Calculate the diameter and velocity of a circular sewer at a slope of 1 in 400 when it is running just full at a discharge of `0.85\ m^3` /sec. The value of n in Manning's formula is 0.011. What will be the discharge and velocity when flowing 0.6 depth of pipe for the same slope.

Solution:

When running just full,

We know,

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

or, `0.85=(pid^2)/4**(1/0.011)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=0.8256\ m`

And the velocity is, `V=Q/A=0.85/((pi**0.8256^2)/4)`

or, `V=1.5875\ m/s`

When flowing at a depth 0.6,

`d/D=0.6`

or, `1/2(1-cos (theta/2))=0.6`

or, `theta=203.07^0`

Now, we know,

`q/Q=q/0.85=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`q=0.571\ m^3/s`

And the velocity is,

`v/1.5875=(1-(360 sin theta)/(2pi theta))^(2/3)`

Solving this equation, we get,

`v=1.89\ m/s`

Solution:

When running just full,

We know,

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

or, `0.85=(pid^2)/4**(1/0.011)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=0.8256\ m`

And the velocity is, `V=Q/A=0.85/((pi**0.8256^2)/4)`

or, `V=1.5875\ m/s`

When flowing at a depth 0.6,

`d/D=0.6`

or, `1/2(1-cos (theta/2))=0.6`

or, `theta=203.07^0`

Now, we know,

`q/Q=q/0.85=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`q=0.571\ m^3/s`

And the velocity is,

`v/1.5875=(1-(360 sin theta)/(2pi theta))^(2/3)`

Solving this equation, we get,

`v=1.89\ m/s`