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Katen [24]
3 years ago
11

A flocculation basin equipped with revolving paddles is 60 ft long (the direction of flow). 45 ft wide, and 14 ft deep and treat

s 10 mgd. The power input to provide paddle-blade velocities of 1.0 and 1.4 fps for the inner and outer blades, respectively, is 930 ft'lb/s. Calculate the detention time, horizontal flow-through velocity, and G (the mean velocity gradient) for a water temperature of 50'F.
Engineering
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

detention time =  40.72 minutes

horizontal velocity v  = 0.295 inch/s

mean velocity gradient, G = 54.38 s^{-1}

Explanation:

given data

long = 60 ft

wide = 45 ft

deep = 14 ft

flow of water to treat = 10 mgd = 15.472 ft³/s         [1 MGD = 1.5472 ft³/s ]

inner velocities = 1.0  fps

outer velocities = 1.4  fps

power required to rotates the paddles = 930 ft-lb/s

solution

we find detention time that is

time = \frac{V}{Q}

here  Q = flow of water and  V is volume of water

so

volume of flocculation basin is = 60 × 45 × 14 = 37800 ft³

so

detention time = \frac{37800}{15.472}

detention time =  40.72 minutes

and

horizontal flow through velocity is

cross-section area of tank = 45 × 14 = 630 ft²

and we know that  Av = Q

so horizontal velocity v =  \frac{Q}{A}

horizontal velocity v =  \frac{15.472}{630}

horizontal velocity v = 0.02456 ft/s

horizontal velocity v  = 0.295 inch/s

and

now we find mean velocity gradient G at  temperature of 50'F

so formula is G = \sqrt{\frac{P}{\mu V} }      

here P is power input  and μ is  viscosity  and V is  volume of flocculation basin

so Volume of flocculation basin is 37800 ft³

and μ for 500F = 8.3197  × 10^{-6} lb-s/ft²

we take from table

so we get

G = \sqrt{\frac{930}{8.3197*10^{-6}*37800} }  

mean velocity gradient, G = 54.38 s^{-1}

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