Chezys Formula - Derivation, Constant Value, Application and Practice Problems (2024)

Chezys Formula

A French hydrologist Antonie chezy gave Chezys formula . He related the rate of flow of water(discharge) through an open channel with the respective dimensions of that channel.

the experiments and their measurements are suggested that frictional resistance experienced through a uniform flow is approximately proportional to the square of the volumetric flow rate. It is generalized in the context of open channel flow to find velocity and discharge.

Here is chezy’s formula

Velocity \(\text{v=C}\sqrt{\text{mi}}\)

where, v = mean velocity,

C = chezy’s coefficient

m= mean hydraulic depth

i= hydraulic gradient.

Chezy’s coefficient is directly related to the frictional factor. Thus it is dependent on surface roughness, Raynold number, and hydraulic mean depth. values of C and another coefficient n (used in manning’s formula) given by chow ( tabulated values).

Also read: Advantages and Disadvantages of Lever Method of Irrigation

Chezy formula derivation

Considering the momentum equation for a uniform flow between sections 1 & 2 as shown in the figure, which is L-distance apart from each other.

\[{{P}_{1}}+w\sin \theta -{{F}_{f}}-{{P}_{2}}={{M}_{2}}-{{M}_{1}}….1\]

As flow is uniform,

\[\begin{align}
& {{P}_{1}}={{P}_{2}}\And {{M}_{1}}={{M}_{2}} \\
& W=\gamma AL\sin \theta \,\text{and }{{\text{F}}_{f}}={{\tau }_{0}}PL \\
\end{align}\]

Where,

A= cross-section area of the channel (remain constant)

γ= unit weight of water

L= length between two sections 1 & 2

\({{\tau }_{0}}\) = average shear stress on the wetted perimeter

P= wetted perimeter of the channel section

Now, \({{P}_{1}}={{P}_{2}}\And {{M}_{1}}={{M}_{2}}\,\text{equation}\,(\text{1)}\,\text{will}\,\text{be}\,\text{written}\,\text{as}\)

\[\begin{align}
& \gamma AL\sin \theta \,-{{\tau }_{0}}PL=0 \\
& \gamma AL\sin \theta \,-{{\tau }_{0}}PL \\
\end{align}\]

Here in uniform flow, frictional forces are balanced by gravity forces. For small values of q like for mild slope (θ is nearer to 0°) \(\sin \theta \approx \tan \theta \) and as we know \(\tan \theta =S{}_{0}\) means bed slope.

\[\begin{align}
& \gamma AS{}_{0}\,={{\tau }_{0}}PL \\
& {{\tau }_{0}}=\gamma \frac{A}{P}S{}_{0} \\
\end{align}\]

We know that, hydraulic radius \(R=\frac{A}{P}\),

Therefore,\({{\tau }_{0}}=\gamma RS{}_{0}\)

_{\({{\tau }_{0}}\)} is average shear stress on a wetted area for a uniform flow condition.

And is known from various experiments that shear stress is directly proportional to dynamic pressure \((\frac{\rho {{v}^{2}}}{2})\) and independent of viscosity.

\[\begin{align}
& {{\tau }_{0}}\propto \frac{\rho {{v}^{2}}}{2} \\
& {{\tau }_{0}}=K\frac{\rho {{v}^{2}}}{2} \\
\end{align}\]

Where,

K= constant independent of the roughness of the channel

ρ= density of water

υ= velocity of flow

\[\begin{align}
& K\frac{\rho {{v}^{2}}}{2}=\gamma RS \\
& {{v}^{2}}=\frac{2\gamma }{K\rho }RS \\
& \\
\end{align}\]

\[v=C\sqrt{RS}……(2)\]

Here \(C=\sqrt{\frac{2\gamma }{K\rho }}RS\) is known as chezy’s coefficient and the above equation 2 is called chezy’s equation.

Chezy formula for head loss

Head loss due to friction for a uniform flow derived by Darcy Weisbach equation.

\[hf=\frac{f}{y}\times \frac{P}{A}\times L{{V}^{2}}\]

As we know hydraulic radius is the ratio of the area of flow to the wetted perimeter. We denoted it as ‘m’.

\(\begin{align}
& m=A/P=\frac{\pi }{4{{d}^{4}}/\pi d}=d/4 \\
& P/A=1/m \\
\end{align}\)

\[\begin{align}
& hf=\frac{f}{y}\times \frac{P}{A}\times L{{V}^{2}} \\
& {{V}^{2}}=\frac{{{h}_{f}}}{L}\times \frac{y}{f}\times m \\
& V=\sqrt{\frac{{{h}_{f}}\times y\times m}{f\times L}} \\
\end{align}\]

Considering the h_f/L = head loss per unit length of pipe and (y/f)^1/2= chezy’s constant,

\[\text{v=C}\sqrt{\text{mi}}\]

Then this is known as chezy’s formula

Relation between the friction factor and chezy’s constant

We know \(v=C\sqrt{RS}\)

And the slope of energy line S for an open channel is, \(S=\frac{{{h}_{f}}}{L}\)

So from darcy Weisbach’s equation

\[{{h}_{f}}=f\frac{L{{v}^{2}}}{D2g}\]

Here we consider the head loss in a pipe, in the open channel related to pipe flow at atmospheric pressure.

So for a pipe with a diameter of D

\[\begin{align}
& R=A/P=\frac{\frac{\pi }{4}{{D}^{4}}}{\pi D} \\
& =D/4 \\
\end{align}\]
\[\begin{align}
& {{h}_{f}}=f\frac{L{{v}^{2}}}{4R2g} \\
& v=\sqrt{\frac{8g}{f}}\sqrt{R}\sqrt{\frac{{{h}_{f}}}{L}} \\
\end{align}\]

From chezy’s equation

\[v=C\sqrt{R}\sqrt{\frac{{{h}_{f}}}{L}}\]

Therefore,

\[C=\sqrt{\frac{8g}{f}}\]

Discharge through the open channel by chezy’s formula

As we know discharge we get is a product of the area of that particular section to velocity.

As per chezy’s formula ,

\[\text{v=C}\sqrt{\text{mi}}\]

\[\text{Discharge = }Q=AV~=AC\sqrt{mi}\]

As per here in open channel flow, bed slope is denoted S and it is equivalent to hydraulic gradient and hydraulic mean depth is equivalent to hydraulic radius R,

Therefore,

\[Q=AC\sqrt{RS}\]

Chezy’s formula for uniform flow

As we know that if the flow is uniform steady-state flow then few conditions remain the same throughout its flow, as a cross-sectional area, and according to that its velocity will also remain the same because of uniform flow.

So the continuity equation states that discharge for the different sections will also be the same.

Therefore,

\[\begin{align}
& Q=vA={{v}_{1}}{{A}_{1}}={{v}_{2}}{{A}_{2}} \\
& ={{A}_{1}}C\sqrt{{{R}_{1}}{{S}_{1}}}={{A}_{2}}C\sqrt{{{R}_{2}}{{S}_{2}}} \\
\end{align}\]

Chezy constant value as per bazin formula

chezy’s constant (C) is dimensionless, it can be determined by different formulas. Here we see how it will be determined through the bazin formula,

\[C=\frac{157.6}{1.81+\frac{K}{\sqrt{m}}}\]

Here K= basin constant

m= hydraulic mean depth/ radius

here K depends on the roughness of the channel surface. More the rough surface gives a higher value of K.

Formula to calculate chezys constant with rugosity coefficient

Another formula to find chezy’s constant is known as Kutter’s formula which is related to the rugosity coefficient,

here we see how it defines,

\[C=\frac{23+\frac{0.00155}{i}+\frac{1}{N}}{1+\left( 23+\frac{0.00155}{i} \right)\frac{N}{\sqrt{m}}}\]

Where N= rugosity coefficient also known as Kutter’s constant,

value of this rugosity coefficient depends on the roughness of the channel surface same as the value of K.

i= bed slope

m= hydraulic mean depth / radius

Manning’s formula

Robert manning 1889 proposed the formula for the average velocity of flow just like chezy’s formula, it became a widely used formula for uniform flow in open channels.

Here is the formula, \(v=\frac{1}{n}{{R}^{\frac{2}{3}}}S{{0}^{\frac{1}{2}}}\)

Where;

v= average velocity of flow

n= roughness coefficient which is a function of the nature of boundary surface, known as manning’s coefficient

R= hydraulic radius

So= slope of the energy line

Also, there is a relation between chezy’s constant and manning’s constant,

Here is manning’s a formula to find chezy’s constant, \(C=\frac{1}{n}{{m}^{\frac{1}{6}}}\)

Where, n= manning’s constant

m= hydraulic mean depth

Application of chezy’s formula

Application of chezy’s formula is as follows:

• It is used to determine the velocity of flow in an open channel,
• Wisely it is used for a determined velocity of flow in pipe also,
• Used for flow over weirs(with different shapes),
• To find discharge through a notch.

Difference between chezy and manning formula

Chezy’s formula practice problems

lets slove and try with a simple example

Question: Find the discharge when the area is 20, the chezy’s constant is 16, the hydraulic radius is 9 and the slope is 21.

As per chezy’s formula

\[\begin{align}
& Q=AC\sqrt{RS} \\
& \,\,\,\,=(20)(16)\sqrt{9\times 21} \\
& \,\,\,\,=\,320\sqrt{189} \\
& \,\,\,\,=4399.27 \\
\end{align}\]

So the discharge of this chezy’s constant is 4399.27 m³/s.

Let’s take another example to find chezy’s constant

Question: discharge given is 38m³/s, the area is 24, the perimeter is 2, given slope is 7. Find chezy’s constant.

First, we find the hydraulic radius.

\[\begin{align}
& R=\frac{A}{P} \\
& \,\,\,\,=24/2 \\
& \,\,\,\,=12m \\
& \text{Now, Q}=AC\sqrt{RS} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,C=\,\frac{Q}{A\sqrt{RS}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\frac{38}{24\sqrt{12\times 7}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\frac{38}{219.96} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.173 \\
& \text{Chezys constant is 0}\text{.173}\text{.} \\
\end{align}\]

FAQ