Example

Chapte.4 iffusion with Chemical eaction Example.4- ------------------------------------------------------------------------------ fluiize coal eacto opeates at 45 K an atm. The pocess will be limite by the iffusion of oxygen countecuent to the cabon ioxie, CO, fome at the paticle suface. ssume that the coal is pue soli cabon with a ensity of 80 g/m 3 an that the paticle is spheical with an initial iamete of.5 0-4 m. i (% O an 79% N ) exists seveal iametes away fom the sphee. The iffusivity of oxygen in the gas mixtue at 45 K is.3 0-4 m /s. If a quasi-steay state pocess is assume, calculate the time necessay to euce the iamete of the cabon paticle to 5.0 0-5 m. (ef. Funamentals of Momentum, Heat, an Mass Tansfe by Welty, Wics, an Wilson, 4 th Eition, 00, pg. 496.) Solution ---------------------------------------------------------------------------------------------- The eaction at the cabon suface is C(s) + O (g) CO (g) We have iffusion of oxygen () towa the suface an iffusion of cabon ioxie (B) away fom the suface. The mola flux of oxygen is given by N, c mix y + y (N, + N B, ) In this equation, is the aial istance fom the cente of the cabon paticle. Since N, N B,, we have N, c mix y The system is not at steay state, the mola flux is not inepenent of since the aea of mass tansfe 4π is not a constant. Using quasi steay state assumption, the mass (mole) tansfe ate, 4π N,, is assume to be inepenent of at any instant of time. W 4π N, 4π c mix y constant y, y,inf -3

t the suface of the coal paticle, the eaction ate is much faste than the iffusion ate to the suface so that the oxygen concentation can be consiee to be zeo: y, 0. Sepaating the vaiables an integating gives y, 4π c mix y 0 W W 4π c mix y, > W 4πc mix y, Since one mole of cabon will isappea fo each mole of oxygen consume at the suface W C W 4πc mix y, Maing a cabon balance gives ρc 4 3 π M C t 3 ρc M C 4π t 4πc mix y, Sepaating the vaiables an integating fom t 0 to t gives t t 0 ρc M c C y mix, f i ρc t M c C y mix, ( i f ) The total gas concentation can be obtaine fom the ieal gas law c P T (0.0806)(45) 0.006 mol/m3 Note: 0.0806 m 3 atm/mol o K The time necessay to euce the iamete of the cabon paticle fom.5 0-4 m to 5.0 0-5 m is then ( ) ( ) 5 5 (80) 7.5 0.5 0 t 4 ()(0.006)(.3 0 )(0.) 0.9 s -4

Example.4- ------------------------------------------------------------------------------ Pulveize coal pellets, which may be appoximate as cabon sphees of aius mm, ae bune in a pue oxygen atmosphee at 450 K an atm. Oxygen is tansfee to the paticle suface by iffusion, whee it is consume in the eaction C(s) + O (g) CO (g). The eaction ate is fist oe an of the fom ɺ " C O whee 0. m/s. This is the eaction ate pe unit suface aea of the cabon pellets. Neglecting change in, etemine the steay-state O mola consumption ate in mol/s. t 450 K, the binay iffusion coefficient fo O an CO is.7 0-4 m /s. (ef. Funamentals of Heat Tansfe by Incopea an ewitt.) Solution ---------------------------------------------------------------------------------------------- We have iffusion of oxygen () towa the suface an iffusion of cabon ioxie (B) away fom the suface. The mola flux of oxygen is given by N, c y + y (N, + N B, ) In this equation, is the aial istance fom the cente of the cabon paticle. Since N, N B,, we have N, c y The system is not at steay state, the mola flux is not inepenent of since the aea of mass tansfe 4π is not a constant. Using quasi steay state assumption, the mass (mole) tansfe ate, 4π N,, is assume to be inepenent of at any instant of time. W 4π N, 4π c y constant y, y,inf The oxygen concentation at the suface of the coal paticle, y,, will be etemine fom the eaction at the suface. The mole faction of oxygen at a location fa fom the pellet is. Sepaating the vaiables an integating gives y, 4π c y y, W W 4π c (y, y, ) > W 4πc ( y, ) -5

The mole of oxygen aive at the cabon suface is equal to the mole of oxygen consume by the chemical eaction W 4π ɺ " 4π C O 4π c y, 4πc ( y, ) 4π c y, ( y, ) y, > y, + " y,.7 0 4 4 3.7 0 + 0. 0.63 The total gas concentation can be obtaine fom the ieal gas law. (Note: 0.0806 m 3 atm/mol K) c P T (0.0806)(450) 0.008405 mol/m3 The steay-state O mola consumption ate is W 4πc ( y, ) 4π(0.008405)(.7 0-4 )( 0.63)(0-3 ) W 6.66 0-9 mol/s Example.4-3 ------------------------------------------------------------------------------ biofilm consists of living cells immobilize in a gelatinous matix. toxic oganic solute (species ) iffuses into the biofilm an is egae to hamless poucts by the cells within the biofilm. We want to teat 0. m 3 pe hou of wastewate containing 0. mole/m 3 of the toxic substance phenol using a system consisting of biofilms on otating is as shown below. Waste wate fee steam biofilm C 0 C (z) Inet soli suface Well-mixe contacto C 0 Teate waste wate biofilm z0 etemine the equie suface aea of the biofilm with mm thicness to euce the phenol concentation in the outlet steam to 0.0 mole/m 3. The ate of isappeaance of phenol (species ) within the biofilm is escibe by the following equation -6

c whee 0.09 s - The iffusivity of phenol in the biofilm at the pocess tempeatue of 5 o C is.0 0-0 m /s. Phenol is equally soluble in both wate an the biofilm. (ef. Funamentals of Momentum, Heat, an Mass Tansfe by Welty, Wics, an Wilson, 4 th Eition, 00, pg. 496.) Solution ---------------------------------------------------------------------------------------------- The ate of phenol pocesse by the biofilms, W, is etemine fom the mateial balance on the pocess unit W 0. m 3 /h(0. 0.0) mol/m 3 8.0 0-3 mol/h W is then equal to the ate of phenol iffuse into the biofilms an can be calculate fom c W S N,z S z 0 In this equation, S is the equie suface aea of the biofilm an N,z is the mola flux of phenol at the suface of the biofilm. The mola flux of (phenol) is given by N,z c y + y (N,z + N B,z ) Since the biofilm is stagnant (o noniffusing), N B,z 0. Solving fo N,z give N,z ( y ) c y The mole faction of phenol in the biofilm, y, is much less than one so that c can be consiee to be constant. Theefoe N,z c y c Biofilm Soli suface N,z Maing a mole balance aoun the contol volume S z gives z -7

S N,z z S N,z z+ z + S z 0 iviing the equation by S z an letting z 0 yiels N, z Substituting N,z c (E-) c into equation (E-) we obtain c c > c c (E-) The solution to the homogeneous equation (E-) has two foms ) c C exp ) c B sinh z z + C exp + B cosh z z The fist exponential fom () is moe convenient if the omain of z is infinite: 0 z while the secon fom using hypebolic functions () is moe convenient if the omain of z is finite: 0 z δ. The constants of integation C, C, B, an B ae to be etemine fom the two bounay conitions. We use the hypebolic functions as the solution to Eq. (E-). c B sinh z + B cosh z (E-3) t z 0, c c s c 0 B t z δ, Theefoe c 0 B cosh δ + B sinh δ sinh B B cosh δ δ sinh c 0 cosh δ δ Equation (E-3) becomes -8

sinh c c 0 cosh δ δ sinh z + c 0 cosh z cosh δcosh z sinh δsinh z c c 0 cosh δ Using the ientity cosh( B) cosh()cosh(b) sinh()sinh(b) we have c c 0 δ cosh δ cosh ( z) c z 0 c 0 sinh ( δ z) cosh δ z 0 c 0 tanh δ The mola flux of phenol at the biofilm suface is given by N,z c c0 δ z 0 tanh δ δ The imensionless paamete δ Fo this poblem we have epesents the atio of eaction ate to iffusion ate. 0.09 0.00 m s 0 m 0 s δ 9.49 This value inicates that the ate of eaction is vey api elative to the ate of iffusion. The flux of phenol into the biofilm is then -9

N,z 0 (0.0)( 0 ) 0.00 (9.49) tanh(9.49) 3.9 0-8 mol/(m s) The equie suface aea of the biofilm is finally S W N, z 8.0 0 3 8 (3.9 0 )(3600) 57.0 m Example.4-4. ---------------------------------------------------------------------------------- Consie a spheical oganism of aius within which espiation occus at a unifom volumetic ate of C. That is, oxygen (species ) consumption is govene by a fist-oe, homogeneous chemical eaction. (a) If a mola concentation of C () C,0 is maintaine at the suface of the oganism, obtain an expession fo the aial istibution of oxygen, C (), within the oganism. (b) Obtain an expession fo the ate of oxygen consumption within the oganism. (c) Consie an oganism of aius 0.0 mm an a iffusion coefficient fo oxygen tansfe of 0-8 m /s. If C,0 5 0-5 mol/m 3 an 0 s -, what is the mola concentation of O at the cente of the oganism? What is the ate of oxygen consumption by the oganism? Solution ------------------------------------------------------------------------------------------ (a) If a mola concentation of C () C,0 is maintaine at the suface of the oganism, obtain an expession fo the aial istibution of oxygen, C (), within the oganism. + Figue E- Illustation of a spheical shell 4π The one-imensional mola flux of is given by the equation " N C (E-) pplying a mole balance on the spheical shell shown in Figue E- yiels fo steay state 4π N " 4π " N + 4π 0 + -30

iviing the equation by the contol volume (4π ) an taing the limit as 0, we obtain ( " N ) + 0 (E-) Fo a fist oe eaction, C an substituting the mola flux fom equation (E-) into the above equation, we have C C 0 C C 0 (E-3) In this equation, an ae constants inepenent of. We want to tansfom this equation into the fom y α y 0 (E-4) Let α, we can tansfom equation (4.6-3) into the fom of equation (E-4) by the following algebaic manipulations C α C 0 C C + α C 0 C C + α C 0 Since becomes ( C ) C C + C + C C +, the above equation ( C ) α C 0 Let y C, the equation has the same fom as equation (E-4) with the solution y B sinh(α) + B cosh(α) o C B sinh(α) + B cosh(α), whee α -3

The two constants of integation B an B can be obtaine fom the bounay conitions t 0, C finite o C 0 t, C C 0 (a nown value) pplying the bounay at 0 yiels 0 B pplying the bounay at yiels C 0 C B sinh(α) B sinh( α) Theefoe the concentation pofile fo species within the oganism is C C 0 sinh( α) sinh( α) (E-5) α t the cente of the oganism, the concentation is given by C ( 0) C 0 sinh( α) (b) Obtain an expession fo the ate of oxygen consumption within the oganism. ate of oxygen consumption within the oganism. 4π ( C ) The oxygen concentation within the oganism is given by equation (E-5) C C 0 sinh( α) sinh( α) (E-5) C C sinh( α) 0 α sinh( α ) + cosh( α) C C sinh( α) 0 α cosh( α) sinh( α) C C 0 [ ] ( α ) coth( α) ) -3

/ Let φ α Thiele moulus fo a fist oe eaction. Ignoing the minus sign, the ate of oxygen consumption within the oganism is then ate of oxygen consumption 4π C 0 (φ cothφ - ) ate of oxygen consumption 4π C 0 (φ cothφ - ) (c) Consie an oganism of aius 0.0 mm an a iffusion coefficient fo oxygen tansfe of 0-8 m /s. If C,0 5 0-5 mol/m 3 an 0 s -, what is the mola concentation of O at the cente of the oganism? What is the ate of oxygen consumption by the oganism? α t the cente of the oganism, the concentation is given by C ( 0) C 0 sinh( α) α / 0 8 0 / 4.47 0 4 m α / 4 ( ) 0 0 8 0 / 4.47 α C ( 0) C 0 sinh( α) 5 0-5 4.47 sinh(4.47) 5. 0-6 mol/m 3 ate of oxygen consumption 4π C 0 (φ cothφ - ) ate 4π(0-4 )(0-8 )( 5 0-5 ) [4.47 coth(4.47) - ].8 0-5 mol/s The following Matlab pogam plots the concentation of oxygen within the oganism as a function of position. % Example.4-4 % alfa4.47e4; % m e-4; % m alfa4.47; C05e-5; % mol/m3 o(:50)/50; *o; -33

CC0*sinh(alfa*)./(o*sinh(alfa)); plot(o,c) gi on xlabel('/');ylabel('c_(mol/m^3)') Figue E.4-4 Oxygen concentation pofile in a spheical oganism. We now consie the iffusion of species into a spheical catalyst paticle whee homogeneous fist oe chemical eaction occus. The concentation pofile fo species within the spheical catalyst paticle is then C C sinh( α) sinh( α) (.4-) In this equation C is the concentation of species at the suface of the catalyst paticle an α is efine by the expession α, whee is the fist oe ate constant an is the iffusivity of in the paticle. t the cente of the bea, the concentation is given by α C ( 0) C sinh( α) (.4-) -34