Question

Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x 1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 15 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter?

Answer 1

Answer:

473,400 L/h

Explanation:  

For an incompressible cake and negligible medium resistance, we can find the  volume of filtrate collected using the following equation:

$V_{f}^{2} = \frac{2A^{2}t_{f}\Delta P}{\mu \alpha \rho}$     (1)          

Where:

$V_{f}$: is the volume of filtrate collected  

A: is the filtration area  

$t_{f}$: is the cake formation time = 20 s  

ΔP: is the vacuum pressure = 80 kPa  

μ: is the viscosity = 5 cP  

α: is the specific cake resistance = 1x10¹¹ cm/g  

ρ: is the cake solid per volume of filtrate = 15 g/L         

First, we need to convert the units:  

$\mu = 5 cP \cdot \frac{0.01 g/cm*s}{1 cP} = 0.05 \frac{g}{cm*s}$

$\rho = 15 \frac{g}{L} \cdot \frac{1 L}{1000 cm^{3}} = 15 \cdot 10^{-3} g/cm^{3}$      

$\Delta P = 80\cdot 10^{3} Pa = 80\cdot 10^{3} \frac{N}{m^{2}} = 80 \cdot 10^{3} \frac{kg}{m*s^{2}} \cdot \frac{1000 g * 1 m}{1 kg * 100 cm} = 8.0 \cdot 10^{5} gcm^{-1}s^{-2}$

Now, we need to find the filtration area (A), which is equal to a cylinder area:    

$A = 2 \pi r^{2} + 2 \pi rh$

where r: is the radius = 8/2 m and h: is the height of the side = 12 m.    

$A = 2\pi (4m)^{2} + 2\pi 4m*12m = 4.02 \cdot 10^{6} cm^{2}$  

Now, we can calculate the volume of filtrate using equation (1):    

$V_{f} = \sqrt {\frac{2(4.02 \cdot 10^{6} cm^{2})^{2}*20 s*8 \cdot 10^{5} gcm^{-1}s^{-2}}{0.05 gcm^{-1}s^{-1}*1 \cdot 10^{11}cmg^{-1}*15 \cdot 10^{-3} gcm^{-3}}} = 2.63 \cdot 10^{6} cm^{3}$

Finally, we can calculate the filtration rate in volumes/hr:  

$rate = \frac{V_{f}}{t_{f}} = \frac{2.63 \cdot 10^{6} cm^{3}}{20 s} \cdot \frac{3600 s * 1 L}{1 h * 1000 cm^{3}} = 473,400 L/h$      

 

I hope it helps you!