Recommended
More Related Content
What's hot
What's hot (20)
Similar to Hydraulic Exponent for Critical flow computation
Similar to Hydraulic Exponent for Critical flow computation (20)
Recently uploaded
Recently uploaded (20)
Hydraulic Exponent for Critical flow computation
- 2. TITLE OF PRESENTATION Hydraulic Exponent for Critical-flow Computation & Computation of Critical-flow By Engr. Zeeshan Ahmed Soomro 2k19-ID-32 Masters Scholar at Faculty of Agricultural Engineering Sindh Agriculture University Tandojam Pakistan
- 3. WHAT IS HYDRAULIC EXPONENT FOR CRITICAL FLOW • Hydraulic exponents are the parameters M and N in the mathematical relations for open-channel flows as expressed in Z²=CyM and K²=CyN, in which Z = the section factor for critical-flow computation; K = the conveyance of the channel flow cross section; y = the flow depth; and C = a coefficient.
- 4. • This clarifies the mathematical derivation of the hydraulic exponents and their implication in application to the computation of flow profiles of gradually varied flow in a prismatic channel. The relationships between these hydraulic exponents, which are assumed to depend only on the average flow depth, and those, which are assumed to vary with the flow depth of a given range, are also explained.
- 5. HYDRAULIC COMPONENT FOR CRITICAL FLOW COMPUTATION • Since the section factor for critical flow computation (Zc) is a function of depth of flow; it may be assume that Where additionally C is coefficient, M is hydraulic exponent for critical flow computation Eq…..1
- 6. Taking log of both sides of equation 1, we get 2 log Z═ Log C + M log y By differentiating Eq 2 2 𝑍 𝑑𝑦 𝑑𝑧 = 0 + 𝑀 𝑦 𝑀 = 2𝑦 𝑍 𝑑𝑍 𝑑𝑦 Eq…..2 Eq…..3
- 7. • But 𝑍 = 𝐴 √𝐷 = 𝐴⅔ √𝑇 𝑍 = 𝐴⅔ × 𝑇−0.5 𝑑𝑍 𝑑𝑦 = 3 2 𝐴0.5 𝑇0.5 𝑑𝐴 𝑑𝑦 − 1 2 𝑇−1.5 𝐴1.5 𝑑𝑇 𝑑𝑦 𝑑𝑍 𝑑𝑦 = 3 2 𝐴0.5 𝑇 𝑇0.5 − 1 2 𝑇−1.5 𝐴1.5 𝑑𝑇 𝑑𝑦
- 8. 𝑑𝑍 𝑑𝑦 = 3 2 𝐴0.5 𝑇0.5 − 1 2 𝑇−1.5 𝐴1.5 𝑑𝑇 𝑑𝑦 Substituting the values in eq:(2); we get 𝑀 = 2𝑦 𝐴1.5 𝑇−0.5 3 2 𝐴0.5 𝑇0.5 − 1 2 𝑇−1.5 𝐴1.5 𝑑𝑇 𝑑𝑦 𝑀 = 3𝑦𝑇 𝐴 − 𝑦 𝑇 𝑑𝑇 𝑑𝑦 𝑀 = 𝑦 𝐴 3𝑇 − 𝐴 𝑇 𝑑𝑇 𝑑𝑦 Eq…..4 Eq…..5
- 9. M FOR RECTANGULAR CHANNEL 𝑀 = 𝑦 𝐴 3𝑇 − 𝐴 𝑇 𝑑𝑇 𝑑𝑦 𝑇 = 𝑏 𝑑𝑇 𝑑𝑦 = 0 𝑀 ≡ 𝑦 𝐴 3𝑇 𝑀 = 3𝑦𝑏 𝑏𝑦 𝑀 = 3
- 10. M FOR TRAPEZOIDAL CHANNEL 𝑀 = 𝑦 𝐴 3𝑇 − 𝐴 𝑇 𝑑𝑇 𝑑𝑦 𝐴 = 𝑏 + 𝑧𝑦 𝑦 𝑇 = 𝑏 + 2𝑧𝑦 𝑑𝑇 𝑑𝑦 = 2𝑧 𝑀 = 𝑦 𝑏 + 2𝑧𝑦 3 +2𝑧𝑦 − 𝑏 + 𝑧𝑦 𝑏 + 2𝑧𝑦 2𝑧 𝑀 = 3 𝑏 + 2𝑧𝑦 𝑏 + 𝑧𝑦 − 𝑏 + 𝑧𝑦 𝑦 𝑏 + 2𝑧𝑦 𝑏 + 𝑧𝑦 2z 𝑀 = 3 1 + 2𝑧𝑦 𝑏 1 + 𝑧𝑦 𝑏 − 2zy 𝑏 1 + 2𝑧𝑦 𝑏 Substituting the values in above equation We get: Eq..6
- 11. • Let zy/b = y • 𝑀 ≡ 3(1+2𝑦) (1 +𝑦) − 2𝑦 1 +2𝑦 • 𝑀 = 3 1 + 2𝑦 2 − 2𝑦 1 +𝑦 1+ 3y + 2y • 𝑀 = 3 +12𝑦 +12𝑦2−2𝑦−2𝑦2 1+ 3y + 2y • 𝑀 = 3 +10𝑦 +10𝑦2 1+ 3y + 2y Eq…..7
- 12. PROBLEM • A trapezoidal channel having bed width of 20 ft, side slope 2:1 and depth of flow 5 ft. caries a discharge od 400 cusecs. Find the hydraulic exponent for critical flow computation using algebraic method and graph / chart? Solution: DATA; Bed width = 20ft Side slope 2:1 Depth of flow = 5ft Discharge = 400 cusecs
- 13. ALGEBRAIC METHOD • 𝑀 = 3 +10𝑦 +10𝑦2 1+3𝑦+2𝑦2 • 𝑦 = 𝑧𝑦 𝑏 • = 2×5 20 • = 0.5 • Therefore • 𝑀 = 3+10𝑥0.5 +10𝑥(0.5)2 (1+3𝑥0.5𝑥+2 (0.52) 𝑀 = 3 + 5 + 2.5 1 + 1.5𝑥 + 0.5 𝑀 = 3.5
- 14. GRAPHICAL METHOD • y/b = 5/20 = 0.25 From figure at y/b = 0.25, the value of M be read against z= 2 thus M= 3.5 Curve for M values
- 15. COMPUTATION OF CRITICAL FLOW • Computation of critical flow involves the determination of critical depth and velocity when the discharge of channel section are known. • On the other hand, if the critical depth and channel section are known, the critical discharge can be determined by method described in example given below.
- 16. • 1. Algebraic method • For a simple gematric channel section, the critical flow can be determined by an algebraic computation using the basic equations. • 2. method of design chart/Graph • The design chart for determining the critical depth can be used with great expediency.
- 17. PROBLEM • Compute the critical depth and velocity of the trapezoidal channel having bed width 20 ft. and side slope 2:1 and carrying a discharge of 400 cusecs using algebraic method and design chart/graph? DATA: Bed width = 20 ft Side slope = 2:1 Discharge = 400 cusecs
- 18. BY ALGEBRAIC METHOD • The hydraulic depth and water area of the trapezoidal section in terms of the depth (y) as: • 𝐷 = 𝐴 𝑇 • 𝐴 = 𝑦 𝑏 + 𝑧𝑦 • 𝑇 = 𝑏 + 2𝑧𝑦 • 𝐷 = 𝑦 𝑏+𝑧𝑦 𝑏+2𝑧𝑦 The velocity is 𝑉 ≡ 𝑄 𝐴 ≡ 400 𝑦 20 + 2𝑦 Substitute the above expression for D and V in Eq: 𝑉2 2𝑔 = 𝐷 2 4002 2𝑔 𝑦 20 + 2𝑦 2 = 𝑦 10 + 𝑦 10 + 2𝑦
- 19. • Simplification of above expression yields; • 2484(5+ y) = 𝑦 10 + 𝑦 3 • Solving this equation for y by a trial and error procedure the value of required parameters is; • Yc =2.15 ft • Ac = 52.2 ft • 𝑉𝑐 = 400 52.2 = 7.66 𝑓𝑡 𝑠𝑒𝑐
- 20. BY DESIGN CHART/GRAPH • 𝑍 = 𝑄 𝑔 • 𝑍 = 400 32.2 • 𝑍 𝑏2.5 = 0.0394 For this value of 𝑍 𝑏2.5 , and z = 2 from graph the value of y/b = 0.108 Then 𝑌𝑐 = 3 𝑄2 𝑏2 𝑔 𝑌𝐶 = 3 4002 (0.1402× 32.2 Yc =2.16 ft