Uniform Flow
The volumetric flow rate in fluid dynamics and hydrometry is the volume of fluid which passes through a given surface per unit time in SI units, or cubic feet per second. It is usually represented by the symbol $Q$.
Key facts
- Chezy Equation:
$v=C \sqrt{m\;i}$
Key facts:
- Manning's formula
$v=\frac{k}{n}\times \;m^{\frac{2}{3}}\;i^{\frac{1}{2}}$
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:
or, $Q=A\;C\;\sqrt{m\;i}$
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
Darcy states that:
where $\displaystyle m=\frac{d}{4}$
This can be written as:
$v=\sqrt{\frac{2g}{f}\times m \times \frac{h}{l}}\; = \;\sqrt{\frac{2g}{f}}\times \sqrt{mi}$
Thus the Darcy and Chezy equations are basically the same, and $\displaystyle C=\sqrt{\frac{2g}{f}}$
The Chezy Constant C
The Chezy constant $C$ 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 is the Manning equation.
The Manning Equation
Manning substituted $M\;m^{\frac{1}{6}}$ for the $C$ of the Chezy equation. Hence: $v=M\;m^{\frac{2}{3}}\;i^{\frac{1}{2}}$
This can be re-written by replacing the $M$ as follows: $v=\frac{k}{n}\times \;m^{\frac{2}{3}}\;i^{\frac{1}{2}}$
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:
$A=b\times d$ and $P=b+2d$. $\therefore \;\;\;\;\;P=\frac{A}{d}+2d$ Differentiating and equating to zero for the minimum wetted area.
$\displaystyle\frac{\mathrm{d} p}{\mathrm{d} d}=-\frac{A}{d^2}+2=0$
$\displaystyle\frac{b\times d}{d^2}=2$
i.e. $b=2d$
Thus for a maximum velocity of flow,$v$ or a maximum rate of flow,$Q$
$b=2d$
A Trapezoidal Channel
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.
$\sin\theta=\frac{d}{\sqrt{d^2n^2+d^2}}=\frac{r}{\displaystyle\frac{b+2nd}{2}}$
But since $d=r$ $\therefore \;\;\;\;\;d\;\sqrt{n^2+1}=\frac{b+2nd}{2}$
$m=$ Hydraulic Mean Depth , $m=\frac{\text{Area of flow}}{\text{Wetted perimeter}}$
$m=\frac{\displaystyle\frac{1}{2}(b+b+2nd)d}{b+2d\sqrt{n^2+1}}=\frac{d(b+nd)}{2(b+nd)}=\frac{d}{2}$
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.
i.e., when: $\frac{d\left ( \displaystyle \frac{A}{P} \right )}{d\theta}=\frac{P \displaystyle \frac{dA}{d\theta}-A \displaystyle\frac{dP}{d\theta}}{P^2}=0$
This occurs when $2\theta=\tan2\theta$ i.e., $2\theta =257\;\displaystyle\frac{1}{2}^0$ And the maximum depth of water is: $0.81\times d$ where $d$ is the diameter
Using the Chezy equation the maximum flow, $Q$ will occur when $\displaystyle \frac{A^3}{P}$ is a maximum.
i.e., when: $\frac{d\left (\displaystyle \frac{A^3}{P} \right )}{d\theta}=\frac{P\times 3A^2\displaystyle\frac{dA}{d\theta}-A^3\displaystyle\frac{dP}{d\theta}}{P^2}=0$
This will occur when,
$2\theta=3\theta\;\cos2\theta-\frac{1}{2}\sin2\theta$ i.e., $2\theta=308^0$ And the maximum depth at the centre is: $0.95\times dc$ where $dc$ is the diameter of the channel
Or using the Manning Equation $Q$ will be a maximum when $\displaystyle \frac{A^5}{P^2}$ is a maximum.
i.e., when: $\frac{d\left ( \displaystyle\frac{A^5}{P^2} \right )}{d\theta}=\frac{P^2\times A^4\displaystyle\frac{dA}{d\theta}-A^5\times 2P\displaystyle\frac{dP}{d\theta}}{P^4}=0$ From which $3\theta=5\theta\cos2\theta-\sin2\theta$ $\therefore \;\;\;\;\;2\theta =302\;\displaystyle\frac{1}{2}^0$ The maximum depth at the centre is: $0.938\times dc$ where $dc$ is the diameter of the channel
Example 1 [imperial]
Water flows down a uniform channel with a slope of $\displaystyle\frac{1}{1600}$. 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.
Using the Chezy Equation:
$Q=A\times 120\;\sqrt{M\times i} = 120\left ( \frac{A^3}{P} \right )^{\frac{1}{2}}\times \sqrt{i}$
From the above equation it can be seen that for a given cross sectional area the flow will be maximum when the wetted area, $P$ is a minimum.
From Equation (3)
$B=\frac{A}{2R}- R\left ( \frac{\pi}{4}+1 \right )$ Substituting in equation(4)
$P=\frac{A}{2R}+R\left ( \frac{\pi}{4} +1\right )+R\left ( \pi+2 \right )$ Thus: $P=\frac{A}{2R}+R\left ( \frac{3\pi}{4}+1 \right )$
Differentiating and equating to zero for minimum value of $P$:
$\frac{dP}{dR}=-\frac{A}{2R^2}+\left ( \frac{3\pi}{4}+1 \right )=0$
Substitute from $A$$\frac{\frac{\pi}{2}R^2+2R^2+2RB}{2R^2}=\frac{3\;\pi}{4}+1$ $\therefore \;\;\;\;\left ( \frac{\pi}{4}+1 \right )+\frac{B}{R}=\frac{3 \pi}{4}+1$ Thus: $\frac{B}{R}=\frac{\pi}{2}$ $m=\frac{A}{P}=\frac{R^2\left ( \displaystyle\frac{\pi}{2}+2 \right )+2R+\displaystyle\frac{\pi R}{2}}{\displaystyle\frac{\pi R}{2}+R(\pi+2)}$ $m=$The Hydraulic Mean Depth $\left ( \frac{\displaystyle\frac{3 \pi}{2}+2}{\displaystyle\frac{3 \pi}{2}+2} \right )\times R=R$ But: $Q=A\times 120\sqrt{m\times i}$ $\therefore \;\;\;\;\;600=\left ( \frac{3 \pi}{2} \right )R^2\times 120\times R^{\frac{1}{2}}\times \frac{1}{\sqrt1600}$
From which: $R=3.89ft.$ And $B=6.11ft.$
Example 2 [imperial]
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 $n=\displaystyle\frac{1}{6}$ .
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.
Let $A$ be the cross sectional area of the water in the conduit and $S$ the length of the wetted area.
From the diagram it can be seen that :
$\sin \theta=\frac{0.5}{1.5}$ And thus: $\theta=19.5^0$ And: $2 \theta=39.0^0$ $FG=1.5\times \cos \theta=1.414$
$\therefore \;\;\;\;\;A=\frac{180+39}{360}\times \pi\times 1.5^2+5\times 1.414=5.027\,ft.^2$ $S=\frac{219}{360}\times \pi\times 3=5.77\,ft.$ Using the Chezy Equation(1) $v=C\sqrt{mi}=Km^{\frac{1}{6}}\;\sqrt{mi}$ $Q=Av=AKm^{\frac{2}{3}}i^{\frac{1}{2}}$
Substituting values into the above equation: $11.5=\frac{5.027\times K\times 0.87^{\frac{2}{3}}}{1500^{\frac{1}{2}}}$ $\therefore \;\;\;\;\;K=97.05$
When the conduit is flowing full:
$m$ now equals $\frac{\pi\times d^2}{4}\times \frac{1}{\pi d}=\frac{d}{4}$ Substituting in the value for $d$: $m=\frac{3}{4}$ $\therefore \;\;\;\;\;C=97.05\times \left ( \frac{3}{4} \right )^{\frac{1}{6}}=92.55$
From Equation (2)
$h_{f}=\frac{flv^2}{2gm}$ And: $v^2=C^2mi$ Where:$i=\frac{h_f}{L}$
re- arranging and substituting in values:
$f=\frac{2g}{C^2}=\frac{2g}{92.55^2}=0.0076$
Using Bernoulli's Equation and neglecting exit losses:
$15=h_f+\frac{v^2}{2g}$ i.e. $15=\frac{4\times 0.0076\times 12,000}{3}\times \frac{v^2}{2g}$ Thus: $v=2.82\;ft./sec.$ And therefore: $Q=Av=\frac{1}{4}\pi\times9\times 2.82=19.9\;ft^2./sec$
$\therefore \;\;\;\;\;K=97.05$ $Q=19.9\;ft^2./sec$
Example 3 [imperial]
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.
From the above diagram it can be seen that:
$A=\frac{\pi+2\theta}{2\pi}\times \pir^2+r^2(\sin\theta\times\cos\theta)$
From the Chezy Equation:
$v=C\sqrt{mi}$
It can be seen that $v$ will be a maximum when $m$ is a maximum i.e. when $A/S$ is a maximum:
When: $S\frac{dA}{d\theta}=A\frac{dx}{d\theta}$
By differentiating equations (5) and (6) and substituting in the above equation.
$r(\pi+2\theta)\times \frac{r^2}{2}\times(2+\cos2\theta)=\frac{r^2}{2}\times \frac{(\pi+2\theta)+\sin2\theta}{2r}$ From which: $\tan2\theta =\pi+2\theta$ $\therefore \;\;\;\;\;\theta=38.7^0$ The depth is: $r+r\sin38.7^0=1.625d=0.812r$ When: $\theta=38.7\,^0$ From equations (5) and (6) $A=2.725r^2$ And: $S=4.474r$ Since: $Q=Av=C\sqrt{i}\;A\;\sqrt{\frac{A}{S}}$
When $v$ is a maximum:
$Q=C\sqrt{i}\times 2.725r^2\sqrt{\frac{2.725r^2}{4.474r}}$ $Q=C\sqrt{i}\times 2.13r^{2.5}$
When the channel is running full: $Q=C\sqrt{i}\times \pi r^2\;\sqrt{\frac{r}{2}}$
$=C\sqrt{i}\times 2.22\;r^{2.5}$
Thus the ratio of the discharges is: $\frac{2.22}{2.13}=1.04$
The ratio of the discharges is $1.04$
The depth is $0.812\;r$