Discussion of “Solution of Specific Energy and Specific Force Equations” by Amlan Das
f ER共y *兲 = 1 − y *R −
Q *R y2 R
=0
共4兲
*
where y *R = y / E and Q*R = Q2 / 共2gB2E3兲. An appropriate arrangement of g2R共y *R兲 and g1R共y *R兲 for calculating y *R2 = y 2 / E and y *R1 = y 1 / E respectively could take the following forms
August 2007, Vol. 133, No. 4, pp. 407–410.
DOI: 10.1061/共ASCE兲0733-9437共2007兲133:4共407兲
Ali R. Vatankhah1 and Salah Kouchakzadeh2 1
Ph.D. Candidate, Irrigation and Reclamation Engineering Dept., University College of Agriculture and Natural Resources, Univ. of Tehran, P.O. Box 4111, Karaj, Iran 31587-77871. E-mail:
[email protected] 2 Professor, Irrigation and Reclamation Engineering Dept., University College of Agriculture and Natural Resources, Univ. of Tehran, P.O. Box 4111, Karaj, Iran 31587-77871. E-mail:
[email protected]
y *R = g2R共y *R兲 = 1 −
y2 R
共5兲
*
y *R = g1R共y *R兲 = The discussers appreciate the author’s presentation of a new methodology for simultaneous determination of alternate and sequent depths. The traditional fixed-point method, however, could be used to present solutions that overcome the difficulty of the traditional computational procedures and converge faster toward the solution than the method proposed by the author. The dimensionless specific energy equation for a trapezoidal cross section takes the form
Q *R
冑
Q *R 1 − y *R
共6兲
Using 90% and 10% of the vertical asymptote value of Eq. 共6兲 共i.e., y *R = 1兲 as initial guesses would guarantee fast convergence to the desired roots y *R2 and y *R1, respectively. Similarly for a triangular cross section Eq. 共1兲 simplifies to f ET共y *兲 = 1 − y *T −
Q *R y4 T
=0
共7兲
*
f E共y *兲 = E* − y * −
Q* 共1 + y *兲2y *2
=0
共1兲
where y * = Zy / B, E* = ZE / B and Q* = 关Q2Z3 / 共2gB5兲兴. Different iterative arrangements of Eq. 共1兲 in the form of y *共n+1兲 = g共y *共n兲兲, n = 0 , 1 , 2 , 3 , . . . might be considered for alternate depths determination. However, the most suitable form that converges to the desired root by using the fixed-point iteration technique should be sought. Two arrangements of f E = 共y *兲 in the forms of Eqs. 共2兲 and 共3兲 are proposed herein for calculating the nondimensional alternate depths y *2 = Zy 2 / B and y *1 = Zy 1 / B, respectively. The subscripts 1 and 2 refer to the small and the large depths, respectively y * = g2共y *兲 = E* −
y * = g1共y *兲 = − 0.5 +
Q*
共2兲
共1 + y *兲2y *2
冑 冑 0.25 +
Q* E* − y *
where y *T = y / E and Q*T = Q2 / 共2gZ2E5兲. For a triangular cross section appropriate arrangements for g2T共y *R兲 and g1T共y *R兲 for calculating y *T2 = y 2 / E and y *T1 = y 1 / E respectively could be presented in the following forms: y *T = g2T共y *T兲 = 1 −
Q *T y4 T
冑 4
y *T = g1T共y *T兲 =
Q *T 1 − y *T
共9兲
For calculating the roots of Eqs. 共8兲 and 共9兲 共i.e., y *T2 and y *T1兲 the condition for initial guesses as previously described might be used for the triangular cross section. The same procedure was pursued for solving the dimensionless specific force equation for a trapezoidal cross section 关Eq. 共10兲兴
共3兲
The general forms of g2共y *兲 and g1共y *兲 are drawn in Fig. 1. The variable g2共y *兲 has two asymptotes 共i.e., z* = E* and y * = 0兲 and intersects with the abscissa at y *g = −0.5+ 关0.25
+ 共Q* / E*兲0.5兴0.5. Also, g1共y *兲 has two asymptotes 共i.e., z* = 0 and y * = E*兲 and intersects with the ordinate at z*g=−0.5+ 关0.25
+ 共Q* / E*兲0.5兴0.5. The slope of g1共y *兲 and g2共y *兲 functions near the root governs their behavior. That is, starting the computation from points having a slope close to zero tend to increase the convergence rate, while using any point with slope greater than +1 would either diverge the iteration process or produce another root. Considering 0.1E* 共10% of the asymptote value兲 as an initial guess for calculating y *1 and 0.9E* 共90% of the asymptote value兲 for calculating y *2 speed up the convergence process. For a rectangular cross section, Eq. 共1兲 simplifies to
共8兲
*
Fig. 1. General presentation of g1共y *兲 and g2共y *兲
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Table 1. Convergence Details of Problem Iteration Problem 1
0 1 2 3 4 5
Problem 2
0 1 2 3 4 5
Problem 3
0 1 2 3 4 5
Problem 4
0 1 2 3 4 5
Problem 5
0 1 2 3 4 5
Problem 6
0 1 2 3 4 5
Problem 7
0 1 2 3 4 5
Fig. 2. General form of h1共y *兲 and h2共y *兲
f F共y *兲 = F* −
3y *2 + 2y *3
−
6
2Q* 共1 + y *兲y *
共10兲
=0
where y * = Zy / B, F* = ZF / B3, and Q* = 共Q2Z3兲 / 共2gB5兲. The sequent depths are obtained by solving Eq. 共10兲. An appropriate arrangement of h2共y *兲 and h1共y *兲 for y *2 = Zy 2 / B and y *1 = Zy 1 / B determination is given by Eqs. 共11兲 and 共12兲, respectively
冑 冉 冑 3
y * = h2共y *兲 =
6y *
3 + 2y *
y * = h1共y *兲 = − 0.5 +
F* −
0.25 +
2Q* 共1 + y *兲y *
冊
2Q* F* − y *2 /2 − y *3 /3
共11兲
共12兲
Fig. 2 presents the general form of h2共y *兲 and h1共y *兲 functions. The function h2共y *兲 has two asymptotes 关i.e., z* = 共3F*兲1/3 and y * = 0兴 and it intersects the abscissa at y *h = −0.5+ 共0.25
+ 2Q* / F*兲0.5. Likewise, h1共y *兲 intersects the ordinate at z*h = −0.5+ 共0.25+ 2Q* / F*兲0.5 and has two asymptotes 共i.e., z* = 0 and y m 兲 one of the following equation roots *
2y *3 + 3y *2 − 6F* = 0 Since the root must be real and positive, it could be determined by the following relationship
冑 + 0.5关12F* + 2冑36F* − 6F* − 1兴
y *m = 0.5关12F* + 2 36F*2 − 6F* − 1兴−1/3 2
1/3
− 0.5
Now h1 and h2 function could be used for calculating y *1 and y *2 provided that an appropriate initial guess is considered. In this case values of 0.1 F*0.45 and F*0.45 共F*0.45 is an approximation of y *m兲 could be used to determine y *1 and y *2, respectively. For a rectangular cross section, Eq. 共10兲 simplifies to f FR共y *兲 = 1 − 0.5y 2 R − *
Q *R y *R
=0
共13兲
y
y
*1
0.0150 0.0336 0.0361 0.0365 0.0366 0.0366 y 1 = 1.830 0.0050 0.0570 — — — — y1 = — 0.1000 0.1162 0.1167 0.1167 0.1167 0.1167 y 1 = 0.700 0.1000 0.1619 0.1678 0.1684 0.1684 0.1685 y 1 = 1.011 0.0124 0.0332 0.0349 0.0351 0.0351 0.0351 y 1 = 1.755 0.1400 0.0691 0.0691 0.0691 0.0691 0.0691 y 1 = 2.020 0.1400 0.3255 0.3404 0.3422 0.3424 0.3424 y 1 = 1.083
*2
0.1350 0.1431 0.1439 0.1440 0.1440 0.1440 y 2 = 7.200 0.0450 −0.0238 −0.2533 0.0454 −0.0223 −0.2939 y2 = — 0.9000 0.9998 0.9998 0.9998 0.9998 0.9998 y 2 = 5.999 0.9000 0.9750 0.9752 0.9752 0.9752 0.9752 y 2 = 5.851 0.1236 0.1183 0.1161 0.1151 0.1147 0.1145 y 2 = 57.25 1.3000 1.4409 1.4411 1.4411 1.4411 1.4411 y 2 = 42.138 1.3000 1.2264 1.2142 1.2120 1.2116 1.2116 y 2 = 3.831
where y *R = y共B / F兲0.5 and Q*R = Q2 / 关g共BF3兲0.5兴. For calculating y *R2 and y *R1 appropriate arrangements of h2R共y *R兲 and h1R共y *R兲 are presented by y *R = h2R共y *R兲 =
冑冉 冊
y *R = h1R共y *R兲 =
2 1−
Q *R y *R
Q *R 1 − 0.5y 2 R
共14兲
共15兲
*
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For evaluating y *R2 and y *R1 initial guesses 0.9冑 2 共90% of asymptote value兲 and 0.1冑 2 might be used respectively. For a triangular cross section Eq. 共10兲 takes the form
Discussion of “Minimum Specific Energy and Critical Flow Conditions in Open Channels” by H. Chanson September/October 2006, Vol. 132, No. 5, pp. 498–502.
DOI: 10.1061/共ASCE兲0733-9437共2006兲132:5共498兲
f ET共y *兲 = 1 −
y3 T *
3
−
Q *R y2 T
=0
共16兲
1
*
where y *T = y共Z / F兲1/3 and Q*T = Q2 / 关g共ZF5兲1/3兴. For this cross section appropriate arrangements of h2T共y *T兲 and h1T共y *T兲 for y *T2 and y *T1 calculation are given by Eqs. 共17兲 and 共18兲, respectively
冑冉 冊 3
y *T = h2T共y *T兲 =
3 1−
Q *T y2 T
共17兲
*
y *T = h1T共y *T兲 = 1/3
冑
Ali R. Vatankhah1; Salah Kouchakzadeh2; and A. H. Hoorfar3
Q *T 1 − y 3 T/3
共18兲
Ph.D. Candidate, Irrigation and Reclamation Engineering Dept., Univ. College of Agriculture and Natural Resources, Univ. of Tehran, P.O. Box 4111, Karaj, Iran 31587-77871. 2 Professor, Irrigation and Reclamation Engineering Dept., Univ. College of Agriculture and Natural Resources, Univ. of Tehran, P.O. Box 4111, Karaj, Iran 31587-77871. E-mail:
[email protected] 3 Assistant Prof., Irrigation and Reclamation Engineering Dept., Univ. College of Agriculture and Natural Resources, Univ. of Tehran, P.O. Box 4111, Karaj, Iran 31587-77871.
The discussers would like to thank the author for presenting nondimensional solutions for the critical flow conditions in open channels. Regarding the S1 solution the author confirmed that the solution does not concur with the experimental data and attributed the discrepancy to the solution instability. It could readily be shown that S1 solution is highly sensitive. Therefore, it might provide nonrealistic answers for some circumstances. Solving Eq. 共8兲 for CD yields
*
1/3
Values 0.9共3兲 and 0.1共3兲 could be used as initial guesses for computing y *T2 and y *T1, respectively. Table 1 presents the convergence details of the proposed equations.
CD =
1.5冑3
冑⌳
dc ⌳ Emin
2
−
dc ⌳ Emin
3
A relative sensitivity index, S, for the dimensionless discharge coefficient,CD, could be defined as follows: S=
Closure to “Solution of Specific Energy and Specific Force Equations” by Amlan Das
冑冉 冊 冉 冊
CD CD
冒
⌳ ⌳ CD = ⌳ CD ⌳
Differentiating CD with respect to ⌳ yields
August 2007, Vol. 133, No. 4, pp. 407–410.
DOI: 10.1061/共ASCE兲0733-9473共2007兲133:4共407兲
Amlan Das1 1
Professor, Dept. of Civil Engineering, National Institute of Technology, Durgapur 713209, West Bengal, India.
The discussers deserve an acknowledgment for taking interest in the writer’s work. The subject matter in the discussion can be put on record as another application of the fixed-point iterative technique for solving the polynomial equations. The faster or slower rate of convergence was never an issue in the writer’s work. The main strength of the writer’s work is its capacity of identifying the data inconsistency in a conclusive manner. The second example of the writer’s work shows the solution results suffice to conclude about the data inconsistency, which no other method can yield. Another advantage of the methodology is that all five roots become available simultaneously, making the methodology applicable in other areas of science and engineering.
Fig. 1. Variation of relative sensitivity and dimensionless critical flow depth at crest for S1 and S3 solutions
882 / JOURNAL OF IRRIGATION AND DRAINAGE ENGINEERING © ASCE / NOVEMBER/DECEMBER 2008
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