Answer :
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:
[tex] V_{f}^{2} = \frac{2A^{2}t_{f}\Delta P}{\mu \alpha \rho} [/tex] (1)
Where:
[tex]V_{f}[/tex]: is the volume of filtrate collected
A: is the filtration area
[tex]t_{f}[/tex]: 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:
[tex] \mu = 5 cP \cdot \frac{0.01 g/cm*s}{1 cP} = 0.05 \frac{g}{cm*s} [/tex]
[tex] \rho = 15 \frac{g}{L} \cdot \frac{1 L}{1000 cm^{3}} = 15 \cdot 10^{-3} g/cm^{3} [/tex]
[tex] \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} [/tex]
Now, we need to find the filtration area (A), which is equal to a cylinder area:
[tex] A = 2 \pi r^{2} + 2 \pi rh [/tex]
where r: is the radius = 8/2 m and h: is the height of the side = 12 m.
[tex]A = 2\pi (4m)^{2} + 2\pi 4m*12m = 4.02 \cdot 10^{6} cm^{2}[/tex]
Now, we can calculate the volume of filtrate using equation (1):
[tex]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}[/tex]
Finally, we can calculate the filtration rate in volumes/hr:
[tex]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[/tex]
I hope it helps you!