Basically, it is a flow with free surface. (Free surface is a surface with atmospheric pressure)

Transcription

1 8 OPEN CHANNEL FLOW 8. Classification & Definition Open channel flows are flows in rivers, streams, artificial channels, irrigation ditches, partiall filled pipe etc. Basicall, it is a flow with free surface. (Free surface is a surface with atmospheric pressure) Open Channel Flow Stead flow Unstead flow Uniform Varied Varied Uniform Graduall varied flow Rapidl varied flow Graduall varied flow Rapidl varied flow 8.. Open Channel Geometr Classifications of Open Channel Flow. Depth of flow, : vertical distance from the bottom to surface. Free surface B Bottom flow A (cross-section) Top width, B: the width of the channel at the free surface Flow area, A: cross-sectional area of the flow 4 Wetted perimeter, P: the length of the channel cross-section in contact with the fluid P.8-

2 5 Hdraulic radius (hdraulic mean depth), R: R = Flow area Wetted perimeter = A P 6 Average depth (hdraulic average depth), ave : ave = Flow area Top width = A B 8.. Rectangular channel B = b A = b* P = b+* b R = b* b + * ave = 8.. Trapezoidal channel B = b + *m* A = *(b+m*) b m P = b+** + m R = ave = *(b + m * ) b + * * + *(b + m* ) b + *m* m P.8-

3 8..4 Triangular channel B = *m* A = m* m P = ** + m R = * m* + m ave = 8..5 Circular channel B = * *(D ) A = D * ( θ sin θ) 8 θ D P = θ*d (θ in radian) D R = sin θ 4 θ ave = ( θ sin θ) D* 8*sin θ P.8-

4 8. Stead Uniform Flow For a stead uniform flow depth is constant along the flow velocit is constant over the cross-section time independent 8.. Manning Equations In 890, Manning, an Irish engineer derived a better and more accurate relationship, Manning equation, based on man field measurement. V = *R *S n (8.) n - Manning s coefficient, s/m / (can be found in most of the hdraulic handbooks) To incorporate the continuit equation, Manning equation becomes Q = A *R *S n (8.) As the flow according to Manning equations is for normal stead uniform flow, the flow is Normal Flow the depth is Normal Depth P.8-4

5 Worked examples:. Water flows in a rectangular, concrete, open channel that is m wide at a depth of.5m. The channel slope is Find the water velocit and the flow rate. (n = 0.0) Answer B Manning equation, V = n *R *S with n = 0.0 S = A = *.5 m = 0 m P = + *.5 m = 7 m R = A/P = 0 / 7 m =.765 m hence V = *(.765) *(0.008) 0.0 = m/s Discharge, Q = A*V = 0*5.945 m /s = 78. m /s P.8-5

6 . Water flows in a rectangular, concrete, open channel that is m wide. The channel slope is If the velocit of the flow is 6 m/s, find the depth of the flow. (n = 0.0) h Answer B Manning equation, m V = n *R *S with V = 6 m/s n = 0.0 S = A = * h m P = + *h m R A *h = = P + *h 6*h 6 + h =.790 h =.55 m = 6*h 6 + h Depth of the flow =.55m P.8-6

7 . A trapezoidal channel with side slopes of /, a depth of m, a bottom width of 8 m and a channel slope of has a discharge of 56 m /s. Find the Manning s n. 4m m m 8m m.5 Answer A = (4+8)*/ m = m P = 8 + * + m = 5. m A/P = / 5. m =.446 m B Manning equation, A Q = n *R *S Q = 56 m /s, S = = *(.446) *(0.0009) n n = P.8-7

8 4. Determine the depth in a trapezoidal channel with side slopes of to.5, a bottom width of 8 m and a channel slope of The discharge is 56 m /s and n = (8+*) m 8m.5 Answer A = (8+8+*)*/ m = (8+.5*)* m P = 8 + ** +.5 m = * m R = A/P = (8+.5*)* / * B Manning equation, A Q = *R *S n Q = 56 m /s, S = (8 +.5* )* (8 +.5* )* 56 = * *(0.0009) * + or 5 [(8 +.5* )* ] [ * ] [( * )* ] [ * ] = = 0 B trial & error, =.7 m. The depth of the trapezoidal channel is.7m. P.8-8

9 5. Water flows in the triangular steel channel shown in the figure below. Find the depth of flow if the channel slope is and the discharge is 0. m /s. (n=0.04) 60 o Answer A = tan0 * / m = *tan0 m P = /cos0 m R = A/P = *tan0 / /cos0 m = sin0 / m B Manning equation, A Q = *R *S n Q = 0. m /s, S = tan 0 sin 0 0. = * * = 8/ * or = 0.68 = 0.67 m 8 m Depth of the channel is 0.67 m. P.8-9

10 8.. Optimum Hdraulic Cross-sections (REFERENCE ONLY) From Manning equation, 5 A * S Q = * n P Hence, Q will be maximum when P is a minimum. For a given cross-sectional area, A of an open channel, the discharge, Q is maximum when the wetted perimeter, P is minimum. Hence if the wetted perimeter, P for a given flow area is minimised, the area, A will give the least expensive channel to be construct. This corresponding cross-section is the optimum hdraulic section or the best hdraulic section Rectangular section width depth = b = b area, A = b P = b+* A = + dp A Hence = + = 0 d i.e. = A or b = Therefore, the optimum rectangular section is P.8-0

11 8... Trapezoidal section B A = b+*m* = (b+m*)* b m P = b+** + m B eliminating b from P, A P = + (* + m m)* For a minimum value of P, δp = 0, dp dp i.e. = 0 and = 0 d dm dp From = 0, d = A dp From = 0, m = dm It implies the side slope of the channel is 60 to horizontal. A b = m = = 4 and P = + = i.e. P = *b The optimum section is given as follow: b 60 o b b P.8-

12 8... Other sections N-side Channel from the conclusion of the previous two sections reflection of the rectangular optimum section about the water surface will form a square of side b. reflection of the trapezoidal optimum section about the water surface will form a regular hexagon of side b. For a N-side channel, the optimum hdraulic section should be in a form of half a N-side regular polgon. b φ b b b b b N φ = * 80 N Triangular Section N =, hence φ = o Circular Section From the result of N-side channel, it can be concluded that the optimum section of a circular channel is a semi-circle. It is the most optimum section for all the possible open-channel crosssection. D P.8-

13 Worked examples. An open channel is to be designed to carr m /s at a slope of The channel material has an n value of 0.0. Find the optimum hdraulic cross-section for a semi-circular section. D Answer The optimum circular section is a semi-circular section with diameter D which can discharge m /s. For a semi-circular section, A = π*d /8 P = π*d/ R = A/P = D/4 As n = 0.0, S = and Q = m /s. A Q = n *R *S π* D D i.e. = * * * D 8/ 8*0.0 = *4 * π D = 0.95 m The diameter of this optimum section is 95mm. P.8-

14 . Find the optimum rectangular section from the last example. Answer A = * P = 4* R = A/P = / B Manning equation, A Q = n *R *S * = * * / *0.0* = * = 0.44 m The optimum rectangular section has dimension of width 0.868m and depth 0.44m. P.8-4

15 . Find the optimum triangular section from the last example. 45 o Answer A = P = * R = A/P = B Manning equation, A Q = *R *S n = * * / 0.0*( ) = = 0.64 m The optimum triangular section is a right angle triangle with depth 0.64 m. P.8-5

16 8. Non-Uniform flow - Specific Energ in Open Channel & Critical Flow energ line E v /g fluid surface hf v /g E z channel bed horizontal datum z In open channel, the solution of man problems are greatl assisted b the concept of specific energ, i.e. E = v g + (8.) In terms of flow rate, Q, E = Q ( ) + (8.4) g A The minimum energ will be given as de d = 0 (8.5) 8.. Rectangular Channel Let q = Q = v* (8.6) b q - the discharge per unit width of a rectangular channel E = q g + (8.7) P.8-6

17 B assuming q is constant de d = - q g = 0 (8.8) or = c q = ( ) (8.9) g - critical depth at which the energ is minimum. c The corresponding energ, E is E min = c (8.0) From (8.6), v = q. Substitute into (8.8), - v g c v c c g c = 0 = (8.) or v c = g c (8.) Since Froude number, Fr is defined as v Fr = g ave (8.) Hence, the minimum energ is occurred when Fr = (8.4) For a given discharge, Q, if the flow is such that E is a min., the flow is critical flow. critical flow - flow with E min critical depth, c - the depth of the critical flow critical velocit - v c = g c P.8-7

18 v /g A c vc /g C subcritical or slow c 45 o B supercritical or fast v /g E If the flow with E > E min, there are two possible depths (, ). - (, ) are called alternate depths. C divides the curve AB into AC and CB regions. - AC - subcritical flow region - CB - supercritical flow region Subcritical Critical Supercritical Depth of flow > c = c < c Velocit of flow v < v c v = v c v > v c Slope Mild S < S c Critical S = S c Steep S > S c Froude number Fr <.0 Fr =.0 Fr >.0 Other v c v < c v = > c g g g P.8-8

19 8.. Non - Rectangular Channel If the channel width varies with, the specific energ must be written Q in the form E = + (8.5) ga The minimum energ also occurs where de = 0 at constant Q d Since A = A(), therefore (8.5) becomes - QA g da d = 0 or da d = ga Q (8.6) Since da d B = ga Q = B - the channel width at the free surface, BQ or A = ( ) (8.7) g v c = Q A ga = ( ) (8.8) B For a given channel shape, A() & B(), and a given Q, (8.7) & (8.8) have to be solved b trial and error to find the A and then v c. If a critical channel flow is also moving uniforml (at constant depth), it must correspond to a critical slope, S c, with n = c. This condition can be analsed b Manning formula. P.8-9

20 Worked examples:. A triangular channel with an angel of 0 made b equal slopes. For a flow rate of m /s, determine the critical depth and hence the maximum depth of the flow. B c 0 o Answer For critical flow, v = g* ave Q = g* ave *A = ga B A ( ave = ) B For critical flow, B = **cot 0 & A = *cot 0 Q = g 5 Hence = ( ) g Q 5 * = ( ) * 98. = m 5 m The maximum depth is m. A The critical depth, c = ave = = B ( B * ) / 0906 = = = 0.45 m. B P.8-0

21 . In the last example, the channel Manning roughness coefficient is 0.0 and the flow rate is m /s. What is the value of the channel slope if the flow is critical, subcritical or supercritical? B 0 o Answer B = * A = *B* P = 4* Using Manning equation, Q = A n * R * S = B *( * B* )*( ) * S n 8 S / c = nq B ( ) B 8 nq * = ( ) * 4 For critical flow, = S / *0.0* *0.906 c = ( ) *(0.906) 4 S c = For flow is critical, S = critical slope subcritical, S < supercritical, S > P.8-

22 8.4 Frictionless Flow over a Bump supercritical approach flow v v subcritical approach flow h When fluid is flowing over a bump, the behaviour of the free surface is sharpl different according to whether the approach flow is subcritical or supercritical. The height of the bump can change the character of the results. Appling Continuit and Bernoulli s equations to sections and, & v * = v * v g + = v + + h g Eliminating v between these two gives a cubic polnomial equation for the water depth over the bump, E * + v * g = 0 (8.9) where E = v + - h (8.0) g This equation has one negative and two positive solutions if h is not too large. P.8-

23 The free surface s behaviour depends upon whether condition is in subcritical or supercritical flow. water depth hmax subcritical bump c h supercritical bump Ec E E specific energ The specific energ E is exactl h less than the approach energ, E, and point will lie on the same leg of the curve as E. A subcritical approach, Fr <, will cause the water level to decrease at the bump. Supercritical approach flow, Fr >, causes a water level increase over the bump. If the bump height reaches h max = E E c, the flow at the crest will be exactl critical (Fr = ). If the bump > h max, there are no phsical correct solution. That is, a bump to large will choke the channel and cause frictional effects, tpicall a hdraulic jump. P.8-

24 Worked example: Water flow in a wide channel approaches a 0 cm high bump at.5 m/s and a depth of m. Estimate (a) the water depth over the bump, and (b) the bump height which will cause the crest flow to be critical. Answer (a) For the approaching flow, v 5. Fr = = g 98. * = subcritical For subcritical approach flow, if h is not too large, the water level over the bump will depress and a higher subcritical Fr at the crest. E = v g 5. + * m =.5 m Hence E = E - h =.5 0. m =.05 m Substitute E into (8.4),.05* = 0 B trial and error, = m, 0.45 m and 0.96 m (inadmissible) The second (smaller) solution is the supercritical condition for E and is not possible for this subcritical bump. Hence = m Checking: v =.745 m/s (B continuit) Fr = 0.60 (> Fr and < ) (OK) P.8-4

25 (b) B considering per m width of the channel, q = v* =.5* m /s For critical flow, E = E min = c q c = ( ) g 5 = (. ) 98. = 0.6 m E = *0.6 = 0.98 m m h max = E - E min = m = 0.97 m P.8-5

26 8.5 Hdraulic Jump in Rectangular Channel A hdraulic jump is a sudden change from a supercritical flow to subcritical flow. Assumptions: the bed is horizontal. the velocit over each cross-section is uniform. the depth is uniform across the width. frictionless boundaries. surface tension effects are neglect. critical depth level v v edd currents Considering the control volume between and, the forces are F = ρg * b * = ρgb* Similarl F = ρgb* (8.a) (8.b) B continuit equation, Q = b* *v = b* *v (8.) B the momentum equation, F = F = 0 hence F F = ρ*q*(v - v ) (8.) Sub. (8.a, b) and (8.) into (8.), then ρgb ( ) = ρq( Q Q b b ) ρq = ( ) b (8.4) P.8-6

27 In a hdraulic jump,, Q * *( + ) = gb + = Q gb Q i.e. ( ) + ( ) = 0 (8.5) gb Solving (8.5), = 8Q [ + + ] (8.6) gb This is the hdraulic jump equation. Using Froude number, then, Fr = v Q = (8.7) g g b = [ + + 8Fr ] (8.8a) or = [ + + 8Fr ] (8.8b) (, ) are called conjugate depths. The energ loss in a jump is given b v g + = v + + g h f v v i.e. h f = ( ) + ( ) g (8.9) Sub. (8.) into above, Q h f = [ gb + ( ) + ]( ) (8.9) P.8-7

28 Using (8.5), (8.9) becomes h f = ( ) 4 (8.0) This is the energ loss equation for the hdraulic jump ( >, h f >0). The power loss in a jump is P = ρgh f *Q This energ loss is useful for getting awa with the unwanted energ of a flow. The energ loss is due to the frictional forces amount the edd currents in the pump. It will increase the temperature of the fluid. P.8-8

29 Worked example: Water flows in a wide channel at q = 0 m /s and =.5 m. If the flow undergoes a hdraulic jump, calculate (a), (b) v, (c) Fr, (d) h f,, and (e) the percentage dissipation of the energ. Answer (a) v = q = 0 m/s = 8 m/s 5. Fr = v g 8 = = * 5. Since = [ + + 8Fr ] = [ + + 8*(. 85) ] =.77 or =.77*.5 m =.46 m (b) B Continuit equation, v = v *( ) = 8* 5. m/s 46. =.89 m/s P.8-9

30 (c) Fr = v g 89. = 98. * 46. = (d) h f = ( ) 4 = ( ) 4* 46. * 5. = 0.65 m (e) E = v g + 8 = + * m = 4.5 m percentage loss = h Ef * 00% = * % = 4 % P.8-0

31 8.6 Graduall Varied Flow It is not alwas possible to have uniform depth across the flow i.e. normal flow with normal depth. The depth of flow can be changed b the conditions along the channel. Examples of Graduall Varied Flow are: backwater curve d dx >0 water surface n dam Downdrop curve water surface n d dx <0 slope change n In a uniform flow, the bod weight effect in balanced out b the wall friction. In graduall varied flow, the weight and the friction effects are unable to make the flow uniform. P.8-

32 δx v/g v THL s fluid surface dhf v/g v z channel bed θ horizontal datum z Basic assumptions are - slowl changing bottom slope - slowl changing water depth (no hdraulic jump) - slowl changing cross section - one dimensional velocit distribution - pressure distribution approximatel hdrostatic Denoting v = v; v = v + dv z = z, z = z + dz =, = + d p = p, p = p Appl Bernoulli s equation between section and, p v z γ + g + + = p v dv d z dz dh f γ + ( + ) + ( + ) + ( + ) + g Neglecting higher order terms, dh f + d + dz + v g du = 0 When lim dx 0, dhf d dz v dv = 0 (8.0) dx dx dx g dx P.8-

33 In (8.0), the four terms are dhf - rate of head loss along the channel dx = S (head loss gradient in Manning equation) d - rate of change of water depth dx - water surface profile s gradient dz - rate of vertical change along channel dx = -sinθ v dv - rate of change of velocit head along the channel g dx B Continuit equation v*a = constant i.e. A dv dx + v da dx = 0 A dv dx + v da d d dx A dv dx + v B d dx dv d or dx = - vb A = - v dx d dx = 0 = 0 Therefore v dv g dx = - v d g dx = -Fr d dx (8.) Hence (8.0) becomes s + d dx - sinθ - d Fr dx = 0 or d dx = [ sin θ S Fr (8.) - general equation of graduall varied flow. P.8-

34 (8.) is a st order non-linear differential equation. Numerical method is used to solve the equation. The equation is rewritten as dx d = [ Fr ] sin θ S x Fr dx = ( ) x sin S d θ Fr x = x + ( ) sin S d θ (8.) The simplest solution is the direct mid-point solution of the integral. Fr i.e. x = x + ( ) sin ( + S *( - ) (8.4) θ ) (8.4) ma be used to calculate the water profile in a step-b-step sequence from a known (x, ) value. P.8-4

35 Worked example: Determine the upstream profile of a backwater curve given: Q = 0 m /s, b = m, sinθ = 0.00, n = 0.0. water surface n θ 5m dam Answer For normal flow, (S sin θ) Q = A n * R * S i.e. 0 = ( n ) n *( )* n =.44 m n The water profile is from.44 m to 5 m along the channel. From Manning equation, v = *R *S n S = n v 4 R Fr Hence x = x + ( n v R 4 / ) *( - ) P.8-5

36 section, I i (m) d (m) ave (m) v (m/s) Fr -Fr*Fr R (m) So - Sf dx (m) x (m) From the table, the water level is not affected b the dam at 5.7 km upstream. 5m 4m m normaldepth dam m m 8km 7km 6km 5km 4km km km km 0 From the graph, the water depth at an location can be obtained. P.8-6

37 8.6. Classifications of Surface Profile of Graduall Varied Flow It is customar to compare the actual channel slope, sinθ or S o with the critical slope S c for the same Q. There are five classes of channel slope giving rise to twelve distinct tpes of solution curves. - S o > S c - Steep (S) - S o = S c - Critical (C) - S o < S c - Mild (M) - S o = 0 - Horizontal (H) - S o < 0 - Adverse (A) There are three number designators for the tpe of profile relates to the position of the actual water surface in relation to the position of the water for normal and critical flow in a channel. - the surface of stream lies above both normal and critical depth - the surface of stream lies between normal and critical depth - the surface of stream lies below both normal and critical depth M S n M c S c M n S Mild slope Steep slope P.8-7

38 Combining the two designators, we have Slope Slope notation Depth Froude Actual depth Profile class class number sinθ > S Steep (S) c > n Fr < > n ; > c S Fr > c > > n S Fr > < n ; < c S sinθ = S Critical (C) c = n Fr < > n = c C Fr > < n = c C sinθ < S Mild (M) c < n Fr < > n ; > c M Fr < n > > c M Fr > < n ; < c M sinθ = 0 Horizontal (H) n = Fr < > c H Fr > < n ; < c H sinθ < 0 Adverse (A) n = Im Fr < > c A Fr > < c A For tpe S For tpe C P.8-8

39 For tpe M For tpe H For tpe A P.8-9

40 Class Exercise 8.: A 500 mm-diameter concrete pipe on a :500 slope is to carr water at a velocit of 0.8 m/s. Find the depth of the flow. (n=0.0) 500mm θ ( = 8 mm) P.8-40

41 Class Exercise 8.: What are the dimensions for an optimum rectangular brick channel (n = 0.05) designed to carr 5 m /s of water in uniform flow with s = 0.00? What will be the percentage increase in flow rate if the channel is a semicircle but retained the same sectional area? (increase = 8.4%) P.8-4

42 Class Exercise 8.: A trapezoidal channel has a bottom width of 6.0 m and side slopes of :. The depth of flow is.5 m at a discharge of 5 m /s. Determine the specific energ and alternate depth. (E =.59 m, = m) P.8-4

43 Class Exercise 8.4: A triangular channel has an apex angle of 60 and carries a flow with a velocit of.0 m/s and depth of.5 m. (a) (b) (c) (d) Is the flow subcritical or supercritical? What is the critical depth? What is the specific energ? What is the alternate depth possible for this specific energ? ( c =.48 m, E =.454 m, =.06 m) P.8-4

44 Class Exercise 8.5: A rectangular channel is 4.0 m wide and carries a discharge of 0 m /s at a depth of.0 m. At a certain section it is proposed to build a hump. Calculate the water surface elevations at upstream of the hump and over the hump if the hump height is 0. m. (Assume no loss of energ at the hump.) Q water surface hump P.8-44

45 Class Exercise 8.6: In a hdraulic jump occurring in a horizontal, rectangular channel it is desired to have an energ head loss equal to 6 times the supercritical flow depth. Calculate the Froude number of the flow necessar to have this jump. (Fr = 4.8 ) P.8-45

46 Tutorial Open Channel Flow. Calculate the normal depth in a concrete trapezoidal channel with side slope of to, a bed slope of 0.000, a bottom width of 4.0 m and a water discharge of 9 m /s. Manning coefficient is m. Determine the critical depth of the trapezoidal channel for a discharge of 5 m /s. The width of the channel bottom, b = 6 m, and the side slope is 45.. Consider a flow in a wide channel over a bump with an approaching velocit, v at the upstream is m/s and the depth, is m. If the maximum bump height is 5 cm, determine (a) the Froude number over the top of the bump, and (b) the depression in the water surface. ( > 0.5 m) 4. Water flows in a trapezoidal channel at a rate of 8.5 m /s. The channel has a bottom width of m and side slope of :. If a hdraulic jump is forced to occur where the upstream depth is 0. m, what will be the downstream depth and velocit? What are the values of Fr and Fr? 5. A wide canal has a bed slope of in 000 and conves water at a normal depth of. m. A weir is to be constructed at one point to increase the depth of flow to.4 m. How far upstream of the weir will the depth be.5 m? (Take n in the Manning equation as 0.0) P.8-46