Best Hydraulic Cross-Section

5. DESIGN OF IRRIGATION CANALS

5.4 Design of lined canals

5.4.2 Best Hydraulic Cross-Section

The best hydraulic (the most efficient) cross-section for a given Q, n, and S0 is the one with a minimum excavation and minimum lining cross-section. A = Amin and P = Pmin. The minimum cross-sectional area and the minimum lining area will reduce construction expenses and therefore that cross-section is economically the most efficient one.

V = Q/A Q/Amin = Vmax

V = 1/n So1/2 R ^2/3 = C*R ^2/3 V = Vmax R = Rmax

R =A/P R = Rmax P= Pmin

The best hydraulic cross-section for a given A, n, and S0 is the cross-section that conveys maximum discharge.

𝑄 = 𝐴1 𝑛𝑅²³𝑆_𝑜¹² Q = C’ x R^2/3 C’ = constant Q = Qmax R = Rmax

R= A/P

R= Rmax P = Pmin

The cross-section with the minimum wetted perimeter is the best hydraulic cross section within the cross-sections with the same area since lining and maintenance expenses will reduce substantially.

Trapezoidal Cross-Sections

67

As can be seen from equation, wetted perimeter is a function of side slope m and water depth y of the cross-section.

For a given side slope m, what will be the water depth y for best hydraulic trapezoidal cross- section?

68

The hydraulic radius R, channel bottom width B, and free surface width L may be found as, Example:

Design the most efficient cross-section of a lined trapezoidal canal to carry a discharge of 15 m³/s when the maximum permissible velocity is 2 m/s. Assume the side slope = 1:1. Also, determine the bed slope for the canal if the \chezy coefficient, C=60

Given

Q = m³ / s V = 2 m/s SS = 1:1 C = 60

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Solution

Step 1: Determine the cross-section area, A A = Q/V = 15/2 = 7.5 m2

Ste 2 = Compute the hydruaic radius, R A = D(B +zD) = D(B + 1D) = D(B + D) P = B + 2D√1 + 𝑧² = 𝐵 + 2𝐷√1 + 1²

P = B + 2√2𝐷 = 𝐵 + 2.828𝐷

→ R = ^{𝐷 (𝐵+𝐷)}

𝐵+2.828𝐷

For most efficient trapezoidal cross-section R = D/2

→ D/2 = ^{𝐷 (𝐵+𝐷)}

𝐵+2.828𝐷

DB + 2.828 D² = 2DB + 2D2

→ DB – 2DB + 2.828 D² – 2D2 = 0 DB + 1.828D² = 0

→ -B + 1.828D = 0 B = 1.828D

→ A = DB + D2

Substitute the value of A = 7.5 m² and B = 1.828D into the above equation

7.5 = D(0.828D) + D²

= 0.828 D² + D² D² = ^7.5

1.828 = 4.1026

D = √4.1028 = 2.03 m = 2.00 m B = 0.828*2 = 1.66 m Determination of bed slope, S

Chezy’s formula V = C√𝑅𝑆

√𝑆 = _𝐶√𝑅^𝑉 → 𝑆 =_𝐶^𝑉₂²_𝑅 S = ²²

60²∗1 = ¹

900

For trapezoidal section the following holds true.

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5.4.3 Practical Lined Canal Sections

The most efficient cross-sections obtained in the preceding derivations need angles in practice. For instance, the type of soil through which the channel is carried may not permit the adoption of a 1:1 side slope for a triangular channel. Secondly, sharp corners in cross-section are virtually ones of stagnation and may lead to the deposition of silt. As such it is desirable to have rounded corners. The Indian practice has been to adopt a triangular section (of the permissible side slope) with a rounded bottom (see Fig below) for discharges less than 55 m³/s. (This limit on the discharge varies from state to state, but a value of 55 m³/s is recommended here as a triangular section for such a discharge is unlikely to result in very high velocities.)

Lined Canal Design

Indian Standard

Q > 55 m3/s

YES

Use Trapizidal canal with rounded

corners

Assume a suitable velocity and use manning equatin

NO Use Triangular

Sectin with rounded corners

Assume suitable Velcity and use mannning equatin

USBR Standard

Use trapizoidal with sharp crners

Find the value f B r B/D ratio from the table and usemanning

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Figure 21 Lined channel section for Q < 55 m³/s

Figure 22 Lined channel for Q > 55 m³/s

To avoid damage to the lining, the maximum velocity in lined channels is restricted to 2.0 m/s. Thus, the design is based on the concept of a limiting velocity.

Table 20 Suitable side slopes for channels excavated through different types of material

Material Side slope (H:V)

Rock Nearly vertical

Muck and peat soil 0.25 : 1

Stiff clay or earth with concrete lining 0.5 : 1 to 1 : 1

Earth with stone lining 1 : 1

Firm clay 1.5 : 1

Loose, sandy aoil 2 : 1

From triangular section, fig 19 Area, A =2 (¹₂ℎ²cot 𝜃) + ¹₂ℎ²(2𝜃)

= ℎ²(𝜃 + cot 𝜃)𝜃

Wetted Perimetre, p = 2hcot 𝜃 + ℎ(2𝜃)

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Hydraulic Radius R =A/P = ^ℎ

2(𝜃+cot 𝜃) 2ℎ(𝜃+cot 𝜃) = ℎ

2 Similarly, for trapezoidal section, fig 2

A= Bh + 2(¹

2ℎ²cot 𝜃) + 2(¹₂ℎ²𝜃)

= Bh + ℎ²(𝜃 + cot 𝜃) P = B + 2h(𝜃 + cot 𝜃)

R = ^𝐴

𝑃

In all these expressions for A, P, and R, the value of θ is in radians. For designing a lined channel, one needs to solve these equations alongwith the Manning’s equation. For given Q, n, S, and A, and R expressed in terms of h for known Q, the Manning’s equation will yield, for triangular section, an explicit relation for h as shown below :

Q = ¹

𝑛[ℎ²(𝜃 + cot 𝜃)] [^ℎ₂]^2/3

H =

[

√𝑆(𝜃+cot 𝜃)^𝑛𝑄22/3

]

^3/8

However, in case of trapezoidal section, Fig. 8.2, the design calculations would start with an assumed value of velocity (less than the maximum permissible velocity of 2.0 m/s) and the expression for h will be in the form of a quadratic expression as can be seen from the following : From the Manning’s equation

R =(𝑈_𝑛

√𝑆

)

^3/2

(𝑈𝑛

√𝑆)^3/2[𝐵 + 2ℎ(𝜃 + cot 𝜃)]

B = _𝑈^𝑄(_𝑈𝑛^√𝑆)^3/2 – 2h (𝜃 + cot 𝜃)

On substituting this value f B in the expression for area of flow A, ne gets, h^𝑄_𝑈(_𝑈𝑛^√𝑆)^3/2− 2ℎ²(𝜃 + cot 𝜃) + ℎ²(𝜃 + cot 𝜃) = 𝐴 = ^𝑄_𝑈

ℎ²(𝜃 + cot 𝜃) − ℎ 𝑄 𝑈(√𝑆

𝑈𝑛)

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+ 𝑄 𝑈 = 0

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h

=

𝑄

𝑈(_𝑈𝑛^√𝑆) ^3/2+ √_𝑈2 ^𝑄2(_𝑈𝑛^√𝑆)^3/2−4(𝜃+cot 𝜃)^𝑄_𝑈 2(𝜃+cot 𝜃)

Therefore, in rder thave a feasible slution, 𝑄²

𝑈² (√𝑆 𝑈𝑛)

3/2

≥ 4(𝜃 + cot 𝜃)𝑄 𝑈 𝑆^3/2

𝑈⁴

≥

^{4𝑛3(𝜃+ cot 𝜃)}_𝑄

U

≤ [

_4𝑛3^𝑄𝑆_{(𝜃+cot 𝜃)}^3/2

]

^1/4

This means that for designing a trapezoidal section for a lined channel, the velocity will have to be suitably chosen so as not to violate the above criterion in order to have a feasible solution (see Example 8.2 for illustration).

A trapezoidal section with rounded corners, as shown below, is used at higher discharges. The side slope is maintained to be the permissible slope of the soil. Brick and concrete tiled linings are commonly used in India and the mean velocity is restricted to 2.0 m/s in these cases to avoid danger to the material forming the lining. Thus the design is based on a limiting velocity rather than on any relation between B, h and z. The procedure of design is illustrated in example 4.4.

The U.S.B.R practice is to specify a width-discharge relationship for trapezoidal channels, thereby placing no direct limit on the velocity. Further, no rounding of the bottom is specified by them. The bottom width recommended by U.S.B.R are as follows:

Table 21 Recommended Bottom Widths for Lined Trapezoidal Canals Q, m³/s 1.0 10.0 20.0 100.0 B, m 2.4 2.9 8.0

The design involves the use of the Manning’s equation and Table 4.1 for the determination of h and B for given values of Q, n and S.

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Example 4.3

Design a lined canal to carry 30 m3/s on a slope of 1 in 1600. The side slope is to be maintained at 1.25H : 1V and the lining is expected to give a value of n equal to 0.014.

Solution

Since Q is less than 55 m³/s, a triangular section with a rounded bottom will be used. Refer to fig

𝐴 = 2 (¹₂ℎ²cot 𝜃) + ¹

2ℎ²2𝜃

= ℎ²(𝜃 + cot 𝜃)

P = 2hcot 𝜃 + 2ℎ𝜃 = 2h (𝜃 + cot 𝜃) R = A/P = h/2

Cot𝜃 = z = 1.25 𝜃 = 38.6⁰ = 0.644 𝑟𝑎𝑑𝑖𝑎𝑛

A = ( 1.25 + 0.644) h² = 1.894 h² From mannings equation

Q = A/p R^2/3 S^1/2 h ^8/3 = 14.1

h = 2.7 m

Example 8.2 Design a lined channel to carry a discharge of 300 m3/s through an alluvium whose angle of repose is 31°. The bed slope of the channel is 7.75 × 10–5 and Manning’s n for the lining material is 0.016.

Solution: Since Q > 55 m3/s, trapezoidal section with rounded corners, Fig. 8.2, is to be designed. Here,

Side slope 𝜃 = 310 = 0.541 radian Cot 𝜃 = 1.664

𝜃 + cot 𝜃 = 2.205

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Side slope 𝜃 = 31⁰ = 0.541 radians Cot 𝜃 = 1.664

𝜃 + cot 𝜃 = 2.205 A = Bh + 2.205 h²

P = B + 4.41 h Adopting U = 2 m/s

A = 300/2 = 150 m² Bh + 2.205h² = 150

∴ 𝐵ℎ + 2.205 ℎ²= 150

And

R = (

^𝑈_√𝑆^𝑛

)

^3/2

= (

_{√7.75 𝑋 10}^2𝑥0.016₋₅

)

^3/2

= 6.93 m

∴ 6.93 (𝐵 + 4.41ℎ) = 150 B = 21.645 – 4.41h

∴ 21.645ℎ − 4.41ℎ² + 2.205ℎ² = 150 Or 2.205h² – 21.645h + 150 = 0

∴ h

=

21645 ± √(21645)²+4𝑥2.205𝑥150 4.41

Obviously, the roots of h are imaginary. Using the criterion, Eq. (8.4), one gets, U ≤ [_4𝑛3^𝑄𝑆_{(𝜃+𝑐𝑜𝑡𝜃)}^3/2 ]^1/4

≤ [300𝑋(7.75𝑋10⁻⁵)^3/2 4(0.016)³(2.205) ]

1/4

≤ 1.543 𝑚/𝑠

∴ Adopt U = 1.5 m/s ∴ A = 200 m²

R =

(

1.5 𝑋 0.016

√7.75 𝑋 10⁻⁵

)

^3/2

= 4.50 𝑚

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4.5x(B + 4.41h) = 200 B = 44.44 – 4.41h

= 14.52 m and h = 6.784 m

Other value of h (= 13.37 m ) gives negative value of B which is meaningless B = 14.52 m and h = 6.784 m.