## Chapter 15

## Q. 15.17

A sluice gate in a 24 ft wide rectangular channel is observed to be operating with an upstream depth y_{1} = 6 ft and a downstream depth y_{2} = 1 ft (Figure 15.46). Find the volume ﬂowrate passing through the channel and all other geometric and ﬂow parameters.

## 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 = V_{1}y_{1 }= V_{2}y_{2}, to write V_{1} = Q/wy_{1} and V_{2} = Q/wy_{2}. Inserting the data, we ﬁnd V_{1} = Q/wy_{1} = 436.8 ft^{3}/s/24 ft(6 ft ) = 3.0 ft/s and V_{2} = Q/wy_{2 }= 436 .8 ft^{3}/s/24 ft(1 ft ) = 18.2 ft/s. The corresponding Froude numbers are found to be Fr_{1} = V_{1}/\sqrt{gy_1} = 3.0 ft/s/\sqrt{(32.2\ \mathrm{ft} /s^2)(6\ \mathrm{ft} )} = 0.22 and Fr_{2} = 18.2 ft/s/\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )} = 3.21. The reservoir depth y_{0} is found by using Eq. 15.59a to write y_{0} = V^2_2 /2g + y_{2} = (18.2 ft/s)^{2}/2(32.2 ft/s^{2}) + 1 ft = 6.14 ft. Note that we could also have used Eq. 15.59b to write y_{0} = V^2_1 /2g + y_{1} = (3.0 ft/s)^{2}/2(32.2 ft/s^{2}) + 6 ft = 6.14 ft. To ﬁnd the gate opening we rearrange Eq. 15.61a to obtain y_{G} = Q /C_{G}w\sqrt{2gy_1}

Q = C_{G}w yw_{G }\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 C_{G} = 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 y_{G} = 1.13 ft and C_{G} = 0.58.