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

  1. E1.1

    A rectangular channel is 3.0m3.0m wide, has a 0.01 slope, flow rate of 5.3m3/s5.3m^{3}/s, and n=0.011n=0.011. Find its normal depth yny_{n} and critical depth ycy_{c}.
    (Answer: yn=0.41my_{n}=0.41m, yc=0.683my_{c}=0.683m)

    Refer to caption
    Figure 1: A rectangular channel section

    For a rectangular channel, we have:






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

    Q=Cbyn(bynb+2yn)1/2S1/2Q=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=1n(byn)5/3(b+2yn)2/3S1/2Q=\frac{1}{n}\frac{(by_{n})^{5/3}}{(b+2y_{n})^{2/3}}S^{1/2} (2)

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

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

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


    And also Fr2=1Fr^{2}=1 then


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


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


    To give

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

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


    Then as q=5.3/3.0=1.767m3/s/mq=5.3/3.0=1.767m^{3}/s/m,

    Critical depthyc=1.76729.81=0.683m\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 1.22m1.22m and discharge of Q=5.66m3/sQ=5.66m^{3}/s. Determine the minimum channel width if the channel is to be sub-critical.
    (Answer: B=1.34mB=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 yc=1.22my_{c}=1.22m and Q=5.66m3/sQ=5.66m^{3}/s and can use Fr=1Fr=1 at critical depth. Equation Eq-13 gives:


    And also Fr2=1Fr^{2}=1 then


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


    We need to solve for BB so


    Which gives

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

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

    1. (a)

      Determine the slope to give a normal depth of yn=2my_{n}=2m when the discharge is 12m3/s12m^{3}/s

    2. (b)

      Determine the critical slope and the critical depth when the discharge is 12m3/s12m^{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=0.00024So=0.00024, (b) Soc=0.0087So_{c}=0.0087, yc=0.61my_{c}=0.61m, (c) Soc=46.03m3/sSo_{c}=46.03m^{3}/s, Q=0.00818Q=0.00818)

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

    Manning’s equation, Eq-2


    For a rectangular channel, we have:






    We have y=yn=2.0my=y_{n}=2.0m, Q=12.0m3/sQ=12.0m^{3}/s and Manning’s n=0.025n=0.025.

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

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

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


    As Q=12.0m/s3Q=12.0m/s^{3} and b=8mb=8m then critical depth yc=0.612my_{c}=0.612m

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




    leading to


    This results in Ac=4.899m2,Pc=9.225mA_{c}=4.899m^{2},P_{c}=9.225m and Rc=0.531R_{c}=0.531, and thus Soc=0.00872So_{c}=0.00872 or 1 in 115

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

    Soc=gn2PcBcRc1/3So_{c}=\frac{gn^{2}P_{c}}{B_{c}R_{c}^{1/3}} (3)

    (where Bc=bB_{c}=b for a rectangular channel)

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


    Which give Q=46.03m3/sQ=46.03m^{3}/s

    Using yc=1.5my_{c}=1.5m we get Ac=12.0m2,Pc=11.0mA_{c}=12.0m^{2},P_{c}=11.0m and Rc=1.091R_{c}=1.091, and, as above, we can cacluate Soc=0.00818So_{c}=0.00818 or 1 in 122

  4. E1.4

    For a trapezoidal channel with a base width b=3.0mb=3.0m, Manning’s n=0.025n=0.025 and side slope s=2s=2 (i.e. 1 vertical: 2 horizontal), calculate the critical depth, critical velocity, and critical slope if its discharge Q=10m3/sQ=10m^{3}/s.

    (Answer:yc=0.855my_{c}=0.855m, vc=2.483m/v_{c}=2.483m/s, Soc=0.00777So_{c}=0.00777)

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


    For a trapezoidal channel, we have:






    Substitution for AA and BB and iterating for yy gives the solution yc=0.855my_{c}=0.855m

    Velocity in any channel is


    Substituting, ycy_{c} to give AcA_{c} and using Q=10m3/sQ=10m^{3}/s, gives v=2.483m./sv=2.483m./s.

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

    Soc=gn2PcBcRc1/3So_{c}=\frac{gn^{2}P_{c}}{B_{c}R_{c}^{1/3}} (4)

    From the data, Pc=6.823mP_{c}=6.823m, Ac=4.027m2A_{c}=4.027m^{2}, bc=6.420mb_{c}=6.420m, Rc=0.590mR_{c}=0.590m and this gives Soc=0.00777So_{c}=0.00777 or 1 in 129

  5. E1.5

    A rectangular channel 9m9m wide carries 7.6m3/s7.6m^{3}/s of water when flowing 1.0m1.0m deep. Work out the flow’s specific energy. Is the flow sub-critical or super-critical?

    (Answer: 1.0361.036m and flow is sub-critical)

    Specific Energy is given by Eq-27,

    Es=y+αV22gE_{s}=y+\frac{\alpha V^{2}}{2g} (5)

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

    Es=1.0+0.84422×9.81=1.036mE_{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=Vgy=0.8441×9.81=0.27Fr=\frac{V}{\sqrt{gy}}=\frac{0.844}{\sqrt{1\times 9.81}}=0.27

    Fr<1.0Fr<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 ycy_{c} is given by Eq-22


    As 1.0>0.1871.0>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=130m3/sQ=130m^{3}/s, flow velocity V=1.6m/sV=1.6m/s, water surface width B=80mB=80m; River 2: flow discharge Q=1530m3/sQ=1530m^{3}/s, flow velocity V=5.6m/sV=5.6m/s, water surface width B=90mB=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


    or in terms of velocity


    We can calculate AA from Q=AVQ=AV.

    For River 1: Q=130.0m3.sQ=130.0m^{3}.s, V=1.6m/sV=1.6m/s, so A=130.0/1.6=81.25m2A=130.0/1.6=81.25m^{2} and as B=80.0mB=80.0m then

    Fr2=(130.02×80.081.253×9.81)=0.507Fr^{2}=\left(\frac{130.0^{2}\times 80.0}{81.25^{3}\times 9.81}\right)=0.507

    For River 2: Q=1530.0m3.sQ=1530.0m^{3}.s, V=5.6m/sV=5.6m/s, so A=1530.0/5.6=273.21m2A=1530.0/5.6=273.21m^{2} and as B=90.0mB=90.0m then

    Fr2=(1530.02×90.0273.213×9.81)=1.027Fr^{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 So=0.0009S_{o}=0.0009 and a Manning roughness factor of n=0.013n=0.013. The bottom width of the channel is b=2.5mb=2.5m, and the side slopes are 1 in 2. Determine the velocity and discharge when the flow is normal at a depth of 1.8m1.8m.
    (Answer: v=2.37m/sv=2.37m/s, Q=26.01m3/sQ=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+1.8×2)=10.98m2A=y(b+ys)=1.8(2.5+1.8\times 2)=10.98m^{2}
    P=b+2y1+s2=2.5+2×1.81+22=10.55mP=b+2y\sqrt{1+s^{2}}=2.5+2\times 1.8\sqrt{1+2^{2}}=10.55m


    Q=1nAR2/3So1/2=10.01310.98×1.042/30.00091/2=26.01m3/sQ=\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

  8. E1.8

    A trapezoidal channel has a bottom slope of So=1S_{o}=1 in 4040 and a Manning roughness factor of n=0.016n=0.016. The bottom width of the channel is b=6.0mb=6.0m, and the side slopes are 1 in 3. Determine the normal depth in this channel for Q=42.3m3/sQ=42.3m^{3}/s.
    (Answer: yn=0.75my_{n}=0.75m). For a rectangular channel, we have:






    Manning’s equation Eq-2 is


    We have Q=42.3m2/sQ=42.3m^{2}/s, Manning’s n=0.016n=0.016, b=6.0mb=6.0m and s=3s=3.

    Iterating gives the solution yn=0.75my_{n}=0.75m.

  9. E1.9

    The discharge in uniform flow in a rectangular channel 4.6m4.6m wide is 11.3m3/s11.3m^{3}/s when the slope is 1:100. Is the flow sub-critical or super-critical? Calculate the slope, SocSo_{c}, that would give critical depth. The Manning roughness coefficient is n=0.012n=0.012.
    (Answer: super-critical, Fr=2.1Fr=2.1, Sc=0.002268S_{c}=0.002268).

    Iterate the Manning’s equation to solve for yny_{n} to give yn=0.521my_{n}=0.521m.

    Then A=byn=2.34m2A=by_{n}=2.34m^{2} and V=Q/A=4.71m/sV=Q/A=4.71m/s

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


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

    Critical depth for a rectangular channel can be calculated from Fr=1Fr=1 or Fr2=2Fr^{2}=2, leading to Eq-22


    From the Manning’s equation SoSo is found


    Ac=byc=4.6×0.85=3.91m2A_{c}=by_{c}=4.6\times 0.85=3.91m^{2}, Pc=b+2yc=6.30mP_{c}=b+2y_{c}=6.30m, Rc=Ac/Pc=0.62mR_{c}=A_{c}/P_{c}=0.62m and thus Soc=0.002268So_{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 So=0.0009S_{o}=0.0009. The Manning roughness factor n=0.025n=0.025 for the main channel and n=0.035n=0.035 for the overbank areas. Determine the normal depth for a discharge of 197m3/s197m^{3}/s. Also, calculate the energy coefficient α\alpha and the momentum coefficient β\beta for the channel with this flow condition.

    Refer to caption
    Figure 2: A Compound section

    (Answer: yn=5.507my_{n}=5.507m, α=1.23\alpha=1.23, β=1.09\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 yny_{n}. If we assume that the normal depth is such that flow is in the flood plains, and our depth is yy, then we have, from the geometry in figure 3, depth on the flood plain yfloody_{flood} as





    A2=(2.0+10.0+2.0)×yflood+4.5×(10.0+2.0)=14yflood+54A_{2}=(2.0+10.0+2.0)\times y_{flood}+4.5\times(10.0+2.0)=14y_{flood}+54

    And also



    Refer to caption
    Figure 3: Section detail for given yy

    n1=n3=0.035n_{1}=n_{3}=0.035, n2=0.025n_{2}=0.025, Q=197m3/sQ=197m^{3}/s, So=0.009S_{o}=0.009 and can calculate the side slope from the section geometry as s2=24.5=0.444s_{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 yy. Then QQ for each of the three regions, and then the total can be calculated and compared against the target of Q=197m3/sQ=197m^{3}/s. The first iteration is shown in figure 4, for an initial y=5.5my=5.5m, equivalent to yflood=1my_{flood}=1m

    Refer to caption
    Figure 4: Iteration table headings and initial calculation with y=5.5my=5.5m

    The aim of the iteration is to reduce the error to be small 0.001m3/s\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(yi)Errorf(y_{i})\equiv\text{Error} in the table of figure 4.

    yi=yi1+f(yi1)(yi1yi2f(yi1f(yi2))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 yy we have taken y=4.5my=4.5m, so that yflood=0.0y_{flood}=0.0. And applying the secant method we get the result shown in he table of figure 5 with yn=5.507my_{n}=5.507m.

    Refer to caption
    Figure 5: Complete iteration table

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

    α=u3𝑑AV¯3A=V13A1+V23A2+V33A3V¯3(A1+A2+A3)\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)


    β=u2𝑑AV¯2A=V12A1+V22A2+V23A3V¯2(A1+A2+A3)\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)


    V¯=QA=V1A1+V2A2+V3A3A1+A2+A3\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 V1=Q1/A1=0.787m/sV_{1}=Q_{1}/A_{1}=0.787m/s, V2=Q2/A2=2.730m/sV_{2}=Q_{2}/A_{2}=2.730m/s, V3=V1V_{3}=V_{1} and so, from equation 10, V¯=2.397m/s\overline{V}=2.397m/s.

    From equations 8 and 9:



  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 0.15m0.15m in the flow depth for the discharge of 197m3/s197m^{3}/s. Assuming normal depth still is present in the channel, determine the minimum allowable channel total width, BB.
    (Answer: B=16.157mB=16.157m.)

    The solution to this question requires finding a width BB such that yny_{n} is the normal depth from E1.10, 5.507m5.507m, plus 0.15m0.15m i.e. yn=5.507+0.15=5.657my_{n}=5.507+0.15=5.657m. We can perform an iteration similar to that in E1.10 (calculating AA, PP, and QQ) for an estimated BB, and the target of the iteration will be the same Q=197m3/sQ=197m^{3}/s, so the error function will be QQiQ-Q_{i}.

    For a given BB we must calculate the floodplain width


    So the formulae that change from question E1.10 are those for A1=A3=Bflood×yfloodA_{1}=A_{3}=B_{flood}\times y_{flood}. As we know that yflood=5.6574.5=1.157my_{flood}=5.657-4.5=1.157m then




    The table of figure 6 shows this implemented in a similar way to that of question E1.10 with BiB_{i} estimated via a secant method and taking B=28mB=28m (full flood plain), and B=14mB=14m (no flood plain) as initial values. This shows the solution of B=16.157mB=16.157m, or the width of each flood plain being 1.079m1.079m.

    Refer to caption
    Figure 6: Complete iteration table for BB
  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: So=2×104S_{o}=2\times 10^{-4}
    Manning’s ns: n1=n2=n3=0.02n_{1}=n_{2}=n_{3}=0.02
    Side slopes: s1=s2=s3=1s_{1}=s_{2}=s_{3}=1
    Bed widths: B1=3mB_{1}=3m, B2=5mB_{2}=5m, B3=4mB_{3}=4m
    Main channel depth: ymain=3.0my_{main}=3.0m


    Normal depth yn=4.5my_{n}=4.5m
    (Answer: Q=69.42m3/sQ=69.42m^{3}/s.)

    Refer to caption
    Figure 7: Idealized river channel with flood plains

    The equations for each geometry element can be written as functions of the depth yy where also y1=y3=yymainy_{1}=y_{3}=y-y_{main}

    BB1\displaystyle BB_{1} =s1y1\displaystyle=s_{1}y_{1}
    BB2\displaystyle BB_{2} =s2ymain\displaystyle=s_{2}y_{main}
    BB3\displaystyle BB_{3} =s3y3\displaystyle=s_{3}y_{3}
    A1\displaystyle A_{1} =(B1+BB1/2)y1\displaystyle=(B_{1}+BB_{1}/2)y_{1}
    A2\displaystyle A_{2} =(B3+BB2)ymain+(B2+2BB2)y1\displaystyle=(B_{3}+BB_{2})y_{main}+(B_{2}+2BB_{2})y_{1}
    A3\displaystyle A_{3} =(B3+BB3/2)y3\displaystyle=(B_{3}+BB_{3}/2)*y_{3}
    P1\displaystyle P_{1} =B1+BB12+y12\displaystyle=B_{1}+\sqrt{BB_{1}^{2}+y_{1}^{2}}
    P2\displaystyle P_{2} =B2+2BB22+ymain2\displaystyle=B_{2}+2\sqrt{BB_{2}^{2}+y_{main}^{2}}
    P3\displaystyle P_{3} =B3+BB32+y32\displaystyle=B_{3}+\sqrt{BB_{3}^{2}+y_{3}^{2}}

    QQ is calculated from Manning’s equation for each of the j=13j=1...3 regions


    And so Q=Q1+Q2+Q3Q=Q_{1}+Q_{2}+Q_{3}.

    Implementing these formulae in a tabular form results in the table of figure 8, which shows Q=69.421m3/sQ=69.421m^{3}/s.

    Refer to caption
    Figure 8: Table calculating QQ for give yny_{n}

    As an extra exercise, the table could be set up to iterate to find a normal depth yny_{n} given a flow QQ. As a test, if the target QQ is set to Q=69.421m3/sQ=69.421m^{3}/s, the solution from an iteration should give yn=4.5my_{n}=4.5m. Such an iteration has been performed using the secant method and is shown in figure 9. With initial estimates of yi=3.0my_{i}=3.0m and 4.0m4.0m this iterates to yn=4.5my_{n}=4.5m in 4 further iterations.

    Refer to caption
    Figure 9: Iteration table for yny_{n} where Q=69.421m3/sQ=69.421m^{3}/s

    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=100m3/sQ=100m^{3}/s is a normal depth yn=5.229my_{n}=5.229m

    Refer to caption
    Figure 10: Iteration table for yny_{n} where Q=100m3/sQ=100m^{3}/s
  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=219.53m3/sQ=219.53m^{3}/s).

    This follows the calculation of geometry of question E1.12, with the calculations for QjQ_{j} changed to take account of So=0.002S_{o}=0.002. If the table has been set up appropriately in the spreadsheet then a simple change of the SoS_{o} value will give a solution of Q=219.53m3/sQ=219.53m^{3}/s, and the table shown in figure 11

    Refer to caption
    Figure 11: Table calculating QQ for S0=0.002S_{0}=0.002

    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 0.5m3/s0.5m^{3}/s. Use the Chezy formula, assuming that C=80m1/2/sC=80m^{1/2}/s.
    (Answer: yn=0.552my_{n}=0.552m, b=0.552mb=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/PR=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,

    Refer to caption
    Figure 12: A trapezoidal channel section

    the area of flow AA and the wetted perimeter PP are given by these familiar equations

    A=by+sy2A=by+sy^{2} (11)
    P=b+2y1+s2P=b+2y\sqrt{1+s^{2}} (12)

    Rearranging the equation for AA to give bb, then substituting this into equation 12 for PP gives

    P=Aysy+2y1+s2P=\frac{A}{y}-sy+2y\sqrt{1+s^{2}} (13)

    To minimize PP, evaluate dP/dydP/dy with AA and ss constant and set dP/dydP/dy equal to zero. Therefore,


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

    A=y2[21+s2s]A=y^{2}\left[2\sqrt{1+s^{2}}-s\right] (14)

    Substituting the expression for AA into Eq. 11, one gets



    b=2y[1+s2s]b=2y\left[\sqrt{1+s^{2}}-s\right] (15)

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



    P=4y1+s22sy=2y[21+s2s]P=4y\sqrt{1+s^{2}}-2sy=2y\left[2\sqrt{1+s^{2}}-s\right] (16)

    Thus, for a minumum wetted perimeter PP,

    R=AP=y2R=\frac{A}{P}=\frac{y}{2} (17)

    Equation 17 indicates that for a trapezoidal section channel and for any side slope ss (or θ\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

    Q\displaystyle Q =1nAR2/3So1/2\displaystyle=\frac{1}{n}AR^{2/3}S_{o}^{1/2}
    Q\displaystyle Q =1n(by+sy2)(y2)2/3So1/2\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

    Q\displaystyle Q =ACR1/2So1/2\displaystyle=ACR^{1/2}S_{o}^{1/2}
    Q\displaystyle Q =C(by+sy2)(y2)1/2So1/2\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, BB and yy. But we know a relationship between bb and yy from equation 15. And using s=3/4=0.75s=3/4=0.75

    b\displaystyle b =2y[1+0.7520.75]\displaystyle=2y\left[\sqrt{1+0.75^{2}}-0.75\right]
    b\displaystyle b =y\displaystyle=y

    So the Chezy equation for this channel section becomes


    This can be solved iteratively with a target of Q=0.5m3/sQ=0.5m^{3}/s and So=1/2000=0.0005S_{o}=1/2000=0.0005. The table of figure 13 shows this iteration using the secant method and that the solution is yn=0.556my_{n}=0.556m and b=y=0.556b=y=0.556

    Refer to caption
    Figure 13: Iteration table calculating yny_{n} for efficient trapezoidal channel s=0.75s=0.75, So=0.0005S_{o}=0.0005 and target Q=0.05m3/sQ=0.05m^{3}/s
  15. E1.15

    An open channel with n=0.011n=0.011 is to be designed to carry 1.0m3/s1.0m^{3}/s of water at a slope of 0.0065. Find the most efficient cross-section for a rectangular section.
    (Answer: b=2y=0.869mb=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, bb, equation 15


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




    and with the usual form of the Manning’s equation


    then an iteration can be performed to solve for yny_{n} and thus bb.

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

    Refer to caption
    Figure 14: Iteration table calculating yny_{n} for efficient rectangular channel s=0s=0, So=0.0005S_{o}=0.0005 and target Q=1.0m3/sQ=1.0m^{3}/s
  16. E1.16

    A rectangular channel has width B=3mB=3m and normal depth y=1my=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 nn and SoS_{o} are the same in the two cases. Compare the two wetted perimeters.
    (Answer: D=2.057mD=2.057m, Prectangular=5.0mP_{rectangular}=5.0m, Pcircular=6.463mP_{circular}=6.463m

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


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

    A=πD24A=\frac{\pi D^{2}}{4}
    P=πDP=\pi D


    R=AP=πD24πD=D4R=\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

    Q\displaystyle Q =1nπD24(D4)2/3So1/2\displaystyle=\frac{1}{n}\frac{\pi D^{2}}{4}\left(\frac{D}{4}\right)^{2/3}S_{o% }^{1/2}
    Q\displaystyle Q =1nπD8/345/3So1/2\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:

    1nby(byb+2y)2/3So1/2=1nπD8/345/3So1/2\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 nn and SoS_{o} are the same for both channels these terms can be cancelled to give

    by(byb+2y)2/3=πD8/345/3by\left(\frac{by}{b+2y}\right)^{2/3}=\frac{\pi D^{8/3}}{4^{5/3}}

    And rearranged to give DD:


    Which results in D=2.057mD=2.057m.

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

  17. E1.17

    What are the dimensions of the most efficient rectangular channel section to carry 5m3/s5m^{3}/s at a slope of 1 in 900. The surface of the channel is of concrete.
    (Answer: y=1.21my=1.21m, b=2y=2.42mb=2y=2.42m using n=0.012n=0.012)

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




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


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


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

    Refer to caption
    Figure 15: Iteration table calculating yny_{n} for an efficient, concrete-lined, rectangular channel
  18. E1.18

    What is the most efficient depth for a brick channel of a trapezoidal section with sides sloping at 4545^{\circ} to the horizontal to carry 3m3/s3m^{3}/s. The bed slope is 0.0009.
    (Answer: y=1.104my=1.104m, using n=0.015n=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=1.104my=1.104m using n=0.015n=0.015 for the brick lining.

    Refer to caption
    Figure 16: Iteration table calculating yny_{n} for an efficient trapezoidal, brick-lined channel