Question

Water flows steadily along a horizontal open channel of uniform width, over a broad-crested weir. The channel bed upstream from the weir (to the left) is d -0.75m above the channel bed downstream from the weir (to the right), as shown in the figure below. The volume flow rate per unit width is Q 6m2 s-1, and the downstream depth of the water is ћ2-d. Take the magnitude of the acceleration due to gravity as g 10ms-2. d - 0.75 m (a) Find the speed u2 of the flow downstream, and show that the flow is supercritical there. (b) By applying Bernoulli's equation along a suitable streamline, show that the speed u1 in the upstream section of the channel satisfies the equation u 64u1 120 0. Verify that ul = 2 ms-1 s a solution of this equation. Hence, by finding also the other positive solution, specify the speed and depth of each of the two possible flows upstream, identifying one as subcritical and the other as supercritical (c) (d) If the flow is subcritical upstream and supercritical downstream, what is the type of flow at the crest of the weir? Find the depth of the water flowing over the top of the weir. Hence, by applying Bernoulli's equation once more, calculate the height D of the crest of the channel bed (above the level of the downstream section). Quote all results to two decimal places

Answer 1

ven d = 0.75 m h_2 = d = 0.75 m Q = 6m^2/s a) The downstream u_2 = Q/d = 6m^2/s/0.75 m u_2 = 8.00 m/s Since f_2 = u_2/squareroot gh_1 = 8 m/s/squareroot 9.8 m/s^2 times 0.75 m = 2.95 f_2 > 1 So the flow is supercritical c) u^3_1 - 64u_1 + 120 = 0 (u_1 - 2) (u^2_1 + 2u_1 - 60) = 0 Solving u_1 - 2 = 0 u_1 = 2 m/s u^2_1 + 2u_1 - 60 = 0 = > u_1 = -2 plusminus squareroot 2^2 - 2 times 1 times (-60)/2 times 1 u_1 = squareroot 61 - 1 u_1 = 6.81 m/s Depth h_1 = Q/u_1 = 6 m^2/s/2 m/s = 3m f_1 = u_1/squareroot gh_1 = 2 m/s/squareroot 9.8 m/s^2 times 3m = 0.369 f_1 < 1 = > flow is subcritical h'_1 = Q/u_1 = 6 m^2/s/6.81 m/s = 0.881 m f'_1 = u_1/squareroot gh_1' = 6.81 m/s/squareroot 9.8 m/s^2