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

1 answer:

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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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A car travels eastwards at 60km/h for 2h, then travels northwards at 20km/h for 8h. Find,

sveta [45]

Answer:

$a) Average\ Speed=28\ km/h\\b) Average\ Velocity= 20\ km/h$

Explanation:

$We\ are\ given\ that,\\Velocity\ of\ the\ car\ Eastwards=60\ km/h\\Time\ taken\ by\ the\ car\ Eastwards=2\ h\\Velocity\ of\ the\ car\ Northwards=20\ km/h\\Time\ taken\ by\ the\ car\ Northwards=8\ h\\Hence,\\As\ we\ know\ that,\\Speed=\frac{Distance}{Time}\\Distance= Speed* Time\\Now,\ lets\ find\ the\ distance\ covered\ by\ the\ car\ in\ both\ the\ cases.$

$Hence,\\Distance\ Covered\ During\ its\ Eastward\ Journey=60*2=120\ km\\Distance\ Covered\ During\ its\ Northward\ Journey=20*8=160\ km\\Now,\\As\ we\ know\ that\ Average\ Speed=\frac{Total\ Distance}{Total\ Time} \\Here,\\Total\ distance\ of\ the\ car=Distance\ Covered\ During\ its\ Northward\ Journey+Distance\ Covered\ During\ its\ Eastward\ Journey\\Hence,\\Total\ distance\ of\ the\ car=120+160=280\ km\\Total\ time\ taken\ by\ the\ car=8+2=10\ hours\\Hence,\\$ $Average\ Speed\ Of\ the\ Car\ throughout\ its\ journey=\frac{280}{10}=28\ km/h$

$Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\$ $We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h$