Question #126580
A suspension of yeast cells is to be filtered at a constant filtration rate of 50 L/min.
The suspension has a solid content of 70 kg/m3 of suspension and the yeast cells
have a bulk density of 800 kg/m3. Laboratory tests indicate that the specific cake
resistance is 40 m/kg and the viscosity of the filtrate is 2.9x10-3 kg/m.s. The filter has
an area of 0.1 m2 and the medium offers negligible resistance. How long can the
filtration rate be maintained before the pressure drop exceeds 1 N/m2? What volume
of cake and filtrate are collected during this time? Note that ρ0 is expressed in mass
of solids per volume of filtrate and NOT total volume of suspension.
1
Expert's answer
2020-07-22T05:50:59-0400

We are given,

Density=800kg/m3=800kg/m^3

Specific cake resistance =40m/kg=40m/kg

Viscosity of the filtrate =2.9 103kg/m.s=2.9\ *10^{-3}kg/m.s

Area=0.1m2=0.1m^2

Pressure=1N/m2=1N/m^2

α\alpha =70kg/m3=70kg/m^3


Formula used.

t =μαρ02ΔP(VA)2t\ =\dfrac{\mu\alpha\rho_0}{2\Delta P}(\dfrac{V}{A})^2


RR c=αρ0(VA)=\alpha\rho_0(\dfrac{V}{A})


40 0.1 =70 800 V4 =56000VV=7.14 10540\ *0.1\ =70\ *800\ *V\newline 4\ =56000V\newline V=7.14\ *10^{-5}


t =2.9103  70  8002 1 (7.141050.1)2t\ =\dfrac{2.9*10^{-3} \ \ *70\ \ *800}{2\ *1}\ (\dfrac{7.14*10^{-5}}{0.1})^2


=81.2  (5.1107)=81.2\ \ *(5.1 *10^{-7})


=4.1 105=4.1\ *10^{-5}






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