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2 years ago

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A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

3 m3/h. The solution leaves the tank at a rate of 2 m3/h. When the tank is full, the solution leave the tank at the same inlet flow of 3 m3/h. Assume perfect mixing. Specific gravity of pure NaOH is 1.22. Determine the time necessary to purge 95% of the NaOH by mass from the tank.

1 answer:

lawyer [7]2 years ago

4 0

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

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Explanation:

Given:-

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Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

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( ma*a - mb*a )*r = 0.5*M*r^2*α

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