Weirs and Notches MCQ Quiz in मराठी - Objective Question with Answer for Weirs and Notches - मोफत PDF डाउनलोड करा

Explanation:

Time required to empty tank due to orifice at bottom is expressed as

\(T=\frac{2A}{{{C}_{d}}a\sqrt{2g}}\left( \sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}} \right)\)

Time required to empty reservoir due to rectangular weir is expressed as

\(T=\frac{3A}{{{C}_{d}}L\sqrt{2g}}\left[ \frac{1}{\sqrt{{{H}_{2}}}}-\frac{1}{\sqrt{{{H}_{1}}}} \right]\)

Time required to empty reservoir due to triangular weir is expressed as

\(T=\frac{5A}{4{{C}_{d}}L.\tan \frac{\theta }{2}\sqrt{2g}}\left[ \frac{1}{{{H_2}^{\frac{3}{2}}}}-\frac{1}{{{H_1}^{\frac{3}{2}}}} \right]\)

Concept:

Weir or notch is a physical structure of masonry constructed across the channel width to calculate the discharge of the channel section.

Rectangular Notch:

The discharge through a rectangular notch weir is,

\(Q = \frac{2}{3}{C_d}L\;\sqrt {2g} \;{H^{3/2}}\)

\(L=\frac{{3Q}}{{2{C_d}\sqrt {2g} \;{H^{3/2}}}}\)

Where, Q = discharge of fluid, Cd = Coefficient of discharge and H = height of water above the notch

Calculation:

Give data;

C_d = 0.6, L = 10 \(\sqrt{10}\) m, H = 1 m, and g = 9.8 m/s2,

The discharge through a rectangular notch weir is,

\(Q = \frac{2}{3}{C_d}L\;\sqrt {2g} \;{H^{3/2}}\)

\(Q = \frac{2}{3}{\times0.6\times}10\sqrt{10}\times\;\sqrt {2\times9.8} \;{1^{3/2}}\) = 56 m³/sec = 56000 litres per sec

Concept:

Discharge over a suppressed rectangular weir is given by,

Q = 1.84 B H^3/2      ----(i)

Where Q = Water flow rate in m3/sec

B = Length of weir in m

H = Head over the Weir in m

Differentiate Eqn (i)

\(\frac{{dQ}}{{dH}} = 1.84 × B × \frac{3}{2} × {H^{1/2}}\)

\(dQ = 1.84 × B × \frac{3}{2} × {H^{1/2}} × dH\)

\(\frac{{dQ}}{Q} = \frac{{1.84 × B × \frac{3}{2} × {H^{1/2}} × dH}}{{1.84 × B × {H^{3/2}}}}\)

\(\frac{{dQ}}{Q} = 1.5\frac{{dH}}{H}\)

Calculation:

Given,

Excess discharge = 3%

So, (dQ/Q) × 100 = 3%

∵ We know that, \(\frac{{dQ}}{Q} = 1.5\frac{{dH}}{H}\)

3% = 1.5 × (dH/H)

dH/H = 2%

So this discrepancy was due to an error in reading the head is 2% excess.

Explanation:

Let us consider that we have channel carrying water and let us think a rectangular notch or weir with this channel as displayed here in the following figure. 

We have the following data from the above figure and these data are as mentioned here. 
H = Head of water over the crest, L = Length of the rectangular notch or weir 

Let us consider one elementary horizontal strip of water of thickness dh and length L as displayed in above figure. 
dh = Thickness of elementary horizontal strip of water flowing over the rectangular notch or weir, h = Depth of elementary horizontal strip of water flowing over the rectangular notch or weir, C_d = Co-efficient of discharge

Area of elementary horizontal strip of water = L x dh 

the value of discharge dQ through the elementary horizontal strip of water is given by,

dQ = C_d × Area of stripe × Theoretical Velocity

dQ = C_d × L × dh × √2gh

Total discharge i.e. Q over a rectangular weir

\(Q = \int_0^HC_d\times L\times \sqrt{2gh}~dh~=~C_d\times L\times \sqrt{2g}\int_0^hh^\frac{1}{2}dh\)

\(Q = {C_d} \times L \times \sqrt {2g} \times {\left[ {\frac{{{h^{{{\frac{1}{2}}} + 1}}}}{{\frac{1}{2} + 1}}} \right]_0}^H = {C_d} \times L \times \sqrt {2g} {\left[ {\frac{{{h^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]^H}_0\)

\(Q~=~\frac{2}{3}\times C_d \times L\times \sqrt{2g}[H]^\frac{3}{2}\)

Concept:

Discharge for V notch is given as:

\(Q = \frac{8}{{15}}{C_d}\tan \frac{\theta }{2}\;\sqrt {2g} \;{H^{\frac{5}{2}}}\)

Calculation:

Given:

Error in the measurement of the head in a V notch is 1 % i.e \(\frac{{dH}}{H} = 1~\% \)

Now differentiating the discharge equation w.r.t 'H' we get:

\(Q = \frac{8}{{15}}{C_d}\tan \frac{\theta }{2}\;\sqrt {2g} \;{H^{\frac{5}{2}}}\)

\(Q = K{H^{\frac{5}{2}}}\)

where: \(K = \;\frac{8}{{15}}{C_d}\tan \frac{\theta }{2}\;\sqrt {2g} \)

\(dQ = K\;\frac{5}{2}\;{H^{\frac{3}{2}}}\;dH\)

\(K = \frac{Q}{{{H^{\frac{5}{2}}}}}\)

\(\frac{{dQ}}{Q} = \;\left( {\frac{{K\;\frac{5}{2}\;{H^{\frac{3}{2}}}\;dH}}{{K{H^{\frac{5}{2}}}}}} \right) = \frac{5}{2}\frac{{dH}}{H}\)

The error in the measurement of discharge

i.e \(\frac{{dQ}}{Q} = 2.5\% \)

Explanation:-

• The velocity of approaches is defined as the velocity with which the water approaches or reaches the weir or notch before it flows over it.
• The whirl velocity is the tangential component of absolute velocity at the blade inlet and outlet. This component of velocity is responsible for the whirling or rotating of the turbine rotor.
• Shear velocity also called friction velocity. It is useful as a method in fluid mechanics to compare true velocities.

Explanation:

Weirs are commonly used for the measurement of open channel flow rate.

Suppressed weir: A rectangular weir whose notch or opening sides are coincident with the sides of the approach channel is known as a Suppressed weir. In this type of Weir crest length (l) is equal to the width of the channel.

If the crest length of the weir is less than the width of the channel, it is known as a Contracted weir.

Francis formula:

When the length of the weir is less than the width of the stream, we find there will be a lateral contraction at each end such a weir is called a contracted weir.

According to Francis each lateral contraction (also called end contraction) is equal to 0.1 H.

If the actual length of the weir is L, then the effective length of the weir will be (L – 0.2 H).

In some cases there may be intermediate obstacles like piers over the weir. In such a case if l is the length of the weir after making deductions for the widths of the obstacles and if there are n lateral contractions the effective length of the weir will be (L – 0.1 nH).

Q =  1.84 (L- nH) H^3/2

If the length of the weir is exactly equal to the width of the approaching stream, there will not be any end contractions(n= 0). Such a weir is called a suppressed weir, and for such a weir Francis formula simplifies to

Explanation:

The length of the channel boundary (sides and the base of the channel) which comes in contact with the flowing liquid is called the wetted perimeter.
The most efficient section must have a minimum wetted perimeter and hence minimum resistance to flow.

Compared to other cross-sections, the semicircular section has the lowest wetted perimeter, and is, therefore, the section of the highest efficiency, but, due to practical limitations in maintaining the section, trapezoidal channels are usually employed.

The hydraulic mean depth or Hydraulic radius is the ratio of the wetted area to the wetted perimeter.

Explanation:

Sheet piles

• Sheet piles are interlocking, vertically driven piles used to create a barrier in the ground.
• They can be used to create cutoff walls that minimize seepage through the foundation.

Additional Information Impervious floors

• Impervious floors refer to the construction of impermeable layers at the base of a structure to prevent or reduce the seepage of water through the foundation.

Inverted filters

• Inverted filters are layers of permeable material (such as gravel or sand) placed above impervious layers to enhance drainage and prevent the development of uplift pressure.

Toe walls

• Toe walls are structures constructed at the base or toe of a dam or embankment to provide additional support and prevent erosion at the foundation.

Triangular weir:

The expression for the discharge over a triangular weir or notch is

Q = \(\frac{8}{{15}} \times {C_d} \times tan\frac{θ }{2} \times \sqrt {2g} \times {H^{\frac{5}{2}}}\)

From this expression flow through the weir depend upon the vertex angle (θ) and coefficient of discharge (Cd) is constant for all the heads, the value of Cd is nearly equal to 0.6.

Rectangular weir:

The expression for the discharge over a rectangular weir or notch is

Q = \(\frac{2}{3} \times {C_d}\; \times L \times \sqrt {2g} \times {H^{\frac{3}{2}}}\)

Advantages of triangular weir over rectangular weir:

A triangular weir is preferred to a rectangular weir due to the following reason.

(1) The expression for discharge for a right-angled V- a notch or weir is very simple.

(2) For measuring low discharge, a triangular weir gives more accurate results than a rectangular weir.

(3) In the case of the triangular weir, only one reading, (H) is required for the computation of discharge.

(4) Ventilation of triangular weir is not necessary.