Question 15.4: Produce a preliminary design for a sea outfall. The dry weat......
Produce a preliminary design for a sea outfall. The dry weather flow through the outfall is 0.1 m^3 \, s^{-1} and the peak flow is 0.5 m^3 \, s^{-1} . Site investigations show that a sea outfall extending to 2 km seaward would not cause pollution of foreshore and recreational waters. The still water depth there is 15 m. The density of sea water is 1026 kg m^{-3} .
If a minimum dilution of 50 is required, estimate the diameter of the ports, number of ports and the length of the diffuser for still water.
Cederwall equations (15.35) and (15.36) are used to obtain the following table (choose FD and compute \left.y_0 / D_{ p } \text { for } S_{ m }=50\right) .
Table 1
S_{\mathrm{m}}=0.54\,F D\left(\frac{y_{0}}{D_{\mathrm{p}}F D}\right)^{7/16}\;\;\mathrm{for}\;\;\;\frac{y_{0}}{D_{\mathrm{p}}}\lt 0.5\,F D (15.35)
S_{\mathrm{m}}=0.54\,F D\left(\frac{0.38y_{0}}{D_{\mathrm{p}}F D}+0.66\right)^{5/3}\ \ \mathrm{for}\ \ \frac{y_{0}}{D_{\mathrm{p}}}≥0.5\,F D. (15.36)
The diameter of the ports is chosen as 175 mm which is large enough to avoid blockage. The exit velocity is about 2.4 m \, s^{-1} which is less than the highest acceptable velocity of 3 m \, s^{-1} to avoid undue headloss at the exit.
The discharge through the port of 175 mm diameter = 0.06 m^3 \, s^{-1}. Number of ports is 0.5/0.06 = 10 (say). The flow through the ports at DWF = 0.1/10 = 0.01 m^3 \, s^{-1}. The velocity through the ports at DWF
Discharge through the ports at peak flow = 0.5/10 = 0.05 m^3 \, s^{-1}. Velocity through the ports at peak flow
\begin{aligned} & =0.05 /\left(\pi \,0.175^2 / 4\right) \\ & =2.1 \,m s ^{-1} \end{aligned}A velocity of 2.1 m \, s^{-1} is expected to be sufficient to avoid blockage of the ports. Densimetric Froude number
F D=2.1 /(9.81 \times 0.026 \times 0.175)^{0.5}= 9.9,
y_0 / D_{ p }=15 / 0.175=85.7From Fig. 15.11, for y_0 / D_{ p }=85.7 \text { and } F D=9.9 ,
w / 2 D_{ p }=15 \text {, }w = 2 × 15 × 0.17 \simeq6\,{\mathrm{m}}.
The diffuser length = 6 × 10 = 60 m.
Table 1
| FD | 1 | 2 | 5 | 10 | 13 | 15 | 20 | 50 |
| y_0 / D_\mathrm{p} | 38 | 49 | 67 | 82 | 88 | 91 | 96 | 100 |
| D_{ \mathrm{p} }( m ) | 0.39 | 0.31 | 0.22 | 0.18 | 0.17 | 0.165 | 0.156 | 0.15 |
| V_{ \mathrm{p} }\left( m s ^{-1}\right) | 0.32 | 0.56 | 1.2 | 2.16 | 2.7 | 3.1 | 4.0 | 9.8 |
| Q_{ \mathrm{p} }\left( m ^3 s ^{-1}\right) | 0.038 | 0.041 | 0.047 | 0.057 | 0.062 | 0.066 | 0.076 | 0.17 |
