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1 year ago

Particle dislodgement in the filtration process is generally undesirable, except during?

1 answer:

4 0

Particle dislodgement in the filtration process is generally undesirable, except during high pressure jets of water applied to filter media during a backwash.

<h3>Define filtration process. </h3>

Filtration is a physical separation procedure that uses a filter medium with a complicated structure that only allows fluid to pass through it to separate solid particles and liquid from a combination. Filtrate is the term for the liquid that passes through the filter medium and is used to describe solid particles that are too large to pass. Larger-than-average particles have the potential to blind the filter by blocking the filter lattice and forming a filter cake on top of the filter. The term "effective pore size" refers to the size of the biggest particles that may successfully pass through a filter.

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Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

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1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

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2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

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