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joja [24]
3 years ago
5

A steady state filtration process is used to separate silicon dioxide (sand) from water. The stream to be treated has a flow rat

e of 50 kg/min and contains 0.22% sand by mass. The filter has a cross-sectional area of 9 square meters and successfully filters out 90% of the input sand by mass. As the filter is used, a cake forms, which we will assume is pure sand (SG of sand = 2.25). The filter needs to be replaced once this cake has a thickness of 0.25 meters.
Required:
How long can a new filter be used before it needs to be replaced, in units of days?
Engineering
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

3.19 days

Explanation:

<em>Given data :</em>

stream flow rate = 50 kg/min

stream contains ; 0.22% sand by mass

Cross sectional area of filter = 9 m^2

Filter successfully filters out 90% of the input sand by mass

SG = 2.25

thickness of cake formed = 0.25 meters

<u>Determine how long a new filter can be used before replacement </u>

Given that for every 1 minute  5.43*10^-5 m thickness of sand layer(cake) forms on the filter and the replacement of filter is done once the cake thickness = 0.25 meters

To determine the number of days ( X ) before replacing filter we apply the relationship below

5.43 * 10^-5 = 1 min

0.25 m = X

hence ; X = 0.25 / (5.43 * 10^-5 )  = 4604.051 minutes ≈ 3.19 days

<u />

Attached below is the beginning part of the detailed solution

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Attached below is the remaining  part of the detailed solution

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Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia
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h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

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h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

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Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the
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Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

where Q = volumetric flowrate

ΔP = Pressure drop across the pipe

μ = fluid viscosity

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If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe

ΔP = μ(8QL/πR⁴)

ΔP = Kμ

K = (8QL/πR⁴) = constant (for this question)

ΔP = Kμ

K = (ΔP/μ)

So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).

μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

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Recall,

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ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

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Hope this Helps!!!

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