Question 10.7: Water flow in a wide channel approaches a 10-cm-high bump at...

Part (a)

First check the approach Froude number, assuming C_0 = \sqrt{gy}:

Fr_1 = \frac{V_1}{\sqrt{gy_1}} = \frac{1.5  m/s}{\sqrt{(9.81  m/s^2)(1.0  m)}} = 0.479 (subcritical)

For subcritical approach flow, if Δh is not too large, we expect a depression in the water level over the bump and a higher subcritical Froude number at the crest. With Δh = 0.1 m, the specific energy levels must be

This physical situation is shown on a specific energy plot in Fig. E10.7. With y_1 in meters, Eq. (10.39) takes on the numerical values

y^3_2 – E_2y^2_2 + \frac{V^2_1 y^2_1}{2g} = 0                             where  E_2 = \frac{V^2_1}{2g} + y_1 – \Delta h                                          (10.39)

There are three real roots: y_2 = +0.859 m, +0.451 m, and -0.296 m. The third (negative) solution is physically impossible. The second (smaller) solution is the supercritical condition for E_2 and is not possible for this subcritical bump. The first solution is correct:

y_2(subcritical) ≈ 0.859 m

The surface level has dropped by y_1 – y_2 – \Delta h = 1.0 – 0.859 – 0.1 = 0.041  m. The crest velocity is V_2 = V_1y_1/y_2 = 1.745  m/s. The Froude number at the crest is Fr_2 = 0.601. Flow downstream of the bump is subcritical. These flow conditions are shown in Fig. E10.7.

Part (b)

For critical flow in a wide channel, with q = Vy = 1.5 m^2/s, from Eq. (10.31),

E_{min} = E(y_c) = \frac{3}{2}y_c                               (10.31)

Therefore the maximum height for frictionless flow over this particular bump is

For this bump, the solution of Eq. (10.39) is y_2 = y_c = 0.612  m, and the Froude number is unity at the crest. At critical flow the surface level has dropped by y_1 – y_2 – \Delta h = 0.191  m.