Centrifugacion PDF

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FLUID MECHANICS II

PROBLEM OF CENTRIFUGATION

DUVAN FELIPE HOSTIA RAMIREZ ARÍN ABRAHAM MÁRQUEZ LARA

JOSÉ ANDRÉS PÉREZ MENDOZA CHEMICAL ENGINEERING UNIVERSITY OF ATLANTIC BARRANQUILLA 2020-1

Example 1. 5 lb/h of a liquid detergent of 100 cps of viscosity and 0.8 g/cm 3 of density is going to be clarified to fine crystals (Na2SO4 ρ=1.46 g/cm3) by centrifugation at 23000 rpm. The length of the inner bowl is 7 ¾ in with R = 7/8 in and (R – ri) = 19/32 in. Determine: 1) The cut-size diameter of the particle. 2) The sigma value.

Solution. Firstly, to begin the development of the exercise, the concept of centrifugation must be taken into account. Centrifuges are separation processes in which centrifugal field forces are used instead of a gravitational field to separate compounds from liquid-liquid or solidliquid systems. Centrifugation can be classified as a unitary operation of sedimentation or filtration, and this will depend on whether the principle is the difference in density of the substances to be removed or a filter medium is used for separation, these two processes are called sedimentation by centrifugation or filtration by centrifugation respectively. For centrifugation calculations, you have the pressure balance

(1)

(2) Assumptions and indicators: There is no slippage between liquids and both phases have the same residence time. q will be the volumetric rate; the subscripts l and h are to identify light and heavy liquids respectively. For the volumetric flow we have q=

V tR

(3)

And for the gravitational sedimentation we have

(4)

(5) Sigma theory Σ (sigma is only a function of the centrifuge ' s characteristics) q=vtΣ (6)

R+r i (7) as the half of liquid layer (R−ri)/2.Then ,the half of the particles 2 will settling to the wall and other half will be inthe liquid when they leave the centrifuge toa critical diameter or cut - size diameter Dcut Taking

r i=

(8) making the respective clearance of the equation

(9) Below are the data provided by the problem in the table: Table 1. centrifugation problem data F(lb/h)= u(cps)= p(g/cm^3)= pNa2SO4(g/cm3)= N(rpm)= R(in)= (R-ri)(in)= H(in)=

5 100 0.8 1.46 23000 0.8750 0.5938 7.75

The first and main thing is to leave all the values of the parameters in the same unit, to facilitate the calculations the units are left in the international system Taking into account the following conversions 1 lb/h = 1.25998*10-4 1 Centipoise [cP] = 0,001 Pa·s [Pas] 1 g/cm3=1000 kg/cm3

he other conversions were done with the excel program TALLER CORTE 3.xlsx, with the function "Convertir" the respective conversions are shown in table 2: Table 2. Data converted to SI of the centrifugation problem F(kg/h)= u(pa*s)= p(Kg/m^3)= pNa2SO4(Kg/m^3 )= ri R(m)= (R-ri)(m)= H(m)=

6.30E-04 0.1 800 1460 0.00714 0.02223 0.01508 0.19685

We calculate the value of the volumetric flow (q) and of w with the following equations kg F h m3 = (10) q= h kg ρ 3 m

( )

( )

w=2 πN (11)

wherein, N (min-1)(s-1)

The value of ri is obtained from the clearance of (R-ri)= 0.01508, then we make the respective replacement of the values of each variable in equation 9 and we obtain the value of cut-size diameter, for the gravitational sedimentation we use equation 4 and finally the value of sigma we make a clearance of equation 6, the results are in table 3 Table 3. Final results of the calculation of the centrifugation problem W(s^-1) q(m3/s)= Dcut(m)= Dcut(um)= vt(m/s)= 

2408.55436 8 7.875E-07 7.4843E-07 0.748 2.01487E-09 390.84

Example 2. In vegetable-oil primary refining, the crude vegetable-oil is partially saponification with caustic soda, and refined oil is separated immediately from soap by a centrifuge. Oil has a density of 920 kg/m3 and viscosity of 20 cps, the soap phase is 980 kg/m3 and 300 cps in density and viscosity respectively. A tubular bowl centrifuge is used to separate the soap formed. The bowl has a length of 0.762 m and 0.0508 m of inner diameter, turning out at 18000 rpm. The radius of weir for the light liquid is 0.0127 m and weir for the heavy liquid is 0.0130 m. Determine: 1) The position of the interface 2) If this centrifuge is fed to volumetric rate of 0.190 m3/h with a 10 % in volume of the soap phase, what is the cut-size diameter of oil drop retained in the soap? Solution. We start by assuming that the process described in the problem is a tubular cup centrifuge to separate two liquid phases. We apply a balance of forces: P

r

∫ ∂ P=∫ ρ w r ∂ r 2

0

Donde

(12)

r0

∂P =ρ w 2 r ∂r

and solving the integral we have 1 2 2 2 P= ρ w (r −r 0 ) 2

(13)

and with this we obtain the pressure balance presented above (equation 1-2), taking into account the following terms: ri radius of the weir for the light liquid; R radius of the bowl rs radius of the interface between of two liquids in the bowl rw radius of the outer weir 2= light phase density (l) 1= density of the heavy phase (h) In the theory, in order for the centrifuge to separate the two liquid phases, the liquid-liquid interface must be located at a radius less than 'R' but greater than that of the upper part of the spill tank (ri). Moreover, it often happens that one of the phases is more difficult to

clarify than the other, and this is compensated for by the fact that the volume of this phase must be greater than the volume of the phase that is easily clarified, and to achieve this the height of the two spill tanks. It should be noted that in one type of centrifuge of liquid clarification, it is only used a tank and the only function of this is to control the volume of liquid kept in the centrifuge. The location of the interface can be determined directly from Equation 2 by clearing the rs radius of the interface between of two liquids in the bowl as follows

r2s ( ρh −ρl ) =−ρl r w2 +r i2 ρh r2s =

−ρl r w2 +r i2 ρh ( ρh− ρl )

−ρl r2w +r2i ρh ) ¿ ¿ ¿ rs = √¿

(14)

rs(m)=

0.0169

again we take into account “Assumptions and indicators" , we clear equation 4 and get equation 15 to find the retention time t R=

V (15) q

expanding the equation for volume we have to 2 2 V =V l +V h =π ( R −r i )∗H

t R=

π ( R2−r 2i )∗H q

(16)

(17)

We found the volume of the soap phase retained in the centrifuge and having the value of the volumetric flow provided by the problem we found the retention time with the above equations, obtaining the following results reported in table 4 Table 4. The position of the interface in the problem of filtration. rs(m)= vol fase jabon=

0.0169 8.569E-04

tR(h)= tR(s)=

4.510E-02 162.35

Then finally the critical drop size, from which 50% of the soap phase will be separated with the residence time of 162 s, is can determine by equation 9, replacing the calculated data given by the problem, the critical diameter gave the following result Dcut(m) =

5.72889E06...