Concept:

Proportionality is the property of the channel and an outlet is said to be proportional when the rate of change of outlet discharge equals to the rate of change of channel discharge.

Proportionality \( = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}}\) = 1 \( = \frac{m}{n} = \frac{H}{y}\)= \( = \frac{{Outlet\;Index}}{{Channel\;Index}}\)

Depending up on the value of Flexibility, outlet can be classified as

1) Proportional outlet (F = 1)

2) Hyper Proportional outlet (F > 1)

3) Sub Proportional outlet (F < 1)

Calculation:

Proportionality = \( = \frac{{0.5}}{{1.5}} = 0.33\) < 1

Hence, the outlet is called Sub Proportional Outlet.

Salinity is due to presence of salt of Na, K, Ca and Mg in soil. It’s a measure of soluble salts in soil.

It is measured by using the following two methods:

1. The electrical conductivity (EC) of a solution or soil and water mix, in the field or laboratory

2. The apparent electrical conductivity of soil using an electromagnetic induction (EM) device

Hence, Electrical conductivity method is should be adopted to measure the salinity in the irrigation field.

Groynes: These are the embankment type structures constructed transverse to the river flow. They extend into the river from the bank and may also be called as transverse dykes.

They are constructed to protect the bank by deflecting the current away from the bank.

As the water is unable to take a sharp embayment, the bank gets protected for certain distance upstream and downstream of the groyne.

However, the nose of the groyne is subjected to a tremendous action of water, and has to be heavily protected by pitching, etc. and the action of eddies reduces from the head towards the bank.

The different of groynes provided are as follows:

Concept:

Cavitation: It is the formation of a void, such as a bubble, within a liquid when a liquid changes its state to a vapor state due to a change in the local pressure while the temperature remains constant.

Cavitation in Ogee spillway:

• When operating head on the Ogee spillway is greater than the designed head, the tailing end of the falling jet may leave the ogee profile, thus generating negative pressure and subsequently formation of cavitation.
• On the other hand, if the operating head is less than the designed head, the falling jet would adhere to the crest of the ogee spillway, creating positive hydrostatic pressure, thus no formation of cavitation.

∴ Statement 1 is incorrect.

Siphon Spillway:

• A siphon spillway consists of one or more siphon units which utilize the siphonic action to discharge the surplus water.
• Instead of allowing water to spill over the crest of a dam or weir, the surplus water is discharged downstream through one or more siphon units which are generally in form of closed duct.

∴ Statement 2 is incorrect.

Concept:

Elementary profile of a gravity dam: It is a right-angled triangle-shaped profile of dam which is subjected only to external water pressure on the upstream side. The shape of this profile is similar to hydrostatic pressure distribution. A diagram showing the elementary profile of gravity dam is shown below:

Width for no tension development: In case of an elementary profile of a gravity dam, for no tension development at the heel with full reservoir, the base width of the profile should be equal to or greater than \({\rm{B}} = \frac{{\rm{H}}}{{\sqrt {{{\rm{S}}_{\rm{c}}} - {\rm{C}}} }}\) where H is the upstream water depth at full reservoir; S_c is specific gravity of dam material and C is seepage coefficient. For, full uplift condition C =1 and for no uplift condition C = 0.

Width for safe in sliding: In case of an elementary profile of a gravity dam, to be safe in sliding with full reservoir, the base width of the profile should be equal to or greater than \({\rm{B}} = \frac{{\rm{H}}}{{{\rm{\mu }}\left( {{{\rm{S}}_{\rm{c}}} - {\rm{C}}} \right)}}\) where H is the upstream water depth at full reservoir; S_c is specific gravity of dam material; µ is coefficient of friction and C is seepage coefficient. For, full uplift condition C = 1 and for no uplift condition C = 0.

Calculation:

Given, upstream water depth at full reservoir condition, H = 55m;

Specific gravity of dam material, S_c = 2.4; coefficient of friction, µ = 0.65.

B₁ is defined as the base width of elementary profile for no tension development with full uplift (C =1).

∴ \({{\rm{B}}_1} = \frac{{\rm{H}}}{{\sqrt {{{\rm{S}}_{\rm{c}}} - 1} }} = \frac{{55}}{{\sqrt {2.4 - 1} }} = 46.5{\rm{\;m}}\) 

B₂ is defined as the base width of elementary profile to be safe in sliding with no uplift (C =0).

∴ \({{\rm{B}}_2} = \frac{{\rm{H}}}{{{\rm{\mu }}\left( {{{\rm{S}}_{\rm{c}}}\; - \;0} \right)}} = \frac{{55}}{{0.65\; \times \;2.4}} = 35.3{\rm{\;m}}\)

Cross regulators & Distributary head Regulators

• Cross regulator and distributary head regulator are provided to control the supplies passing down the parent channel and the offtaking channel respectively.
• A cross regulator is provided on the parent channel at the d/s of the offtake to head up the parent channel at the channel to draw the required supply.
• A distributary head regulator is provided at the head of the offtaking channel (or distributary) to control the supplies entering the offtaking channel.

The best alignment of offtake is when the offtake channel makes zero angle with the parent channel initially and then separates out along transition curves s shown in figure.

Function of cross regulators

1. Cross regulator enable effective regulation of the entire canal system.
2. They help to raise the water level in the parent channel is low.
3. They help in closing the supply to the d/s of the parent channel for the purposes of repairs and construction works.
4. In conjunction with escapes they help water escapes they help water to escape from the channels.
5. They facilitate communication, since a road can be taken over them with a little extra cost.
6. They help to absorb fluctuations in the various sections of the canal system and hence to prevent possibilities of breaches in the tail reaches.
7. They help to control discharge at an outfall of canal into another canal or lake.
8. They help to control water surface slope for bringing the canals to regime slope and section.

Functions of distributary head regulators

1. They regulate or control the supplies to the offtaking channel from the parent channel.
2. They control the entry of silt in the offtaking channel.
3. They serve as a meter for measurement of discharge entering the offtaking channel.
4. They help in shutting off the supplies when not needed in the offtaking channel or when the offtaking channel is required to be closed for repairs.

Concept:

Leaching Requirement (L.R) \(= \frac{{{{\left( {E.C} \right)}_i}}}{{{{\left( {E.C} \right)}_d}}}\)

Where (E.C)_i = electrical Conductivity value of irrigation water

(E.C)_d = electrical Conductivity value of leaching water

\(L.R = 1 - \frac{{{C_u}}}{{{D_i}}}\)

D_d = -C_u + D_i

Calculations:

Given: D_i = 5.94 mm

(E.C)_i = 0.9 m.mho/cm

(E.C)_d = 2 × 13

= 26 m

L.R = 75% = 0.075

\(0.075 = \frac{{{{\left( {E.C} \right)}_i}}}{{26}}\)

(E.C)_i = 1.95 m.mho/cm

LR = 1 - \(\frac{{{C_u}}}{{{D_i}}}\)

0.075 = 1 - \(\frac{{{C_u}}}{{{D_i}}}\)

\(\begin{array}{l} 0.075 = 1 - \frac{{{C_u}}}{{5.94 }}\\ \frac{{{C_u}}}{{5.94 }} = 1 - 0.075 \end{array}\)

C_u = 5.4945 mm

Concept:

Uplift pressure on a dam:

• It is the pressure extracted on the base of the dam by seeping water through the pores, cracks, and fissures of the foundation material and water seeping through dam body.
• An uplift force usually reduces the downward weight of the body of the dam and hence, acts against dam stability.

Drainage Gallery: It is a gallery in a masonry gravity dam that runs parallel to the top of the dam and primarily provided to intercept seepage from the upstream face and conduct it away from the downstream face. Apart from this drainage gallery also facilitates inspection of the dam body.

When drainage galleries are provided to relieve the uplift, the recommended uplift at the face of the gallery is equal to hydrostatic pressure at the toe \(\left( {{{\rm{\gamma }}_{\rm{w}}}{\rm{H'}}} \right)\) plus 1/3^rd the difference of the hydrostatic pressure between heel and toe i.e. \(\left[ {{{\rm{\gamma }}_{\rm{w}}}{\rm{H'}} + \frac{1}{3}\left( {{{\rm{\gamma }}_{\rm{w}}}{\rm{H}} - {{\rm{\gamma }}_{\rm{w}}}{\rm{H'}}} \right)} \right]\). Diagram for uplift pressure when drainage gallery provided is shown below:

Calculation:

Given, Upstream water depth, H = 75m; Tailwater depth, H’ = 7m; Unit weight of water, \({{\rm{\gamma }}_{\rm{w}}} = 9.81{\rm{\;kN}}/{{\rm{m}}^3}\)

∴ Hydrostatic pressure at heel face of the dam = \({{\rm{\gamma }}_{\rm{w}}}{\rm{H}} = 9.81 \times 75 = 735.75{\rm{\;kN}}/{{\rm{m}}^2}{\rm{\;}}\)

∴ Hydrostatic pressure at toe face of the dam = \({{\rm{\gamma }}_{\rm{w}}}{\rm{H'}} = 9.81 \times 7 = 68.67{\rm{\;kN}}/{{\rm{m}}^2}\)

∴ Hydrostatic pressure at centerline face of drainage gallery = \(\left[ {{{\rm{\gamma }}_{\rm{w}}}{\rm{H'}} + \frac{1}{3}\left( {{{\rm{\gamma }}_{\rm{w}}}{\rm{H}} - {{\rm{\gamma }}_{\rm{w}}}{\rm{H'}}} \right)} \right]\)

\(= \left[ {68.67 + \frac{1}{3}\left( {735.75 - 68.67} \right)} \right] = 291.03{\rm{\;kN}}/{{\rm{m}}^2}\)

Concept:

Sliding Failure: Sliding (or shear) failure will occur when the net horizontal force above any plane in the dam or at the base of the dame exceeds the frictional resistance developed at that level.

Shear Friction Factor (SFF): In low dams, the safety against sliding should be checked only for friction. But in high dams the shear strength of the joint, which is an additional shear resistance, is also be considered from an economical point of view. If this shear resistance of the joint is also considered then sliding failure is characterized by shear friction factor (SFF) instead of a plain factor of safety.

It is given by,

\({\rm{SFF}} = {\rm{\;}}\frac{{{\rm{\mu }}\sum {\rm{V}} + {\rm{Bq}}}}{{\sum {\rm{H}}}}\)  

Where, B = width of the dam at the joint; q is the Average shear strength of the joint; µ is the frictional coefficient between soil and dam surfaces; \(\sum {\rm{V}}\) is total vertical forces and \(\sum {\rm{H\;}}\) is the net total horizontal force.

Calculation:

Given, Width of the dam at the joint, B = 45 m;

Frictional coefficient between soil and dam surfaces, µ = 0.68;

Required shear friction factor, SFF = 4.0;

Total downward force, ΣV = 54500 kN and Net horizontal force, ΣH = 32000 kN;

\(\therefore 4 = {\rm{\;}}\frac{{0.68\; \times \;54500 \;+ \;45\; \times \;{\rm{q}}}}{{32000}}{\rm{\;\;}} \Rightarrow {\rm{q}} = 2020.89{\rm{\;kN}}/{{\rm{m}}^2}\)

Important Point:

Sliding failure for high gravity dam is measured by shear friction factor (SFF) where SFF = \(\frac{{{\rm{\mu }}\sum {\rm{V}} + {\rm{Bq}}}}{{\sum {\rm{H}}}}\)

Sliding failure for low gravity dam is measured by a factor of safety against sliding (FSS) where FSS = \(\frac{{{\rm{\mu }}\sum {\rm{V}}}}{{\sum {\rm{H}}}}\)

Concept:

Flexibility of a module/ canal outlet: Flexibility is defined as the ratio of the rate of change of discharge of the outlet to the rate of change of discharge of the distributary channel.  Thus \({\rm{F}} = \frac{{{\rm{dq}}/{\rm{q}}}}{{{\rm{dQ}}/{\rm{Q}}}}\)

where F = Flexibility of the outlet; q = Discharge passing through the outlet; Q = Discharge in the distributary channel.

If H is the head acting on the outlet, the discharge through the outlet may be expressed as \({\rm{q}} = {\rm{C}}{{\rm{H}}^{\rm{m}}}\)

where, C and m are constants depend upon the type of outlet.

Similarly, the discharge passing down the distributary channel may be expressed as \({\rm{Q}} = {\rm{K}}{{\rm{y}}^{\rm{n}}}\) where K and n are constants and y is the depth of water in the distributary.

By differentiating it can be obtained that, \(\frac{{{\rm{dq}}}}{{\rm{q}}} = {\rm{m}}\frac{{{\rm{dH}}}}{{\rm{H}}}{\rm{\;\;\;and\;\;}}\frac{{{\rm{dQ}}}}{{\rm{Q}}} = {\rm{n}}\frac{{{\rm{dy}}}}{{\rm{y}}}{\rm{\;\;\;\;\;}}\therefore {\rm{F}} = \frac{{\rm{m}}}{{\rm{n}}}\frac{{\rm{y}}}{{\rm{H}}}\frac{{{\rm{dH}}}}{{{\rm{dy}}}}\)

But since a change in the head working on the outlet (dH) would result in an equal change in the water depth of the distributary (dy), hence dy = dH.  

\(\therefore {\bf{F}} = \frac{{\bf{m}}}{{\bf{n}}}\frac{{\bf{y}}}{{\bf{H}}}\)

Calculation:

Given, flow is proportional to \({{\rm{y}}^{5/3}}\) in the channel. Hence n = 5/3.

Also, for a weir type outlet flow is proportional to \({{\rm{H}}^{3/2}}\). Hence m = 3/2.

Flexibility of the module, F = 1.15; water depth in the channel, y = 3.5 m.

\(\therefore {\rm{F}} = \frac{{\rm{m}}}{{\rm{n}}}\frac{{\rm{y}}}{{\rm{H}}}{\rm{\;}} \Rightarrow 1.15 = \frac{{\frac{3}{2}}}{{\frac{5}{3}}} \times \frac{{3.5}}{{\rm{H}}}{\rm{\;\;\;}}\therefore {\rm{H}} = 2.74{\rm{\;m}}\)

Important point:

For a weir type outlet flow is proportional to \({{\bf{H}}^{3/2}}\).

For an orifice type outlet flow is proportional to \({{\bf{H}}^{1/2}}\).