Open Channel Flow Practice Questions

A hydraulic jump occurs when there is a sudden change in the flow characteristics of the fluid. This phenomenon typically occurs when the flow transitions from a high velocity (steep slope) to a lower velocity (mild slope). The change in slope causes a sudden rise in water level and a loss of kinetic energy, resulting in a turbulent flow downstream. Thus, the correct answer is (B) STEEP slope to MILD slope .

Given: rectangular channel.

For Froude number at upstream

Now, $\Delta E =\frac{(1.68-0.5)^{3}}{4 \times 1.68 \times 0.5}=0.49$ From above equation

Discharge $(\mathrm{Q})=30 \mathrm{~m}^{3} / \mathrm{s}$ Width of channel $(B)=2 \mathrm{~m}$ Bed slope $=1$ in 300 Energy correction factor $(\alpha)=1.1$ $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Discharge per unit width $(q)=\frac{Q}{B} =\frac{30}{2}=15 \mathrm{~m}^{3} / \mathrm{sec} / \mathrm{m}$ Critical depth does not depend on slope or the roughness of channel. Critical depth $\left(y_{c}\right)=\left(\frac{q^{2} \cdot \alpha}{g}\right)^{1 / 3} =\left(\frac{1.1 \times 15^{2}}{10}\right)^{1 / 3}=2.914 \mathrm{~m} \simeq 2.91 \mathrm{~m}$

Unit discharge $(q)=7.5 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m}$

Tail water depth, $y_{2}=4.54 \mathrm{~m}$

width of channel $(B)=4 \mathrm{~m}$ discharge $(Q)=2 \mathrm{~m}^{3} / \mathrm{s}$ depth of $\operatorname{flow}(\mathrm{y})=0.4 \mathrm{~m}$ Specific energy at section (1) - (1) (E)

When channel section is contracted to minimum width and for constant discharge Q. The flow over contracted section will be critical flow.

For critical flow condition, $\frac{\mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA}^{3}}=1$ $\frac{\mathrm{Q}^{2} \mathrm{~B}_{\min }}{\mathrm{gB}_{\min \mathrm{y}_{\mathrm{c}}^{3}}^{3}}=1$

Degree of freedom in a mobile boundary channel are bed width, depth, slope and plain form.

(I) Solubility of bitumen: It is defined as weight of bitumen soluble in $100 \mathrm{ml}$ of carbon disulphide. (II) Softening point of bitumen: It is the temperature at which bitumen attains a particular degree of softneess. (III) Los angles abrasion test: It is reported as percentage of material finer than $1.7 \mathrm{~mm}$ with respect to total weight. (IV) Flash point of bitumen: It is the lowest temperature at which bitumen catch five momentarily. (V) Ductility of bitumen: Ductility is measured by stretching a standard briquette of bitumen. The distance in $\mathrm{cm}$ that the briquette can be stretched before breaking is the ductility. (VI) Specific gravity of bitumen: It is the ratio of unit weight of bitumen to unit weight of water. (VII) Thin film oven test: It measure mass charge a sample as a percent of initial mass.

Given,

The value of specific force

For horizontal, frictionless rectangular channel $\therefore \quad \frac{2 q^{2}}{g}=y_{1} y_{2}\left(y_{1}+y_{2}\right)$

Now, specific force $=0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}$

When super critical stream enters a mild (m) channel section, the type of flow. Pofile would be $M_{3}$ or $S_{1}$ .

$\eta=0.01$ Critical depth, $y_{C}=\left(\frac{q^{3}}{g}\right)^{1 / 3}=\left(\frac{70^{2}}{9.81}\right)^{1 / 3}=7.934 \mathrm{~m}$ For normal depth of flow for wide rectangular channel. For bed slope $(0.001)$

The profile would be $\mathrm{M}_{2}$ and $S_{2}$ .

Currently no explanation available

Specific Energy $(E)=y+\frac{v^{2}}{2 g}$ $\Rightarrow \quad E=y+\alpha \cdot \frac{Q^{2}}{2 g A^{2}}\left[\right.$ as $\left.\frac{Q}{A}=v\right]$ For critical flow condition,

'dA' can be written as 'Tdy' $\Rightarrow \frac{d A}{d y}=T$ $\Rightarrow \quad \frac{\mathrm{dE}}{\mathrm{dy}}=1-\frac{\alpha \mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA} \mathrm{A}^{3}}$ $\Rightarrow 1-\frac{\alpha Q^{2} \mathrm{~T}}{g A^{3}}=0$ $\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{\mathrm{A}^{3}}{\mathrm{~T}}$ At critical flow, $A=A_{C}$ ; $\mathrm{T}=\mathrm{T}_{\mathrm{C}}$ $\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}$

$Q=30m^3/sec$ $B=6m$ $y_1=0.5m$ $E_L=1.6m$ $q= \frac{Q}{B} =\frac{30}{6}=5m^3/sec/m$ We know that,

Hence, $y_2=2.5m$ Post jump Froude's No. $(F_2)=\frac{V_2}{\sqrt{\sqrt{gy_2}}}=\frac{\left ( \frac{30}{6 \times 2.5} \right )}{\sqrt{10 \times 2.5}}=0.4$ $$

Currently no explanation available

Currently no explanation available

Given: $B = 3m, y = 2m, Q = 12 m^3 /sec$ $Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec$ Specific energy at section (1-1) $E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m$ When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place. $E=E_c=\frac{108}{49}$ We have that, $E_c=\frac{3}{2}y_c$ (for rectangular crosssection) $y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}$ For critical flow condition,

Minimum specific energy will correspond to a critical flow condition. The critical depth

Efficient channel section

Assuming channle bed to be horizontal and frictionless. $q=\frac{12}{6}=2 m^3/s/m$ Initial Froude No.

From Belenger's Momentum equation for a rectangular channel

Heal loss in the jump

Power lost in the jump