Question

A rotary vacuum filter is available with an area of 200 m^2 and vacuum pressure of 75 kPa. Filter leaf tests have been performed on a cell broth with a viscosity of 5 cP.

The leaf tests gave a specific cake resistance of 1×10^11 cm/g and medium resistance of 1×10^8 cm-1. The cake solids (dry basis) per volume of filtrate was 15 g/L. It is desired to operate the large filter with cycle time of 45 s and a cake formation time of 10 s. What is the filtration rate expected for the rotary vacuum filter? How significant is the resistance medium?

Answer 1

From the given information, we get the following data:

$\Delta\rho=75Kpa; \mu_\omicron=5cp=0.05g/cms; a=1*10^{11}cm/g;A'=200m^2; R_m=1*10^8cm^{-1}; \rho_c=15g/litre; t=10s$

$\therefore\implies \frac{t}{V/A}=\frac{\mu_\omicron a\rho_c}{2\Delta p}(\frac{V}{A})+\frac{\mu_\omicron R_m}{\Delta p}$

and the area A can be calculated from;

$A=A'/(\frac{45s}{10s})=200m^2/(\frac{45s}{10s})=44.44m^2$

$So, \frac{10}{V/A}=\frac{\frac{0.05g}{cms}*1*10^{11}\frac{cs}{g}*15\frac{g}{Litre}*\frac{Litre}{10^3cm^3}}{2*75*10^4\frac{g}{cms^2}}(\frac{V}{A})+\frac{0.05g/cms*1*10^8cm^{-1}}{75*10^4\frac{g}{cms^2}}$

$\therefore \frac{V}{A}=3.855*10^{-3}m$

$V=3.855*10^{-3}m*44.44m^2=0.1713m^3$

• The filtration rate expected for the rotary vacuum filter is;

$$\frac{0.1713m^3}{10s}=17.13L/s$$

• To answer how significant is the resistance medium is;

$R_c=a\rho_c(\frac{V}{A})=1*10^{11}\frac{cm}{g}*15\frac{g}{L}*3.855*10^{-3}m= 5.78*10^{10}m^{-1}$

So, at the end of the filtration,

$\frac{R_m}{R_c+R_m}=\frac{1*10^8cm^{-1}}{5.78*10^{10}m^{-1}+1*10^8cm^{-1}}=0.1475$

Therefore the filter medium is contributing very little of the resistance to filtration