A 16 ft wide irrigation channel is equipped with a sharp-edged, full-width rectangular weir as shown in Figure 15.49. The weir height is 3 ft, and at the maximum design ﬂow the upstream water depth is 4.5 ft. What is the discharge under these conditions? What is the discharge of a 90° V-notch weir

Question

A 16 ft wide irrigation channel is equipped with a sharp-edged, full-width rectangular weir as shown in Figure 15.49. The weir height is 3 ft, and at the maximum design ﬂow the upstream water depth is 4.5 ft. What is the discharge under these conditions? What is the discharge of a 90° V-notch weir with the same weir head?

Accepted Answer

We can calculate the ﬂowrate of a full-width rectangular weir by using Eqs. 15.62a and 15.62b, noting that the weir head y₁ − y_W = 4.5 ft − 3.0 ft = 1.5 ft exceeds the minimum required value of 0.2 ft. To calculate the discharge coefficient we use Eq. 15.62b: C_W = 0.59 + 0.08[(y₁ − y_W)/y_w]. Inserting the data, we have C_W = 0.59 + 0.08[(4.5 − 3)/3] = 0.63. The ﬂowrate is then found by using Eq. 15.62a:

[latex]Q=C_Ww\left(\frac{L}{w} ight)\sqrt{g}(y_1-y_W)^{3/2}[/latex]

[latex]=0.63(16\ \mathrm{ft})\left(\frac{16\ \mathrm{ft}}{16\ \mathrm{ft}} ight)\sqrt{32.2\ \mathrm{ft}/s^2}(4.5\ \mathrm{ft}-3\ \mathrm{ft})^{3/2}=105\ \mathrm{ft}^3/s[/latex]

The 90° V-notch weir is described by Eqs. 15.63. Combining these equations and noting that θ = 90°, we have

[latex]Q=C_W( an \frac{ heta }{2})\sqrt{g}(y_1-y_W)^{5/2}[/latex] (15.63a)

[latex]C_W=0.44[/latex] (15.63b)

Q = 0.44[latex]\sqrt{g} [/latex](y₁ − y_W)^5/2 = 0.44[latex]\sqrt{32.2\ \mathrm{ft}/s^2}[/latex](4.5 ft − 3 ft )^5/2 = 6.9 ft³/s

Question 15.18: A 16 ft wide irrigation channel is equipped with a sharp-edg...

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