A sluice gate in a 24 ft wide rectangular channel is observed to be operating with an upstream depth y1 = 6 ft and a downstream depth y2 =1 ft (Figure 15.46). Find the volume ﬂowrate passing through the channel and all other geometric and ﬂow parameters.

Question

A sluice gate in a 24 ft wide rectangular channel is observed to be operating with an upstream depth y₁ = 6 ft and a downstream depth y₂ = 1 ft (Figure 15.46). Find the volume ﬂowrate passing through the channel and all other geometric and ﬂow parameters.

Accepted Answer

Applying Eq. 15.58e, [latex]Q =[2gw^2y^2 _1y^2 _2/(y_1 +y_2)]^{1/2} [/latex], the ﬂowrate is found to be

[latex]Q =\left[\frac{2(32.2\ \mathrm{ft} /s^2)(24\ \mathrm{ft} )^2(6\ \mathrm{ft} )^2(1\ \mathrm{ft} )^2 }{6\ \mathrm{ft} +1\ \mathrm{ft} } \right]^{1/2}=436.8\ \mathrm{ft} ^3/s[/latex]

The velocities upstream and downstream can now be determined by using Eq. 15.58a, Q/w = V₁y₁= V₂y₂, to write V₁ = Q/wy₁ and V₂ = Q/wy₂. Inserting the data, we ﬁnd V₁ = Q/wy₁ = 436.8 ft³/s/24 ft(6 ft ) = 3.0 ft/s and V₂ = Q/wy₂= 436 .8 ft³/s/24 ft(1 ft ) = 18.2 ft/s. The corresponding Froude numbers are found to be Fr₁ = V₁/[latex]\sqrt{gy_1}[/latex] = 3.0 ft/s/[latex]\sqrt{(32.2\ \mathrm{ft} /s^2)(6\ \mathrm{ft} )}[/latex] = 0.22 and Fr₂ = 18.2 ft/s/[latex]\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )}[/latex] = 3.21. The reservoir depth y₀ is found by using Eq. 15.59a to write y₀ = [latex]V^2_2[/latex] /2g + y₂ = (18.2 ft/s)²/2(32.2 ft/s²) + 1 ft = 6.14 ft. Note that we could also have used Eq. 15.59b to write y₀ = [latex]V^2_1[/latex] /2g + y₁ = (3.0 ft/s)²/2(32.2 ft/s²) + 6 ft = 6.14 ft. To ﬁnd the gate opening we rearrange Eq. 15.61a to obtain y_G = Q /C_Gw[latex]\sqrt{2gy_1}[/latex]

Q = C_Gw yw_G[latex]\sqrt{2gy_1}[/latex] (15.61a)

and note that the discharge coefficient is given by Eq. 15.61b:
[latex]C_G=\frac{0.61 }{\sqrt{1+0.61(y_G/y_1)} } [/latex]

Assuming C_G = 0.61, a ﬁrst iteration yields

[latex]y_G=\frac{Q}{C_Gw\sqrt{2gy_1}}=\frac{436.8\ \mathrm{ft} ^3/s }{0.61(24\ \mathrm{ft} )\sqrt{2(32.2\ \mathrm{ft} /s^2)(6\ \mathrm{ft} )}}=1.52\ \mathrm{ft} [/latex]

However the 1.52 ft corresponds to a discharge coefficient of

[latex]C_G=\frac{0.61 }{\sqrt{1+0.61(y_G/y_1)} }=\frac{0.61 }{\sqrt{1+0.61(1.52\ \mathrm{ft} /6\ \mathrm{ft} )}}=0.57[/latex]

so further iteration is required. The ﬁnal results are y_G = 1.13 ft and C_G = 0.58.

Question 15.17: A sluice gate in a 24 ft wide rectangular channel is observe...

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