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Chapter 15

Q. 15.17

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.

15.46

Step-by-Step

Verified Solution

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

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

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/\sqrt{gy_1} = 3.0 ft/s/\sqrt{(32.2\ \mathrm{ft} /s^2)(6\ \mathrm{ft} )} = 0.22 and Fr2 = 18.2 ft/s/\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )} = 3.21. The reservoir depth y0 is found by using Eq. 15.59a to write y0 = V^2_2 /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 = V^2_1 /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 yG = Q /CGw\sqrt{2gy_1}

Q = CGw yw\sqrt{2gy_1}        (15.61a)

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

Assuming CG = 0.61, a first iteration yields

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}

However the 1.52 ft corresponds to a discharge coefficient of

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

so further iteration is required. The final results are yG = 1.13 ft and CG = 0.58.