Question 12.13: Water flows from a reservoir under a sluice gate and into th......
Water flows from a reservoir under a sluice gate and into the 1.5-m-wide unfinished concrete rectangular channel, which is horizontal, Fig. 12–28a. At the control point 0, the measured flow is 2 m³/s and the depth is 0.2 m.
Show how to determine the variation of the water depth along the channel, measured downstream from this point. Take n = 0.014.
Fluid Description. Assuming the reservoir maintains a constant level, then the flow will be steady. Just after point 0, it will be nonuniform because its depth is changing. As usual, we assume water to be an ideal fluid, and use a mean velocity profile.
Analysis. First we will classify the water surface profile. Since q = Q/b = (2 m³/s)/(1.5 m) = 1.333 m²/s, the critical depth is
y_c = (\frac{q^2}{g})^{1/3} = [\frac{(1.333 m^2/s)^2}{9.81 m/s^2}]^{1/3} = 0.5659 m
Here y = 0.2 m < y_c = 0.5659 m, so the flow is supercritical. Since the channel is horizontal, S_0 = 0. Using Table 12–2, the water surface profile is H3, as shown in Fig. 12–28a. This indicates that the water depth will increase as x increases from the control point. (Note that if the channel had a slope, then the Manning equation would have to be used to find y_n since this value is also needed for a profile classification.)
To determine the downstream depths, we will divide the calculations into increments of depth, choosing Δy = 0.01 m. At station 0, y = 0.2 m, and the velocity is
The hydraulic radius is therefore
R_{h0} = \frac{A_0}{P_0} = \frac{(1.5 m)(0.2 m)}{[1.5 m + 2(0.2 m)]} = 0.1579 mThese results are entered into the first line of the table shown in Fig. 12–28b.
At station 1, y_1 = 0.2 m + 0.01 m = 0.21 m, thus, V_1 = Q/A_1 = 6.349 m/s and R_{h1} = A_1/P_1 = 0.1641 m (third line of table). The intermediate second line reports the values of
V_m = \frac{V_0 + V_1}{2} = \frac{6.667 m/s + 6.349 m/s}{2} = 6.508 m/sR_{hm} = \frac{R_{h0} + R_{h1}}{2} = \frac{0.1579 m + 0.1641 m}{2} = 0.1610 m
S_{fm} = \frac{n^2V_m^2}{k^2R_{hm}^{4/3}} = \frac{(0.014)^2(6.508 m/s)^2}{(1)^2 (0.1610 m)^{4/3}} = 0.09479
Δ x = \frac{(y_1 – y_0) + (V_1^2 – V_0^2)}{S – S_{fm}}
= \frac{(0.21 m – 0.2 m) + [(6.349 m/s)^2 – (6.667 m/s^2)]/[2(9.81 m/s^2)]}{(0 – 0.09479)}
= 2.116 m
The calculations are repeated for the next two stations as shown in the table, and the results are plotted in Fig. 12–28c. They appear satisfactory, since each increase Δy = 0.01 m produces somewhat uniform changes Δx. Had larger changes occurred, then smaller increments of Δy should be selected to improve the accuracy of the finite-difference method.
TABLE 12–2 Surface Profile Classification
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