A 60 cm diameter sewer is to discharge `0.07` `m^3`/sec at a velocity as self cleansing as sewer flowing full at 0.85 m/sec. Find depth and velocity of flow and required slope. Take uniform value of manningā??s coefficients (n) = 0.015

Solution:

When sewer is running full,

`d=0.6\ m`

`v=0.85=1/0.015(d/4)^(2/3)**(S)^(1/2)`

or, `S=1/490.269` and the discharge is,

`Q=pi**0.6^2/4**0.85=0.24\ m^3/s`

For partial depth self cleansing flow

`q_s=0.07\ m^3/s`

or, `q_s/Q=0.07/0.24=0.2912`

or, `0.2912=(theta/360-(sin theta)/(2pi))(1-(360 sin theta)/(2 pi theta))^(1/6)`

or, `theta=143.63^0`

Also, we have,

`d/D=1/2(1-cos theta/2)=0.3439`

or, `d=0.2063\ m`

Also,

`r/R=(1-(260 sin theta)/(2 pi theta))`

`=0.763`

Now,

`V_s/V=(r/R)^(1/6)`

or, `V_s=0.956**0.85=0.8162\ m/s`

Also,

`s=(R/r)S=1/0.763**2.093**10^-3`

`=1/374.20`

Solution:

When sewer is running full,

`d=0.6\ m`

`v=0.85=1/0.015(d/4)^(2/3)**(S)^(1/2)`

or, `S=1/490.269` and the discharge is,

`Q=pi**0.6^2/4**0.85=0.24\ m^3/s`

For partial depth self cleansing flow

`q_s=0.07\ m^3/s`

or, `q_s/Q=0.07/0.24=0.2912`

or, `0.2912=(theta/360-(sin theta)/(2pi))(1-(360 sin theta)/(2 pi theta))^(1/6)`

or, `theta=143.63^0`

Also, we have,

`d/D=1/2(1-cos theta/2)=0.3439`

or, `d=0.2063\ m`

Also,

`r/R=(1-(260 sin theta)/(2 pi theta))`

`=0.763`

Now,

`V_s/V=(r/R)^(1/6)`

or, `V_s=0.956**0.85=0.8162\ m/s`

Also,

`s=(R/r)S=1/0.763**2.093**10^-3`

`=1/374.20`