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Heat transfer in agitated vessal

Heat transfer in agitated vessal. Shawala Azhar 06-chem-63 Qudsia Ramzan 06- chem-62 Samina 06-chem-90 Humera 06-chem-100. Agitation of liquids. Purpose of agitation

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Heat transfer in agitated vessal

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  1. Heat transfer in agitated vessal Shawala Azhar 06-chem-63 Qudsia Ramzan 06- chem-62 Samina 06-chem-90 Humera 06-chem-100 engineering-resource.com

  2. Agitation of liquids • Purpose of agitation liquids are agitated for number of purposes depending upon the objective of the processing step. 1- Suspending solid particles. 2- Blending miscible liquids. 3- Dispersing a gas through a liquid in the form small bubbles. 4- promoting a heat transfer between the liquid and a coil or jacket. engineering-resource.com

  3. Agitation of liquid • Often one agitator serves dual purposes example: In the catalytic hydrogenation of a liquid agitator is used to 1- Disperse gas through the liquid in which the solid particles of catalyst are suspended. 2-Promoting the removal of heat of reaction by a cooling coil and the jacket. engineering-resource.com

  4. Time required for heating and cooling • It is frequently necessary to heat or cool the contents of a large batch reactor or storage tank. In this case physical constants of the liquor may alter and the overall heat transfer coefficient may change during the process • In the calculations average value of overall heat transfer coefficient is assumed so as to simplify the calculations of time required for heating and cooling. engineering-resource.com

  5. Heating or cooling time The heating or cooling time can be reduced by improving the rate of heat transfer to the fluid. By agitating the fluid. By reducing the heat losses from the vessal by insulation. engineering-resource.com

  6. Problem statement • A vessal contains 1ton (1Mg) of a liquid of a specific heat capacity 4.0kJ/kg K. The vessal is heated by steam at 393K which is fed to the coil immersed in the agitated liquid and the heat lost to the surrounding at 293K from the outside of the vessal. (a) How long does it take to heat the liquid from 293 to 353K and what is the maximum temperature to which the liquid can be heated. (b)When the liquid temperature has been reached to 353K, the steam supply is turned off for 2 hours(7.2 ks) and the vessal cools. How long will it take to reheat the material to 353K. Data given: The surface area of the coil is 0.5m2 and the overall heat transfer coefficient for the liquid may be taken as 600W/m2 K. The outside area of the vessal is 6m2and the coefficient of heat transfer to the surroundings may be taken as 10 W/m2 K. engineering-resource.com

  7. Solution • Part (a) If T(K) is the temperature of the liquid at the time t(sec), then a heat balance an the vessal gives (1000 *4000)dT/dt = (600*0.5)(393-T)-(10*6)(T-293) 4 * 10^6dT/dt = 135,480-360T 11,111dT/dt = 376.3-T engineering-resource.com

  8. CONTINUE • The equilibrium temperature occurs when dT/dt=0 so, T=376.3K • In heating from 293 to 353K, the time taken is t=11,111 293 ∫353dT/ (376.3-T) =11,111 ln(83.3/23.3) =14,155 sec (or 3.93 hrs) engineering-resource.com

  9. Continue • Part (b) The steam is turned off for 7200sec and during this time a heat balance gives: (1000*4000)dT/ dt = -(10*6)(T-293) 66,700dT/dt = 293-T engineering-resource.com

  10. CONTINUE • The temperature change is then given by: 353 ∫Tdt / (293-T) = (1/66,700) 0∫7200 dt ln(-60/293-T) = (7200/66,700) = 0.108 T= 346.9K engineering-resource.com

  11. CONTINUE • The time taken to reheat the liquid to353K is then given by: t =11,111 346.9∫353 dt/(376.3-T) =11,111 ln (29.4/23.3) t =2584 sec (0.72hrs) engineering-resource.com

  12. THE END engineering-resource.com

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