Question 5.5: Sketch possible water-surface profiles in a long channel wit......

The method presented in the previous section to determine the channel discharge from a reservoir is followed. First, the channel bottom slope corresponding to the critical flow (critical slope) is calculated. If the actual channelbottom slope is steeper than the critical slope, then the flow is supercritical. The flow is subcritical if the channel bottom slope is less than the critical slope. Since it is possible to have more than one critical slope for a compound channel, it is necessary to consider them in the analysis. The following discussion will illustrate this point.
\quadFigure 5-15 shows the discharge and Froude number versus depth for the given reservoir level, H = 1.2 m. The discharge, Q, is calculated from the following energy equation
\quad\quad\quad\quad Q=\left[\frac{2(H-y)gA^{2}}{\alpha}\right]^{\frac{1}{2}} (5-19)
in which y is the flow depth in the channel. The entrance losses and the velocity of approach are neglected in this equation.
\quadCritical conditions occur when F_{r_{c}}= 1. It is clear from the discharge versus depth diagram (Fig. 5-15) that the discharge is not necessarily a local maximum at critical conditions. For example, it is actually minimum for y_{c_{2}}.
As can be seen, there are three critical depths: y_{c_{1}}= 0.8 m, y_{c_{2}}= 1.03 m and y_{c_{3}}= 1.10 m. Corresponding to these critical depths, there are three critical discharges, Q_{c_{1}} = 2.241 m³ /s, Q_{c_{2}} = 1.960 m³ /s and Q_{c_{3}}= 2.0 m³ /s. Critical bottom slopes, S_{c}, corresponding to these critical discharges are determined from the Manning equation as S_{c_{1}} = 0.0064, S_{c_{2}}= 0.0024 and S_{c_{3}} = 0.0015, respectively.
\quadDepending upon the channel bottom slope, S_{o}, the following four cases are possible.

This bottom slope is steeper than all three critical slopes and the flow in the channel is supercritical. Flow depth varies rapidly from the reservoir level to the critical depth, y_{c_{3}}, passes through the other two critical depths, y_{c_{2}} and y_{c_{1}} and approaches the normal depth as shown in Fig. 5-16a. The discharge in the channel is equal to the discharge corresponding to y_{c_{1}}, i.e., Q = Q_{c_{1}} =2.24 m³/s. The discharge is the same for all values of the bottom slope, S_{o} > S_{c_{1}}.

For this bottom slope, y_{n} is greater than y_{c_{1}} but less than y_{c_{2}}. Therefore, as indicated by Fig. 5-15, the Froude number is less than 1 and the flow is subcritical. The discharge in the channel depends on the channel bottom slope and may be determined by solving the energy equation simultaneously with the Manning equation for uniform flow. For S_{o} = 0.0035, the discharge is 2.08 m³ /s; the corresponding depth is 0.96 m and the flow profile is shown in Fig. 5-16b.

For this bottom slope, the normal depth, y_{n} is greater than y_{c_{2}} but less than y_{c_{3}}. Therefore, as indicated by Fig. 5-15, the Froude number is greater than 1 and the flow is supercritical. Flow depth passes through the critical depth y_{c_{3}} and then approaches the normal depth. The channel discharge is equal to the discharge corresponding to y_{c_{3}}, i.e., 2.0 m³/s. The flow profile is shown in Fig. 5-16c.

In this case, the channel bottom slope is less than all critical slopes and the flow is subcritical. The channel discharge may be calculated as discussed previously for Case 2. The flow profile is shown in Fig. 5-16d.