Uniform Flow — Channel Flow

Introduces the Chezy and Manning formulae for open channel flow

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Introduction

Uniform Flow occurs in long inclined channels of uniform cross section when the terminal velocity is reached. This occurs when the loss of potential energy equals the work done against the Channel Surface Friction. In this condition the Water Surface is parallel to the bed of the Channel.

The Chezy Equation

The Chezy Equation is analogous to the Darcy Equation used in the flow of liquid through Pipes. Chezy states that:-

$$\mathbf{v=C\;\sqrt{m\;i}}\;\;\;\;or\;\;\;\;Q=A\;c\;\sqrt{m\;i$$

(1)

Where

v is the velocity of flow
C is the Chezy constant.
i is the slope of the channel or the slope of the hydraulic gradient
A is the area of flow.
P is the wetted perimeter of the channel
m is the hydraulic mean depth which equals $\displaystyle =\frac{A}{P}$

A comparison between the Chezy and Darcy equations

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Darcy states that:-

$$\text{Head Lost}\;=\frac{4flv^2}{2dg}\;=\frac{flv^2}{2gm}\;\;\;\;\;\text{Where}\;\; m=\frac{d}{4}$$

(2)

This can be written as:-

$$v=\sqrt{\frac{2g}{f}\times m \times \frac{h}{l}}\; = \;\sqrt{\frac{2g}{f}}\times \sqrt{mi}$$

(3)

Thus the Darcy and Chezy equations are basically the same and $\displaystyle C=\sqrt{\frac{2g}{f}}$

The Chezy Constant C

The Chezy "C" varies with the nature of the channel walls and with "m" the Hydraulic mean depth. There are a number of Empirical equations which seek to eliminate the variations of C with m. Amongst these the Manning equation

The Manning Equation

Manning substituted $\displaystyle M\;m^{\frac{1}{6}}$ for the C of the Chezy equation. Hence:-

$$v=M\;m^{\frac{2}{3}}\;i^{\frac{1}{2}}$$

(4)

This can be re-written by replacing the "M" as follows:-

$$v=\frac{k}{n}\times \;m^{\frac{2}{3}}\;i^{\frac{1}{2}}$$

(5)

In this form

k is a conversion constant and equals 1.486 in the Imperial system and 1.0 in SI units.
n is the Manning Coefficient and is independent of units. It depends only on the nature of the channel surface and varies from 0.01 (smooth) to 0.035 (rough)

The conditions for the most Economic Channel Section

Three cross sectional areas will be considered .

Rectangular Section

The maximum discharge (or maximum velocity) for a given area will occur when the wetted area, P, is a minimum. For a unit length:-

$$\text{Area}\;\;A=b\times d$$

(6)

$$P=b+2d$$

(7)

$$\therefore \;\;\;\;\;P=\frac{A}{d}+2d$$

(8)

Differentiating and equating to zero for the minimum wetted area.

$$\frac{\mathrm{d} p}{\mathrm{d} d}=-\frac{A}{d^2}+2=0$$

(9)

$$\text{Or}\;\;\;\;\;\frac{b\times d}{d^2}=2\;\;\;\;\;i.e.\;\;\;\;\;b=2d$$

(10)

Thus for a maximum velocity of flow,v or a maximum rate of flow,Q

$$b=2d$$

(11)

A Trapezoidal Channel

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The maximum Flow (Q,) or the maximum velocity (v,) of flow for a given area and slope $\theta$ occurs when the wetted area P is a minimum. It can be shown that this happens when the sloping sides and the base are tangential to a semi-circle described on the water surface.

$$\sin\theta=\frac{d}{\sqrt{d^2n^2+d^2}}=\frac{r}{\frac{b+2nd}{2}}$$

(12)

But since $d=e$

$$\therefore \;\;\;\;\;d\;\sqrt{n^2+1}=\frac{b+2nd}{2}$$

(13)

$$\text{And the Hydraulic Mean Depth (m)}=\frac{\text{Area of flow}}{\text{Wetted perimeter}}$$

(14)

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$$m=\frac{\frac{1}{2}(b+b+2nd)d}{b+2d\sqrt{n^2+1}}=\frac{d(b+nd)}{2(b+nd)}=\frac{d}{2}$$

(15)

Circular Section

Using either the Chezy or Manning formulae, the maximum velocity for a given radius R will occur when $\displaystyle \frac{A}{P}$ is a maximum.

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$$i.e.\;\;\;\;\;\;\text{When}\;\;\;\frac{d\left ( \frac{A}{P} \right )}{d\theta}=\frac{P\frac{dA}{d\Theta}-A\frac{dP}{d\theta}}{p^2}=0$$

(16)

$$\text{This occurs when}\;\;\;\;2\theta=\tan2\theta$$

(17)

$$i.e.\;\;\;\;\;2\theta =257\;\tfrac{1}{2}^0$$

(18)

$$\text{And the maximum depth of water is }\;0.81\times \text{The diameter}$$

(19)

Using the Chezy equation the maximum flow, Q will occur when $\displaystyle \frac{A^3}{P}$ is a maximum.

$$\text{i.e. When}\;\frac{d\left ( \frac{A^3}{P} \right )}{d\theta}=\frac{P\times 3A^2\frac{dA}{d\theta}-A^3\frac{dP}{d\theta}}{P^2}=0$$

(20)

This will occur when:-

$$2\theta=3\theta\;\cos2\theta-\frac{1}{2}\sin2\theta$$

(21)

$$i.e.\;\;\;\;2\theta=308^0$$

(22)

$$\text{And the maximum depth at the centre } = 0.95\times \text{The diameter of the channel}$$

(23)

Or Using the Manning Equation Q will be a maximum when $\displaystyle \frac{A^5}{P^2}$ is a maximum

$$\text{i.e. When}\;\;\;\frac{d\left ( \frac{A^5}{P^2} \right )}{d\theta}=\frac{P^2\times A^4\frac{dA}{d\theta}-A^5\times 2P\frac{dP}{d\theta}}{P^4}=0$$

(24)

$$\text{From which}\;\;\;3\theta=5\theta\cos2\theta-\sin2\theta$$

(25)

$$\therefore \;\;\;\;\;2\theta =302\;\tfrac{1}{2}^0$$

(26)

$$\text{This gives a maximum depth at the centre of }\;0.938\times \text{The diameter of the channel}$$

(27)

Worked Examples

The workings to the following questions can be seen by clicking on the red buttons.

Example 1

Water flows down a uniform channel with a slope of $\frac{1}{1600}$. The cross section shown in the diagram.

If the channel is to convey 600 Cusec under the condition of maximum discharge for a given area find the values of R and B. C the Chezy constant is 120.

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More...

Example 2

A 3 ft. diameter conduit 12,000 ft. long is laid at a uniform slope of 1 in 1,500 and connects two reservoirs. When the levels in the reservoirs are low the conduit runs partly full and it is found that the normal depth of 2 ft. gives a rate of flow of $11.5 ft.^3/sec.$. The Chezy coefficient C is given by $Km^n$ where K is a constant, m is the hydraulic mean depth and $n=\frac{1}{6}$ . Neglecting the losses of head at entry and exit, obtain:-

The value of K.

The discharge when the conduit is flowing full and the difference in levels between the two reservoirs is 15 ft.

(B.Sc. Part 1)

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To see the solution please click on the red button

Example 3

Find an expression for the theoretical depth for maximum velocity in a closed circular channel in terms of the diameter d.

Compare the discharge at maximum velocity with that when the channel is running full, assuming that the Chezy constant is unaltered and that the pressure remains atmospheric. (B.Sc. Part 1)

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Example 4

A Special sewer consists of a semicircular top and bottom of radius R joined by parallel sides of length R so that the overall height is 3R.

a) Show that for maximum flow fro a given cross sectional area the angle subtended by the water surface at the centre of curvature of the upper semi circle is approximately $64^0$

b) If the water surface now rises until it reaches the top of the sewer, find the percentage decrease in flow.

The Chezy coefficient = $Km^{12.6}$ (B.Sc. Part2)

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To see the calculations please click on the red button

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