Best Hydraulic Sections

Faculty of Engineering & TechnologyINTI International University

Best hydraulic section (or) Most economical sectionIt is known that the conveyance of a channel section increases with increase in the hydraulic radius or with decrease in the wetted perimeter.

From a hydraulic view point, therefore the channel section having the least wetted perimeter for a given area has the maximum conveyance; such a section is known as the best hydraulic section.

Best hydraulic section is also called asMost economical section, orOptimum cross section, orOptimum shape, orOptimum depth

Best hydraulic section for a rectangular channelRectangular channelWidth = B,Depth = D, wetted perimeter P = B + 2D D Area = BD,B = A/D To obtain best hydraulic section, minimize the wetted perimeter by differentiating P with respect to the controlling variable and equate to zero.P = B + 2D = A/D + 2DFor a given value of A, surface roughness and channel slope, P will be minimum when dP/dD = 0,dP/dD = -AD-2 + 2 = 0A/D2 = 2 B = 2DSo, for the maximum discharge, the width B should be twice the depth D.B

Best hydraulic section for a rectangular channelRectangular channelFor a given value of area, surface roughness and channel slope,B = 2DFor the maximum discharge, the width B should be twice the depth D.

Hydraulic radius: R = A/P = BD/(B + 2D) = 2D2/4D = D/2R = D/2Hydraulic radius is half the depth of water

Best hydraulic section for a trapezoidal channelTrapezoidal channelP = B + 2D(1+z2) A = (B + 2zD + B) D = D (B + zD) B = A/D zD 1 D Substituting B into P Z P = B + 2D(1+z2) = A/D zD + 2D(1+z2) (1) BIf A and z are fixed, P will be a minimum when dP/dD = 0.

Differentiating equation (1) with respect to D,dP/dD = -AD-2 z + 2(1+z2) = 0A = D2(2(1+z2) z)BD + zD2 = D2(2(1+z2) z)BD = 2D2(1+z2) 2zD2B = 2D((1+z2) z)B + 2zD = 2D(1+z2)Top width is twice the wetted slope length B + 2zD

Best hydraulic section for a trapezoidal channelTrapezoidal channelFor a given value of area, surface roughness and channel slope,B + 2zD = 2D(1+z2)For maximum discharge, the top width is twice the wetted slope length

Hydraulic radius: R = A/P = (B + zD)D/(B + 2D(1+z2)) = (B + zD)D/(B + B + 2zD) R = (B + zD)D/2(B + zD) R = D/2Hydraulic radius is half the depth of water

Hydraulically efficient trapezoidal channel O R 1 T z Le Be OS = ye OT = OR sin = OR/(z2 +1)OR = 1/2Be +zye = (2ye 1+z2 - 2zye ) + zye = ye (z2 +1) OT = OS = ye The proportions of a hydraulically efficient trapezoidal channel will be such that a semicircle. S

The side slope m was held constant. If m is allowed to vary, the optimum value of m to make Pe most efficient is obtained by dPe /dz = 0

Hydraulically efficient trapezoidal channelSetting dPe /dz = 0,

4z2 = 1 + z2zem = 1/3 = cotm = 60Pem = 2/3 yem Bem = 2/3 yem A = 3 yem 2 Lem = 2/3 yem = Bem

Hydraulically efficient trapezoidal channel

Table 1 Proportion of some most efficient sections

NoChannel shapeAPemBemRemTem1Rectangular2yem24yem2yemYem/22yem2Trapezoidal(half hexagon)3 yem223 yem2/3 yem yem/24/3 yem3Circular (semicircle)/2 yem2 yem-yem/22yem4Triangle (vertex angle = 90)yem223 yem-yem/222yem

Example 15.2A trapezoidal channel has side slopes of 3 horizontal to 4 vertical and the slope of its bed is 1 in 2000. Determine the optimum dimensions of the channel if it is to carry water at 0.5 m3 /s.Use the Chezy formula, assuming that C = 80 m1/2 s-1

1 D 4 1 z 3 3/4 BB + 2zD

Example 15.2For optimum dimensions: Top width (B+2zD) is twice the wetted slope lengthB + 2zD = 2D(1+z2)B = 2D (1+z2) 2zD = = 2D [(1+z2) z] = 2D [(1+9/16) 3/4] = 2D[(25/16) 3/4]B = 2D(5/4 3/4) = 2D(1/2)B = D

Area A = (B + zD))D = (D + zD)D = D2(1 + z) = D2 (1 + 3/4) = 7/4 D2

Hydraulic radius R = D/2

0.5 = (7/4D2) (80)(D/2)1/2(1/2000)1/2 = 7/4[80D5/2 / (2x2000)1/2]D = 7/4[(80D5/2/(63.25)1/2] D = 0.552 mTherefore, B = D = 0.552 m

Best hydraulic section for a circular channelFor circular channels which are not flowing full, there will be optimum depths of flow for maximum velocity and for maximum discharge.Chezys formula,V = C RS = C(A/P)SIf the free surface subtends an angle 2 at the center O for any depth y,Area A= sector OSTU triangle OSU = r.(r2) (r sin)(r cos)2 = r.(r2) r2sin.cos = r2 - r2 sin2 = r2( - sin 2)Wetted perimeter, P = 2r

yrrOSTUP

Best hydraulic section for a circular channel(a) For maximum velocity:V = C RS = C(A/P)SFor constant values of C and S, the maximum value of v will occur when A/P is maximum. Differentiating (A/P) with respect to and equal to zero,(1)By substituting A, P, dA/d and dP/d into (1), we obtain2 = tan 2giving 2 = 257.5 Depth of flow, y = r r cos = 1.62 r y = 0.81 DDepth of flow = 0.81 Diameter of pipe

Best hydraulic section for a circular channel(b) For maximum discharge:Q = ACRS = AC(A/P)S = C(A3/P)1/2CS1/2For given values of C and S, the discharge Q will be a maximum when A3/P is a maximum. Differentiating (A3/P) with respect to and equating to zero.(1)By substituting A, P, dA/d and dP/d into (1), we obtain2 = 308, = 154 = 2.68 radiansDepth for maximum discharge, y = r(1 cos ) = 1.9 r y = 0.95DDepth of flow = 0.95 Diameter of pipe

Example on circular sectionA sewer diameter D = 0.6 m, has a slope of 1 in 200.What will be the maximum velocity of flow that can occur and what is the discharge at this velocity?Take Chezy C = 55 SI units.

Example on circular sectionFor max. velocity, depth of flow y = 1.62r = 1.62(D/2) = 1.62x0.3 = 0.486 m

Velocity V = CRSFor max. velocity, 2 = 257.5 = 4.5 radians(or) = 128.75 = 2.247 rad. Area A = r2( - sin 2)A = 0.32 (2.247 sin257.5) = 0.246 m2 P = 2r = 2x0.3x2.247 = 1.35 mR = A/P = 0.246/1.35 = 0.1825 m

V = 55(0.1825x1/200) = 1.66 m/s

Discharge at max. velocity, Q = AV = 0.246x1.66 = 0.408 m3/s

Most economical triangular channel sectionA = 2( y tan * y) = y2 tan P = 2y sec y = (A/tan )0.5

y=

2tan sec tan = sec3 Sec (2 tan2 - sec2 ) = 02 tan2 = sec22 tan = sec 2 sin /cos = 1/cos sin = 1/2 = 45

= 0

Compound sectionsSome channel sections may be formed as a combination of elementary sections. Typically natural channels, such as rivers, have flood plains which are wide and shallow compared to the deep main channel.A simplified section of a stream with flood banks are known as compound sections.

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Compound sectionsmethod of discharge estimationThe discharge is calculated as the sum of the partial discharges in the sub-areas; for e.g. units 1, 2 and 3 in the figure QP = Qi = Vi Ai The discharge is also calculated by considering the whole section as one unit, (portion ABCDEFGH in the figure), say Qw The larger of the above two discharges Qi and Qw is adopted as the discharge at the depth y.

Poseys methodIn this method, while calculating the wetted perimeter for the sub-area, the imaginary divisions (FJ and CK) are considered as boundaries for the deeper portion only and neglected completely in the calculation relating to the shallower portion. This way the shear stress that occurs at the interface of the deeper and shallower parts is empirically accounted for.Zero shear method The interfaces are not counted as perimeter either for the deep portion or for the shallow portion.

ExampleFor the compound channel shown in the figure below determine the discharge for a depth of flow of 1.2 m. Calculate the discharge.

17 m1327 m0.9 m3 myn = 0.02 So = 0.0002

SolutionPartial area discharge by Poseys methodSubarea 1: A1 = 7x0.3 = 2.1 m2 P1 = 0.3 + 7 = 7.3 m R = 2.1/7.3 = 0.288 m = 0.647 m3 /s Q3 = 0.647 m3 /sSubarea 2: A2 = 3x1.2 = 3.6 m2 P2 = 3 + 1.2 + 1.2 = 5.4 m R = 3.6/5.4 = 0.667 m Q2 = 1.943 m3 /s Q = Q1 + Q2 + Q3 = 0.647 + 1.943 + 0.647 = 3.237 m3 /s

SolutionPartial area discharge by zero shear methodSubarea 1: A1 = 7x0.3 = 2.1 m2 P1 = 0.3 + 7 = 7.3 m R = 2.1/7.3 = 0.288 m = 0.647 m3 /s Q3 = 0.647 m3 /sSubarea 2: A2 = 3x1.2 = 3.6 m2 P2 = 3 + 0.9 + 0.9 = 4.8 m R = 3.6/4.8 = 0.75 m Q2 = 2.1 m3 /s Q = Q1 + Q2 + Q3 = 0.647 + 2.1 + 0.647 = 3.395 m3 /s

SolutionTotal section methodA = 2.1 + 2.1 + 3.6 = 7.8 m2 P = 0.3 + 7 + 0.9 + 3 + 0.9 + 7 + 0.3 = 19.4 mR = 7.8/19.4 = 0.402 m = 3.005 m3 /s

ExampleA circular channel of 1 meter diameter has a bed slope of 1 in 1500. (i) Find the maximum discharge through the channel. Assume maximum discharge when the wetted perimeter subtends an angle of 308 degrees at the centre and take C = 50 in the Chezy formula. (ii) Using that maximum discharge, find the optimum dimension of a rectangular channel while the bed slope and C is same as the circular section.

Optimum section for a circular channel: A = r2 ( - sin cos ) 2 = 308, = 2.688 rad = 0.52x2.688 0.52 sin 154 cos 154 = 0.77 m2P = 2r = 2x0.5x2.688 = 2.688 mR = A/P = 0.286 m

= 50 x 0.77 x 0.2861/2 x (1/1500)1/2 = 0.532 m3/s

For optimum section for rectangular section, B = 2D R = D/2A = BD = 2D2P = B + 2D = 4DQ = 2D2x50x(0.5D)1/2x(1/1500)1/20.532 = 1.826 D5/2D = 0.611 mB = 2D = 1.221 m

The End