E2 Questions on Rapidly Varied Flow - Sudden transitions and Hydraulic Jumps

  1. E2.1

    For a trapezoidal channel with a base width=3.0m\text{base width}=3.0m, and side slope 1 vertical 2 horizontal, calculate the critical depth if the discharge is Q=10m3/sQ=10m^{3}/s.
    (Answer: yc=0.86my_{c}=0.86m)

    The question asks for the critical depth, that is the depth when Froude number Fr=1Fr=1.

    We have for the Froude number this general equation Eq-11


    Written in terms of Q(=VA)Q\;(=VA) gives Eq-12:


    And at critical flow, when the depth is critical, ycy_{c} then


    Or, more conveniently for calculation


    We are considering the trapezoidal channel as shown in figure 17, the area of flow section AA and the wetted perimeter PP are given as

    Refer to caption
    Figure 17: A trapezoidal channel section

    This can be written in terms of critical depth, ycy_{c}:


    To solve for ycy_{c} (given QQ and bb) requires an iterative solution method. The iteration table shown in figure 18 solves this using the secant method. The solution is yc=0.856my_{c}=0.856m.

    Refer to caption
    Figure 18: Iteration table calculating ycy_{c} for a trapezoidal channel
  2. E2.2

    Water is flowing at a normal depth in a 3m3m wide rectangular channel with a bed slope of 1:500. If Manning’s n=0.025n=0.025 and the discharge is 5m3/s5m^{3}/s. Calculate the height of a bump that would produce the critical flow without causing backwater upstream (i.e. without raising the upstream water level). (Answer: yn=1.215my_{n}=1.215m, Bump height =0.326m=0.326m)

    First of all Manning’s equation needs to be solved for normal depth yny_{n} in the rectangular channel. An iteration table can be set up to do this. The table of figure 19 demonstrates this, and a solution of yn=1.215my_{n}=1.215m is calculated.

    Refer to caption
    Figure 19: Iteration table calculating yny_{n} for a rectangular channel

    If we consider the flow over a bump. The depth at point 1, just upstream of the bump, will be yn=1.215my_{n}=1.215m.

    Refer to caption
    Figure 20: Steady flow over a raised bump

    We can apply the specific energy concept from point 1 to point 2, just on the bump.

    Es1=Es2+ΔzE_{s1}=E_{s2}+\Delta z


    y1+V122g=y2+V222g+Δzy_{1}+\frac{V_{1}^{2}}{2g}=y_{2}+\frac{V_{2}^{2}}{2g}+\Delta z


    y1+Q2b1y12g=y2+Q2b2y22g+Δzy_{1}+\frac{Q^{2}}{b_{1}y_{1}2g}=y_{2}+\frac{Q^{2}}{b_{2}y_{2}2g}+\Delta z

    y2=ycy_{2}=y_{c}, critical depth. We can write the Froude number squared, at critical depth in a rectangular channel, from Eq-14 as


    so, using q=Q/Bq=Q/B



    yc=(5.029.81×3.02)1/3=0.657my_{c}=\left(\frac{5.0^{2}}{9.81\times 3.0^{2}}\right)^{1/3}=0.657m

    Applying Q=5m3/sQ=5m^{3}/s, b1=b2=B=3mb_{1}=b_{2}=B=3m with y2=ycy_{2}=y_{c} in the specific energy equation we get Δz=0.326m\Delta z=0.326m.

  3. E2.3

    Water is flowing at a velocity of 3.4m/s3.4m/s and a depth of 3.4m3.4m in a channel of rectangular section with a width of 3.4m3.4m. Find the changes in depth produced by

    1. (a)

      A smooth contraction to a width of 3.0m

    2. (b)

      The smallest allowable contraction for the flow to be possible upstream as described.

    (Answer: (a) y2=3.048my_{2}=3.048m, (b) y2=2.89my_{2}=2.89m)

    We have Q=AV=(3.4×3.4)3.4=39.304m3/sQ=AV=(3.4\times 3.4)3.4=39.304m^{3}/s and thus q=QB=39.3043.4=11.56m3/s/mq=\frac{Q}{B}=\frac{39.304}{3.4}=11.56m^{3}/s/m.

    Apply the specific energy equation from point 1 just upstream to point 2 in the throat.





    y1+Q2B12y122g=y2+Q2B22y222gy_{1}+\frac{Q^{2}}{B_{1}^{2}y_{1}^{2}2g}=y_{2}+\frac{Q^{2}}{B_{2}^{2}y_{2}^{2}% 2g}

    or using flow per unit width, q1=QB1q_{1}=\frac{Q}{B_{1}} and q2=QB2q_{2}=\frac{Q}{B_{2}}, then




    Part E2.3a: Solve the above for y2y_{2} given y1=3.4m,Q=3.4m3/s,B1=3.4m,B2=3.0my_{1}=3.4m,Q=3.4m^{3}/s,B_{1}=3.4m,B_{2}=3.0m. We have the cubic

    0=y23y22(y1+q12y122g)+q222g0=y_{2}^{3}-y_{2}^{2}\left(y_{1}+\frac{q_{1}^{2}}{y_{1}^{2}2g}\right)+\frac{q_% {2}^{2}}{2g}

    We can calculate Es1=3.99mE_{s1}=3.99m so the cubic equation to solve becomes


    This must be solved using a numerical tool or trial and error to give these 3 solutions y2=3.048,1.288,2.229y_{2}=3.048,-1.288,2.229.

    We know the negative solution is not physically possible, so can discount that. We need to check whether the approach flow is sub or super-critical. We need to calculate the critical depth upstream


    Also, we need to calculate the critical depth in the contraction


    The critical depth in the contraction cannot be passed so as the upstream flow is subcritical as y1>yc1y_{1}>y_{c1}, and depth will drop from 3.4m3.4m to 3.048m3.048m (it cannot physically pass yc2y_{c2} and drop down to 2.229m2.229m.

    We can see this on the specific energy curves and depths shown in figure 21. The solid horizontal green line intersects with the blue q1q_{1}, Es1E_{s1} curve at Es1=3.99mE_{s1}=3.99m, y1=3.4my_{1}=3.4m. The vertical dark blue Es1E_{s1} line drawn down from that point intersects the orange q2q_{2}, Es2E_{s2} curve at y2=3.048my_{2}=3.048m.

    Refer to caption
    Figure 21: Specific Energy curves Es1E_{s1}, Es2E_{s2}, and depths y1y_{1}, y2y_{2} and yc1y_{c1}, yc2y_{c2}

    Part E2.3b: The upstream flow depth will not be affected by the contraction width unless it goes narrower than the critical width, i.e. the width when the depth in the contraction is critical. We know that Escritical{E_{s}}_{critical} (or Esc{E_{s}}_{c}) can be expressed in terms of ycy_{c} as Eq-33


    And as Es2=EscE_{s2}={E_{s}}_{c} and because of conservation of energy Es1=Es2E_{s1}=E_{s2} this gives


    Using the expression for Fr2=1Fr^{2}=1 at critical depth we can calculate BB the minimum width to not affect the upstream depth

  4. E2.4

    The normal depth of flow in a rectangular channel (with a 5m wide bases and 2m high side walls) is 1m. It is laid to a slope of 1m/km with Manning’s n=0.02n=0.02. Some distance downstream there is a hump of height 0.5m on the stream bed. If critical flow occurs on the bump, determine the depth of flow (y1y_{1}) immediately upstream of the bump and the depth of flow (y2y_{2}) above the bump. If the bump is reduced to 0.1m, what values will y1y_{1} and y2y_{2} be?
    (Answer: Q=6.32m3/sQ=6.32m^{3}/s, y1=1.25my_{1}=1.25m, yc=0.54my_{c}=0.54m, y1=1.0my_{1}=1.0m, y2=0.87my_{2}=0.87m)

    We first calculate the flow QQ in the channel using Manning’s equation for a rectangular channel given, yn=1my_{n}=1m, b=5mb=5m, n=0.02n=0.02 and So=1/1000=0.001S_{o}=1/1000=0.001

    Q=1nAR2/3So1/2=1nbyn(bynb+2yn)2/3So1/2=6.317m3/sQ=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}by_{n}\left(\frac{by_{n}}{b+2y_{n}% }\right)^{2/3}S_{o}^{1/2}=6.317m^{3}/s

    Apply the specific energy equation from point 1 just upstream to point 2 above the bump with Δz\Delta z representing the height of the bump.

    Es1=Es2+ΔzE_{s1}=E_{s2}+\Delta z


    y1+V122g=y2+V222g+Δzy_{1}+\frac{V_{1}^{2}}{2g}=y_{2}+\frac{V_{2}^{2}}{2g}+\Delta z

    or (and considering that B1=B2=BB_{1}=B_{2}=B)

    y1+Q2B2y122g=y2+Q2B2y222g+Δzy_{1}+\frac{Q^{2}}{B^{2}y_{1}^{2}2g}=y_{2}+\frac{Q^{2}}{B^{2}y_{2}^{2}2g}+\Delta z

    or using flow per unit width, q=QBq=\frac{Q}{B}, then

    y1+q2y122g=y2+q2y222g+Δzy_{1}+\frac{q^{2}}{y_{1}^{2}2g}=y_{2}+\frac{q^{2}}{y_{2}^{2}2g}+\Delta z

    First, we need to know whether the flow arriving at the bump is sub or super-critical. To do this calculate the Froude number, Eq-13:


    So the flow is sub-critical.

    We could have calculated the critical depth Eq-22:


    And as this is less than the normal depth this confirms that flow is sub-critical.

    We have two possibilities here: 1) that the bump is less than the critical height, y1y_{1} will stay the same and will drop over the bump so we need to calculate y2y_{2}, or 2) the bump is greater than the critical height and so depth over the bump will be critical y2=ycy_{2}=y_{c} and the depth will increase upstream, so we need to calculate y1y_{1}. (of course, if the bump is equal to critical height, then y1y_{1} would not change.)

    So, we need to calculate the critical bump height. If we set y2=ycy_{2}=y_{c} and use the other given values then we have

    y1+q2y122g=yc+q2yc22g+Δzcy_{1}+\frac{q^{2}}{y_{1}^{2}2g}=y_{c}+\frac{q^{2}}{y_{c}^{2}2g}+\Delta z_{c}

    As q=Q/B=1.263m3/s/mq=Q/B=1.263m^{3}/s/m (or m2/sm^{2}/s) and yc=0.546my_{c}=0.546m

    1.081=0.819+Δzc1.081=0.819+\Delta z_{c}
    Δzc=0.262m\Delta z_{c}=0.262m

    As the bump height, 0.5m0.5m is greater than this Δzc=0.262m\Delta z_{c}=0.262m, then y2=yc=0.546my_{2}=y_{c}=0.546m and we need solve for y1y_{1}

    0=y13y22(yc+q2yc22g+Δz)+q22g0=y_{1}^{3}-y_{2}^{2}\left(y_{c}+\frac{q^{2}}{y_{c}^{2}2g}+\Delta z\right)+% \frac{q^{2}}{2g}

    Which, using the known values, gives


    This can be solved using a numerical tool or trial and error to give these 3 solution values y1=1.268,0.229,0.280y_{1}=1.268,-0.229,0.280.

    We know that the depth upstream must increase (from 1.0m1.0m), so the solution is y1=1.268my_{1}=1.268m, and y2=yc=0.546my_{2}=y_{c}=0.546m

    If the bump height is Δz=0.1m\Delta z=0.1m, then as this is less than the critical bump height (of 0.262m0.262m), the depth upstream will not change from the normal depth (of y1=yn=1.0my_{1}=y_{n}=1.0m), but the depth over the bump will fall.

    We simply take the same specific energy equation as above and now solve for y2y_{2} using y1=yn=1.0my_{1}=y_{n}=1.0m and Δz=0.1m\Delta z=0.1m.

    0=y23y22(y1+q2y122gΔz)+q22g0=y_{2}^{3}-y_{2}^{2}\left(y_{1}+\frac{q^{2}}{y_{1}^{2}2g}-\Delta z\right)+% \frac{q^{2}}{2g}

    Which, using the known values, gives


    This can be solved using a numerical tool or trial and error to give these 3 solution values y2=0.875,0.256,0.363y_{2}=0.875,-0.256,0.363.

    We know that the depth must drop from 1.0m1.0m and that the depth cannot be less than yc=0.546my_{c}=0.546m, and as the negative depth is not physically possible, the solution is y2=0.875my_{2}=0.875m.

  5. E2.5

    Water is flowing at a rate of 10m3/s10m^{3}/s through a rectangular channel 4m4m wide, at a depth of 0.5m0.5m. A weir downstream causes the water to back up the channel and a hydraulic jump occurs. Find the sequent depth and the loss of energy at the jump.
    (Answer: y2=1.37my_{2}=1.37m, ΔE=0.238m\Delta E=0.238m)

    We will use the following equation that relates depths on each side of a hydraulic jump, Eq-39:


    First calculate the velocity and Froude numbers using Q=10m3/sQ=10m^{3}/s, B=4mB=4m, y1=0.5my_{1}=0.5m


    Froude number for the rectangular channel


    It is useful to calculate the specific energy of the flow, as we need to calculate how much is lost in the jump, so


    Applying the hydraulic jump equation gives the height of the jump (the sequent depth):


    Then V2=Q/(By2)=1.830m/sV_{2}=Q/(By_{2})=1.830m/s and


    So loss of energy in the jump is

    ΔE=Es1Es2=0.237m\Delta E=E_{s1}-E_{s2}=0.237m

    Alternatively, this formula can be used to calculate the energy loss in a jump

    ΔE=(y2y1)34y1y2=0.237m\Delta E=\frac{(y_{2}-y_{1})^{3}}{4y_{1}y-2}=0.237m
  6. E2.6

    Water flows in a rectangular channel at a depth of 30cm and with a velocity of 16m/s16m/s. If a downstream sill forces a hydraulic jump, what will be the depth and velocity downstream of the jump? What head loss is produced by the jump?
    (Answer: y2=3.81my_{2}=3.81m, V2=1.26m/sV_{2}=1.26m/s, ΔE=9.46m\Delta E=9.46m)

    This is very similar to the previous question, however this time we do not know the width, so we work with flow per unit width, qq.

    We will again use the following equation that relates depths on each side of a hydraulic jump:


    First calculate the flow per unit width (q=QB=ByVB=Vyq=\frac{Q}{B}=\frac{ByV}{B}=Vy) and Froude numbers using V1=16m/sV_{1}=16m/s, y1=0.3my_{1}=0.3m

    q=V1y1=4.8m3/s/m(or m2/s)q=V_{1}y_{1}=4.8m^{3}/s/m\qquad(\text{or }m^{2}/s)

    Froude number for the rectangular channel Eq-11


    Calculate the specific energy of the flow, LABEL:eqn:specific_energy


    Applying the hydraulic jump equation gives the height of the jump (the sequent depth):


    Then V2=q/y2=1.26m/sV_{2}=q/y_{2}=1.26m/s and


    So loss of energy in the jump is

    ΔE=Es1Es2=9.457m\Delta E=E_{s1}-E_{s2}=9.457m

    Alternatively, the formula Eq-41 can be used to calculate the energy loss in a jump

    ΔE=(y2y1)34y1y2=9.457m\Delta E=\frac{(y_{2}-y_{1})^{3}}{4y_{1}y-2}=9.457m
  7. E2.7

    Water passes under a sluice gate in a horizontal channel of width 2m. The depths of flow on either side of the sluice gate are y1=1.8my_{1}=1.8m and y2=0.3my_{2}=0.3m. A hydraulic jump occurs a short distance downstream. Assuming no energy loss at the gate, calculate:

    1. (a)

      The force on the gate

    2. (b)

      The depth of flow downstream of the hydraulic jump

    3. (c)

      The fraction of the fluid energy that is dissipated in the jump

    (Answer: F=15.7kNF=15.7kN, y3=1.22my_{3}=1.22m, Fraction of energy lost (dissipated) =28%=28\%)

    We will consider the arrangement and labelling shown in figure 22 for the depths in all subsequent equations.

    Refer to caption
    Figure 22: Sluice and hydraulic jump arrangements

    We first need to determine the flow rate using the given depths, y1=1.8my_{1}=1.8m and y2=0.3my_{2}=0.3m. This can be done using the specific energy equation. In terms of flow per unit width, this can be written as equation Eq-32:


    This can be rearranged in terms of qq:


    And applying the known depths


    and thus q=1.651m3/s/mq=1.651m^{3}/s/m             (or m2/sm^{2}/s)

    We can now calculate the velocities:




    For E2.7a to find the force on the gate we need to apply the force equation


    Where, RR is the force on the gate, and MM is the total force (force due to the rate of change in momentum), given by:

    M\displaystyle M =M2M1\displaystyle=M_{2}-M_{1}
    =ρQV2ρQV1\displaystyle=\rho QV_{2}-\rho QV_{1}
    =ρQ(V2V1)\displaystyle=\rho Q(V_{2}-V_{1})

    or in terms of flow per unit width qq

    M=ρqb(V2V1)M=\rho qb(V_{2}-V_{1})

    And PP is the pressure force (force due to changes in pressure), given by:

    P\displaystyle P =P1P2\displaystyle=P_{1}-P_{2}
    =ρgz1¯A1ρgz2¯A2\displaystyle=\rho g\bar{z_{1}}A_{1}-\rho g\bar{z_{2}}A_{2}
    =ρg(z1¯A1z2¯A2)\displaystyle=\rho g(\bar{z_{1}}A_{1}-\bar{z_{2}}A_{2})

    For a rectangular channel, where z¯=y2\bar{z}=\frac{y}{2} and A=byA=by, so

    P=ρgb12(y12y22)P=\rho gb\frac{1}{2}(y^{2}_{1}-y^{2}_{2})

    Putting these together gives

    ρqb(V2V1)=ρgb12(y12y22)R\rho qb(V_{2}-V_{1})=\rho gb\frac{1}{2}(y^{2}_{1}-y^{2}_{2})-R


    R\displaystyle R =ρgb12(y12y22)ρqb(V2V1)\displaystyle=\rho gb\frac{1}{2}(y^{2}_{1}-y^{2}_{2})-\rho qb(V_{2}-V_{1})

    Or, force on the gate R=15.7kNR=15.7kN.

    For E2.7b we know that the hydraulic jump occurs soon after the gate, so we can assume the conditions from the above calculation: i.e. just upstream of the jump, depth, y2=0.3my_{2}=0.3m and velocity, V2=5.503m/sV_{2}=5.503m/s.

    And the Froude number here is equation Eq-16

    Fr2V2gy2=5.5039.81×0.03=3.208Fr_{2}\frac{V_{2}}{\sqrt{gy_{2}}}=\frac{5.503}{\sqrt{9.81\times 0.03}}=3.208

    We will apply the hydraulic jump equation as previously, using the upstream location as 22 and the downstream as 33, following the notation in figure 22 The hydraulic jump equation Eq-39 is thus

    y3=0.32(1+8×3.20821)=1.219my_{3}=\frac{0.3}{2}\left(\sqrt{1+8\times 3.208^{2}}-1\right)=1.219m

    For E2.7c we need to calculate the specific energy on both sides of the jump.

    Es2=0.3+5.50322×9.81=1.843mE_{s2}=0.3+\frac{5.503^{2}}{2\times 9.81}=1.843m

    Velocity downstream of the jump is calculated as



    Es3=1.219+1.35422×9.81=1.312mE_{s3}=1.219+\frac{1.354^{2}}{2\times 9.81}=1.312m

    So loss of energy in the jump is

    ΔE=Es2Es3=0.531m\Delta E=E_{s2}-E_{s3}=0.531m

    The fraction of energy loss is

    ΔEEs2=0.5311.843=0.2881\frac{\Delta E}{E_{s2}}=\frac{0.531}{1.843}=0.2881

    Which in percentage terms is an energy loss of 28.8%28.8\%.