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Assoli18 [71]
3 years ago
14

A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of 10 m ´ 8 m, an effective filtration rate of 7.70 m/h.

A complete filter cycle duration is 52 h and the filter is rinsed for 20 minutes at the start of each cycle. a. What flow rate (m3 /s) does the filter handle during production? b. What volume of water is used for backwashing plus rinsing the filter in each filter cycle?
Engineering
1 answer:
galben [10]3 years ago
8 0

Answer:

Explanation:

given data

loading rate = 8.00 m/h

filtration rate = 7.70 m/h

dimensions = 10 m × 8 m

filter cycle duration = 52 h

time = 20 min

to find out

flow rate  and  volume of water is used for back washing plus rinsing the filter  

solution  

we consider here production efficiency is 96%

so here flow rate will be  

flow rate = area × rate of filtration  

flow rate = 10 × 8 × 7.7  

flow rate = 616 m³/h

and  

we know back washing generally 3 to 5 % of total volume of water per cycle so  

volume of water is = 616 × 52

volume of water is  32032 m³

and  

volume of water of back washing is = 4% of 32032  

volume of water of back washing is 1281.2 m³

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Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
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Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

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Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

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6 0
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A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
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Answer:

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2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

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The defective rate of the device = 3%

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The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

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Therefore, the probability of not exactly 1 defective = q = 1 - p

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The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

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