Irrigation Engineering Test 3

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.

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:

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:

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 \({{\bf{H}}^{1/2}}\).