch 13.) Repeated steps 9 to 12 for different values of feed flow rate viz: 3, 5, 7.5 LPH

Observations and Calculations
Feed density = (26.28/25)*1000 = 1051.2 kg/m3 Feed concentration = (1.0512 – 0.9832)/0.0114 = 5.965% Heat of Solution (5% Na2CO3, Feed) = 242 kJ/kg ρw = 1000, kg/m3 Volume of sample = 25 ml Weight of empty bottle=24.32 gram 6

Weight of feed solution (Mf) = 50.6 – 24.32 = 26.28 gram Specific Gravity of feed = 26.28 / 25 = 1.0512 gm/ml

Table 1: Temperature observations for various feed Flow

Flow Rate(L/h) 3.0000 5.0000 7.5000

T1 105.0000 107.1000 105.6000

T2 98.4000 98.3000 98.5000

T3 33.0000 32.9000 32.8000

T4 102.2000 102.5000 101.6000

T5 97.1000 96.5000 96.6000

T6 36.5000 41.5000 48.6000

T7 55.2000 70.1000 81.7000

Table 2 : Concentrated product observations

Vp 60.0000 70.0000 230.0000

tp 60.0000 60.0000 120.0000

Mp 24.4400 24.7600 25.3100

ρp 977.6000 990.4000 1012.4000

Wp 0.0010 0.0012 0.0019

Conc (wt %) 0.4912 0.6316 2.5614

hp 171.3444 172.0286 160.2050

Table 3: Feed observations

Ff 3.0000 5.0000 7.5000

Wf 0.0021 0.0020 0.0048

hf 4.8161 4.4074 3.9987

Fc 100.0000 100.0000 100.0000

Table 4: Condensed vapour observations

Vv tv 68.0000 60.0000 52.0000 60.0000 345.0000 120.0000

Wv 0.0011 0.0009 0.0029

hv 2679.4320 2679.9000 2678.4960

Table 5: Calculation of Heat transfer coefficient

C 4.0800 3.1200 10.3500

ΔT 3.8000 4.2000 3.1000

Q 3.1940 2.5124 7.9923

U 4.4893 3.1950 13.7700

Table 6: Mass Balance Calculations

Ff 3.0000 5.0000 7.5000

Wf Wf Theoritical Calculated 0.0009 0.0021 0.0014 0.0020 0.0022 0.0048

Error % 59.0702 28.7881 55.1442


Sample Calculations
Feed flow rate = 3 LPH Weight of product (Mp) =24.44 gram ρf = (Mf /25) ×1000 = (26.28/25)*1000 = 1051.2 Kg/m3 ρp = (Mp /25)×1000= 977.6 Kg/m3 Wv = (Vv × 10-6 × ρw) / tv= 0.001133 kg/sec Wp = (Vp × 10-6 × ρp) / tp= 0.000978 kg/sec Wf = (Ff × ρf × 10-3) / 3600 = 0.002111 kg/sec C = Wv × 3600 = 4.08 kg/h Ws*(Hfg + Cp(100-T2)) = Q Ws = 0.002667 Kg/s E = Wv / Ws = 0.425 Mass Balance : Wf = Wp + Wv = 0.002111 kg/sec Error = (WF T – WF C )/WF c x 100 = 59.07 % Q=Wp*hp+Wv*hv-Wf*hf 3.194029 kW A= n (Pi *d*l)tube+(Pi*D*L)downcomer = 0.18723 m2 U=Q/(A*(T4-T3)) = 4.489313 KW/m2 degC Heat Transfer Coefficient = Uaverage = 7.151439 kW/0C-m2.

Results and Discussion
1. Average heat transfer coefficient of the evaporator (taking the inner surface area of the evaporator) came out to be 7.151 kW/0C-m2. 2. Near about 50% error is involved in mass balance calculations. 3. Average steam economy is 1.8673 4. Average evaporator capacity is 5.8500 kg/h

1) Calandria Evaporator is useful in concentrating non-volatile solute from a solution by removing volatile solvent. 2) Careful considerations and measures are required in order to maintain steady state. 3) A good steam economy (above 1) can be achieved through Calandria evaporator. 4) Due to high power consumption approx. 3.2 KW Calandria evaporator are less in use now a days.


Sources of Error
   Steady state is assumption is made in the experiment which is difficult to maintain throughout the experiment due to fluctuations in flow rates etc. The deposition of salt can take place on the side of the wall causing the concentration of solution of product to show a value lesser than the actual concentration inside the evaporator. Difficult to get perfect insulation

1. Never run the apparatus if power supply is less than 180 V and above 230 V 2. Never switch ON mains power supply before ensuring that all the ON/OFF switches given on the panel are at OFF position 3. Operator selectors switch OFF temperature indicator gently 4. Always keep the apparatus free from dust

Wv = Rate of condensation of vapor from condenser, kg/sec. Wp = Flow rate of product, kg/sec. Wf = Flow rate of feed, kg/sec ρf = density of feed stream (g/ml) ρp = density of product stream(g/ml) ρv = density of vapor stream (g/ml) hi = enthalpy of i-th stream(J/kg) xf = mole fraction of solute in feed xp = mole fraction of solute in bottom product Vf = volume of feed collected (ml) Vp = volume of product collected (ml) Vw = volume of top product collected (ml)

  Kern, D. Q. , “Process Heat Transfer, 16th edition, McGraw Hill, ND, 2007, Page 401-403


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