Chapter 13, Problem 167P

Consider water flow in the range of 10 to 15 m³/s through a horizontal section of a 5-m-de rectangular channel. A rectangular or triangular thin-plate weir is to be installed to measure the flow rate. If the water depth is to remain under 2 m at all times, specify the type and dimensions of an appropriate was. What uid your response be if the flow range were 0 to15 m³/s?

To determine

The appropriate dimension for the rectangular weir for the varying flow rate.

The appropriate dimension for the triangular weir for the varying flow rate.

Answer to Problem 167P

The appropriate dimensions for the rectangular weir are $5\text{m}$ base and $2.65\text{m}$ height for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

The appropriate dimensions for the triangular weir are $5\text{m}$ base and $2.65\text{m}$ height for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given Information:

The water flow rate range is $10{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$, the width of the weir is $5\text{m}$, the water depth is $2\text{m}$ and the changed water flow rate range is $10{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

Write the expression for the weir head.

  $H=y_1-P_w$...... (I)

Here, the flow depth at upstream is $y_1$ and the height of the rectangular weir is $P_w$.

Write the expression for the discharge coefficient of the weir.

  $C_D=0.598+0.0897\frac{H}{P_w}$...... (II)

Write the expression for the water flow rate of the channel per meter width.

  $\dot{Q}=C_D\frac{2}{3}b\sqrt{2g}{H}^{3/2}$...... (III)

Here, the width of the channel is $b$ and the acceleration due to gravity is $g$.

Calculation:

Substitute $2\text{m}$ for $y_1$ in Equation (I).

  $\begin{array}{c}H=2\text{m}-p_w\\ p_w=2\text{m}-H\end{array}$

Substitute $2\text{m}-H$ for $P_w$ in Equation (II).

  $C_D=0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)$

Case I.

Substitute $10{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $5\text{m}$ for $b$, $0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)$ for $C_D$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (III).

  $\begin{array}{l}10{\text{m}}^{\text{3}}/\text{s}=\left(0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)\right)\frac{2}{3}\left(5\text{m}\right)\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{\left(H\right)}^{3/2}\\ 10{\text{m}}^{\text{3}}/\text{s}=\left(0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)\right)\left(3.33\text{m}\right)\left(4.4294\sqrt{\text{m}}/\text{s}\right){\left(H\right)}^{3/2}\\ 10{\text{m}}^{\text{3}}/\text{s}\left(2\text{m}-H\right)=\left[0.598\left(2\text{m}-H\right)+0.0897H\right]\left(3.33m\times 4.4294\sqrt{\text{m}}/\text{s}\right){\left(H\right)}^{3/2}\\ \left(2\text{m}-H\right)=\left[1.1978-0.5092H\right]\left(1.476\right){\left(H\right)}^{3/2}\end{array}$

  $\begin{array}{l}\left(2\text{m}-H\right)=\left[1.768-0.7516H\right]\left(H^{3/2}\right)\\ \left(2\text{m}-H\right)=-0.7516\left(\sqrt{H}-1.53373\right)\left(\sqrt{H}+1.53373\right)\left(H^{3/2}\right)\\ \left(2\text{m}-H\right)=\left(1.768H^{3/2}-0.7516H^{3/2}\right)\\ H=2.526\text{m}\end{array}$

The head of the weir plate for the different values of the volume flow rate is shown in below Table 1.

Table-1

S.No.	$\dot{Q}$${\text{m}}^{\text{3}}/\text{s}$	$H$$\text{m}$
1	$10{\text{m}}^{\text{3}}/\text{s}$	$2.526\text{m}$
2	$11{\text{m}}^{\text{3}}/\text{s}$	$2.549\text{m}$
3	$12{\text{m}}^{\text{3}}/\text{s}$	$2.574\text{m}$
4	$13{\text{m}}^{\text{3}}/\text{s}$	$2.598\text{m}$
5	$14{\text{m}}^{\text{3}}/\text{s}$	$2.625\text{m}$
6	$15{\text{m}}^{\text{3}}/\text{s}$	$2.655\text{m}$

Case II.

Substitute $0{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $5\text{m}$ for $b$, $0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)$ for $C_D$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (III).

  $\begin{array}{l}0{\text{m}}^{\text{3}}/\text{s}=\left(0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)\right)\frac{2}{3}\left(5\text{m}\right)\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{\left(H\right)}^{3/2}\\ 0{\text{m}}^{\text{3}}/\text{s}=\left(0.598+0.0897\left(\frac{H}{2\text{m}-H}\right)\right)\left(3.33\text{m}\right)\left(4.4294\sqrt{\text{m}}/\text{s}\right){\left(H\right)}^{3/2}\\ 0{\text{m}}^{\text{3}}/\text{s}\left(2\text{m}-H\right)=\left[0.598\left(2\text{m}-H\right)+0.0897H\right]\left(3.33m\times 4.4294\sqrt{\text{m}}/\text{s}\right){\left(H\right)}^{3/2}\\ 0=\left[1.1978-0.5092H\right]\left(1.476\right){\left(H\right)}^{3/2}\end{array}$

  $\begin{array}{l}\left[17.68-7.516H\right]\left(H^{3/2}\right)=0\\ -7.516\left(\sqrt{H}-1.53373\right)\left(\sqrt{H}+1.53373\right)\left(H^{3/2}\right)=0\\ \left(17.68H^{3/2}-7.516H^{5/2}\right)=0\\ H=0\text{m}\end{array}$

The head of the weir plate for the different values of the volume flow rate is shown in below Table 2.

Table-2

S.No.	$\dot{Q}$${\text{m}}^{\text{3}}/\text{s}$	$H$$\text{m}$
1	$0{\text{m}}^{\text{3}}/\text{s}$	$0\text{m}$
2	$1{\text{m}}^{\text{3}}/\text{s}$	$2.367\text{m}$
3	$2{\text{m}}^{\text{3}}/\text{s}$	$2.38\text{m}$
4	$3{\text{m}}^{\text{3}}/\text{s}$	$2.39\text{m}$
5	$4{\text{m}}^{\text{3}}/\text{s}$	$2.41\text{m}$
6	$5{\text{m}}^{\text{3}}/\text{s}$	$2.42\text{m}$
7	$6{\text{m}}^{\text{3}}/\text{s}$	$2.44\text{m}$
8	$7{\text{m}}^{\text{3}}/\text{s}$	$2.46\text{m}$
9	$8{\text{m}}^{\text{3}}/\text{s}$	$2.48\text{m}$
10	$9{\text{m}}^{\text{3}}/\text{s}$	$2.50\text{m}$
11	$10{\text{m}}^{\text{3}}/\text{s}$	$2.52\text{m}$
12	$11{\text{m}}^{\text{3}}/\text{s}$	$2.54\text{m}$
13	$12{\text{m}}^{\text{3}}/\text{s}$	$2.57\text{m}$
14	$13{\text{m}}^{\text{3}}/\text{s}$	$2.59\text{m}$
15	$14{\text{m}}^{\text{3}}/\text{s}$	$2.62\text{m}$
16	$15{\text{m}}^{\text{3}}/\text{s}$	$2.65\text{m}$

The Table 2 shows that the height of the weir plate is increase with increase in the flow rate of the water. Therefore, consider the maximum height of the weir plate which is suitable for all volume flow rate varying from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

The rectangular weir of $5\text{m}$ width and $2.65\text{m}$ height is used for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

The triangular weir of $5\text{m}$ base and $2.65\text{m}$ height is used for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

Conclusion:

The appropriate dimensions for the rectangular weir are $5\text{m}$ base and $2.65\text{m}$ height for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.

The appropriate dimensions for the triangular weir are $5\text{m}$ base and $2.65\text{m}$ height for the flow rate varies from $0{\text{m}}^{\text{3}}/\text{s}$ to $15{\text{m}}^{\text{3}}/\text{s}$.