Question Q2. A filter with an area of 1 m² operates at a constant pressure drop of 100 kPa to filter a mineral slurry. A test gives the following results with an incompressible cake: Volume of filtrate collected (litre) 1 10 50 100 200 Time (min) 0.01 0.11 0.75 2 6 If you run the above filtration test at a constant pressure drop of 50 kPa, predict the experimental outcomes: a) What time is required to collect 0.1 m² of filtrate? b) What are the filtration rates (volume flow rate of filtrate) at the start and end of the filtration, respectively? Comment on the change in the filtration rate as a function of time and briefly explain why. c) Immediately following the completion of the filtration, the resulting cake is washed using water. Estimate the time required to wash the cake with 0.1 m at a constant pressure drop of 25 kPa.

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Transcribed Image Text: Q2. A filter with an area of 1 m² operates at a constant pressure drop of 100 kPa to filter a mineral slurry. A test gives the following results with an incompressible cake: Volume of filtrate collected (litre) 1 10 50 100 200 Time (min) 0.01 0.11 0.75 2 6 If you run the above filtration test at a constant pressure drop of 50 kPa, predict the experimental outcomes: a) What time is required to collect 0.1 m² of filtrate? b) What are the filtration rates (volume flow rate of filtrate) at the start and end of the filtration, respectively? Comment on the change in the filtration rate as a function of time and briefly explain why. c) Immediately following the completion of the filtration, the resulting cake is washed using water. Estimate the time required to wash the cake with 0.1 m at a constant pressure drop of 25 kPa.

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plot grapb t / v e V\begin{array}{l}t /=1 \times b^{-4} \gamma+0.01 \\P=100 \mathrm{kPa} .\end{array}i)\begin{array}{l}t / v=k v+B \\k \propto \frac{1}{\Delta P} \text { and } B \propto \frac{1}{\Delta p} \\\text { So when } \Delta p=50 \mathrm{kPa} \\t / v=\frac{1 \times 10^{-4}}{0.5} v+\frac{0.01}{0.5}\end{array}put V=100 \mathrm{ml}t=45b) filteration rate =\frac{d v}{d t} \cdot 1 / B+2 k x (from the table) at end\begin{aligned}d v /\left.a t\right|_{F} & =\frac{1}{\frac{0.01}{0.5}+\frac{2 \times 1 \times 10^{-4} \times}{0.5}} 200 \\& =20\end{aligned} at start (from the table)\begin{aligned}\left.d \frac{a t}{}\right|_{F} & =\frac{1}{\frac{0.01}{0.5}+\frac{2 \times 1 \times 10^{-4} \times 1}{0.5}} \\& =49.02\end{aligned}(ii) New \frac{1}{V} V_{S} V equation.\begin{array}{l}\Delta p=25 k p a \\f / v=\frac{1 \times 10^{-4}}{0.25} \times v+\frac{0.01}{.25} \\\frac{d v}{d t}=\frac{1}{\frac{0.01}{0.25}+\frac{2 \times 1 \times 10^{-4}}{0.25} \times 200}=5\end{array}time: \left.\frac{0.1 \times 1000}{d y / a b}\right|_{F}\begin{array}{l}=\frac{0.1 \times 1000}{5} \\=205\end{array} -------------------------------------------------------------------------------------------- Really iam trying my best to answer your questions. So, Please dont dislike because my cf score is less. If you really didnt get the answer for your question, comment below so that i will check and rectify it. Once again iam requesting you please dont dislike my answers, as i am trying my best for you. ...