A river, transporting sediment, flows into a tidal estuary. The maximum freshwater flow into the estuary is 4000 m³ s^-1. It is required to make a preliminary design of a scale model in a laboratory where the space available dictates the horizontal scale, Ml=250; the pumping capacity available for

Question

A river, transporting sediment, flows into a tidal estuary. The maximum freshwater flow into the estuary is 4000 ${m}^{3}\,{s}^{-1}$ . It is required to make a preliminary design of a scale model in a laboratory where the space available dictates the horizontal scale, $M_{l}=250;$ the pumping capacity available for the model is 271 ${s}^{-1}$ and it is desirable to use it reasonably fully to avoid viscous effects on the model. Establish

1. a suitable vertical scale for the model,
2. the discharge rate,
3. the tidal period scale,
4. the scale of the fall velocity of suspended sediment,
5. the probable scale of the bed material, and
6. the scale for the density of the bed load.

Accepted Answer

1. Utilizing fully the discharge capacity gives a discharge scale 4 × [latex] 10^6 / 27=148150 [/latex]. From equation (16.7),

[latex] M_Q=M_A M_v=M_h^{3 / 2} M_l [/latex] (16.7)

[latex] M_h=\left(M_Q / M_l ight)^{2 / 3}=(148\,150 / [/latex][latex] 250)^{2 / 3}=70.57 [/latex]. Therefore let us choose [latex] extstyle{M_{h}}[/latex] = 75, giving a distortion of 3.33, which is probably quite acceptable in this case.
2. [latex] M_Q=M_h^{3 / 2} M_l=75^{3 / 2} imes 250=162380 [/latex] . The maximum model discharge will be [latex] 4 imes 10^6 / 162380=24.61 \, s ^{-1} [/latex].
3. [latex] M_t=M_l / M_v=M_l M_h^{-1 / 2}=250 / 75^{1 / 2}=[/latex]28.87 (equation (16.3)).

[latex] \frac{M_l}{M_v M_t}=\frac{M_g M_l}{M_v^2}=\frac{M_p}{M_ ho M_v^2}=\frac{M_\mu}{M_ ho M_v M_l}=1 [/latex] (16.3a–d)

4. [latex] M_{w_s}=M_h^{3 / 2} / M_l=75^{3 / 2} / 250=2.6 [/latex](equation (16.13)).

[latex] M_w=M_h / M_t=M_h^{3 / 2} / M_l [/latex] (16.13)
5. From the Manning–Strickler equation for a wide channel with [latex]R\simeq y,[/latex]

[latex] M_v=M_n^{-1} M_h^{2 / 3} M_s^{1 / 2}=M_d^{-1 / 6} M_h^{2 / 3} M_h^{1 / 2} M_l^{-1 / 2}=M_d^{-1 / 6} M_h^{7 / 6} M_l^{-1 / 2}=M_h^{1 / 2}[/latex],

[latex] M_d=M_h^4 M_l^3=75^4 / 250^3=2.025 \simeq 2 . [/latex]

6. [latex] M_{\Delta}=\frac{M_h^2}{M_l M_d}=\frac{M_h^2 M_l^3}{M_h^4 M_l}=\frac{M_l^2}{M_h^2}=\left(\frac{250}{75} ight)^2=11.11 ext { (equation (16.12)). } [/latex]

[latex] M_{\Delta}=\frac{M_{U *}^2}{M_d}=\frac{M_R M_S}{M_d}=\frac{M_R M_h}{M_l M_d} . [/latex] (16.12)

Note that the answer to 5 assumes that there is no effect of bedforms (i.e. that the bed is flat) which may be unrealistic; this has to be checked by further computation, and the final value of [latex] extstyle{M_{d}}[/latex] may influence the whole model design.

Question 16.1: A river, transporting sediment, flows into a tidal estuary. ......