E4 Questions on Gradually varied flow: Flow Transitions.

  1. 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 7.5m3/s/m7.5m^{3}/s/m and determine whether a hydraulic jump occurs and if one occurs which reach of the channel it occurs.
    (Answers: yn1=1.406my_{n1}=1.406m, yn2=3.467my_{n2}=3.467m, yc=1.79my_{c}=1.79m. Jump in reach 1.)

    For all flow transition problems you must first calculate the normal depth yny_{n} for reach and the critical depth ycy_{c}, for that flow (it will be the same for each reach).

    In this example, we have a wide channel which implies that byb\gg y and can be assumed rectangular. This results in A=byA=by, PbP\approx b and so RyR\approx y.

    The Manning’s equation for a wide channel is equation Eq-3

    q=1nyn5/3So1/2q=\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=Cyn3/2So1/2q=Cy_{n}^{3/2}S_{o}^{1/2}

    Then yny_{n} can be calculated by this explicit expression

    yn=(qCSo1/2)2/3y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}

    For reach 1, So=1/100=0.01S_{o}=1/100=0.01 and

    yn1=(7.545×0.011/2)2/3=1.406my_{n_{1}}=\left(\frac{7.5}{45\times 0.01^{1/2}}\right)^{2/3}=1.406m

    For reach 2, So=1/1500=0.0006671S_{o}=1/1500=0.0006671 and

    yn2=(7.545×0.00066711/2)2/3=3.467my_{n_{2}}=\left(\frac{7.5}{45\times 0.0006671^{1/2}}\right)^{2/3}=3.467m

    For a wide channel, Fr2Fr^{2} can be written from equation Eq-17 as

    Fr2=Q2b2gy3=qgy3Fr^{2}=\frac{Q^{2}}{b^{2}gy^{3}}=\frac{q}{gy^{3}}

    As ycy_{c} is the depth when Fr=Fr2=1Fr=Fr^{2}=1 then ycy_{c} can be calculated using equation Eq-23:

    yc=(q2g)1/3=1.790my_{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 Eq-11:

    Fr=VgAB=Qbygy=qygyFr=\frac{V}{\sqrt{g\frac{A}{B}}}=\frac{Q}{by\sqrt{gy}}=\frac{q}{y\sqrt{gy}}

    Which gives Frreach 1=1.44Fr_{reach\;1}=1.44 and Frreach 2=0.37Fr_{reach\;2}=0.37.

    We can compare the normal and critical depths for each reach. We see that for reach 1, yn1<ycy_{n_{1}}<y_{c} so the reach is steep and the flow super-critical. Also confirmed as Frreach 1>1Fr_{reach\;1}>1.

    For reach 2, yn2>ycy_{n_{2}}>y_{c} so the reach is mild and the flow sub-critical. Also confirmed as Frreach 2<1Fr_{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.

    Refer to caption
    Figure 45: Sketch of slopes and relative critical and normal depths

    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.

    Refer to caption
    Figure 46: Sketch of possible positions of hydraulic jump joining the two normal depths

    We calculate the conjugate depth to yn1y_{n_{1}} in reach 1 using the following, where y2y_{2} is the conjugate depth in reach 1 using equation Eq-39:

    yn1conjugate=y2=y12(1+8Fr121)y_{n1\;conjugate}=y_{2}=\frac{y_{1}}{2}\left(\sqrt{1+8F_{r1}^{2}}-1\right)

    From earlier Fr1=1.44F_{r1}=1.44 and y1=yn1=1.406y_{1}=y_{n1}=1.406, this results in the conjugate depth yn1conjugate=y2=2.239my_{n1\;conjugate}=y_{2}=2.239m.

    And for the conjugate depth to yn2y_{n_{2}} in reach 2 using the following, where y1y_{1} is the conjugate depth in reach 2 using equation Eq-40:

    yn2conjugate=y1=y22(1+8Fr221)y_{n2\;conjugate}=y_{1}=\frac{y_{2}}{2}\left(\sqrt{1+8F_{r2}^{2}}-1\right)

    From earlier Fr2=0.26F_{r2}=0.26 and y2=yn2=3.467y_{2}=y_{n2}=3.467, this results in the conjugate depth yn2conjugate=y1=0.779my_{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

    Refer to caption
    Figure 47: Sketch of conjugate depths in relation to normal and critical depths

    We can first consider if a jump is possible in reach 1. We see that yn2y_{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 yn2y_{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.

    Refer to caption
    Figure 48: Sketch of conjugate depths and identification of hydraulic jump position
  2. 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 3m3/s/m3m^{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 free-hand sketch of the final profile and label each of the gradually varied profiles.
    (Answer: yn1=0.609my_{n1}=0.609m, yn2=3.041my_{n2}=3.041m, yc=0.972my_{c}=0.972m, yn1conjugate=1.459my_{n1\;conjugate}=1.459m, yn1conjugate=0.187my_{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 yny_{n} for reach and the critical depth ycy_{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=Cyn3/2So1/2q=Cy_{n}^{3/2}S_{o}^{1/2}

    And the yny_{n} can be calculated by this explicitly from equation LABEL: expression

    yn=(qCSo1/2)2/3y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}

    For reach 1, q=3m3/s/mq=3m^{3}/s/m, So=0.025S_{o}=0.025 and C=40C=40, giving yn1=0.608my_{n1}=0.608m For reach 2, qq is the same at q=3m3/s/mq=3m^{3}/s/m, So=0.0002S_{o}=0.0002 and C=40C=40, giving yn2=3.041my_{n2}=3.041m

    For a wide channel, setting Fr2=1Fr^{2}=1 and solving for critical depth ycy_{c} gives

    yc=(q2g)1/3=0.972my_{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 FrFr in a wide channel is:

    Fr=qygyFr=\frac{q}{y\sqrt{gy}}

    Which gives Frreach 1=2.02Fr_{reach\;1}=2.02 and Frreach 2=0.18Fr_{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.

    Refer to caption
    Figure 49: Sketch of slopes and relative critical and normal depths

    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 yn1y_{n_{1}} in reach 1 from equation Eq-39 to give

    yn1conjugate=y2=y12(1+8Fr121)=1.459my_{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 Eq-40 to give

    yn2conjugate=y1=y22(1+8Fr221)=0.187my_{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 yn2y_{n2} is greater than the conjugate depth, and above critical, so it is possible to jump down to the yn1y_{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 yn2y_{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

    Refer to caption
    Figure 50: Sketch of conjugate depths and identification of hydraulic jump position
  3. 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 1m3/s/m1m^{3}/s/m. The upstream reach has a slope of 1-in-500, the middle reach has a slope of 1-in-50 and the downstream reach has a slope of 1-in-2,000. Determine which flow profile occurs.
    (Answer: yn1=0.787my_{n1}=0.787m, yn2=0.366my_{n2}=0.366m, yn3=1.250my_{n3}=1.250m, yc=0.467my_{c}=0.467m, yn2conjugate=0.586my_{n2\;conjugate}=0.586m, yn3conjugate=0.119my_{n3\;conjugate}=0.119m )

    For all flow transition problems you must first calculate the normal depth yny_{n} for reach and the critical depth ycy_{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 Eq-7,

    q=Cyn3/2So1/2q=Cy_{n}^{3/2}S_{o}^{1/2}

    And the yny_{n} can be calculated by the explicit equation Eq-10

    yn=(qCSo1/2)2/3y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}

    For reach 1, q=1.0m3/s/mq=1.0m^{3}/s/m, So=1/500=0.002S_{o}=1/500=0.002 and C=32C=32, giving yn1=0.787my_{n1}=0.787m.
    For reach 2, qq is the same at q=1.0m3/s/mq=1.0m^{3}/s/m, So=1/50=0.02S_{o}=1/50=0.02 and C=32C=32, giving yn2=0.366my_{n2}=0.366m.
    For reach 3, qq is still the same at q=1.0m3/s/mq=1.0m^{3}/s/m, So=1/2000=0.0005S_{o}=1/2000=0.0005 and C=32C=32, giving yn3=1.250my_{n3}=1.250m

    For a wide channel, setting Fr2=1Fr^{2}=1 and solving for critical depth ycy_{c} gives equation Eq-23

    yc=(q2g)1/3=0.467my_{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 FrFr in a wide channel is is equation Eq-18

    Fr=qygyFr=\frac{q}{y\sqrt{gy}}

    Which gives Frreach 1=0.46Fr_{reach\;1}=0.46, Frreach 2=1.44Fr_{reach\;2}=1.44 and Frreach 3=0.23Fr_{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.

    Refer to caption
    Figure 51: Sketch of slopes and relative critical and normal depths

    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 sub-critical, super-critical, sub-critical 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 mild-sloped reach 1 and the second region of the steep-loped 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 sub-critical to super-critical 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 yn3y_{n3} will be an S1 (steep-sloped in region 1). If it occurs in reach 3, then the surface profile that joins it to yn2y_{n2} will be an M1 (mild-sloped 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, yn3y_{n3} to critical.

    Refer to caption
    Figure 52: Sketch of possible flow profiles joining the normal depths of the three reaches

    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 yn2y_{n_{2}} in reach 2 from equation Eq-39 to give

    yn2conjugate=y2=y12(1+8Fr121)=0.586my_{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 Eq-40) to give

    yn3conjugate=y1=y22(1+8Fr221)=0.119my_{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 yn3y_{n3} is greater than the conjugate depth, and above critical, so it is possible to jump down to the yn2y_{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 yn3y_{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

    Refer to caption
    Figure 53: Sketch of conjugate depths and identification of hydraulic jump position for reaches 2 and 3
  4. E4.4

    A wide channel has two reaches with different bed slopes changing from so=0.01s_{o}=0.01 to So=0.0005S_{o}=0.0005. Both reaches have the same cross-section shape and bed roughness of n=0.03n=0.03. Sketch the critical flow depth ycy_{c} line, uniform flow depth yny_{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=0.8m3/s/mq=0.8m^{3}/s/m.
    (Answer: yn1=0.425my_{n1}=0.425m, yn2=1.043my_{n2}=1.043m, yc=0.403my_{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 Eq-3 is:

    q=1nyn5/3So1/2q=\frac{1}{n}y_{n}^{5/3}S_{o}^{1/2}

    Then yny_{n} can be calculated by this explicit equation Eq-6

    yn=(qnSo1/2)3/5y_{n}=\left(\frac{qn}{S_{o}^{1/2}}\right)^{3/5}

    For reach 1, So=0.01S_{o}=0.01 and

    yn1=(0.8×0.030.011/2)3/5=0.425my_{n_{1}}=\left(\frac{0.8\times 0.03}{0.01^{1/2}}\right)^{3/5}=0.425m

    For reach 2, So=0.0005S_{o}=0.0005 and

    yn2=(0.8×0.030.00051/2)2/3=1.043my_{n_{2}}=\left(\frac{0.8\times 0.03}{0.0005^{1/2}}\right)^{2/3}=1.043m

    For a wide channel, Fr2Fr^{2} can be written from equation Eq-17 as

    Fr2=Q2b2gy3=qgy3Fr^{2}=\frac{Q^{2}}{b^{2}gy^{3}}=\frac{q}{gy^{3}}

    As ycy_{c} is the depth when Fr=Fr2=1Fr=Fr^{2}=1 then ycy_{c} can be calculated using equation Eq-23:

    yc=(q2g)1/3=0.403my_{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. FrFr in a wide channel can be calculated from equation Eq-18:

    Fr=VgAB=Qbygy=qygyFr=\frac{V}{\sqrt{g\frac{A}{B}}}=\frac{Q}{by\sqrt{gy}}=\frac{q}{y\sqrt{gy}}

    Which gives Frreach 1=0.92Fr_{reach\;1}=0.92 and Frreach 2=0.24Fr_{reach\;2}=0.24.

    We can compare the normal and critical depths for each reach. We see that for reach 1, yn1>ycy_{n_{1}}>y_{c} so the reach is mild and the uniform flow sub-critical. Also confirmed as Frreach 1>1Fr_{reach\;1}>1.

    Similarly for reach 2, yn2>ycy_{n_{2}}>y_{c} so the reach is also mild and the uniform flow also sub-critical. As again confirmed as Frreach 2<1Fr_{reach\;2}<1.

    We can sketch normal and critical depths to help understand the physical situation. This is shown in figure 54

    Refer to caption
    Figure 54: Sketch of normal depth for reach 1 and 2 and the critical depth

    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 yn1y_{n1} to yn2y_{n2}, again joining the normal depth asymptotically. This is sketched in figure 55

    Refer to caption
    Figure 55: Sketch of the M1 water surface profile connecting the normal depths of reach 1 and reash 2
  5. 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 0.75m3/s/m0.75m^{3}/s/m.
    (Answer: yn1=0.728my_{n1}=0.728m, yn2=0.338my_{n2}=0.338m, yn3=1.105my_{n3}=1.105m, yc=0.386my_{c}=0.386m, yn2conjugate=0.438my_{n2\;conjugate}=0.438m, yn3conjugate=0.087my_{n3\;conjugate}=0.087m. Jump in reach 2 (middle). )

    For all flow transition problems you must first calculate the normal depth yny_{n} for reach and the critical depth ycy_{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 Eq-7

    q=Cyn3/2So1/2q=Cy_{n}^{3/2}S_{o}^{1/2}

    And the yny_{n} can be calculated by this explicit equation Eq-10

    yn=(qCSo1/2)2/3y_{n}=\left(\frac{q}{CS_{o}^{1/2}}\right)^{2/3}

    For reach 1, q=0.75m3/s/mq=0.75m^{3}/s/m, So=0.002S_{o}=0.002 and C=27C=27, giving yn1=0.728my_{n1}=0.728m.
    For reach 2, qq is the same at q=0.75m3/s/mq=0.75m^{3}/s/m, So=0.02S_{o}=0.02 and C=27C=27, giving yn2=0.338my_{n2}=0.338m.
    For reach 3, qq is still the same at q=0.75m3/s/mq=0.75m^{3}/s/m, So=0.00057S_{o}=0.00057 and C=27C=27, giving yn3=1.105my_{n3}=1.105m

    For a wide channel, setting Fr2=1Fr^{2}=1 and solving for critical depth ycy_{c} gives equation Eq-23

    yc=(q2g)1/3=0.386my_{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. FrFr in a wide channel is given by equation Eq-18:

    Fr=qygyFr=\frac{q}{y\sqrt{gy}}

    Which gives Frreach 1=0.39Fr_{reach\;1}=0.39, Frreach 2=1.22Fr_{reach\;2}=1.22 and Frreach 3=0.21Fr_{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.

    Refer to caption
    Figure 56: Sketch of slopes and relative critical and normal depths

    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 sub-critical, super-critical, sub-critical.

    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 mild-sloped reach 1 and the second region of the steep-loped 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 sub-critical to super-critical 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 yn3y_{n3} will be an S1 (steep-sloped in region 1). If it occurs in reach 3, then the surface profile that joins it to yn2y_{n2} will be an M1 (mild-sloped in region 3).

    Refer to caption
    Figure 57: Sketch of possible flow profiles joining the normal depths of the three reaches

    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 yn2y_{n_{2}} in reach 2 from equation Eq-39 to give

    yn2conjugate=y2=y12(1+8Fr121)=0.438my_{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 Eq-40) to give

    yn3conjugate=y1=y22(1+8Fr221)=0.087my_{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 yn3y_{n3} is greater than the conjugate depth, and above critical, so it is possible to jump down to the yn2y_{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 yn3y_{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

    Refer to caption
    Figure 58: Sketch of conjugate depths and identification of hydraulic jump position for reaches 2 and 3