Rapidly Varied MCQ Quiz in తెలుగు - Objective Question with Answer for Rapidly Varied - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Classification of jump according to the Froude numbers-

\(F_{r1}^2 = \frac{{{q^2}}}{{gy_1^3}}\)

Where F_r1 = Froude’s Number

q = Discharge per metre width

g = acceleration due to gravity

y₁ = depth of flow

Calculation:

Given, Unit discharge = 1.019 m³/sec/m

Depth = 101.9 mm = 0.1019

F_r1² = (1.019²) / (9.81 × 0.1019³) = 100

F_r1 =√100 = 10 (>9)

So it is a strong jump.

Explanation:

Classification of Jump

I. Undular Jump → 1 < F₁ ≤ 1.7

II. Weak Jump → 1.7 < F₁ ≤ 2.5

III. Oscillating Jump → 2.5 < F₁ ≤ 4.5

IV. Steady Jump → 4.5 < F₁ ≤ 9.0

V. Strong Jump F₁ > 9

F₁ > 9,

Concept-

A hydraulic jump occurs when the flow changes from a supercritical flow (unstable) to a sub-critical flow (stable).

There is a sudden rise in the water level at the point where the hydraulic jump occurs.

Rollers (eddies) of turbulent water form at this point. These rollers cause dissipation of energy.

A hydraulic jump occurs in practice at the toe of a dam or below a sluice gate where the velocity is very high.

Important Points

Supercritical velocity is greater than critical velocity and subcritical velocity is less than the critical velocity. In hydraulic jump the flow is from supercritical flow (unstable) to a sub-critical flow.

We have critical velocity,\({{\rm{V}}_{\rm{c}}} = \sqrt {{\rm{g}} \times {{\rm{Y}}_{\rm{c}}}} \)

For supercritical flow depth is less, so area of flow will be less and velocity will be more. Similarly for subcritical flow depth is more so area of flow will be more so velocity will be less to maintain a constant discharge.so accordingly v1>vc and v2<vc.

Concept:

The flow in the open channel can be classified based on Froude's number.

Froude Number > 1 ⇒ Supercritical flow

Froude Number = 1 ⇒ Critical flow

Froude Number < 1 ⇒ Subcritical flow

Given:

y = 1 m, B = 5 m, and v = 3.13 m/s

\({\rm{Froude\;number\;}}\left( {{{\rm{F}}_{\rm{R}}}} \right) = \frac{{\rm{v}}}{{\sqrt {\frac{{{\rm{gA}}}}{{\rm{T}}}} }}\)

Where, T= Top width, and A= Area of cross-section of the flow.

∴ Top width (T) = 5 + 2 × 2 = 9 m

\(\therefore {\rm{A}} = \frac{1}{2} \times \left( {9 + 5} \right) \times 1 = 7{\rm{\;}}{{\rm{m}}^2}\)

\({{\rm{F}}_{\rm{R}}} = \frac{{3.13}}{{\sqrt {9.81 \times \frac{7}{9}} }}\)

⇒ F_R = 1.133 > 1

∴ The flow is supercritical.

Concept:

Hydraulic jump: Whenever supercritical flow merges into subcritical flow then to reduce the energy a jump is formed, called a hydraulic jump.

For a rectangular frictionless channel, both depths are related as:

\(\frac{{{{\rm{y}}_2}}}{{{{\rm{y}}_1}}} = \frac{1}{2}\left[ {\sqrt {1 + 8{{\rm{F}}_1}^2} - 1} \right]\)

Froude number (F) = \(\sqrt {\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}}}\)

Where,

Y₁ = Pre jump depth, Y₂ = Post jump depth, F₁= Froude Number before jump, Q = Discharge, T = top width, and A = area of flow

Length of hydraulic jump (L) is calculated empirically by: L = 7 × (Y₂ - Y₁)

Concepts:

The energy loss due to hydraulic jump is given as:

\( E = \frac{(y_2 - y_1 )^3}{4y_1y_2} \)

Where

y₁ is the pre-jump depth

y₂ is the post jump depth

Calculation:

Given:  y₁ = 0.25 m; y₂ = 1.25 m

\( E = \frac{(y_2 - y_1 )^3}{4y_1y_2} \)

\( E = \frac{(1.25-0.25)^3}{4\times 0.25\times 1.25} \)

∴ Energy Loss = E = 0.8 m

Additional InformationThe relation between pre and post jump depth is given as

\(\frac{y_2}{y_1} = \frac{1}{2} (-1 +\sqrt{ 1 + 8F_1^2} ) \)

where, F₁ is the Froude number before the jump.

Concept: 

The energy loss due to the hydraulic jump is given as:

\( E = \frac{(y_2 - y_1 )^3}{4y_1y_2} \)

Where

y1 is the pre-jump depth

y2 is the post-jump depth

Calculations:

The sequent depths given in the question are as follows:

y1 = 0.25 m

y2 = 1.25 m

\( E = \frac{(y_2 - y_1 )^3}{4y_1y_2} \)

\( E = \frac{(1.25 - 0.25 )^3}{4\times 1.25\times0.25} \)

Energy Loss (EL) = 0.80 m

Concept:

Critical Flow: It is defined as the flow for which the specific energy is minimum and Froude’s number is equal to unity.

Froude’s no. (F_r) = \(\sqrt {\frac{{{\rm{Inertial\;force}}}}{{{\rm{gravity\;force}}}}} \)

Where,

Inertial force = ρ × Q × v

(ρ = Density, Q = Discharge, v = velocity)

Gravity force = ρ × g × V;

(ρ = Density, g = Gravity acceleration, V = Volume)

Now,

\({{\rm{F}}_{\rm{r}}} = \sqrt {\frac{{{\rm{\rho Qv}}}}{{{\rm{\rho gV}}}}} = \sqrt {\frac{{{\rm{Qv}}}}{{{\rm{gAL}}}}} = \sqrt {\frac{{{\rm{Q}}\left( {\frac{{\rm{Q}}}{{\rm{A}}}} \right)}}{{{\rm{gA}}\left( {\frac{{\rm{A}}}{{\rm{T}}}} \right)}}} = \sqrt {\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}}} \)

 (∵ A = L × T, ν = Q/A)

A = Area, L = Length, and T = Top width

Now, for a critical flow F_r = 1

\(\therefore {{\rm{F}}_{{{r}}}} = 1\)

\( \Rightarrow \frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}} = 1\)

Important points:

For a critical flow:

Specific energy is minimum for a given discharge.

Discharge is maximum for a given specific energy.

Specific force is minimum for a given discharge.

Froude’s number= \(\frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \left( {\frac{{\rm{A}}}{{\rm{T}}}} \right)} }}\)

Where,

A = Area of cross-section, and T = Top width

For a triangular channel as stated above

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {\frac{{{\rm{gA}}}}{{\rm{T}}}} }}\)

T = 2y + 2y = 4y

\({\rm{A}} = \frac{1}{2} \times 4{\rm{y}} \times {\rm{y}} = 2{y^2}\)

\(\therefore {{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \frac{{2{{\rm{y}}^2}}}{{4{\rm{y}}}}} }}\)

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}}\left( {\frac{{\rm{y}}}{2}} \right)} }}\)