Assignment 2 Solution to Problem 1 on Fluidized Beds

<strong>Assignment</strong> 2<strong>Solution</strong> <strong>to</strong> <strong>Problem</strong> 1 on Fluidized Beds

Data givenParameterValueParticle Size (m) 350 x 10 -6Particle Density (kg m -3 ) 2400Gas Density (kg m -3 ) 1.01Specific heat of particles (J kg -1 K -1 ) 800Specific heat of gas (J kg -1 K -1 ) 1050Conductivity of particles (W m -1 K -1 ) 18.9Conductivity of gas (W m -1 K -1 ) 0.027Gas viscosity (kg m -1 s -1 ) 1.9 x 10 -5Bed voidage (-) 0.4Bed Diameter (m) 0.1Bed Height (m) 0.15

<strong>Solution</strong> <strong>to</strong> Q11. What is the Geldart type of this particle? What are its expected flow characteristics?Answer: From the data given the particle diameter is 350 µm and particle density is 2400 kg/m 3 .From the above figure which divides area in four different regions based on theparticle properties, the said particles are of type “B”. These particles will have goodfluidization quality, vigorous bubbling, slug at high velocity, small bed expansion, goodsolids mixing in bubbling

<strong>Solution</strong> <strong>to</strong> Q22. What is the minimum fluidization velocity?Answer: For whole range of Reynolds numberAll properties are known, hence simply finding the value of minimum fluidization velocityHence U mf = 0.089 m s -1 for this case Reynolds number isRe p,mf = 1.65

<strong>Solution</strong> <strong>to</strong> Q33. What is the terminal settling velocity?Answer: Single particle settling velocity, We assume that 0.4

<strong>Solution</strong> <strong>to</strong> Q3Answer: Hindered particle settling velocity, We assume that Rep>500,( )( )1/2⎡ 3.1 ρ ⎤p-ρg gdpurel,t= ⎢⎥ εf ( ε ),⎢⎣ρg ⎥⎦where0.4f ε =ε =0.693.Hence,urel,t=1.4 m s-1Checking validity of assumption by recalculating Reynolds numberρRe=whereρd uave p rel,tμeffave f g=25637>500, which justify our assumption=ερ +(1-ε)ρ =1440 kg mμμeff= =2.74 × 100.4εkg m-3s-5 -1 -1

<strong>Solution</strong> <strong>to</strong> Q4, 5 and 64. What will be the bed condition if the bed height is doubled?5. What will be the bed condition if the velocity is twice minimum fluidization velocity6. What is the recommended operating velocityAnswer: Before we answer the question, we need <strong>to</strong> know the bed condition with existingcondition. First we calculate Froude number, Reynolds number and bed height atminimum fluidizing condition [2,3] ,mf p gRep,mf= =1.7,mfpu d ρμ2umfFrmf= =2.3,d gH =1.3d0.175t=1.3mNext, we determine fluidization regime by using⎛ ρs-ρg ⎞⎛ L ⎞mfFr Re =12072>100⎜⎟⎜ρ ⎟ ⎟⎝ g ⎠ ⎝ dt⎠( mf )( p,mf )[2]A. S. Mujumdar, 2010, Heat Transfer in Fluidized Beds- An overview. Available online at: www.serve.me.nus.edu.sg/arun[3]J. P. Constantineau, J.R. Grace, C.J. Lim, G.G. Richards, 2007, Generalized bubbling–slugging fluidized bed reac<strong>to</strong>r model, Chemical Engineering Science Vol 62 pp. 70 – 81

<strong>Solution</strong> <strong>to</strong> Q4, 5 and 6This implies that fluidization condition is bubbling or aggregative fluidization.If the bed height is doubled, the bed condition will not change except that it will increasepressure drop across the bed, for which higher pumping power is required.Now we check slugging criteriauums mf tms=u +0.07 gd=0.31m shence,2u

<strong>Solution</strong> <strong>to</strong> Q4, 5 and 6In fluidized bed operation, velocity has <strong>to</strong> be higher than minimum fluidization velocity.However, we do not want velocity in which the particle is flashed out from the chamber; hence,it should not exceed hindered settling velocity.Maximum allowable velocity = 1.4 m s -1Minimum allowable velocity = 0.089 m s -1As such, recommended velocity is the velocity in between both value. Moreover, <strong>to</strong> achieveslugging condition, recommended velocity is chosen as 4 u mf = 0.35 m s -1Of course the choice of velocity can be different depending on the conditions of fluidizationone is interested in, but for our next calculations, we have used this velocity

<strong>Solution</strong> <strong>to</strong> Q77. What is particle <strong>to</strong> gas heat transfer coefficient?Answer: Single Particle <strong>to</strong> gas heat transfer coefficient can be calculated as [1] ,h d =2+0.6Prks p 1/3g( Re ) 1/2pThe recommended velocity (0.35 m s -1 ) will be usedcpgμPr= =0.74,kgp gRep= =6.5,hencehsd p1/3 1/2=2+0.6 0.74 6.5 ,kghsd p=3.38,ksgud ρμh =261 W m K( ) ( )-2 -1[1]D. Kunii, O. Levenspiel, 1969, Fluidization Engineering, Wiley, New York

<strong>Solution</strong> <strong>to</strong> Q77. What is particle <strong>to</strong> gas heat transfer coefficient?Answer: For fluidized Bed, Particle <strong>to</strong> gas heat transfer coefficient can be calculated as [1] ,The recommended velocity (0.35 m s -1 ) will be usedcpgμPrg= =0.74,While usingkgudpρforgRep= =6.5,μhsdp h( ) 1.3 sdp=0.03 6.5 or =0.34,hencekkhsd p1/3 1.4=0.0282 0.74 6.5 ,kghsd p=0.351,ksgh =27.06 W m K( ) ( )-2 -1forforsgh =26.37 W m K-2 -1It can be seen that the particle-gas heat transfercoefficient for fluidized bed is much lower than forsingle particle – gas heat transfer coefficient.At higher particle Reynolds number, however, heattransfer coefficient in fluidized bed increases, mainlyfor particles of larger diameterg[1]D. Kunii, O. Levenspiel, 1969, Fluidization Engineering, Wiley, New York

<strong>Solution</strong> <strong>to</strong> Q88. What if operating velocity is one half of the minimum fluidizing velocity? Estimate the heattransfer coefficient for this conditionAnswer: At this condition the bed will be in a fixed bed regime,p gRep= =0.83,hencehsd p1/3 0.6=2+1.1 0.74 0.83 ,kghsd p=2.89,ksgud ρμh =222.9 W m K( ) ( )-2 -1

<strong>Solution</strong> <strong>to</strong> Q99. Estimate the heat transfer coefficient for operating velocity of the fluidized bed which is 6times the minimum fluidizing velocity. What will be the fluidized bed condition in this case?Answer: the bed condition is slugging fluidization and the heat transfer coefficient is,h d =0.0282 Prks p 1/3gp gRep= =9.9,hence( Rep )1.4hsd p1/3 1.4=0.0282 0.74 9.9 ,kghsd p=0.63,ksgud ρμh =48.78 W m K( ) ( )-2 -1

<strong>Solution</strong> <strong>to</strong> Q1010. Estimate heat transfer coefficient if a vertical tube of 5cm diameter is immersed in the bed.Answer: When a vertical tube is immersed in the bed, the heat transfer coefficient is [2] ,h ⎛ ρ ⎞ ⎛ ⎞ ⎛− ρ ⎞cdp 4 0.23Cpg gCps s= 3.51× 10 CR( 1 −ε) Rep⎜ ⎟ ⎜ ⎟ ⎜ ,k ρ ⎟g ⎝ kg ⎠ ⎝ Cpg ⎠ ⎝ g ⎠⎛ r ⎞ ⎛ r ⎞C =1.07+3.04⎜ ⎟-3.29 ⎜ ⎟ , where R =0.5 d⎝ Rb⎠ ⎝ Rb⎠R b tfor 10