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Alecsey [184]
3 years ago
12

4. Design an aerated grit chamber installation (2 tanks in parallel) for an average flowrate of 0.3 m3/s and a peak flowrate of

1.0 m3/s. The average depth is 3 m, the width-to-depth ratio is 1.5:1, and the detention time at peak flow is 3.5 min. The aeration rate is 0.4 m3/min per m of the tank length. Determine the dimensions of the grit chambers and the total air required
Physics
1 answer:
Sergio [31]3 years ago
6 0

Answer

given,

average flow rate = 0.3 m³/s

peak flow rate = 1 m³/s

width to depth ratio = 1.5 : 1

detention time = 3.5 min

average depth = 3 m

aeration rate = 0.4 m³/min/m

dimensions of grit chamber = ?

volume of grit chamber = \dfrac{1 \times 3.5 \times 60}{2}

                                       = 105 m³

average depth = 3 m

width-to-depth ratio = 1.5:1,

width = 1.5 × 3 = 4.5 m

length of the grit chamber = \dfrac{105}{3\times 4.5}

                                            = 7.78 m

increase length by 20% to account for inlet and outlet of the tank

length = 1.2 × 7.78 = 9.34 m

dimension of grit chamber = 9.34 m × 4.5 m × 3 m

air required = aeration rate × length

                    = 0.4 m³/min /m × 9.34

                    = 3.736 m³/min

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Explanation:

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    The current on the copper is  I  = 20 \ A

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          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

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substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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