Water flowing in a rectangular channel of width 3 m at a supercritical velocity of 4.823 m/sec and a supercritical depth of 0.933 m encounters a subcritical flow, as illustrated in Figure EP 5.22. As the flow transitions from supercritical to subcritical flow, a hydraulic jump occurs, where there

Question

Water flowing in a rectangular channel of width 3 m at a supercritical velocity of 4.823 m/sec and a supercritical depth of 0.933 m encounters a subcritical flow, as illustrated in Figure EP 5.22. As the flow transitions from supercritical to subcritical flow, a hydraulic jump occurs, where there is a major head loss due the intense agitation and turbulence, where the depth of flow is at critical depth of 1.273m. (a) Determine the subcritical depth of flow and the subcritical velocity of flow after the jump. (b) Determine the head loss due to the flow turbulence at the jump.

Accepted Answer

(a) In order to determine the subcritical depth of flow after the jump, the hydraulic jump equation (see Equation 9.248 [latex]y_{2} = \frac{y_{1}}{2} \left( - 1 + \sqrt{1 + 8 F_{1}^{2}} \right) [/latex]), which is derived by applying the integral form of the momentum equation, is applied. Given that the upstream conditions (depth and velocity) are known, the hydraulic jump equation (Equation 9.248), [latex]y_{2} = \frac{y_{1}}{2} \left( - 1 + \sqrt{1 + 8 F_{1}^{2}} \right) [/latex] requires the definition of the Froude number for the supercritical depth, y 1 upstream of the jump, [latex] F_{1} = \frac{v_{1}}{\sqrt{gy_{1}} } [/latex] Furthermore, the continuity equation is applied to determine the subcritical velocity of flow.

[latex] V_{1}: = 4.823 \frac{m}{sec} [/latex] [latex] Y_{1}: = 0.933 m[/latex] [latex] b: = 3 m[/latex] [latex] g: = 9.81 \frac{m}{sec^{2}} [/latex]

[latex] F_{1}: = \frac{V_{1}}{\sqrt{g. Y_{1}} } = 1.594[/latex] [latex] Y_{2}: = \frac{Y_{1}}{2} ( - 1 + \sqrt{ 1 + 8 . F^{2}_{1}} ) = 1.688 m[/latex]

[latex] Q: = V_{1}. b. Y_{1} = 13.5 \frac{m^{3}}{s} [/latex] [latex] V_{2}: = \frac{Q}{b. Y_{2}} = 2.666 \frac{m}{s} [/latex]

(b) In order to determine the major head loss due to the jump, the energy equation is applied between the upstream point 1 and the downstream point 2 as follows:

[latex] Z_{1} : = 0 m[/latex] [latex] Z_{2} : = 0 m[/latex]

Guess value: [latex] h_{f}: = 1 m[/latex]

Given

[latex] Y_{1} + Z_{1} + \frac{ V^{2}_{1}}{2.g} - h_{f} = Y_{2} + Z_{2} + \frac{ V^{2}_{2}}{2.g} [/latex]
[latex] h_{f}: = Find (h_{f}) = 0.068 m [/latex]

Question 5.22: Water flowing in a rectangular channel of width 3 m at a sup...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions