E3 Questions on Gradually varied flow - including integration of the backwater curve.

  1. E3.1

    A Rectangular channel is 3.0m wide, has a 0.01 slope, discharge of 5.3m3/s5.3m^{3}/s, and n=0.011. Find yny_{n} and ycy_{c}. If the actual depth of flow is 1.7m, what type of profile exists?
    (Answer: yny_{n} = 0.4m, ycy_{c} = 0.683m, S1 Curve)

    We need to solve Manning’s equation for yny_{n} for a rectangular channel.

    Refer to caption
    Figure 23: A rectangular channel section

    For a rectangular channel, we have:

    A=byA=by

    and

    P=b+2yP=b+2y

    so

    R=AP=byb+2yR=\frac{A}{P}=\frac{by}{b+2y}

    The Manning’s equation Eq-2 is thus:

    Q=1nAR2/3So1/2=1n(byn)5/3(b+2yn)2/3So1/2Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}

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

    This gives for Q=5.3m3/sQ=5.3m^{3}/s, b=3.0mb=3.0m, So=0.01So=0.01 and n=0.011n=0.011:

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

    At critical depth ycy_{c}, Fr=1Fr=1 and also Fr2=1Fr^{2}=1, so we can use equation Eq-20

    Fr2=1=Q2BgA3Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}

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

    1=Q2gB2yc31=\frac{Q^{2}}{gB^{2}y_{c}^{3}}

    This can be formulated to be solved explicitly for ycy_{c}, eqn Eq-22:

    yc=Q2gB23=(Q2gB2)1/3y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}=\left(\frac{Q^{2}}{gB^{2}}\right)^{1/3}

    To give

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

    As yn<ycy_{n}<y_{c} then the flow is super-critical.

    We can calculate the Froude number of normal depth (eqn Eq-16) to confirm this:

    Fr=Vgyn=2.13Fr=\frac{V}{\sqrt{gy_{n}}}=2.13

    As Fr>1Fr>1 this confirms that uniform flow is super-critical.

    So we have a steep channel, and all surface profiles will be "S" curves.

    Figure 24 shows the relative positions of the critical and normal depth lines for a steep slope.

    Refer to caption
    Figure 24: Critical and normal depth lines for a steep slope

    If at a position in the channel, the depth is 1.7m1.7m, as this is above the critical depth then the profile is in region 1. The surface profile will be an S1S1 profile. This takes the form as shown in figure 25

    Refer to caption
    Figure 25: The S1S1 profile
  2. E3.2

    A rectangular channel with a bottom width of 4.0m4.0m and a bottom slope of 0.00080.0008 has a discharge of 1.50m3/s1.50m^{3}/s. In a gradually varied flow in this channel, the depth at a certain location is found to be 0.30m0.30m assuming n=0.016n=0.016, determining the type of GVF profile, the critical depth, and the normal depth.
    (Answer: M2, yc=0.24my_{c}=0.24m, yn=0.43my_{n}=0.43m)

    Following example E3.1, the Manning’s equation Eq-2 is:

    Q=1nAR2/3So1/2=1n(byn)5/3(b+2yn)2/3So1/2Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}

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

    This gives for Q=1.5m3/sQ=1.5m^{3}/s, b=4.0mb=4.0m, So=0.0008So=0.0008 and n=0.016n=0.016:

    Normal depthyn=0.426m\text{Normal depth}\quad y_{n}=0.426m

    At critical depth ycy_{c}, Fr=1Fr=1 and also Fr2=1Fr^{2}=1, so we can use equation Eq-14 equated to 1

    Fr2=1=Q2BgA3Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}

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

    1=Q2gB2yc31=\frac{Q^{2}}{gB^{2}y_{c}^{3}}

    This can be formulated to be solved explicitly for ycy_{c} (eqn Eq-22):

    yc=Q2gB23y_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}

    To give

    Critical depthyc=0.243m\text{Critical depth}\quad y_{c}=0.243m

    As yn>ycy_{n}>y_{c} then the flow is sub-critical.

    We can calculate the Froude number of normal depth to confirm this:

    Fr=Vgyn=0.43Fr=\frac{V}{\sqrt{gy_{n}}}=0.43

    As Fr<1Fr<1 this confirms that uniform flow is sub-critical.

    So we have a mild channel, and all surface profiles will be "M" curves.

    Figure 26 shows the relative positions of the critical and normal depth lines for a mild slope.

    Refer to caption
    Figure 26: Critical and normal depth lines for a mild slope

    If at a position in the channel, the depth is 0.3m0.3m, as this is above the critical depth and below the normal depth the profile is in region 2. The surface profile will be an M2M2 profile. This takes the form as shown in figure 27

    Refer to caption
    Figure 27: The M2M2 profile
  3. E3.3

    The figure below shows a backwater curve in a long rectangular channel. Determine using a numerical integration method, the profile for the following high flow conditions: Q=10m3/sQ=10m^{3}/s, b=3mb=3m, n=0.022n=0.022, and a bed slope of 0.0010.001. Take the depth just upstream of the dam as the control point equal to 5m5m. At what distance is the water level not affected by the dam? perform your integration using a) 2-steps and b) 10-steps

    (Answer: yn=2.437my_{n}=2.437m.
    Using y0y_{0} (Euler) 2-step x=3490mx=-3490m, 10-step x=4724mx=-4724m.
    Using y1/2y_{1/2} 2-step x=4480mx=-4480m, 10-step x=5906mx=-5906m.
    Using a forth-order Runge Kutta method gave 2-step x=3893mx=-3893m, 10-step x=5461mx=-5461m)

    Refer to caption
    Figure 28: The backwater curve for question E3.3

    We will integrate using a numerical method that solves to obtain a distance from depth change. There are various methods. We will demonstrate 3 here.

    We must solve (integrate) this backwater, or gradually varied flow, equation Eq-42:

    dydx=SoSfFr2\frac{dy}{dx}=\frac{S_{o}-S_{f}}{\-Fr^{2}}

    Solving for distance, xx, given a change in depth yy. So in total differential form, we will solve:

    Δx=Δy(1Fr2SoSf)\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)

    There are options now as to the way the right-hand side is approximated (or discretised). Some options are

    1. i)

      As in equation i), the first-order Euler method (where the subscript o{}_{o} indicates at the known, or initial, point), i.e. y0y_{0}:

      Δx=Δy(1Fr2SoSf)o\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{o}
    2. ii)

      Or equation ii), at the averaged depth y1/2=(y0+y1)/2y_{1/2}=(y_{0}+y_{1})/2

      Δx=Δy(1Fr2SoSf)mean=Δy(1Fry1/22SoSfy1/2)=Δy(1Fr2SoSf)1/2\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{mean}=\Delta y% \left(\frac{1-Fr^{2}_{y{{}_{1/2}}}}{S_{o}-S_{f_{y_{1/2}}}}\right)=\Delta y% \left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{{1/2}}
    3. iii)

      or, as in equation iii), the whole function averaged between the initial and subsequent point:

      Δx=Δy(1Fr2SoSf)mean\displaystyle\Delta x=\Delta y\left(\frac{1-Fr^{2}}{S_{o}-S_{f}}\right)_{mean} =Δy[(1Fr02SoSfy0)+(1Fr12SoSfy1)]/2\displaystyle=\Delta y\left.\left[\left(\frac{1-Fr_{0}^{2}}{S_{o}-S_{f_{y_{0}}% }}\right)+\left(\frac{1-Fr_{1}^{2}}{S_{o}-S_{f_{y_{1}}}}\right)\right]\middle/% 2\right.
      =Δy[(1FrSoSf)0+(1FrSoSf)1]/2\displaystyle=\Delta y\left.\left[\left(\frac{1-Fr}{S_{o}-S_{f}}\right)_{0}+% \left(\frac{1-Fr}{S_{o}-S_{f}}\right)_{1}\right]\middle/2\right.

    For all methods, we calculate Fr2Fr^{2} and SfS_{f} in the same way using the appropriate depth yy and other channel properties. The equations to use are, for the Froude number, in a rectangular channel (equation Eq-14):

    Fr2=Q2bA3g=Q2b(by)3g=Q2b2y3gFr^{2}=\frac{Q^{2}b}{A^{3}g}=\frac{Q^{2}b}{(by)^{3}\;g}=\frac{Q^{2}}{b^{2}y^{3% }\;g}

    And remembering that SfS_{f} is determined from the Manning’s equation (Eq-2), assuming that over the short distance Sf=SoS_{f}=S_{o} then:

    Q=1nAR2/3So1/2=1nAR2/3Sf1/2Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}AR^{2/3}S_{f}^{1/2}

    rearranging for SfS_{f} and substituting appropriately for a rectangular channel we get from equation Eq-45:

    Sf=Q2n2A2R4/3=Q2n2(by)2(byb+2y)4/3S_{f}=\frac{Q^{2}n^{2}}{A^{2}R^{4/3}}=\frac{Q^{2}n^{2}}{(by)^{2}\left(\frac{by% }{b+2y}\right)^{4/3}}

    The question asks to find the distance to where the depth is not affected by the dam - this is asking, when does the depth reach normal depth?. So we must first calculate yny_{n}. This is done via a straightforward iteration of the Manning’s equation for a rectangular channel (see previous examples).

    In this case using Q=10m3/sQ=10m^{3}/s, b=3mb=3m, n=0.022n=0.022 and So=0.001S_{o}=0.001 then the solution is, normal depth yn=2.437my_{n}=2.437m.

    The level just upstream of the dam is 5m5m, so we need to integrate to find xx when yy has dropped to yn=2.437y_{n}=2.437. That is a drop, (so negative), of dy=2.4375.0=2.563dy=2.437-5.0=-2.563

    We are asked to calculate the distance using a 2-step integration which would give Δy=2.563/2=1.2815m\Delta y=-2.563/2=-1.2815m

    and also for a 10-step integration which would give Δy=2.563/10=0.2563m\Delta y=-2.563/10=-0.2563m

    We can now take each of the three discretisation methods and construct an iteration table (e.g. in Excel) for each to calculate the two solutions using the two Δy\Delta y values.

    Integration method E3.3i) - First-order Euler

    Figure 29 shows the integration table for the 2-step integration and a distance of 3490m3490m upstream of the dam to normal depth.

    Refer to caption
    Figure 29: Method E3.3i) Euler integration. 2-steps, Δy=1.2815m\Delta y=-1.2815m

    Figure 30 shows the integration table for the 10-step integration and a distance of 4724m4724m upstream of the dam to normal depth.

    Refer to caption
    Figure 30: Method E3.3i) Euler integration. 10-steps, Δy=0.2563m\Delta y=-0.2563m

    Integration method E3.3ii) - Average yy

    Figure 31 shows the integration table for the 2-step integration and a distance of 4480m4480m upstream of the dam to normal depth.

    Refer to caption
    Figure 31: Method E3.3ii) Average yy integration. 2-steps, Δy=1.2815m\Delta y=-1.2815m

    Figure 32 shows the integration table for the 10-step integration and a distance of 5906m5906m upstream of the dam to normal depth.

    Refer to caption
    Figure 32: Method E3.3ii) Average yy integration. 10-steps, Δy=0.2563m\Delta y=-0.2563m

    Integration method E3.3iii) - Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f})

    Figure 33 shows the integration table for the 2-step integration and a distance of 1712420m1712420m upstream of the dam to normal depth. This seems incorrect (unlikely), however, let’s have a look a the 10-step solution.

    Refer to caption
    Figure 33: Method E3.3iii) Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f}) integration. 2-steps, Δy=1.2815m\Delta y=-1.2815m

    Figure 34 shows the integration table for the 10-step integration and a distance of 346510m346510m upstream of the dam to normal depth. This again seems incorrect.

    Refer to caption
    Figure 34: Method E3.3iii) Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f}) integration. 10-steps, Δy=0.2563m\Delta y=-0.2563m

    The reason why this method does not give expected answers is that it fails when depth is equal to yny_{n}. The reason is that (SoSf)(S_{o}-S_{f}) would is zero at yny_{n}. In the numerical calculation, it is close to zero, 107\approx 10^{-7}, and gives a very long last Δx\Delta x. We can say that this method should not be used when the depth of yny_{n} is to be encountered.

    A hack that could allow us to use this method would be to perform an Euler step for the last integration step (i.e. avoid the use of yny_{n}). [We might call this a hybrid method.]. This would give, for the 2-step x=2718mx=-2718m and for the 10-step x=4570mx=-4570m.

    A fourth-order Runge-Kutta method is usually considered as accurate as practically possible for a numerical method. Using this on the above problem gave for the 2-step x=3893mx=-3893m, and for the 10-step x=5461mx=-5461m.

    In summary, the solutions are shown in the table 1 below

    Table 1: Summary of distances upstream in metres to normal depth from 4 methods
    Step Δy(m)\Delta y\;(m) E3.3i) Euler E3.3ii) y1/2y_{1/2} E3.3iii) - Average GVF Runge Kutta
    funct. (hybrid) 4th Order
    2 -1.2815 3490 4480 2718 3893
    10 -0.2563 4724 5906 4570 5461
    100 -0.0256 6719 7916 6703 7512
  4. E3.4

    A trapezoidal, concrete-lined, channel has a constant bed slope of 0.00150.0015, a bed width of 3m3m, and side slopes of 1:11:1. A control gate increased the depth immediately upstream to 4m4m. When the discharge is 19m3/s19m^{3}/s compute the water surface profile upstream and identify the distance when the water depth is 1.8m1.8m. (n=0.017n=0.017)
    (Answer: Using y0y_{0} (Euler) 2-step x=1540mx=-1540m, 10-step x=1695mx=-1695m.
    Using y1/2y_{1/2} 2-step x=1670mx=-1670m, 10-step x=1809mx=-1809m.
    Using mean GVF function: 2-step x=2729mx=-2729m, 10-step x=1933mx=-1933m.
    Using a forth-order Runge Kutta method gave 2-step x=2023mx=-2023m, 10-step x=1850mx=-1850m)

    This solution follows similarly to that of example question E3.3. We have been given an initial depth of y0=4my_{0}=4m and asked to find the distance upstream to where the depth is y=1.8y=1.8.

    We should first determine the normal and critical depths. Although these are not directly used in the question, they do inform what solutions is expected and whether the depths rise or fall. The calculations must be done iteratively, as seen in previous examples.

    Undertaking the iterative calculation we get yn=1.725my_{n}=1.725m and yc=1.364my_{c}=1.364m. So flow is sub-critical, and the slope is mild. The depths given for the calculation of the surface are above the normal depth so the surface profile will be a M1M1 profile. It will slope downwards from 4m toward the target of 1.8m.

    The depth fall is a drop, (so negative), of dy=4.01.8=2.2mdy=4.0-1.8=-2.2m

    We are asked to calculate the distance using a 2-step integration which would give Δy=2.2/2=1.1m\Delta y=-2.2/2=-1.1m

    And also for a 10-step integration which would give Δy=2.1/10=0.22m\Delta y=-2.1/10=-0.22m

    Here we demonstrate results from the three integration methods identified in example E3.3 above.

    Integration method E3.3i) - First-order Euler

    Figure 35 shows the integration table for the 2-step integration and a distance of 1540m1540m upstream of the dam to the target depth.

    Refer to caption
    Figure 35: Method E3.3i) Euler integration. 2-steps, Δy=1.1m\Delta y=-1.1m

    Figure 36 shows the integration table for the 10-step integration and a distance of 1696m1696m upstream of the dam to the target depth.

    Refer to caption
    Figure 36: Method E3.3i) Euler integration. 10-steps, Δy=0.22m\Delta y=-0.22m

    Integration method E3.3ii) - Average yy

    Figure 37 shows the integration table for the 2-step integration and a distance of 1670m1670m upstream of the dam to the target depth.

    Refer to caption
    Figure 37: Method E3.3ii) Average yy integration. 2-steps, Δy=1.1m\Delta y=-1.1m

    Figure 38 shows the integration table for the 10-step integration and a distance of 1809m1809m upstream of the dam to the target depth.

    Refer to caption
    Figure 38: Method E3.3ii) Average yy integration. 10-steps, Δy=0.22m\Delta y=-0.22m

    Integration method E3.3iii) - Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f})

    Figure 39 shows the integration table for the 2-step integration and a distance of 2279m2279m upstream of the dam to the target depth. This seems incorrect (unlikely), however, let’s have a look a the 10-step solution.

    Refer to caption
    Figure 39: Method E3.3iii) Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f}) integration. 2-steps, Δy=1.1m\Delta y=-1.1m

    Figure 40 shows the integration table for the 10-step integration and a distance of 1934m1934m upstream of the dam to the target depth. This again seems incorrect.

    Refer to caption
    Figure 40: Method E3.3iii) Average (1Fr2)/(SoSf)(1-Fr^{2})/(S_{o}-S_{f}) integration. 10-steps, Δy=0.22m\Delta y=-0.22m

    A fourth-order Runge-Kutta method is usually considered as accurate as practically possible for a numerical method. Using this on the above problem gave for the 2-step x=2020mx=-2020m, and for the 10-step x=1850mx=-1850m. A 100-step Runge Kutta give x=1844mx=-1844m, so the 10-step can probably be considered sufficiently accurate.

    In summary, the solutions are shown in the table 1 below

    Table 2: Summary of distances upstream in metres to target depth from 4 methods
    Step Δy(m)\Delta y\;(m) E3.3i) Euler E3.3ii) y1/2y_{1/2} E3.3iii) - Average GVF Runge Kutta
    funct. 4th Order
    2 -1.1 1540 1670 2729 2023
    10 -0.22 1696 1809 1933 1851
    100 -0.022 1822 1843 1845 1844
  5. E3.5

    Using the figure below, determine the profile for the channel conditions using a step length of Δx=100m\Delta x=100m. Q=600m3/sQ=600m^{3}/s, n=0.04n=0.04, the bed slope of the rectangular channel is So=0.002S_{o}=0.002 and has a width of B=50mB=50m. The sill height of the weir is 2.5m2.5m and the water depth over the weir is 4m4m. Compare results from each method.

    Refer to caption
    Figure 41: The backwater curve for question E3.5

    The question is asking to integrate upstream from a depth of 6.5m(=4.0+2.5)6.5m(=4.0+2.5) at the weir enough distance to understand and compare the flow profile for each method. It suggests an integration step of Δx=100m\Delta x=100m which should be used for the depth from distance integration. It does not specify that this should be the method so the method integration distance from depth will also be used and the results compared.

    We should first understand the flow regime by calculation normal and critical depths and by comparison examine the expected flow profile.

    We have a rectangular channel and Manning’s n, so will use the following uniform flow equation:

    Q=1nAR2/3So1/2=1n(byn)5/3(b+2yn)2/3So1/2Q=\frac{1}{n}AR^{2/3}S_{o}^{1/2}=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{% 2/3}}So^{1/2}

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

    This gives for Q=600.0m3/sQ=600.0m^{3}/s, b=50.0mb=50.0m, So=0.002So=0.002 and n=0.04n=0.04:

    Normal depthyn=4.434m\text{Normal depth}\quad y_{n}=4.434m

    At critical depth ycy_{c}, Fr=1Fr=1 and also Fr2=1Fr^{2}=1, so we can use the equation Eq-19, written for a rectangular channel

    Fr2=1=Q2BgA3=Q2gB2yc3Fr^{2}=1=\frac{Q^{2}B}{gA^{3}}=\frac{Q^{2}}{gB^{2}y_{c}^{3}}

    And formulated to solve for critical depth ycy_{c}, giving from equation Eq-22

    yc=Q2gB23=2.448my_{c}=\sqrt[3]{\frac{Q^{2}}{gB^{2}}}=2.448m

    As yn>ycy_{n}>y_{c} then uniform flow is sub-critical. The slope of the channel is mild. The depth at the weir is higher that normal depth and moving upstream must fall gradually toward normal depth. This is the region 1 (above normal fro a mild slope), so the profile must be an M1 profile.

    The first integration will be the following depth from distance, using Δx=100m\Delta x=100m and we will use am Euler method i.e. use the conditions at the known point to calculate the derivative. Equation Eq-53

    Δy=Δx(SoSf1Fr2)i\Delta y=\Delta x\left(\frac{S_{o}-S_{f}}{1-Fr^{2}}\right)_{i}

    Where the friction slope, SfS_{f}, term is, as previously, equation Eq-45:

    Sf=n2Q2R4/3A2S_{f}=\frac{n^{2}Q^{2}}{R^{4/3}A^{2}}

    And the Froude number squared, equation Eq-14

    Fr2=Q2BA3gFr^{2}=\frac{Q^{2}B}{A^{3}g}

    See previous questions for a greater description of the Euler integration methodology. Tabulating the calculation and produces the integration table in figure 42. The integration has been performed for 20 steps and depth has fallen to 4.64m4.64m. Which indicates that normal depth has still not been achieved 2km2km upstream of the weir.

    Refer to caption
    Figure 42: Integration table for fixed Δx=100m\Delta x=100m

    As we wish to compare results from a distance from depth integration we should select appropriate parameters. The integration above used 20 steps and reached as depth of y=4.636my=4.636m. Using these as parameters would give a Δy=(6.54.636)/20=0.0932m\Delta y=(6.5-4.636)/20=0.0932m and 20 steps of integration.

    Refer to caption
    Figure 43: Integration table for fixed Δy=0.0932m\Delta y=-0.0932m and 20 steps

    Results in an integration table using a derivative approximated at y1/2y_{1/2} is shown in figure 43. We see from the table that there is a slight difference in the distance to that same depth of 75m75m. To investigate any further differences by plotting a graph of the two profiles. Such a graph is shown in figure 44. It is clear that the two integrations are very similar - differing by only millimeters along the whole profile length. The shape of the profile is as expected that of an M1 profile.

    Refer to caption
    Figure 44: Integrated Water Profile from the weir upstream for the two integration methods