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7nadin3 [17]
2 years ago
15

Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib

le medium resistance. The cake solids (dry basis) per volume of filtrate was 20 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.
Engineering
1 answer:
insens350 [35]2 years ago
4 0

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m ,  Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr  expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( \frac{60}{20} ) * 1706.0670

   = 5118.201 liters  ≈ 5.118 m^3/hr

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Answer:

<em>the % recovery of aluminum product is 80.5%</em>

<em>the % purity of the aluminum product is 54.7%</em>

<em></em>

Explanation:

feed rate to separator = 2500 kg/hr

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of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

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x 100% = <em>54.7%</em>

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2 years ago
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wi
Eduardwww [97]

Answer: 255

255 turns are required to create 25 ohms of  secondary impedance.

Explanation:

Given that,

Number of turns in primary wire N₁ = 900

impedance on Primary wire Z₁ = 400 ohms

Number of turns in Secondary wire N₂ = ?

impedance on Secondary wire Z₂ = 25 ohms

we know that, the relationship between turn and impedance is

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there fore

Z₁ / Z₂ = ( N₁ / N₂ )²

Now we substitute

( 400 / 25 ) = ( 900 / N₂ )²

400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

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N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

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N₂ = 225

Therefore 255 turns are required to create 25 ohms of  secondary impedance.

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