Calculate the diameter and velocity of a circular sewer at a slope of 1 in 400 when it is running just full at a discharger of `2` `m^3`/sec. The Manning's coefficient n = 0.013. What will be the discharge and velocity when flowing one third full?

Solution:

When running just full,

We know,

`Q=A/nR^(2/3)S^(1/2)`

or, `2=(pid^2)/4**(1/0.013)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=1.2115\ m`

And the velocity is, `V=Q/A=2/((pi**1.2115^2)/4)`

or, `V=1.735\ m/s`

When flowing at a one third depth,

`d/D=1/3`

or, `1/2(1-cos (theta/2))=1/3`

or, `theta=141.057^0`

Now, we know,

`q/Q=q/2=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`q=0.4794\ m^3/s`

And the velocity is,

`v/1.735=(1-(360 sin theta)/(2pi theta))^(2/3)`

Solving this equation, we get,

`v=1.425\ m/s`

Solution:

When running just full,

We know,

`Q=A/nR^(2/3)S^(1/2)`

or, `2=(pid^2)/4**(1/0.013)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=1.2115\ m`

And the velocity is, `V=Q/A=2/((pi**1.2115^2)/4)`

or, `V=1.735\ m/s`

When flowing at a one third depth,

`d/D=1/3`

or, `1/2(1-cos (theta/2))=1/3`

or, `theta=141.057^0`

Now, we know,

`q/Q=q/2=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`q=0.4794\ m^3/s`

And the velocity is,

`v/1.735=(1-(360 sin theta)/(2pi theta))^(2/3)`

Solving this equation, we get,

`v=1.425\ m/s`