Chapter 13, Problem 31P

To determine

(a)

The critical depth.

Answer to Problem 31P

The critical depth is $0.7415\text{m}$.

Explanation of Solution

Given information:

The temperature of the water is $10°\text{C}$, the width of the rectangular channel is $6\text{m}$, the flow depth is $0.55\text{m}$ and the volume flow rate is $12{\text{m}}^{\text{3}}/\text{s}$.

Write the expression for the critical depth.

   $y_c={\left(\frac{{\dot{Q}}^2}{g{b}^2}\right)}^{1/3}$     ...... (I)

Here, the flow rate is $\dot{Q}$, the acceleration due to gravity is $g$ and the width of the rectangular channel is $b$.

Calculation:

Substitute $12{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $6\text{m}$ for $b$ in Equation (I).

   $\begin{array}{c}{y}_c={\left(\frac{{\left(12{\text{m}}^{\text{3}}/\text{s}\right)}^2}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(6\text{m}\right)}^2}\right)}^{1/3}\\ ={\left(\frac{\left(144{\text{m}}^6/{\text{s}}^2\right)}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(36{\text{m}}^2\right)}\right)}^{1/3}\\ ={\left(0.4077{\text{m}}^{\text{3}}\right)}^{1/3}\\ =0.7415\text{m}\end{array}$

Conclusion:

Thus, the critical depth is $0.7415\text{m}$.

To determine

(b)

Whether the flow is subcritical or supercritical.

Answer to Problem 31P

The flow is supercritical.

Explanation of Solution

Given information:

The temperature of the water is $10°\text{C}$, the width of the rectangular channel is $6\text{m}$, the flow depth is $0.55\text{m}$ and the volume flow rate is $12{\text{m}}^{\text{3}}/\text{s}$.

Write the expression for the cross- sectional flow area.

   $A_c=b\times y$     ...... (II)

Write the expression for the velocity of the flow.

   $V=\frac{\dot{Q}}{A_c}$     ...... (III)

Write the expression for the Froude number.

   $Fr=\frac{V}{\sqrt{gy}}$     ...... (IV)

Calculation:

Substitute $6\text{m}$ for $b$ and $0.55\text{m}$ for $y$ in Equation (II).

   $\begin{array}{c}{A}_c=6\text{m}\times 0.55\text{m}\\ =\text{3}\text{.3}{\text{m}}^2\end{array}$

Substitute $3.3{\text{m}}^{\text{2}}$ for $A_c$ and $12{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ in Equation (III).

   $\begin{array}{c}V=\frac{12{\text{m}}^{\text{3}}/\text{s}}{3.3{\text{m}}^{\text{2}}}\\ =3.63\text{m}/\text{s}\end{array}$

Substitute $3.63\text{m}/\text{s}$ for $V$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.55\text{m}$ for $y$ in Equation (IV).

   $\begin{array}{c}Fr=\frac{3.63\text{m}/\text{s}}{\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\left(0.55\text{m}\right)}}\\ =\frac{3.63\text{m}/\text{s}}{\sqrt{5.3955{\text{m}}^2/{\text{s}}^{\text{2}}}}\\ =\frac{3.63\text{m}/\text{s}}{2.32\text{m}/\text{s}}\\ =1.56\end{array}$

Since, the Froude number is greater than one hence the flow is super critical.

Conclusion:

Thus, the flow is supercritical.

To determine

(c)

The alternate flow depth.

Answer to Problem 31P

The alternate flow depth is $1.033\text{m}$.

Explanation of Solution

Given information:

The temperature of the water is $10°\text{C}$, the width of the rectangular channel is $6\text{m}$, the flow depth is $0.55\text{m}$ and the volume flow rate is $12{\text{m}}^{\text{3}}/\text{s}$.

Write the expression for the initial specific energy.

   $E_{c1}=y_1+\frac{{\dot{Q}}^2}{2g{b}^2{y}_1^2}$     ...... (V)

Here, the initial flow depth is $y_1$.

Write the expression for the final specific energy.

   $E_{c2}=y_2+\frac{{\dot{Q}}^2}{2g{b}^2{y}_2^2}$     ...... (VI)

Here, the final flow depth is $y_2$.

Calculation:

The flow depth of the channel is same as the initial flow depth of the channel.

Substitute $0.55\text{m}$ for $y_1$, $12{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $6\text{m}$ for $b$ in Equation (V).

   $\begin{array}{c}{E}_{c1}=0.55\text{m}+\frac{{\left(12{\text{m}}^{\text{3}}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(6\text{m}\right)}^2{\left(0.55\text{m}\right)}^2}\\ =0.55\text{m}+\frac{\left(144{\text{m}}^6/{\text{s}}^2\right)}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(36{\text{m}}^2\right)\left(0.3025{\text{m}}^2\right)}\\ =0.55\text{m}+0.6739\text{m}\\ =\text{1}\text{.2239}\text{m}\end{array}$

Since, the specific energy at final and initial position is same hence, $E_{c2}=1.2239\text{m}$.

Substitute $1.2239\text{m}$ for $E_{c1}$, $12{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $6\text{m}$ for $b$ in Equation (V).

   $\begin{array}{l}1.2239\text{m}=y_2+\frac{{\left(12{\text{m}}^{\text{3}}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(6\text{m}\right)}^2{\left(y_2\right)}^2}\\ \left(1.2239\text{m}\right)\left(706.32{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right){\left(y_2\right)}^2=y_2\left(706.32{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right){\left(y_2\right)}^2+\left(144{\text{m}}^6/{\text{s}}^2\right)\\ y_2^3-\left(1.2239\text{m}\right)y_2^2+0.204{\text{m}}^3=0\\ y_2=0.55m,1.033\text{m},-0.359\text{m}\end{array}$

Since, the alternate depth must be greater than the critical depth hence $y_2=1.033\text{m}$.

Conclusion:

Thus, the alternate flow depth is $1.033\text{m}$.