Calculate diameter and velocity of circular sewer at a slope of 1 in 400 when it is running just full at a discharge of `1\ m^3`/sec. The value of n in manning's coefficient is 0.012. Will self cleansing velocity be maintained in sewer when flow drops to `0.6\ m^3`/sec?

Solution:

When running just full,

We know,

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

or, `1=(pid^2)/4**(1/0.012)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=0.9066\ m`

A=pi d^2/4=0.65\ m^2`

And the velocity is, `V=Q/A=1/((pi**0.9066^2)/4)`

or, `V=1.5489\ m/s`

Now, when flows drops to `0.6\ m^3/s`,

`q/Q=0.6/1=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`theta=193.38^0`

Also,

`v/1.5489=(1-(360 sin theta)/(2pi theta))^(2/3)`

or, `v=1.618\ m/s`

Since  `1.618 > 0.6`, Self cleansing velocity will be maintained in the sewer.

Solution:

When running just full,

We know,

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

or, `1=(pid^2)/4**(1/0.012)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=0.9066\ m`

A=pi d^2/4=0.65\ m^2`

And the velocity is, `V=Q/A=1/((pi**0.9066^2)/4)`

or, `V=1.5489\ m/s`

Now, when flows drops to `0.6\ m^3/s`,

`q/Q=0.6/1=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`theta=193.38^0`

Also,

`v/1.5489=(1-(360 sin theta)/(2pi theta))^(2/3)`

or, `v=1.618\ m/s`

Since  `1.618 > 0.6`, Self cleansing velocity will be maintained in the sewer.