Chapter 10, Problem 10.4CP

To determine

The feasibility of installing such a weir which will be accurate and yet not cause the water to overflow to the side of the channel.

Answer to Problem 10.4CP

The first two of these are plausible although H/Y is large than 2 . The third result is not recommended because Y is too far below 9 cm. We can say that reasonable designs are possible.

Explanation of Solution

Figure:

Given:

For asphalt take $n=0.016$, $b=1.5m$, $g=9.81m/s^2$

$S_o=0.0036$, Depth $y=0.7m$

Concept Used:

Annuity problem requires the use of the formula as follows:

$Q=\frac{1}{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}$

Here

Area = $A$, Slope = $S_o$, Hydraulic radius $=R_h$

Annuity problem requires the use of the formula as follows:

$R_h=\frac{by}{b+2y}$

Here,

Hydraulic radius $=R_h$, Depth $=y$, width $=b$

Annuity problem requires the use of the formula as follows:

$A=by$

Here,

Area= $A$, Depth $=y$, width $=b$

Annuity problem requires the use of the Weir formula as follows:

$Q_{weir}=0.581\left(b-0.1H\right)g^{1/2}{H}^{3/2}$

Here, width $=b$, acceleration gravity $=g$

Calculation:

As per the given problem

$b=1.5m$, $y=0.7m$

Annuity problem requires the use of this formula:

$A=by$

Substitute these values in the formula

$A=1.5\times 0.7=1.05m^2$

$A=1.05m^2$

Annuity problem requires the use of this formula:

$b=1.5m$, $y=0.7m$

$R_h=\frac{by}{b+2y}$

Substitute these values in the formula:

$R_h=\frac{1.5\times 0.7}{1.5+2\times 0.7}=0.362m$

$R_h=0.362m$

Annuity problem requires the use of this formula

$R_h=0.362m$, $A=1.05m^2$, $S_o=0.0036$, $n=0.016$

$Q=\frac{1}{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}$

Substitute these values in the formula

$Q=\frac{1}{0.016}\left(1.05\right){\left(0.362\right)}^{2/3}{\left(0.0036\right)}^{1/2}=2.0m^3/s$

$Q=2.0m^3/s$

The weir discharge must equal this flow rate. Let us being by making b equal to the full available width of 1.5 m and holding Y to the minimum height of 9 cm.

Annuity problem requires the use of the Weir formula as follows

$b=1.5m$, $g=9.81m/s^2$, $Q=2.0m^3/s$

$Q_{weir}=0.581\left(b-0.1H\right)g^{1/2}{H}^{3/2}$

Substitute these values in the formula

$2.0m^3/s=0.581\left(1.5m-0.1H\right){\left(9.81m/s\right)}^{1/2}{H}^{3/2}$

$\Rightarrow$

$H=0.845m$

$H=0.845m$

$\to$ H is independent of Y ,

To find out the ratio H/Y

$\frac{H}{Y}=\frac{0.845}{0.09}=9.40$

$\to$ Which far exceeds Sturm’s recommendation H/Y $\le$ 2.0. If we rise Y to H/2=0.423 m,

The total upstream water depth

$H+Y=1.27m$ which overflow the channel walls?

If we back down to $Y=0.35m$,

The upstream depth

$Y=0.35m$, $H=0.845m$

$H+Y=?$

$0.845+0.35=1.195m$

So, a wide weir design is possible with $\frac{H}{Y}=2.4$ not bad

Similarly,

To let shorter values of b

$*$ Upstream depth will exceed 1.2 m

$*$ The ratio $\frac{H}{Y}$ will exceed 2.0

Annuity problem requires the use of this formula:

$b=1.4m$, $y=0.7m$

$A=by$

Substitute these values in the formula:

$A=1.4\times 0.7=0.98m^2$

$A=0.98m^2$

Annuity problem requires the use of this formula:

$b=1.4m$, $y=0.7m$

$R_h=\frac{by}{b+2y}$

Substitute these values in the formula:

$R_h=\frac{1.4\times 0.7}{1.4+2\times 0.7}=0.35m$

$R_h=0.35m$

Annuity problem requires the use of this formula:

$R_h=0.35m$, $A=0.98m^2$, $S_o=0.0036$, $n=0.016$

$Q=\frac{1}{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}$

Substitute these values in the formula:

$Q=\frac{1}{0.016}\left(0.98\right){\left(0.35\right)}^{2/3}{\left(0.0036\right)}^{1/2}=1.837m^3/s$

$Q=1.837m^3/s$

Annuity problem requires the use of the Weir formula as follows:

$b=1.4m$, $g=9.81m/s^2$, $Q=1.837m^3/s$

$Q_{weir}=0.581\left(b-0.1H\right)g^{1/2}{H}^{3/2}$

Substitute these values in the formula

$1.8375m^3/s=0.581\left(1.4m-0.1H\right){\left(9.81m/s\right)}^{1/2}{H}^{3/2}$

$\Rightarrow$

$H=0.89m$

$H=0.89m$

$H+Y\approx 1.2m$ if $Y=0.31m$

To find out the ratio H/Y

$\frac{H}{Y}=\frac{0.89}{0.31}=2.9$

Annuity problem requires the use of this formula:

$b=1.25m$, $y=0.7m$

$A=by$

Substitute these values in the formula:

$A=1.25\times 0.7=0.875m^2$

$A=0.875m^2$

Annuity problem requires the use of this formula:

$b=1.25m$, $y=0.7m$

$R_h=\frac{by}{b+2y}$

Substitute these values in the formula:

$R_h=\frac{1.25\times 0.7}{1.25+2\times 0.7}=0.33m$

$R_h=0.33m$

Annuity problem requires the use of this formula:

$R_h=0.33m$, $A=0.875m^2$, $S_o=0.0036$, $n=0.016$

$Q=\frac{1}{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}$

Substitute these values in the formula:

$Q=\frac{1}{0.016}\left(0.875\right){\left(0.33\right)}^{2/3}{\left(0.0036\right)}^{1/2}=1.575m^3/s$

$Q=1.575m^3/s$

Annuity problem requires the use of the Weir formula as follows:

$b=1.25m$, $g=9.81m/s^2$, $Q=1.575m^3/s$

$Q_{weir}=0.581\left(b-0.1H\right)g^{1/2}{H}^{3/2}$

Substitute these values in the formula

$1.575m^3/s=0.581\left(1.4m-0.1H\right){\left(9.81m/s\right)}^{1/2}{H}^{3/2}$

$\Rightarrow$

$H=0.97m$

$H=0.97m$

$H+Y\approx 1.2m$ if $Y=0.23m$

To find out the ratio H/Y:

$\frac{H}{Y}=\frac{0.97}{0.23}=4.2$

Conclusion:

The first two of these are plausible although H/Y is large than 2.0.The third result is not recommended because Y is too far below 9 cm .We conclude that reasonable designs are possible.