Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

Holooly Tables

All the data tables that you may search for.

Holooly Help Desk

Need Help? We got you covered.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Help Desk

Need Help? We got you covered.

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.

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.