Water at 20^◦ C flows at a depth of 3.7 m in a rectangular channel of width 4 m at a mild slope. A weir is inserted in the channel as illustrated in Figure EP 5.17 in order to measure the flowrate. The depth of flow downstream of the weir is 1.5 m. The weir discharge coefficient, Cd is 0.66. (a)

Question

Water at [latex] 20^{\circ} C [/latex] flows at a depth of 3.7 m in a rectangular channel of width 4 m at a mild slope. A weir is inserted in the channel as illustrated in Figure EP 5.17 in order to measure the flowrate. The depth of flow downstream of the weir is 1.5 m. The weir discharge coefficient, [latex] C_{d}[/latex] is 0.66. (a) Determine the ideal flow rate. (b) Determine the actual flowrate. (c) Determine the reaction force acting on the weir due to the thrust force of flow (anchoring force required to keep the weir from moving).

Accepted Answer

(a) In order to determine the ideal velocity at either point 1 or point 2, the Bernoulli equation is applied between points 1 and 2. However, in order to determine the relationship between the ideal velocities at points 1 and 2, the continuity equation is applied between points 1 and 2; thus, we have two equations and two unknowns. Furthermore, in order to determine the ideal flowrate in the pipe, the continuity equation is applied at either points 1 or 2.

[latex]Y_{1}: = 3.7 m[/latex] [latex]Y_{2}: = 1.5 m[/latex] [latex] b: = 4m [/latex] [latex]g: = 9.81 \frac{m}{sec^{2}} [/latex]

[latex]A_{1}: = b.Y_{1} = 14.8 m^{2}[/latex] [latex]A_{2}: = b.Y_{2} = 6 m^{2}[/latex] [latex]Z_{1}: = 0 m[/latex] [latex]Z_{2}: = 0 m[/latex]

Guess value: [latex]V_{1}: = 1 \frac{m}{sec} [/latex] [latex]V_{2}: = 1 \frac{m}{sec} [/latex] [latex]Q: = 1 \frac{m^{3}}{sec} [/latex]

Given

[latex]Y_{1} + Z_{1} + \frac{V^{2}_{1}}{2. g} = Y_{2} + Z_{2} + \frac{V^{2}_{2}}{2. g}[/latex]
[latex] V_{1}.b .Y_{1} = V_{2} .b .Y{2} [/latex]
[latex] Q = V_{1} .A_{1} [/latex]
[latex] \left ( \begin{matrix} V_{1} \ V_{2} \ Q \end{matrix} ight ) : = Find (V_{1}, V_{2}, Q ) [/latex]

[latex]V_{1}: = 2.914 \frac{m}{s} [/latex] [latex]V_{2}: = 7.187 \frac{m}{s} [/latex] [latex]Q: = 43.122 \frac{m^{3}}{s} [/latex]

(b) The actual flowrate, Q a is computed from Equation 5.51b [latex] Q_{a} = \frac{\sqrt{2g\left(\Delta y - \frac{ au _{w}L}{\gamma R_{h}} ight) } }{\sqrt{2g\Delta y} } \frac{A_{a}}{A_{i}}Q_{i} = C_{d} Q_{i} [/latex] as follows:

[latex] Q_{i}: = Q = 43.122 \frac{m^{3}}{s} [/latex] [latex] C_{d}: = 0.66 [/latex] [latex] Q_{a}: = C_{d}. Q_{i} = 28.461 \frac{m^{3}}{s} [/latex]

(c) In order to determine the reaction force acting on the weir due to the thrust force of flow, the integral form of the momentum equation is applied in the x direction for the control volume illustrated in Figure EP 5.17. The flow enters at control face 1 and leaves at control face 2. The viscous forces have already been accounted for indirectly through the discharge coefficient,[latex] C_{d}[/latex] in the continuity equation, and there isno gravitational force component along thex-axis because the channel bottom slope is mild and is assumed to be horizontal [latex] (z{1} = z_{2} = 0)[/latex] .

[latex] \sum{F_{x}} = \underbrace{(p_{1}A_{1})}_{F_{H1}} + \underbrace{(p_{2}A_{2})}_{F_{H2}} + R_{x} = ho Q(v_{2x}-v_{1x}) [/latex]

[latex] ho : = 998 \frac{kg}{m^{3}} [/latex] [latex] \gamma : = ho .g = 9.79 imes 10^{3} \frac{kg}{m^{2}s^{2}} [/latex]

[latex] h_{ca1}: = \frac{Y_{1}}{2} = 1.85 m [/latex] [latex] h_{ca2}: = \frac{Y_{2}}{2} = 0.75 m [/latex]

[latex] P_{1}: = \gamma . h_{ca1} = 1.811 imes 10^{4} \frac{N}{m^{2}} [/latex] [latex] P_{2}: = \gamma . h_{ca2} = 7.434 imes 10^{3} \frac{N}{m^{2}} [/latex]

[latex] P_{1x}: = P_{1} = 1.811 imes 10^{4} \frac{N}{m^{2}} [/latex] [latex] P_{2x}: = - P_{2} = - 7.343 imes 10^{3} \frac{N}{m^{2}} [/latex]

[latex] F_{H1}: = P_{1x}.A_{1} = 2.681 imes 10^{5} N [/latex] [latex] F_{H2}: = P_{2x}.A_{2} = -4.406 imes 10^{4} N [/latex]

[latex] V_{1x}: = V_{1} = 2.914 \frac{m}{s} [/latex] [latex] V_{2x}: = V_{2} = 7.187 \frac{m}{s} [/latex]

Guess value: [latex] R_{x}: = 1000 N [/latex]

Given

[latex] F_{H1}+ F_{H2} - R_{x} = ho .Q(V_{2x}- V_{1x} ) [/latex]
[latex] R_{x}: = Find (R_{x}) = 4.01 imes 10^{4} N [/latex]

Thus, the reaction force acting on the weir,[latex] R_{x} [/latex] acts to the left as assumed and is illustrated in Figure EP 5.17.

Question 5.17: Water at 20^◦ C flows at a depth of 3.7 m in a rectangular c...

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