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Solution:

+ ^ ( 2 ) + ^ ( 2 ) + ^ ( 2 ) + ^ ( 2 ) + «-p ( 2 ) + ^ ( 2 ) + 445+51 ( 2 ) + 51 + 45 ( 2 ) + 440+51 ( 2 ) + 40+51 ( 2 )

- 37.7 m3/hr Ans Example 3.9 Using the data in Example 3.8, design the equalization basin. Solution:

Vbasn = 2>s of (Qm)(t, - t-2) = pos of ^^^^T2-- 37.7^(2)

+ (188+2- - 37.7)( 2) + (119+18 - 37.7) 2) + (27+1- - 3^ 2)

+ (39(27 - 37.72) + (- 37.72) + (62 + 52 - 37.7)(2) + (51ir2 - 37.7)(2) + (45 + 51 - 37.3 2) + (5' + 45 - 37.7)(2) + (440+51-37.7)( 2) + (^+-40 - 37.7)(2)} = (- 37.72) + (^p - 37.7)(2) + (51+6- - 37.7)(2)

+ (45 + ^ - 37.3 2) + (5- + 45 - 37.7 2) + (440+5-- - 37.72)}

= 15.6 + 38.6 + 37.6 + 20.6 + 20.6 + 15.6 = 148.6 m3 Use a circular basin at a height of 4 m. Therefore,

Therefore, dimensions: height = 4 m, diameter = 10 m; use two tanks, one for standby. Ans

Example 3.10 The following table shows the BOD values read from Figure 3.9 at intervals of 2 h. Along with the data in Example 3.8, calculate each equalized value of the BOD at every time interval when the tank is filling.

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