Fluid Mechanics Test 7

Concept:

Hydraulic Jump: It is a phenomena associated with sudden rise in water depth when a flow with high velocity and low depth (supercritical flow) strikes another flow with low velocity and high depth (subcritical flow).

For any hydraulic jump, the following relation always holds good:

\({{\rm{y}}_1}{{\rm{y}}_2}\left( {{{\rm{y}}_1} + {{\rm{y}}_2}} \right) = \frac{{2{{\rm{q}}^2}}}{{\rm{g}}}\)

Where, y₁ and y₂ are the depths before and after the hydraulic jump respectively.

q is discharge per unit width and g is acceleration due to gravity.

Calculation:

Given, upstream depth which is depth before hydraulic jump = y₁ = 0.4 m.

Downstream depth which is depth after hydraulic jump = y₂ = 5 m.

Acceleration due to gravity = g = 9.81 m/s².

\(\therefore {{\rm{q}}^2} = \frac{{{{\rm{y}}_1}{{\rm{y}}_2}\left( {{{\rm{y}}_1} + {{\rm{y}}_2}} \right) \times {\rm{g}}}}{2} = \frac{{0.4 \times 5 \times \left( {0.4 + 5} \right) \times 9.81}}{2} = 52.974{\rm{\;}}\therefore {\rm{q}} = 7.278\frac{{{{\rm{m}}^2}}}{{\rm{s}}} \approx 7.3{\rm{\;}}\frac{{{{\rm{m}}^2}}}{{\rm{s}}}{\rm{\;}}\)

∴ The discharge per unit width is nearly 7.3 m²/s. 

Concept:

The flow is said to be critical, if it satisfies the following equation irrespective of the type of channel:

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

Where,

Q is the discharge

T is the top width of channel

A is the wetted area of channel

Calculation:

Let y_c is the critical depth of flow

Wetted area, A = By_c

Given: Q = 20 m³/s, B = 10 m, g = 10 m/s²

Top width, T = B = 10 m

We know that: \(\frac{{{Q^2}T}}{{g{A^3}}} = 1\)

\( \Rightarrow \frac{{{{\left( {20} \right)}^2}\; \times \;10}}{{10\; \times\; {{\left( {10\; \times\; {y_c}} \right)}^3}}} = 1 \Rightarrow {y_c} = 0.736\;m\) 

Critical velocity of flow, \({V_c} = \frac{Q}{A} = \left( {\frac{{20}}{{10\; \times \;0.736}}} \right)\)

⇒ V_c = 2.72 m/s

Concept:

For a triangular notch

\({\rm{Q\;}} = {\rm{\;}}\frac{8}{{15}}{{\rm{c}}_{\rm{d}}}\sqrt {2{\rm{g}}} \tan \frac{{\rm{\theta }}}{2} \times {{\rm{H}}^{3/2}}\)

Where,

Q = Dischange, Cd = co-eff of discharge, H = Head above the bottom of notch, and θ = enclosed angle

Calculation:

Differentiating equation w.r.t. H, we have

\(\frac{{{\rm{dQ}}}}{{{\rm{dH}}}}{\rm{\;}} = {\rm{\;}}\frac{8}{{15}}{{\rm{c}}_{\rm{d}}}\sqrt {2{\rm{g}}} \tan \frac{{\rm{\theta }}}{2} \times \frac{5}{2}{{\rm{H}}^{3/2}}\)

\({\rm{dQ\;}} = {\rm{\;}}\frac{8}{{15}}{{\rm{c}}_{\rm{d}}}\sqrt {2{\rm{g}}} \tan \frac{{\rm{\theta }}}{2} \times \frac{2}{2}{{\rm{H}}^{\frac{3}{2}}} \times {\rm{dH}}\)

\(\frac{{{\rm{dQ}}}}{{\rm{Q}}}{\rm{\;}} = {\rm{\;}}\frac{{\frac{8}{{15}}{{\rm{c}}_{\rm{d}}}\sqrt {2{\rm{g}}} \tan \frac{{\rm{\theta }}}{2} \times \frac{3}{2}{{\rm{H}}^{3/2}} \times {\rm{dH}}}}{{\frac{8}{{15}}{{\rm{c}}_{\rm{d}}}\sqrt {2{\rm{g}}} \tan \frac{{\rm{\theta }}}{2} \times {{\rm{H}}^{3/2}}}}{\rm{\;}} = {\rm{\;}}\frac{5}{2}\frac{{{\rm{dH}}}}{{\rm{H}}}\)

\(\therefore \frac{{{\rm{dQ}}}}{{\rm{Q}}}{\rm{\;}} = {\rm{\;}}\frac{5}{2}\frac{{{\rm{dH}}}}{{\rm{H}}}{\rm{\;}} = {\rm{\;}}2.5{\rm{\;}}\frac{{{\rm{dH}}}}{{\rm{H}}}\)

Given error in head \( = {\rm{}}\frac{{{\rm{dH}}}}{{\rm{H}}}{\rm{}} = {\rm{}}2{\rm{\% }}\)

So, \(\frac{{{\rm{dQ}}}}{{\rm{Q}}}{\rm{}} = {\rm{}}2.5 \times 2{\rm{}} = {\rm{}}5{\rm{\% }}\)

Important Points:

Discharge over a rectangular notch

\({\rm{Q}} = {\rm{}}{{\rm{C}}_{\rm{d}}}\frac{2}{3}{\rm{B\;}}\sqrt {2{\rm{g}}} {{\rm{H}}^{3/2}}\)

Where,

Concept:

The Froude’s number is given as:

\({F_r} = \frac{{{V^2}}}{{\sqrt {gD} }}\)

Where,

D is the hydraulic depth and it is given as D = A/T

Where A is the wetted area and T is the top width

For rectangular channel, A = By and T = B ⇒ D = (By)/B = y i.e. depth of flow

Now,

If F_r < 1 then flow is sub-critical.

If F­_r > 1 then flow is supercritical.

If F_r = 1 then flow is critical.

As per, Manning’s formula, the velocity of flow for normal depth y_n is given as:

\(V = \frac{1}{n}{R^{2/3}}{S^{1/2}}\)

Where,

R is the Hydraulic radius and it is given as R = A/P

(A is wetted area and P is wetted perimeter)

S is the bed slope or energy line slope.

Calculation:

Given: B = 4 m, S \(= \frac{1}{{5000}},\) Q = 30 m³/s, n = 0.012

\(V = \frac{1}{n}{\left( R \right)^{\frac{2}{3}}}\;{S^{\frac{1}{2}}}\)

\(R = \frac{A}{P} = \left( {\frac{{B{y_n}}}{{B + 2{y_n}}}} \right) = {y_n}\)                                   (For wide rectangular channel)

Q = A V

\(Q = \left( {B{y_n}} \right)\frac{1}{n}{\left( {{y_n}} \right)^{\frac{2}{3}}}{S^{\frac{1}{2}}}\)

\( \Rightarrow Q = \left( {\frac{B}{n}{S^{\frac{1}{2}}}{\rm{\;}}} \right)\left( {y_n^{\frac{5}{3}}} \right)\)

\( \Rightarrow 30 = \left( {\frac{4}{{0.012}}} \right){\left( {\frac{1}{{5000}}} \right)^{\frac{1}{2}}}\left( {y_n^{\frac{5}{3}}} \right)\) ⇒ y_n = 3.04 m

Now, velocity of flow, V

\(V = \frac{1}{n}{\left( {{y_n}} \right)^{\frac{2}{3}}}{S^{\frac{1}{2}}}\)

\( \Rightarrow V = \frac{1}{{0.012}} \times {\left( {3.04} \right)^{\frac{2}{3}}}{\left( {\frac{1}{{5000}}} \right)^{{V_2}}}\)

⇒ V = 2.47 m/s

D = y_n (for rectangular channel)

\({F_r} = \frac{V}{{\sqrt {g{y_n}} }} = \frac{{2.47}}{{\sqrt {10\; \times \;3.04} }}\) ⇒ F_r = 0.447

F_r < 1 ⇒ Flow is sub-critical.

Concept

The specific force for a channel is given as,

\(F = A\bar Z + \;\frac{{{Q^2}}}{{Ag}}\)

Where,

A is the wetted area

Q is the discharge in the channel

Z̅ is the centroid of the channel section measured from the top of the channel.

Calculation

Given:

Q = 10 m³/s; B = 2 m

tan 45° = y/1 ⇒ y = 1 m

\(\bar Z = \frac{1}{3}y = \frac{1}{3} \times 1 = 0.33\;m\)

Wetted area, A \(= \frac{1}{2} \times 2 \times 1 = 1\;{m^2}\)

\(\Rightarrow {F_s} = A\bar Z + \frac{{{Q^2}}}{{Ag}}\)  \(= \left( {1 \times 0.67} \right) + \frac{{{{\left( {10} \right)}^2}}}{{1\; \times \;10}}\) 

⇒ F_s = 10.33 kN m³/kN

Concept

For a constant specific energy, the discharge will be maximum when flow is critical and it satisfies the following condition:

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

Where,

Q is the discharge

T is the top width of channel

A is the wetted area of channel

Specific energy under critical conditions for a given critical depth y_c is given as:

\({E_c} = {y_c} + \;\frac{{{V^2}}}{{2g}}\)

Where,

V is the velocity of flow

y_c is the depth flow at a particular section

 Calculation:

Given: E_c = 3 m, B = 2 m

\({E_c} = {y_c} + \frac{{{V^2}}}{{2g}}\) 

\({E_c} = {y_c} + \frac{{{{\left( {\frac{Q}{{B{y_c}}}} \right)}^2}}}{{2g}}\) 

\({E_c} = {y_c} + \frac{{{Q^2}}}{{{B^2}y_c^2\left( {2g} \right)}}\)     …1)

For critical flow,

\(\frac{{{Q^2}T}}{{g{A^3}}} = 1\) ⇒ \(\frac{{{Q^2}B}}{{g{{\left( {B{y_c}} \right)}^3}}} = 1\) 

\(\Rightarrow \frac{{{Q^2}}}{{y_c^2}} = g{B^2}{y_c}\)      …2)

Substitute the value of \(\frac{{{Q^2}}}{{y_c^2}}\) in equation 1)

\({E_c} = {y_c} + \frac{{g{B^2}{y_c}}}{{2g{B^2}}}\) 

\( \Rightarrow {E_c} = {y_c} + \frac{{{y_c}}}{2} = \frac{3}{2}{y_c}\) 

\( \Rightarrow 3 = \frac{3}{2} \times {y_c} \Rightarrow {y_c} = 2\;m\) 

Substitute y_c = 2 m in equation 2)

\(\frac{{{Q^2}}}{{{{\left( 2 \right)}^2}}} = 10 \times {\left( 2 \right)^2} \times {\left( 2 \right)^3}\) 

⇒ Q = 17.89 m³/s

∴ The maximum discharge is 17.89 m³/s

Concept

The energy loss due to hydraulic jump is given as:

\(E = \;\frac{{{{\left( {{y_2} - {y_1}} \right)}^3}}}{{4{y_1}{y_2}}}\)

Where,

y₁ is the pre jump depth

y₂ is the post jump depth

The relation between pre and post jump depth is given as,

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

Where,

F₁ is the Froude number before jump.

Calculation:

Given: F₁ = 10, B = 2 m

Let y₂ = Ky₁

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

\(\frac{{{y_2}}}{{{y_1}}} = K = \frac{1}{2}\left( { - 1 + \sqrt {1 + 8 \times {{10}^2}} } \right)\) 

⇒ K = 13.65

Energy loss is,

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

\( \Rightarrow 2 = \frac{{{{\left( {K{y_1} - {y_1}} \right)}^3}}}{{\left( {4{y_1}} \right)\; \times \;\left( {K{y_1}} \right)}}\) 

\( \Rightarrow 2 = \frac{{{{\left( {K - 1} \right)}^3}{y_1}}}{{4\;K}}\) 

Putting K = 13.65 in above,

\( \Rightarrow 2 = \left( {\frac{{{{\left( {13.65\; - \;1} \right)}^3}}}{{4\; \times \;13.65}}} \right) \times {y_1}\) 

⇒ y₁ = 0.054 m

Now, y₂ = K y₁ = 13.65 × 0.054 = 0.736 m

y₂ = 0.74 m