For a inflow of 25 $m^{3} / d a y$ and percolation rate of 12 minutes, the absorption area required (in $m^{2}$ , rounded off to nearest integer) for dispersion trench is
420

Open in App

Solution

The correct option is A 420
Absorption area required= $\frac{Q (\frac{L}{d a y})}{(\frac{204}{\sqrt{t}}) \frac{\frac{l}{m^{2}}}{d a y}}$
$w h e r e Q = 25 \frac{m^{3}}{d a y} = 25000 \frac{l}{d a y}$
$t = 12 m i n u t e s$

$A = \frac{25000}{\frac{204}{\sqrt{12}}} = 424.52 m^{2}$

Suggest Corrections

Similar questions

What do you understand by water absorption rate?
It took 60 minutes for 360 mL of water to percolate through a soil sample. What is the rate of percolation for the soil sample? [5 MARKS]

Q. A municipal corporation is required to treat

1000

m^{3} / d a y

of water. It is found that an overflow rate of 20 m/day will produce a satisfactory removal of the discrete suspended particles at a depth of

3 m

. The diameter (in meters, rounded to the nearest integer) of a circular settling tank designed for the removal of these particles would be.

Q. The rates of rainfall for successive 10 minutes period of a 60 minute duration storm, are shown in the below figure,

If the value of

ϕ

- index is 3 cm/hour, the runoff will be_____mm. (rounded off to nearest integer)

An army camp of 250 men has provision for 350 days. After 25 days, 50 men died due to epidemic. The food will last for

___ days. (Round off the answer to the nearest integer)

An army camp of 250 men has provision for 350 days. After 25 days, 50 men left the camp The food will last for

___ days. (Round off the answer to the nearest integer)

Join BYJU'S Learning Program

For a inflow of 25 m3/day and percolation rate of 12 minutes, the absorption area required (in m2, rounded off to nearest integer) for dispersion trench is420

The correct option is A 420Absorption area required=Q(Lday)(204√t)lm2day where Q=25m3day=25000lday t=12 minutes A=25000204√12=424.52m2

For a inflow of 25 $m^{3} / d a y$ and percolation rate of 12 minutes, the absorption area required (in $m^{2}$ , rounded off to nearest integer) for dispersion trench is
420

The correct option is A 420
Absorption area required= $\frac{Q (\frac{L}{d a y})}{(\frac{204}{\sqrt{t}}) \frac{\frac{l}{m^{2}}}{d a y}}$
$w h e r e Q = 25 \frac{m^{3}}{d a y} = 25000 \frac{l}{d a y}$
$t = 12 m i n u t e s$

$A = \frac{25000}{\frac{204}{\sqrt{12}}} = 424.52 m^{2}$