1 / 48

Specific Energy

Specific Energy . Channel transitions Steps, cross-section Slope change Local loss locations Gate Piers Hydraulic Jump. EGL. v 2 /2g. y. y c. D z. datum. 1. 2. If we assume no energy loss through the transition we can say:.

jaden
Download Presentation

Specific Energy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Specific Energy • Channel transitions • Steps, cross-section • Slope change • Local loss locations • Gate • Piers • Hydraulic Jump CJM

  2. EGL v2/2g y yc Dz datum 1 2 • If we assume no energy loss through the transition we can say: which says for a constant total energy E, there is a specific energy loss between 1-2 of Dz. CJM

  3. CJM

  4. Open Channel Hydraulics CJM

  5. Open Channel Hydraulics • Three basic relationships • Continuity equation • Energy equation • Momentum equation CJM

  6. 3a Inflow 3 A Change in Storage 3b Outflow 1 2 A Section AA Continuity Equation Inflow – Outflow = Change in Storage CJM

  7. Area of the cross-section (ft2) or (m2) Avg. velocity of flow at a cross-section (ft/s) or (m/s) Flow rate (cfs) or (m3/s) General Flow Equation Q = vA CJM

  8. 3 A 3 Section AA 1 2 A Q1 Q2 Q1 – Q2 = Change in storage rate CJM

  9. elevation head velocity head Energy loss between sections 1 and 2 pressure head Energy Equation (Bernoulli’s) CJM

  10. Open Channel Energy Equation • In open channel flow (as opposed to pipe flow) the free water surface is exposed to the atmosphere so that p/g is 0 (on the surface; or p/g is y at the invert), leaving: CJM

  11. Q Channel Bottom z1 z2 Datum 2 1 CJM

  12. HGL y1 y2 Channel Bottom z1 z2 Datum 1 2 CJM

  13. EGL HGL y1 y2 Channel Bottom z1 z2 Datum 2 1 CJM

  14. hL1-2 EGL HGL y1 y2 Channel Bottom z1 z2 Datum 1 2 CJM

  15. A E y Datum and channel bottom A Eqn. 2.2 Specific Energy E is the specific energy. CJM

  16. y Section AA q = vy Where q is the flow per unit width. (Eqn. 2.4) Specific Energy (cont.) Eqn. 2.2 becomes: CJM

  17. y y=E y2 yc y1 Ec E Eo Specific Energy Diagrams *Note q is constant. CJM

  18. or Critical Depth, yc • The depth of flow corresponding to the minimum E is the critical depth, yc CJM

  19. Channel Transitions CJM

  20. Channel transitions occur where there is a change in width, shape, slope, roughness, bottom elevation of the channel. • For changes in slope and roughness we can use backwater curves to evaluate the effects of the changes. • For other types of smooth transitions (transitions were energy loss is minimal), we can use energy relationships to evaluate the impact of the transition. CJM

  21. EGL v2/2g y yc Dz datum 1 2 • If we assume no energy loss through the transition we can say: which says for a constant total energy E, there is a specific energy loss between 1-2 of Dz. CJM

  22. y y=E y1 y2 Dz yc E2 E1 E Specific Energy Representation of a Transition CJM

  23. If the flow must pass through critical depth, the assumption of no energy loss may not be valid. • This is especially true when going from supercritical to subcritical flow. • A hydraulic jump accompanied by considerable energy loss occurs. CJM

  24. Other Minimum Specific Energy Relations • Froude Number: Minimum Specific Energy • Emin:yc • Wave speed of small disturbance CJM

  25. DERIVATIONS • E = y + v2/2g • E = y + q2/2gy2 • dE/dy = 0 = 1- q2/gyc3 • yc = (q2/g) 0.333……(eqtn 2.6)……..SO…..!!!!!…… • Emin = yc + q2/2gyc2 = 3/2 yc • Emin = 3/2 yc CJM

  26. is known as the Froude Number, F Froude Number, F • Compare to (2.6) : q2/gyc3 = 1 • Then, vc2/gyc = 1 at critical conditions • So, at critical conditions, the Froude number =1! CJM

  27. ……..Tie in with wave speed • Speed of a small disturbance on water can be shown to be: vw = (gy)1/2 • So, at critical conditions…surface wave has same velocity as the river (vw = vc) CJM

  28. Flow classification based on Froude number • If F = 1, y = yc and flow is critical. • If F < 1, y > yc and flow is subcritical. • If F > 1, y < yc and flow is supercritical. • F is independent of the slope of the channel, yc dependent only on Q. CJM

  29. Critical Step Height • Figure 2.6: Approach Froude number vs non-dimensional step height (Dz/y1) CJM

  30. Example 2.1 For an approach flow in a rectangular channel with depth of 2.0 m (6.6 ft) and velocity of 2.2 m/s (7.2 ft/s), determine the depth of flow over a gradual rise in the channel bottom of Dz = 0.25 m (0.82 ft). Repeat the solution for Dz = 0.50 m (1.64 ft). CJM

  31. EGL v2/2g V1 = 2.2 Y1= 2.0 y2 Dz = 0.25 datum 1 2 • If we assume no energy loss through the transition we can say: which says for a constant total energy (TE), there is a specific energy loss between 1-2 of Dz (meters). CJM

  32. Steps • Determine U/S Condition (sub- or super-) • Calculate critical depth, compare to u/s depth • Alternately, calculate F1 • Solve specific energy equation for correct root for y2, v2 • Water surface elevation = y2 + Dz CJM

  33. Repeat for a step of 0.50 m • Check value of Emin • Determine value of E2 = (E1-Dz) • Find E2<Emin (CAN’T BE!!!!) • Flow “backs-up” to allow passage of q • Y1 increases; conditions at step are critical, with y2 = yc; E2 = Emin CJM

  34. Contraction/Expansion in Width CJM

  35. Contraction/Expansion in Width width1 width2 Plan View 1 2 • If we assume no energy loss through the transition we can say: V12/2g + y1 = V22/2g + y2 OR q12/2gy12 + y1 = E1 = E2 = q22/2gy22 + y2 q2 and q1 related by continuity .. So two equations; two unknowns CJM

  36. Continuity • q1w1 = q2w2 = Q • q2 = q1 w1/w2 Now, find y2 (3 roots, 2 real…find one that matches with upstream flow condition) Ensure that flow isn’t “choked” by contraction condition CJM

  37. q1 y q2>q1 y1 y2 E CJM

  38. Choking Condition q1 y q2 y1 E E1< E2min CJM

  39. Revised Upstream Condition q1 y q2 y1 y1 THIS IS THE CONTROL! y2 E E1 = E2 = Emin (q2) CJM

  40. yc, General Form • E = y + Q2/2gA2 • Differentiate with respect to y and set DE/dy = 0; find minimum point; ie condition where E = Emin; y = yc • dE/dy = 0 =1 – Q2/gA3 (dA/dy) • 1 = Q2/gA3 (dA/dy) • (if you can determine dA/dy, and can write A(y), can find yc) CJM

  41. General Channel Shape B(y) dy dA=Bdy y A(y) CJM

  42. dA/dy = B (topwidth, B(y)) • So, Q2Bc/gAc3 = 1 (2.18) • Define hydraulic depth D = A/B • V = Q/A • F = V/(gD)1/2 or V/(gD/a)1/2 • Minimum specific energy, Emin = yc + Dc/2 • (From 2.16 and 2.18) CJM

  43. Solve for yc in a non-rectangular channel: • aQ2Bc/gAc3 = 1 • Find value of yc that satisfies this equality • Geometric Elements in Table 2-1 CJM

  44. Example 2.2 • Find the critical depth in a trapezoidal channel with a 20 ft bottom width and 2:1 side slopes if Q = 1000 cfs. Use the bisection technique and compare the solution with that from Figure 2.13 1 2 20’ CJM

  45. F(y) = Q – (g1/2A3/2)/B1/2 • YOYC provides solution OR • Graphical technique (Figure 2.13) • yc = 3.740 feet CJM

  46. Overbank Flow Flood Stage Normal Stage CJM

  47. Overbank Flow: calculate yc • Can’t neglect a • Must consider variation of a withy • dE/dy = 1 – aQ2B/gA3 + Q2/2gA2 da/dy • (2.16 differentiated with A(y) and a(y)) CJM

  48. Overbank Flow (Compound Channel) continued • Fc=(aQ2B/gA3 – Q2/2gA2 da/dy)0.5 • Use Ycomp! CJM

More Related