# E1 Questions on Uniform Flow and Critical Flow. - Calculation of normal and critical depth.

1. E1.1

A rectangular channel is $3.0m$ wide, has a 0.01 slope, flow rate of $5.3m^{3}/s$, and $n=0.011$. Find its normal depth $y_{n}$ and critical depth $y_{c}$.
(Answer: $y_{n}=0.41m$, $y_{c}=0.683m$)

For a rectangular channel, we have:

$A=by$

and

$P=b+2y$

so

$R=\frac{A}{P}=\frac{by}{b+2y}$

The Chezy equation, eqn Eq-1, for a rectangular channel (writing $y=y_{n}$ as we are now using a uniform flow equation) becomes:

$Q=Cby_{n}\left(\frac{by_{n}}{b+2y_{n}}\right)^{1/2}S^{1/2}$ (1)

The Manning equation, eqn Eq-2, for a rectangular channel becomes:

$Q=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{2/3}}S^{1/2}$ (2)

This equation must be solved for $y_{n}$ by a suitable numerical solution (e.g. trial and error or secant method). This gives:

$\text{Normal depth}\quad y_{n}=0.412m$

At critical depth $y_{c}$, $Fr=1$, so using equation Eq-13:

$Fr=1=\left(\frac{Q^{2}B}{gA^{3}}\right)^{1/2}$

And also $Fr^{2}=1$ then

$Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}$

For a rectangular channel this simplifies (using $A=By$) to

$1=\frac{Q^{2}}{gB^{2}y_{c}^{3}}$

This can be formulated to be solved explicitly for $y_{c}$:

$y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}=\left(\frac{Q^{2}}{gB^{2}}\right)^{1/3}$

To give

$\text{Critical depth}\quad y_{c}=0.683m$

You could also note that this can be formulated using the unit flow concept $q=\frac{Q}{B}$, so Which again can be simplified to

$y_{c}=\sqrt[3]{\frac{q^{2}}{g}}=\left(\frac{q^{2}}{g}\right)^{1/3}$

Then as $q=5.3/3.0=1.767m^{3}/s/m$,

$\text{Critical depth}\quad y_{c}=\sqrt{\frac{1.767^{2}}{9.81}}=0.683m$
2. E1.2

Water flows in a long rectangular channel at a depth of $1.22m$ and discharge of $Q=5.66m^{3}/s$. Determine the minimum channel width if the channel is to be sub-critical.
(Answer: $B=1.34m$)

The maximum depth will give the minimum width for a given flow. As the channel is to flow sub-critical then the critical depth is the limiting maximum depth. We have then $y_{c}=1.22m$ and $Q=5.66m^{3}/s$ and can use $Fr=1$ at critical depth. Equation Eq-13 gives:

$Fr=1=\left(\frac{Q^{2}B}{gA^{3}}\right)^{1/2}$

And also $Fr^{2}=1$ then

$Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}$

For a rectangular channel, this simplifies (using $A=By$) to Eq-21

$1=\frac{Q^{2}}{gB^{2}y_{c}^{3}}$

We need to solve for $B$ so

$B=\left(\frac{Q^{2}}{gy_{c}^{3}}\right)^{1/2}$

Which gives

$\text{Minimum width }\quad B=1.34m$
3. E1.3

A rectangular channel has a bottom width of $B=8m$ and Manning’s $n=0.025$

1. (a)

Determine the slope to give a normal depth of $y_{n}=2m$ when the discharge is $12m^{3}/s$

2. (b)

Determine the critical slope and the critical depth when the discharge is $12m^{3}/s$

3. (c)

Determine the critical slope to give a the critical depth of 1.5m and compute the corresponding discharge.

(Answer:(a) $So=0.00024$, (b) $So_{c}=0.0087$, $y_{c}=0.61m$, (c) $So_{c}=46.03m^{3}/s$, $Q=0.00818$)

(a) Using Manning’s equation and the formulae for a rectangular channel in terms of $y$, iterate to solve for slope $So$

Manning’s equation, Eq-2

$Q=\frac{1}{n}AR^{2/3}So^{1/2}$

For a rectangular channel, we have:

$A=by$

and

$P=b+2y$

so

$R=\frac{A}{P}$

We have $y=y_{n}=2.0m$, $Q=12.0m^{3}/s$ and Manning’s $n=0.025$.

Iterating gives the solution $So=0.00024$ or 1 in 4167.

(b) Critical slope is the slope when the normal depth, $y_{n}$, is equal to the critical depth $y_{c}$

For a rectangular channel, the critical depth is given by Eq-22

$y_{c}=\left(\frac{Q^{2}}{b^{2}g}\right)^{1/3}$

As $Q=12.0m/s^{3}$ and $b=8m$ then critical depth $y_{c}=0.612m$

Critical Slope, $So_{c}$, can be calculated by rearranging Manning’s equation and substituting in the geometric parameters calculated at critical depth i.e.

$A_{c}=by_{c}$
$P_{c}=b+2y_{c}$

and

$R_{c}=\frac{A_{c}}{P_{c}}$

$So_{c}=\left(\frac{Qn}{A_{c}R_{c}^{2/3}}\right)^{2}$

This results in $A_{c}=4.899m^{2},P_{c}=9.225m$ and $R_{c}=0.531$, and thus $So_{c}=0.00872$ or 1 in 115

Alternatively, the formula for critical slope, Eq-24, could be used directly.

$So_{c}=\frac{gn^{2}P_{c}}{B_{c}R_{c}^{1/3}}$ (3)

(where $B_{c}=b$ for a rectangular channel)

(c) For this part we must calculate a new value of $Q$ that will give critical depth $y_{c}=1.5m$, so using equation Eq-22

$y_{c}=\left(\frac{Q^{2}}{b^{2}g}\right)^{1/3}$

Which give $Q=46.03m^{3}/s$

Using $y_{c}=1.5m$ we get $A_{c}=12.0m^{2},P_{c}=11.0m$ and $R_{c}=1.091$, and, as above, we can cacluate $So_{c}=0.00818$ or 1 in 122

4. E1.4

For a trapezoidal channel with a base width $b=3.0m$, Manning’s $n=0.025$ and side slope $s=2$ (i.e. 1 vertical: 2 horizontal), calculate the critical depth, critical velocity, and critical slope if its discharge $Q=10m^{3}/s$.

(Answer:$y_{c}=0.855m$, $v_{c}=2.483m/$s, $So_{c}=0.00777$)

For any channel, the critical depth can be found by solving equation Eq-14 equated to 1.

$Fr^{2}=1=\left(\frac{Q^{2}B}{A^{3}g}\right)$

For a trapezoidal channel, we have:

$A=y(b+ys)$

and

$B=b+2ys$

and

$P=b+2y\sqrt{1+s^{2}}$

Substitution for $A$ and $B$ and iterating for $y$ gives the solution $y_{c}=0.855m$

Velocity in any channel is

$V=\frac{Q}{A}$

Substituting, $y_{c}$ to give $A_{c}$ and using $Q=10m^{3}/s$, gives $v=2.483m./s$.

The formula for critical slope, Eq-24, could be used directly.

$So_{c}=\frac{gn^{2}P_{c}}{B_{c}R_{c}^{1/3}}$ (4)

From the data, $P_{c}=6.823m$, $A_{c}=4.027m^{2}$, $b_{c}=6.420m$, $R_{c}=0.590m$ and this gives $So_{c}=0.00777$ or 1 in 129

5. E1.5

A rectangular channel $9m$ wide carries $7.6m^{3}/s$ of water when flowing $1.0m$ deep. Work out the flow’s specific energy. Is the flow sub-critical or super-critical?

(Answer: $1.036$m and flow is sub-critical)

Specific Energy is given by Eq-27,

$E_{s}=y+\frac{\alpha V^{2}}{2g}$ (5)

Taking, as usual, $\alpha=1$ then $V=\frac{Q}{by}=0.844m/s$

$E_{s}=1.0+\frac{0.844^{2}}{2\times 9.81}=1.036m$ (6)

Calculate the Froude number to determine is sub or super-critical flow. For a rectangular channel Eq-16 gives

$Fr=\frac{V}{\sqrt{gy}}=\frac{0.844}{\sqrt{1\times 9.81}}=0.27$

$Fr<1.0$, so flow is sub-critical.

Alternatively, we could calculate the critical depth and compare this to the current depth. For a rectangular channel critical depth $y_{c}$ is given by Eq-22

$y_{c}=\left(\frac{Q^{2}}{gb^{2}}\right)^{1/3}=0.187m$

As $1.0>0.187$, i.e. the depth of flow is greater than the critical depth, the flow is sub-critical.

6. E1.6

Two engineers observed two rivers and recorded the following flow parameters: River 1: flow discharge $Q=130m^{3}/s$, flow velocity $V=1.6m/s$, water surface width $B=80m$; River 2: flow discharge $Q=1530m^{3}/s$, flow velocity $V=5.6m/s$, water surface width $B=90m$. Decide the flow regime of two rivers, i.e. sub-critical or super-critical.

(Answer: River 1 is sub-critical and River 2 is super-critical)

For any channel, the Froude number (squared) can be found from either Eq-20

$Fr^{2}=\left(\frac{Q^{2}B}{A^{3}g}\right)$

or in terms of velocity

$Fr^{2}=\left(\frac{V^{2}}{A^{2}g\frac{A}{B}}\right)$

We can calculate $A$ from $Q=AV$.

For River 1: $Q=130.0m^{3}.s$, $V=1.6m/s$, so $A=130.0/1.6=81.25m^{2}$ and as $B=80.0m$ then

$Fr^{2}=\left(\frac{130.0^{2}\times 80.0}{81.25^{3}\times 9.81}\right)=0.507$

For River 2: $Q=1530.0m^{3}.s$, $V=5.6m/s$, so $A=1530.0/5.6=273.21m^{2}$ and as $B=90.0m$ then

$Fr^{2}=\left(\frac{1530.0^{2}\times 90.0}{273.21^{3}\times 9.81}\right)=1.027$

So River 1 is flowing sub-critical, River 2 is flowing super-critical.

7. E1.7

A concrete, trapezoidal channel has a bottom slope of $S_{o}=0.0009$ and a Manning roughness factor of $n=0.013$. The bottom width of the channel is $b=2.5m$, and the side slopes are 1 in 2. Determine the velocity and discharge when the flow is normal at a depth of $1.8m$.
(Answer: $v=2.37m/s$, $Q=26.01m^{3}/s$)

The solution is a straightforward use of Manning’s equation Eq-2 applied for a trapezoidal channel.

$A=y(b+ys)=1.8(2.5+1.8\times 2)=10.98m^{2}$
$P=b+2y\sqrt{1+s^{2}}=2.5+2\times 1.8\sqrt{1+2^{2}}=10.55m$
$R=\frac{A}{P}=1.04m$

So

$Q=\frac{1}{n}AR^{2/3}So^{1/2}=\frac{1}{0.013}10.98\times 1.04^{2/3}0.0009^{1/2% }=26.01m^{3}/s$

And the velocity of flow in the channel is

$v=\frac{Q}{A}=\frac{26.01}{10.98}=2.37m/s$
8. E1.8

A trapezoidal channel has a bottom slope of $S_{o}=1$ in $40$ and a Manning roughness factor of $n=0.016$. The bottom width of the channel is $b=6.0m$, and the side slopes are 1 in 3. Determine the normal depth in this channel for $Q=42.3m^{3}/s$.
(Answer: $y_{n}=0.75m$). For a rectangular channel, we have:

$A=by$

and

$P=b+2y$

so

$R=\frac{A}{P}$

Manning’s equation Eq-2 is

$Q=\frac{1}{n}AR^{2/3}So^{1/2}$

We have $Q=42.3m^{2}/s$, Manning’s $n=0.016$, $b=6.0m$ and $s=3$.

Iterating gives the solution $y_{n}=0.75m$.

9. E1.9

The discharge in uniform flow in a rectangular channel $4.6m$ wide is $11.3m^{3}/s$ when the slope is 1:100. Is the flow sub-critical or super-critical? Calculate the slope, $So_{c}$, that would give critical depth. The Manning roughness coefficient is $n=0.012$.
(Answer: super-critical, $Fr=2.1$, $S_{c}=0.002268$).

Iterate the Manning’s equation to solve for $y_{n}$ to give $y_{n}=0.521m$.

Then $A=by_{n}=2.34m^{2}$ and $V=Q/A=4.71m/s$

For any channel, the Froude number can be calculated from Eq-11

$Fr=\frac{V}{\sqrt{g\frac{A}{B}}}=2.1$

As $Fr>1$ the flow is super-critical.

Critical depth for a rectangular channel can be calculated from $Fr=1$ or $Fr^{2}=2$, leading to Eq-22

$y_{c}=\left(\frac{Q^{2}}{gb^{2}}\right)^{1/3}=0.85m$

From the Manning’s equation $So$ is found

$So_{c}=\frac{Q^{2}n^{2}}{A_{c}^{2}R_{c}^{4/3}}$

$A_{c}=by_{c}=4.6\times 0.85=3.91m^{2}$, $P_{c}=b+2y_{c}=6.30m$, $R_{c}=A_{c}/P_{c}=0.62m$ and thus $So_{c}=0.002268$ or 1 in 440

Compound Channels

10. E1.10

The cross-section of a stream can be approximated by the compound channel shown in figure 2. The bottom slope is $S_{o}=0.0009$. The Manning roughness factor $n=0.025$ for the main channel and $n=0.035$ for the overbank areas. Determine the normal depth for a discharge of $197m^{3}/s$. Also, calculate the energy coefficient $\alpha$ and the momentum coefficient $\beta$ for the channel with this flow condition.

(Answer: $y_{n}=5.507m$, $\alpha=1.23$, $\beta=1.09$.)

We will apply Manning’s equation for each of the three regions and the total flow will be the sum of the three calculated flows.

We will need to perform an iterative solution to solve Manning’s equation for $y_{n}$. If we assume that the normal depth is such that flow is in the flood plains, and our depth is $y$, then we have, from the geometry in figure 3, depth on the flood plain $y_{flood}$ as

$y_{flood}=y-4.5$

and

$A=A_{1}+A_{2}+A_{3}$

where

$A_{1}=A_{3}=7.0y_{flood}$
$A_{2}=(2.0+10.0+2.0)\times y_{flood}+4.5\times(10.0+2.0)=14y_{flood}+54$

And also

$P=P_{1}+P_{2}+P_{3}$

where

$P_{1}=P_{3}=y_{flood}+7$
$P_{2}=10+2\times\sqrt{2^{2}+4.5^{2}}=19.85m$

$n_{1}=n_{3}=0.035$, $n_{2}=0.025$, $Q=197m^{3}/s$, $S_{o}=0.009$ and can calculate the side slope from the section geometry as $s_{2}=\frac{2}{4.5}=0.444$

An iteration table can be set up in, say, Excel and each of the above geometric and physical parameters calculated for a given $y$. Then $Q$ for each of the three regions, and then the total can be calculated and compared against the target of $Q=197m^{3}/s$. The first iteration is shown in figure 4, for an initial $y=5.5m$, equivalent to $y_{flood}=1m$

The aim of the iteration is to reduce the error to be small $\approx 0.001m^{3}/s$. This can be done by various methods including trial and error. Here we will use the secant method to estimate the solution. The secant method uses the following formula, where $f(y_{i})\equiv\text{Error}$ in the table of figure 4.

$y_{i}=y_{i-1}+f(y_{i-1})\left(\frac{y_{i-1}-y_{i-2}}{f(y_{i-1}-f(y_{i-2})}\right)$ (7)

For the second $y$ we have taken $y=4.5m$, so that $y_{flood}=0.0$. And applying the secant method we get the result shown in he table of figure 5 with $y_{n}=5.507m$.

The equations for the Energy coefficient $\alpha$ (Eq-28) and Momentum coefficient $\beta$ (Eq-36) are:

$\alpha=\frac{\int u^{3}\ dA}{\overline{V}^{3}A}=\frac{V_{1}^{3}A_{1}+V_{2}^{3}% A_{2}+V_{3}^{3}A_{3}}{\overline{V}^{3}(A_{1}+A_{2}+A_{3})}$ (8)

and

$\beta=\frac{\int u^{2}\ dA}{\overline{V}^{2}A}=\frac{V_{1}^{2}A_{1}+V_{2}^{2}A% _{2}+V_{2}^{3}A_{3}}{\overline{V}^{2}(A_{1}+A_{2}+A_{3})}$ (9)

where

$\overline{V}=\frac{Q}{A}=\frac{V_{1}A_{1}+V_{2}A_{2}+V_{3}A_{3}}{A_{1}+A_{2}+A% _{3}}$ (10)

From the iteration table of figure 5 then we have $V_{1}=Q_{1}/A_{1}=0.787m/s$, $V_{2}=Q_{2}/A_{2}=2.730m/s$, $V_{3}=V_{1}$ and so, from equation 10, $\overline{V}=2.397m/s$.

From equations 8 and 9:

$\alpha=1.230$

and

$\beta=1.093$
11. E1.11

The total width of the channel considered in Question E1.10 is to be decreased by reducing the overbank portions symmetrically; however, this reduction must not cause an increase of more than $0.15m$ in the flow depth for the discharge of $197m^{3}/s$. Assuming normal depth still is present in the channel, determine the minimum allowable channel total width, $B$.
(Answer: $B=16.157m$.)

The solution to this question requires finding a width $B$ such that $y_{n}$ is the normal depth from E1.10, $5.507m$, plus $0.15m$ i.e. $y_{n}=5.507+0.15=5.657m$. We can perform an iteration similar to that in E1.10 (calculating $A$, $P$, and $Q$) for an estimated $B$, and the target of the iteration will be the same $Q=197m^{3}/s$, so the error function will be $Q-Q_{i}$.

For a given $B$ we must calculate the floodplain width

$B_{flood}=\frac{B}{2}-\frac{10}{2}-2=\frac{B}{2}-7$

So the formulae that change from question E1.10 are those for $A_{1}=A_{3}=B_{flood}\times y_{flood}$. As we know that $y_{flood}=5.657-4.5=1.157m$ then

$A_{1}=A_{3}=1.157B_{flood}$

and

$P_{1}=P_{3}=1.157+B_{flood}$

The table of figure 6 shows this implemented in a similar way to that of question E1.10 with $B_{i}$ estimated via a secant method and taking $B=28m$ (full flood plain), and $B=14m$ (no flood plain) as initial values. This shows the solution of $B=16.157m$, or the width of each flood plain being $1.079m$.

12. E1.12

The cross-section of a river with flood plains flowing in uniform flow may be idealized as shown in Fig. 7. Determine the discharge carried by the river when its dimensions and roughness parameters are:

Bed slope: $S_{o}=2\times 10^{-4}$
Manning’s ns: $n_{1}=n_{2}=n_{3}=0.02$
Side slopes: $s_{1}=s_{2}=s_{3}=1$
Bed widths: $B_{1}=3m$, $B_{2}=5m$, $B_{3}=4m$
Main channel depth: $y_{main}=3.0m$

and

Normal depth $y_{n}=4.5m$
(Answer: $Q=69.42m^{3}/s$.)

The equations for each geometry element can be written as functions of the depth $y$ where also $y_{1}=y_{3}=y-y_{main}$

 $\displaystyle BB_{1}$ $\displaystyle=s_{1}y_{1}$ $\displaystyle BB_{2}$ $\displaystyle=s_{2}y_{main}$ $\displaystyle BB_{3}$ $\displaystyle=s_{3}y_{3}$
 $\displaystyle A_{1}$ $\displaystyle=(B_{1}+BB_{1}/2)y_{1}$ $\displaystyle A_{2}$ $\displaystyle=(B_{3}+BB_{2})y_{main}+(B_{2}+2BB_{2})y_{1}$ $\displaystyle A_{3}$ $\displaystyle=(B_{3}+BB_{3}/2)*y_{3}$
 $\displaystyle P_{1}$ $\displaystyle=B_{1}+\sqrt{BB_{1}^{2}+y_{1}^{2}}$ $\displaystyle P_{2}$ $\displaystyle=B_{2}+2\sqrt{BB_{2}^{2}+y_{main}^{2}}$ $\displaystyle P_{3}$ $\displaystyle=B_{3}+\sqrt{BB_{3}^{2}+y_{3}^{2}}$

$Q$ is calculated from Manning’s equation for each of the $j=1...3$ regions

$Q_{j}=\frac{1}{n_{j}}A\left(\frac{A_{j}}{P_{j}}\right)^{2/3}S_{o}^{1/2}$

And so $Q=Q_{1}+Q_{2}+Q_{3}$.

Implementing these formulae in a tabular form results in the table of figure 8, which shows $Q=69.421m^{3}/s$.

As an extra exercise, the table could be set up to iterate to find a normal depth $y_{n}$ given a flow $Q$. As a test, if the target $Q$ is set to $Q=69.421m^{3}/s$, the solution from an iteration should give $y_{n}=4.5m$. Such an iteration has been performed using the secant method and is shown in figure 9. With initial estimates of $y_{i}=3.0m$ and $4.0m$ this iterates to $y_{n}=4.5m$ in 4 further iterations.

Using this iteration table, we could answer the question, what is the normal depth for different flows? E.g. the table of figure 10, shows that the solution for a flow of $Q=100m^{3}/s$ is a normal depth $y_{n}=5.229m$

13. E1.13

For the channel of question E1.12 calculate the flow, if all dimensions, including the normal depth were the same, but the slope of the channel is 0.002.
(Answer $Q=219.53m^{3}/s$).

This follows the calculation of geometry of question E1.12, with the calculations for $Q_{j}$ changed to take account of $S_{o}=0.002$. If the table has been set up appropriately in the spreadsheet then a simple change of the $S_{o}$ value will give a solution of $Q=219.53m^{3}/s$, and the table shown in figure 11

Efficient Channels

14. E1.14

A trapezoidal channel has side slopes of 1:3/4 and the slope of the bed is 1 in 2000. Determine the optimum dimensions of the channel if it is to carry water at $0.5m^{3}/s$. Use the Chezy formula, assuming that $C=80m^{1/2}/s$.
(Answer: $y_{n}=0.552m$, $b=0.552m$).

Let us review the concept and equations for efficient channels. We would want to minimise the excavation and materials used to construct the channel. This suggests that the designed channel should have a maximum hydraulic radius for a given area or minimum perimeter for the given area since hydraulic radius $R=A/P$. A semi-circular channel would be the most optimal. However, we must consider channel shapes that are to be built.

In this example, we will consider a trapezoidal channel section, as shown in figure 12,

the area of flow $A$ and the wetted perimeter $P$ are given by these familiar equations

$A=by+sy^{2}$ (11)
$P=b+2y\sqrt{1+s^{2}}$ (12)

Rearranging the equation for $A$ to give $b$, then substituting this into equation 12 for $P$ gives

$P=\frac{A}{y}-sy+2y\sqrt{1+s^{2}}$ (13)

To minimize $P$, evaluate $dP/dy$ with $A$ and $s$ constant and set $dP/dy$ equal to zero. Therefore,

$\frac{dP}{dy}=-\frac{A}{y^{2}}-s+2\sqrt{1+s^{2}}=0$

and rearranging we can obtain an expression for the area to give this optimum $P$:

$A=y^{2}\left[2\sqrt{1+s^{2}}-s\right]$ (14)

Substituting the expression for $A$ into Eq. 11, one gets

$by+sy^{2}=y^{2}\left[2\sqrt{1+s^{2}}-s\right]$

$\therefore$

$b=2y\left[\sqrt{1+s^{2}}-s\right]$ (15)

Likewise, substituting the value of $A$ from Eq. 14 in Eq. 13, one gets

$P=y\left[2\sqrt{1+s^{2}}-s\right]-sy+2y\sqrt{1+y^{2}}$

or

$P=4y\sqrt{1+s^{2}}-2sy=2y\left[2\sqrt{1+s^{2}}-s\right]$ (16)

Thus, for a minumum wetted perimeter $P$,

$R=\frac{A}{P}=\frac{y}{2}$ (17)

Equation 17 indicates that for a trapezoidal section channel and for any side slope $s$ (or $\theta$), the most efficient uniform flow channel would be the one for which the hydraulic radius is half the depth of flow.

So the Manning’s equation Eq-2 for uniform flow in the most efficient trapezoidal section channel would be

 $\displaystyle Q$ $\displaystyle=\frac{1}{n}AR^{2/3}S_{o}^{1/2}$ $\displaystyle Q$ $\displaystyle=\frac{1}{n}(by+sy^{2})\left(\frac{y}{2}\right)^{2/3}S_{o}^{1/2}$

And for the Chezy equation Eq-1

 $\displaystyle Q$ $\displaystyle=ACR^{1/2}S_{o}^{1/2}$ $\displaystyle Q$ $\displaystyle=C(by+sy^{2})\left(\frac{y}{2}\right)^{1/2}S_{o}^{1/2}$

So we have an equation (either Manning’s or Chezy) that has two unknowns, $B$ and $y$. But we know a relationship between $b$ and $y$ from equation 15. And using $s=3/4=0.75$

 $\displaystyle b$ $\displaystyle=2y\left[\sqrt{1+0.75^{2}}-0.75\right]$ $\displaystyle b$ $\displaystyle=y$

So the Chezy equation for this channel section becomes

$Q=1.75y^{2}C\left(\frac{y}{2}\right)^{1/2}S_{o}^{1/2}$

This can be solved iteratively with a target of $Q=0.5m^{3}/s$ and $S_{o}=1/2000=0.0005$. The table of figure 13 shows this iteration using the secant method and that the solution is $y_{n}=0.556m$ and $b=y=0.556$

15. E1.15

An open channel with $n=0.011$ is to be designed to carry $1.0m^{3}/s$ of water at a slope of 0.0065. Find the most efficient cross-section for a rectangular section.
(Answer: $b=2y=0.869m$).

See question solution E1.14 for the following equations for efficient trapezoidal channels. The equation of the only variable we can change in this question is that for the bed width, $b$, equation 15

$b=2y\left[\sqrt{1+s^{2}}-s\right]$

For a square channel the $s=0$. Using this with the familiar equations for the geometry of a trapezoidal channel

$A=by+sy^{2}$

and

$P=b+2y\sqrt{1+s^{2}}$

and with the usual form of the Manning’s equation

$Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}$

then an iteration can be performed to solve for $y_{n}$ and thus $b$.

The table of figure 14 shows this iteration using the secant method and that the solution is $y_{n}=0.434m$ and $b=2y=0.869m$

16. E1.16

A rectangular channel has width $B=3m$ and normal depth $y=1m$. What is the diameter of a semicircular channel that will have the same discharge as in the rectangular channel, when flowing just full in uniform flow? Assume that $n$ and $S_{o}$ are the same in the two cases. Compare the two wetted perimeters.
(Answer: $D=2.057m$, $P_{rectangular}=5.0m$, $P_{circular}=6.463m$

For this, we need to first calculate the $Q$ for the rectangular channel for the given parameters using Manning’s equation Eq-2

$Q=\frac{1}{n}by\left(\frac{by}{b+2y}\right)^{2/3}S_{o}^{1/2}$

Then second, apply the manning equation for a circular channel and equate the $Q$. When the channel (pipe) is just full, ie. depth is equal to the diameter or $y_{n}=D$ we should realise that the area, $A$, will be the area of the circle and the wetted perimeter, $P$, the circumference of the same circle. i.e.

$A=\frac{\pi D^{2}}{4}$
$P=\pi D$

So

$R=\frac{A}{P}=\frac{\pi D^{2}}{4\pi D}=\frac{D}{4}$

The Manning’s equation can then be written for a ’just-full’ pipe as

 $\displaystyle Q$ $\displaystyle=\frac{1}{n}\frac{\pi D^{2}}{4}\left(\frac{D}{4}\right)^{2/3}S_{o% }^{1/2}$ $\displaystyle Q$ $\displaystyle=\frac{1}{n}\frac{\pi D^{8/3}}{4^{5/3}}S_{o}^{1/2}$

Equating this with the Manning’s equation for a rectangular channel gives:

$\frac{1}{n}by\left(\frac{by}{b+2y}\right)^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{% \pi D^{8/3}}{4^{5/3}}S_{o}^{1/2}$

As $n$ and $S_{o}$ are the same for both channels these terms can be cancelled to give

$by\left(\frac{by}{b+2y}\right)^{2/3}=\frac{\pi D^{8/3}}{4^{5/3}}$

And rearranged to give $D$:

$D=\left[\frac{4^{5/3}}{\pi}by\left(\frac{by}{b+2y}\right)^{2/3}\right]^{3/8}$

Which results in $D=2.057m$.

Comparing the wetted perimeters we can simply calculate these as
$P_{rectangular}=5.0m$, $P_{circular}=6.463m$.

17. E1.17

What are the dimensions of the most efficient rectangular channel section to carry $5m^{3}/s$ at a slope of 1 in 900. The surface of the channel is of concrete.
(Answer: $y=1.21m$, $b=2y=2.42m$ using $n=0.012$)

From the earlier section on Efficient uniform flow channels we saw that for a trapezoidal channel, the most efficient section is when

$P_{min}=2y\left[2\sqrt{1+s^{2}}-s\right]$

and

$A=y^{2}\left[2\sqrt{1+s^{2}}-s\right]$

For a rectangular channel then $s=0$ and so

$P_{min}=4y$
$A=2y^{2}$
$R=\frac{2y\;y}{4y}=\frac{y}{2}$

So Manning’s equation Eq-2 becomes, for an efficient rectangular channel

$Q=\frac{1}{n}(2y^{2})\left(\frac{y}{2}\right)^{2/3}S_{o}^{1/2}$

This requires an iterative solution. A table showing this solved using the secant method is shown in figure 15. The result is that $y=1.143m$ and $b=2y=2.286m$

18. E1.18

What is the most efficient depth for a brick channel of a trapezoidal section with sides sloping at $45^{\circ}$ to the horizontal to carry $3m^{3}/s$. The bed slope is 0.0009.
(Answer: $y=1.104m$, using $n=0.015$)

This requires an iterative solution. A table showing this solved using the secant method is shown in figure 16. The result is that $y=1.104m$ using $n=0.015$ for the brick lining.