written 5.2 years ago by |

As discussed earlier, that for a most economical section, the discharge slope of bed and resistance co-efficient is maximum, but in case of circular channels, the area of flow cannot be maintained constant, with change of flow in circular channels of any radios, the wetted area and wetted perimeter changes, thus in case of circular section there are two different conditions.

1] Condition for maximum velocity,

2] Condition for maximum discharge.

**1] Condition for maximum velocity for circular section:**

Let, d = depth of water

2 $\theta$ = angle subtended at the center by water surface

R = radius of channel

i = slope of the bed

The velocity of flow according to Chezy's formula,

$v = c \sqrt{mi}$

$c = \sqrt \frac{A}{P} i$

$\because m = \frac{A}{P}$

The velocity of flow through a circular channel will be maximum when the hydraulic mean depth 'm' or (A/P) is maximum for a given value of c and i, In case of circular pipe, the variable is $\theta$ only, hence, for maximum value of $\frac{A}{P}$ we have the condition.

$\frac{d(\frac{A}{P})}{d \theta} = 0$ - - - - (1)

where A and P both are functions of $\theta$

The value of wetted area A, is given by equation.

$A = R^2 ( \theta - \frac{sin 2 \theta}{2})$ - - - -(2)

The value of wetted perimeter, P is given by equation,

P = 2 R $\theta$ - - - - (3)

Differentiating equation (1)

$\frac{P \frac{dA}{d\theta} - A \frac{dp}{d \theta}}{p^2} = 0$

$p \frac{dA}{d \theta} - A \frac{dp}{d \theta} = 0$ - - - - (4)

from equation (2) , $\frac{dA}{d \theta} = R^2 ( 1 - \frac{cos 2 \theta}{2} \times 2)$

$R^2 (1- cos 2 \theta)$

from equation (3), $\frac{dp}{d \theta} = 2R$

Substituting the values of A, $\frac{PdA}{d \theta}$ and $\frac{dP}{d \theta}$ in equation (4),

$2R \theta [ R^2 (1- cos 2 \theta)] - R^2 (\theta - \frac{sin 2 \theta}{2}) (1R)$ = 0

$2 R^3 \theta (1- cos 2 \theta ) - 2R^3 ( \theta - \frac{sin 2 \theta}{2})$ = 0

$\theta (1- cos 2\theta) - (\theta - \frac{sin 2 \theta}{2} )$ = 0

$\theta - \theta cos 2 \theta - \theta + \frac{sin 2\theta}{2}$ = 0

$\frac{sin 2\theta}{coz 2\theta} = 2\theta$

$tan 2\theta = 2\theta$

The solution of this equation by hit and trail, gives

$2\theta = 257˚ 30'$

$\theta = 128˚ 45'$

The depth of flow for maximum velocity

d = OD - OC

= R - Rcos$\theta$

= R [ 1- cos $\theta$]

= R [1 - cos 128˚ 45']

= R [1- cos(180˚ - 51˚ 15')]

= R [ 1 - 0 + cos 51˚ 15')]

= R [ 1 + cos 51˚ 15']

= R [1 + 0.62]

= 1.62 R

$= 1.62 \times \frac{D}{2}$

= 0.81 D

D = Diameter of circular channel

Thus, for maximum velocity of flow, the depth of water in circular channel should be equal to 0.81 times the diameter of the channel.

Hydraulic mean depth for maximum velocity is,

$m = \frac{A}{p}$

$= \frac{R^2 (\theta - \frac{sin 2 \theta}{2})}{2 R \theta}$

$\frac{R}{2 \theta} [ \theta - \frac{sin 2 \theta}{2}]$

where, $\theta$ = 128˚ 45' = 128.75˚

$\therefore$ $128.75 \times \frac{\pi}{180}$

= 2.247 radius

$\therefore$ m = $\frac{R}{2 \times 2.247} [ 2.247 - \frac{sin 257˚ 30'}{2}]$

$= \frac{R} {4.494} [ 2.247 + \frac{sin 87.5˚}{2}]$

= 0.611 B

$m = 0.611 \times \frac{D}{2} = 0.3D$

**2] condition for maximum discharge for circular section,**

$Q = Ac \sqrt mi$

$= Ac \sqrt{ \frac{A}{p}}i$

$c \sqrt{\frac{A^3}{p}} i$

The discharge will be maximum for constant values of c and i, when $\frac{A^3}{p}$ is maximum.

$\frac{A^3}{p}$ will be maximum when $[ \frac{d}{d \theta} (\frac{A^3}{p}) = 0 ]$

Differentiating this equation with respect to $\theta$ and equation the same to zero, we get,

$\frac{p \times 3A^2 \frac{dA}{d \theta} - A^3 \frac{dp}{d \theta}}{p^2} = 0$

$3PA^2 \frac{dA}{d \theta} - A^3 \frac{dp}{d \theta} = 0$

Dividing by $A^2, 3p \frac{dA}{d \theta} - A \frac{dp}{d \theta} = 0$

but from the eqution ,$ (p = 2R \theta)$

$\therefore$ $\frac{dp}{d \theta} = 2R$

from equation, $A = R^2 (\theta - \frac{sin 2 \theta}{2})$

$\therefore$ $\frac{dA}{d \theta } = R^2 (1- cos 2 \theta)$

Substituting the values of $P, A \frac{dp}{d \theta}$ and $\frac{dA}{d \theta}$ in equation (1),

$3 \times 2 R \theta \times R^2 (1- cos 2 \theta) - R^2 (\theta - \frac{sin 2 \theta}{2}) \times 2R = 0$

$6 R^3 \theta (1- cos 2\theta) - 2 R^3 ( \theta - \frac{sin 2 \theta}{2})= 0$

Dividing by $ 2 R^3$, we get

$3 \theta (1 - cos 2 \theta) - ( \theta - \frac{sin 2 \theta}{2}) = 0$

$3 \theta - 3 \theta cos 2 \theta - \theta + \frac{sin 2 \theta}{2} = 0$

$2 \theta - 3 \theta cos 2 \theta + \frac{sin 2 \theta}{2} = 0$

$4 \theta - 6 \theta cos 2 \theta + sin 2 \theta = 0$

The solution of this equation hit and trail goes,

$2 \theta = 308˚$

$ \theta = \frac{308˚}{2}$

$\therefore$ $ \theta$ = 154

Depth of flow for maximum discharge,

d = OD - OC

$= R - R cos \theta$

$= R [ 1 - cos \theta]$

$= R [ 1 - cos 154˚]$

$R [ 1- cos (180˚ - 26˚)]$

= R [1 + cos 26˚]

= 1.898 B

$= 1.898 \frac{D}{2}$

[ d = 0.948 D = 0.95 D]