E4 Questions on Gradually varied flow: Flow Transitions.

E4.1
A wide channel consists of two long reaches. The upstream reach has a bed slope of 1 in 100 and the downstream reach has a bed slope of 1 in 1500. The Chezy C for both lengths can be assumed constant and equal to 45. Determine the profile that occurs when the flow per unit width is $7.5m^{3}/s/m$ and determine whether a hydraulic jump occurs and if one occurs which reach of the channel it occurs.
(Answers: $y_{n1}=1.406m$, $y_{n2}=3.467m$, $y_{c}=1.79m$. Jump in reach 1.)For all flow transition problems you must first calculate the normal depth $y_{n}$ for reach and the critical depth $y_{c}$, for that flow (it will be the same for each reach).
In this example, we have a wide channel which implies that $b\gg y$ and can be assumed rectangular. This results in $A=by$, $P\approx b$ and so $R\approx y$.
The Manning’s equation for a wide channel is equation Eq3
$q=\frac{1}{n}y_{n}^{5/3}S_{o}^{1/2}$And, applicable to this example, the Chezy equation for a wide channel becomes
$q=Cy_{n}^{3/2}S_{o}^{1/2}$Then $y_{n}$ can be calculated by this explicit expression
$y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}$For reach 1, $S_{o}=1/100=0.01$ and
$y_{n_{1}}=\left(\frac{7.5}{45\times 0.01^{1/2}}\right)^{2/3}=1.406m$For reach 2, $S_{o}=1/1500=0.0006671$ and
$y_{n_{2}}=\left(\frac{7.5}{45\times 0.0006671^{1/2}}\right)^{2/3}=3.467m$For a wide channel, $Fr^{2}$ can be written from equation Eq17 as
$Fr^{2}=\frac{Q^{2}}{b^{2}gy^{3}}=\frac{q}{gy^{3}}$As $y_{c}$ is the depth when $Fr=Fr^{2}=1$ then $y_{c}$ can be calculated using equation Eq23:
$y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}=1.790m$It is useful to calculate the Froude number of the normal depths as they will be used later. The equation for Fr in a wide channel the Froude number can be calculated from Eq11:
$Fr=\frac{V}{\sqrt{g\frac{A}{B}}}=\frac{Q}{by\sqrt{gy}}=\frac{q}{y\sqrt{gy}}$Which gives $Fr_{reach\;1}=1.44$ and $Fr_{reach\;2}=0.37$.
We can compare the normal and critical depths for each reach. We see that for reach 1, $y_{n_{1}}<y_{c}$ so the reach is steep and the flow supercritical. Also confirmed as $Fr_{reach\;1}>1$.
For reach 2, $y_{n_{2}}>y_{c}$ so the reach is mild and the flow subcritical. Also confirmed as $Fr_{reach\;2}<1$.
It is always useful to sketch out the normal and critical depths so that you can start to understand what is physically being calculated and what may be physically possible. These are shown in figure 45.
We see that the depths in the two reaches change from below critical depth to above critical depth. The only way that a water surface can join these (and cross critical) is by a hydraulic jump. The hydraulic jump could be in either the upstream reach or the downstream reach, see figure 46. To determine which, we must calculate the conjugate (or sequent) depths for each of the two normal depths.
We calculate the conjugate depth to $y_{n_{1}}$ in reach 1 using the following, where $y_{2}$ is the conjugate depth in reach 1 using equation Eq39:
$y_{n1\;conjugate}=y_{2}=\frac{y_{1}}{2}\left(\sqrt{1+8F_{r1}^{2}}1\right)$From earlier $F_{r1}=1.44$ and $y_{1}=y_{n1}=1.406$, this results in the conjugate depth $y_{n1\;conjugate}=y_{2}=2.239m$.
And for the conjugate depth to $y_{n_{2}}$ in reach 2 using the following, where $y_{1}$ is the conjugate depth in reach 2 using equation Eq40:
$y_{n2\;conjugate}=y_{1}=\frac{y_{2}}{2}\left(\sqrt{1+8F_{r2}^{2}}1\right)$From earlier $F_{r2}=0.26$ and $y_{2}=y_{n2}=3.467$, this results in the conjugate depth $y_{n2\;conjugate}=y_{1}=0.779m$.
To understand what these two conjugate depths tell us about the position of the hydraulic jump it is again useful to sketch the depths together with the normal and critical depths. This is shown in figure 47
We can first consider if a jump is possible in reach 1. We see that $y_{n2}$ is greater than the conjugate depth, and above critical, so it could be possible to jump down to that level in reach 1. This is depicted in figure 48. Considering reach 2, we see that the conjugate depth is lower than the normal depth of reach 1. So for a jump to occur the depth would have to drop down to the conjugate, then jump up to $y_{n2}$. This cannot happen (as there is nothing to cause the drop in level). So the result is a hydraulic jump in reach 1. The surface profile will be an S1. It is in the steep channel and in region 1.

E4.2
A long wide channel consists of an upstream section with a bed slope of 0.025 and a downstream section with a bed slope of 0.0002. The Chezy C is constant for the whole of the channel and equal to 40 in SI units. When the discharge in the channel is $3m^{3}/s/m$. width demonstrates that a hydraulic jump occurs. Determine in which reach of the channel the jump occurs. Complete your answer with a freehand sketch of the final profile and label each of the gradually varied profiles.
(Answer: $y_{n1}=0.609m$, $y_{n2}=3.041m$, $y_{c}=0.972m$, $y_{n1\;conjugate}=1.459m$, $y_{n1\;conjugate}=0.187m$ ) This is very similar to example problem E4.1 so some of the more detailed explanations are not shown here  see that example to see how and why calculations are performed.For all flow transition problems you must first calculate the normal depth $y_{n}$ for reach and the critical depth $y_{c}$, for that flow (it will be the same for each reach). It is also useful to calculate the Froude numbers for each normal depth.
In this example, we have a wide channel and so the Chezy equation for uniform flow in this form is used:
$q=Cy_{n}^{3/2}S_{o}^{1/2}$And the $y_{n}$ can be calculated by this explicitly from equation LABEL: expression
$y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}$For reach 1, $q=3m^{3}/s/m$, $S_{o}=0.025$ and $C=40$, giving $y_{n1}=0.608m$ For reach 2, $q$ is the same at $q=3m^{3}/s/m$, $S_{o}=0.0002$ and $C=40$, giving $y_{n2}=3.041m$
For a wide channel, setting $Fr^{2}=1$ and solving for critical depth $y_{c}$ gives
$y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}=0.972m$It is useful to calculate the Froude number of the normal depths as they will be used later. The equation for $Fr$ in a wide channel is:
$Fr=\frac{q}{y\sqrt{gy}}$Which gives $Fr_{reach\;1}=2.02$ and $Fr_{reach\;2}=0.18$.
A sketch showing the normal and critical depths helps to understand what is physically being calculated and what may be physically possible. These are shown in figure 49.
We see that the depths in the two reaches change from below critical depth to above critical depth. The only way that a water surface can join these (and cross critical) is by a hydraulic jump. The hydraulic jump could be in either the upstream reach or the downstream reach, see figure 46 (from the previous example). To determine which, we must calculate the conjugate (or sequent) depths for each of the two normal depths.
We calculate the conjugate depth to $y_{n_{1}}$ in reach 1 from equation Eq39 to give
$y_{n1\;conjugate}=y_{2}=\frac{y_{1}}{2}\left(\sqrt{1+8F_{r1}^{2}}1\right)=1.4% 59m$and for reach 2, from equation Eq40 to give
$y_{n2\;conjugate}=y_{1}=\frac{y_{2}}{2}\left(\sqrt{1+8F_{r2}^{2}}1\right)=0.1% 87m$We now need to consider where the hydraulic jump occurs. Considering reach 1, we note that $y_{n2}$ is greater than the conjugate depth, and above critical, so it is possible to jump down to the $y_{n1}$ level in reach 1. This is depicted in figure 50. Considering reach 2, we see that the conjugate depth is lower than the normal depth of reach 1. So for a jump to occur the depth would have to drop down to the conjugate, then jump up to $y_{n2}$. This cannot happen (as there is nothing to cause the drop in level).
So the result is a hydraulic jump in reach 1. The surface profile will be an S1. It is in the steep channel and in region 1. As shown in figure 50

E4.3
A wide channel has three reaches each with a different slope. Each can be considered to be long and at the downstream end of the three reaches there is a free outfall. For a specific case, the Chézy C value for all reaches is 32 and the flow per unit width is $1m^{3}/s/m$. The upstream reach has a slope of 1in500, the middle reach has a slope of 1in50 and the downstream reach has a slope of 1in2,000. Determine which flow profile occurs.
(Answer: $y_{n1}=0.787m$, $y_{n2}=0.366m$, $y_{n3}=1.250m$, $y_{c}=0.467m$, $y_{n2\;conjugate}=0.586m$, $y_{n3\;conjugate}=0.119m$ )For all flow transition problems you must first calculate the normal depth $y_{n}$ for reach and the critical depth $y_{c}$, for that flow (it will be the same for each reach). It is also useful to calculate the Froude numbers for each normal depth.
It is a wide channel and we are given a Chezy C so we will use the Chezy equation for uniform flow to calculate normal depths,, equation Eq7,
$q=Cy_{n}^{3/2}S_{o}^{1/2}$And the $y_{n}$ can be calculated by the explicit equation Eq10
$y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}$For reach 1, $q=1.0m^{3}/s/m$, $S_{o}=1/500=0.002$ and $C=32$, giving $y_{n1}=0.787m$.
For reach 2, $q$ is the same at $q=1.0m^{3}/s/m$, $S_{o}=1/50=0.02$ and $C=32$, giving $y_{n2}=0.366m$.
For reach 3, $q$ is still the same at $q=1.0m^{3}/s/m$, $S_{o}=1/2000=0.0005$ and $C=32$, giving $y_{n3}=1.250m$For a wide channel, setting $Fr^{2}=1$ and solving for critical depth $y_{c}$ gives equation Eq23
$y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}=0.467m$It is useful to calculate the Froude number of the normal depths as they will be used later. The equation for $Fr$ in a wide channel is is equation Eq18
$Fr=\frac{q}{y\sqrt{gy}}$Which gives $Fr_{reach\;1}=0.46$, $Fr_{reach\;2}=1.44$ and $Fr_{reach\;3}=0.23$.
A sketch showing the normal and critical depths helps to understand what is physically being calculated and what may be physically possible. These are shown in figure 51.
We see that the depths in the three reaches change from above critical depth to below critical depth to above. The slopes are thus mild, steep, mild and at uniform flow would be subcritical, supercritical, subcritical respectively.
It is now useful to sketch the possible flow profiles joining the three reaches of uniform flow. Figure 52 shows such a sketch.
Starting from reach 1, we have a normal depth greater than critical which joins to the normal depth lower than critical. To connect these the flow must cross through critical. It does this smoothly at the point where the channel changes slope. This is shown on figure 52 as surface profiles M2 and S2 as profile is in the second region of the mildsloped reach 1 and the second region of the steeploped reach 2.
For the change from reach 2 to reach 3 we have to increase depth from below critical to above depth i.e. from subcritical to supercritical flow. This is only possible via a hydraulic jump. There are two options as show, a jump that occurs in reach 2 or a jump that occurs in reach 3. If it occurs in reach 2, then the surface profile that joins it to $y_{n3}$ will be an S1 (steepsloped in region 1). If it occurs in reach 3, then the surface profile that joins it to $y_{n2}$ will be an M1 (mildsloped in region 3).
As the end of the three reaches is a free outfall the flow will be forced to be critical at that outfall, so the depth critical. There will in that case be, at the end of reach 3, an M2 profile as the depth drops from normal, $y_{n3}$ to critical.
We now need to consider whether the jump is in reach 2 or reach 3. To determine which, we must calculate the conjugate (or sequent) depths for each of the two normal depths.
We calculate the conjugate depth to $y_{n_{2}}$ in reach 2 from equation Eq39 to give
$y_{n2\;conjugate}=y_{2}=\frac{y_{1}}{2}\left(\sqrt{1+8F_{r1}^{2}}1\right)=0.5% 86m$and for reach 3 (from equation Eq40) to give
$y_{n3\;conjugate}=y_{1}=\frac{y_{2}}{2}\left(\sqrt{1+8F_{r2}^{2}}1\right)=0.1% 19m$We now need to consider where the hydraulic jump occurs. Considering reach 2, we note that $y_{n3}$ is greater than the conjugate depth, and above critical, so it is possible to jump down to the $y_{n2}$ level in reach 2. This is depicted in figure 53. Considering reach 3, we see that the conjugate depth is lower than the normal depth of reach 2. So for a jump to occur the depth would have to drop down to the conjugate, then jump up to $y_{n3}$. This cannot happen (as there is nothing to cause the drop in level).
So the result is a hydraulic jump in reach 2. The surface profile will be an S1. It is in the steep channel and in region 1. As shown in figure 53

E4.4
A wide channel has two reaches with different bed slopes changing from $s_{o}=0.01$ to $S_{o}=0.0005$. Both reaches have the same crosssection shape and bed roughness of $n=0.03$. Sketch the critical flow depth $y_{c}$ line, uniform flow depth $y_{n}$ lines, and water surface profile in the channel and identify the type of water surface profile, if the flow discharge per unit width is $q=0.8m^{3}/s/m$.
(Answer: $y_{n1}=0.425m$, $y_{n2}=1.043m$, $y_{c}=0.403m$, M1 profile )This is very similar to example E4.1 and E4.2 but this time uses then Manning’s n to define roughness, so the normal depth calculations will be slightly different.
The Manning’s equation for a wide channel, equation Eq3 is:
$q=\frac{1}{n}y_{n}^{5/3}S_{o}^{1/2}$Then $y_{n}$ can be calculated by this explicit equation Eq6
$y_{n}=\left(\frac{qn}{S_{o}^{1/2}}\right)^{3/5}$For reach 1, $S_{o}=0.01$ and
$y_{n_{1}}=\left(\frac{0.8\times 0.03}{0.01^{1/2}}\right)^{3/5}=0.425m$For reach 2, $S_{o}=0.0005$ and
$y_{n_{2}}=\left(\frac{0.8\times 0.03}{0.0005^{1/2}}\right)^{2/3}=1.043m$For a wide channel, $Fr^{2}$ can be written from equation Eq17 as
$Fr^{2}=\frac{Q^{2}}{b^{2}gy^{3}}=\frac{q}{gy^{3}}$As $y_{c}$ is the depth when $Fr=Fr^{2}=1$ then $y_{c}$ can be calculated using equation Eq23:
$y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}=0.403m$It is useful to calculate the Froude number of the normal depths as may often be used later. $Fr$ in a wide channel can be calculated from equation Eq18:
$Fr=\frac{V}{\sqrt{g\frac{A}{B}}}=\frac{Q}{by\sqrt{gy}}=\frac{q}{y\sqrt{gy}}$Which gives $Fr_{reach\;1}=0.92$ and $Fr_{reach\;2}=0.24$.
We can compare the normal and critical depths for each reach. We see that for reach 1, $y_{n_{1}}>y_{c}$ so the reach is mild and the uniform flow subcritical. Also confirmed as $Fr_{reach\;1}>1$.
Similarly for reach 2, $y_{n_{2}}>y_{c}$ so the reach is also mild and the uniform flow also subcritical. As again confirmed as $Fr_{reach\;2}<1$.
We can sketch normal and critical depths to help understand the physical situation. This is shown in figure 54
We need to decide which profile joins the two normal depths. They are both mild so it will be an M profile. One option would be for the depth to rise in reach 3 with an M2 profile. However, an M2 falls in the direction of flow and we need it to rise. So the only option is for an M1 profile in reach 1 which rises asymptotically from $y_{n1}$ to $y_{n2}$, again joining the normal depth asymptotically. This is sketched in figure 55

E4.5
A wide channel consists of three long reaches. The upstream reach has a bed slope of 1 in 500, the middle reach a slope of 1 in 50, and the downstream reach slope of 1 in 1,750. The Chezy C for all three lengths can be assumed constant and equal to 27. Determine the profiles that occur if the flow per unit width is $0.75m^{3}/s/m$.
(Answer: $y_{n1}=0.728m$, $y_{n2}=0.338m$, $y_{n3}=1.105m$, $y_{c}=0.386m$, $y_{n2\;conjugate}=0.438m$, $y_{n3\;conjugate}=0.087m$. Jump in reach 2 (middle). )For all flow transition problems you must first calculate the normal depth $y_{n}$ for reach and the critical depth $y_{c}$, for that flow (it will be the same for each reach). It is also useful to calculate the Froude numbers for each normal depth.
It is a wide channel and we are given a Chezy C so we will use the Chezy equation for uniform flow to calculate normal depths, equation Eq7
$q=Cy_{n}^{3/2}S_{o}^{1/2}$And the $y_{n}$ can be calculated by this explicit equation Eq10
$y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}$For reach 1, $q=0.75m^{3}/s/m$, $S_{o}=0.002$ and $C=27$, giving $y_{n1}=0.728m$.
For reach 2, $q$ is the same at $q=0.75m^{3}/s/m$, $S_{o}=0.02$ and $C=27$, giving $y_{n2}=0.338m$.
For reach 3, $q$ is still the same at $q=0.75m^{3}/s/m$, $S_{o}=0.00057$ and $C=27$, giving $y_{n3}=1.105m$For a wide channel, setting $Fr^{2}=1$ and solving for critical depth $y_{c}$ gives equation Eq23
$y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}=0.386m$It is useful to calculate the Froude number of the normal depths as they will be used later. $Fr$ in a wide channel is given by equation Eq18:
$Fr=\frac{q}{y\sqrt{gy}}$Which gives $Fr_{reach\;1}=0.39$, $Fr_{reach\;2}=1.22$ and $Fr_{reach\;3}=0.21$.
A sketch showing the normal and critical depths helps to understand what is physically being calculated and what may be physically possible. These are shown in figure 56.
We see that the depths in the three reaches change from above critical depth to below critical depth to above. The slopes are thus mild, steep, mild and uniform flow would be subcritical, supercritical, subcritical.
It is now useful to consider and sketch the possible flow profiles joining the three reaches of uniform flow. Figure 57 shows such a sketch.
Starting from reach 1, we have a normal depth greater than critical which joins to the normal depth lower than critical. To connect these the flow must cross through critical. It does this smoothly at the point where the channel changes slope. This is shown on figure 57 as surface profiles M2 and S2 as the profile is in the second region of the mildsloped reach 1 and the second region of the steeploped reach 2.
For the change from reach 2 to reach 3 we have to increase depth from below critical to above depth i.e. from subcritical to supercritical flow. This is only possible via a hydraulic jump. There are two options as shown, a jump that occurs in reach 2 or a jump that occurs in reach 3. If it occurs in reach 2, then the surface profile that joins it to $y_{n3}$ will be an S1 (steepsloped in region 1). If it occurs in reach 3, then the surface profile that joins it to $y_{n2}$ will be an M1 (mildsloped in region 3).
We now need to consider whether the jump is in reach 2 or reach 3. To determine which, we must calculate the conjugate (or sequent) depths for each of the two normal depths.
We calculate the conjugate depth to $y_{n_{2}}$ in reach 2 from equation Eq39 to give
$y_{n2\;conjugate}=y_{2}=\frac{y_{1}}{2}\left(\sqrt{1+8F_{r1}^{2}}1\right)=0.4% 38m$and for reach 3 (from equation Eq40) to give
$y_{n3\;conjugate}=y_{1}=\frac{y_{2}}{2}\left(\sqrt{1+8F_{r2}^{2}}1\right)=0.0% 87m$We now need to consider where the hydraulic jump occurs. Considering reach 2, we note that $y_{n3}$ is greater than the conjugate depth, and above critical, so it is possible to jump down to the $y_{n2}$ level in reach 2. This is depicted in figure 58. Considering reach 3, we see that the conjugate depth is lower than the normal depth of reach 2. So for a jump to occur the depth would have to drop down to the conjugate, then jump up to $y_{n3}$. This cannot happen (as there is nothing to cause the drop in level).
So the result is a hydraulic jump in reach 2. The surface profile will be an S1. It is in the steep channel and in region 1. As shown in figure 58