CHAPTER 10 Flow in Open Channels

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1 CHAPTER 10 Flow in Open Channels Chapter 10 / Flow in Open Channels Introdution 1 α os (1 0.) rad or QB Qdsinα ga g d /4( α sinα os α) 4 sin(1.159) ( d / 64) sin1.159os1.159) 5 This equation redues to: 19.1 d d.86 m Uniform Flow Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

2 Use Chez-Manning equation in the form Q 5 / A 1 / S0. Substitute in n P appropriate expressions for A and P, and solve for 0 b trial and error. (a) (.5 +.5) / ( ) 5/ This redues to (4.5 + ) 8.06 ( ) 5/ 0 0 / Solving, 0.15 m, A m, P m () 5/ 1.8 / 4 ( α sinαos α) / 0.01(1.8 α ) ( α sinαos α) This redues to.6 / α 1.8 Solving, α 1.94 rad, 0 (1 os1.94) 1.55 m 5/ 1.8 (1.94 sin1.94 os1.94) m A 4 P m 10.8 b 0, m 1 8, m A b + ( m + m ) m ( ) ( ) ( ) 1 1 P b+ 1+ m + 1+ m 1+ m / / S / Q AR n (a) 8/ 0.1 m, Q.956(0.1) m /s (b) /8 Q m /s, 0.44 m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

3 Energ Conepts q g ( E ), E onstant, g ( E ) dq 14 g( E ) g g( E ) g d q Setting dq / d 0 and noting that q 0, so that the numerator is zero, one an solve for at the ondition q q max (note: 0 is a trivial solution): g[ ( E ) ] 0, E 0, E/ 10.1 q V m /s q E m g1 q / g 7.5 / m, E m (a) E E1 h m, E E > is subritial q E +, or g 981. Solving,.1 m, and + h m () Set E E.68 m and maximum h is hmax E1 E E1 E m + hmax m (a) q V 9 m /s, Q b q 9 7 m /s q E m, 9 / m 1 g E m Without hange in width at lo., E E1 h m Sine E E, width must hange to prevent hoking. Set < 1 E.76 m m, q g m /s b Q/ q 7 / m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

4 q 5.5 E m 1 1 g q 5.5 q 5.5 Fr m g g E m (a) The maximum height of the raised bottom at loation will be one for whih the energ is a minimum: E1 E + h, h, h m (b) () (d) Sine Fr 1 < 1, if h > 0.0 m, subritial nonuniform flow will our upstream of the transition. At the anal entrane, the ondition of ritial flow is Fr QB/ ga 1, to be solved for the unknown width b. (a) Retangular hannel: B b, A b, Qb Q Q 18 1 b 5.75 m gb gb g Objetive is to determine the zero of the funtion f( ) Q B( )/ g[ A( )] 1, in whih eqns. for B() and A() representing either a irular or trapezoidal setion area an be substituted. The false position algorithm is explained in Example 10.10; this was used to determine the roots. The solutions are: (a) 1.00 m, () 1.81 m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

5 In the lower setion ( 1 m), the area is 0 0 / Q B Q Q A bdη ηdη Assume < 1 m; then Fr ga g8 8g 9/ If Q 1 m, then 1, or Q / 5.1 m /s Q 5.1 m /s, ritial depth will be > 1 m. Cross-setional area inluding when > upper region ( > 1 m) is / A 1 ( ) + ( 1) 1, and B m (a) QB 55 ga 981. ( 1) (b) 1/ (709) m , ( 1) 709, QB.5 4 1, 0.468, 0.8 m ga (a) m /s, q m E m, 0 E m E E E h m > E no hoking. Sine losses are negleted throughout the transition, 1 0, and an be omputed b writing the energ equation between loations 1 and : h , 0.95 m (b) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

6 B H (a) B 1 m, H 0.6 m B.81 and H (1.5 B) (1.5.18) Q 0.11BH 0.11 (.81)(1.18) 0.4 m /s (b) H (m) Q m /s Q b g h Q b g h / / 1 ( 1 ), ( ) given Q 1, 1, Q,, solve for b and h 10.0 (a) / / 1 Q1 1 ( 1/ ) or h / 1 ( 1/ ) h Q Q h Q Q Q / (0.15/0) h 1.0 m / 1 (0.15/0) Q 0 b 8.8 m / / g ( h) 9.81 ( ) Momentum Conepts M b + q g M b 1 q + g q + g( / ) But q / g 1 M b ( / ) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

7 (a) Conditions at loation 1: q g E m q 1 / g 1.5 / m. d/ q 1 q b 1 d/ b The smallest onstrition at loation is one that establishes ritial flow, i.e., where minimum energ exists: E E, or sine E E, 1 E E m 10.4 b q g( ) m /s qb m q 4. Hene the maximum diameter is d b1 b m The momentum eqn. annot be used to determine the drag sine no information is given with regard to loation downstream of offerdam. But we do have available the drag relation, whih will provide the required drag fore: Frontal area: A 1d m Approah veloit: V1 q1/ 1 1.5/ m/s 1 F CD AρV / / 6 N, a rather insignifiant drag fore! Asill 1 wh m 1 A m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

8 M Q 1 Q Q 1 A ga , 7.58 M Q 1 Q A ga sill F CDA Q Q Q γ ga 9.81 ( 0.5 ) 10. Substitute into momentum eqn: M1 M + F/ γ Q, Q Q Q Q ( ) Q 7.0 m /s 1 Given ondition: V 0.4V 1 1 Continuit eqn: 1( V1+ w) ( V + w) 10.8 Momentum eqn: 1 ( V1 + w) g 1 Unknowns are V, and w. Eliminate V and w, and solve for. (a) V ( 1+ w) V + w w, or w 0.9/( 1.5) /( ) 1 or Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

9 Solving, 1.77 m V 0.9 m/s w. m/s (a) To ompute the disharge, write the energ equation between loations 1 and, and solve for q: g ( 1) 9.81(0.10.5) q m /s ( 1 ) ( ) Q bq m /s (b) To ompute depth downstream, first ompute the Froude number at : Fr q g ( 1 8Fr ) ( ) m () To alulate power lost, first ompute the head loss in the jump: ( ) ( ) h j m W γ Qh j 4 watt, or 5.0 kw Use Eqn with F 0: M 1 Q 1 m 1 b 6 ( + ) + gb ( + m ) ( ) m ( ) M M 1 or Q m b gb + m ( + ) ( ) 60 + ( + 5) (5 ) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

10 Energ loss aross jump is h j : Solving,.55 m W j Q Q j 1 1+ ga1 ga 1 + A h j m h E E A m, m, Power dissipated is W j : 5 γqh W j First find Q, then ompute 1 : V Fr g m/s Q VA m /s q Q/ b 11.51/ 5.0 m /s q g /.0 / m To ompute, use momentum eqn, M1 F/ γ M or q CDAρV1 / q + + g 1 bγ g Substitute in known data, plus the relation V1 q / 1: whih redues to Solving, m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

11 Nonuniform Graduall Varied Flow (a) q Q/ b 0.5 / m /s Use Chez-Manning eqn. to ompute 0 : q / g / m Qn S 1 0 5/ (1.8 0) (1.8 + ) 0 / Solving, m E + h + h m E q g m Sine E0 < E + h, normal onditions annot exist at lo. 1, and hoking will our at lo.. Compute alternate depths at los. 1 and : E E E + h Solving, m, m. (Note: lo. is a ritial ontrol, with subritial flow upstream and superritial flow downstream.) A jump is loated upstream of lo. 1. Find the depth onjugate to 0 : 0 q go Fr / /( ) j ( 1+ 8Fr0 1) ( ) 0.4 m q Q/ b /4 8.5 m /s Use Chez-Manning eqn. to ompute 0 : q / g 8.5 / m / Qn 0.01 (4 0) 1.4 S (4 + ) / Solving, 0.98 m Sine 0 >, mild slope onditions prevail. With entrane <, an M profile exists downstream of the entrane. For free outfall onditions, exit, andan M profile exists upstream of the exit. The M and M profiles are separated b a hdrauli jump loated approximatel 60 m downstream of the entrane (determined b numerial analsis) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

12 A spreadsheet solution is shown below. An M profile is situated upstream, with a hdrauli jump at approximatel 150 m followed b an M 1 profile. Q 0 n S L 00 g 9.81 b 0 m 1.5 u 0.50 d m.5 Depth Residual E Station A V E m S( m ) Δx x j FM Residual E E E E E E E E E E E E E E E E E E E E E E E E E q Q/ b 15/4.75 m /s q / g.75 / m upstream reah has a steep slope. 1.4 m, < 0 0 downstream reah has a mild slope. Compute depth onjugate to 01 : 0.9 m, > q g01 Fr /( ).75 /( ) j 1+ 8Fr m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

13 Hdrauli jump ours upstream, followed b an S 1 urve up to the transition q m. Sine 01 < and 0 >, there will be a (a) g 9.81 hdrauli jump between A and C. (a) Compute using Fr 1: Q B Q dsinα ga g ( d /4)( α sinα os α) sinα ( α sinαos α).5.5sinα 9.81 (.5 /4)( α sinα os α) Solving d.5 α rad and (1 os α) (1 os1.115) 0.70 m Compute 0 using Qn ( d /4)( α sinα os α) / S ( αd) 1 0 5/, (.5 /4)( α sinαos α) / ( α.5) 5/ whih redues to ( α sinαos α) 1.09 / α Solving, α1. 61 rad, and d.5 0 (1 os α) (1 os1.61) 0.99 m Sine 0 >, a mild slope ondition exists. The water surfae onsists of an M urve beginning at the inlet, followed b a hdrauli jump to an M urve, whih terminates at ritial depth at the outlet. The numeriall-predited water surfae and energ grade line are provided in the following table. 5/ Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

14 x E x E (m) (m) (m) (m) (m) (m) * 0.49 * 1.18 * * 0.97 * 1.07 * * hdrauli jump (a) 0 q 6 m /s m E m 9.81 E m m m m m 1.9 m, A m, P m, R 9.5 / m Qn S ( m) A R / 4/ m m x E E 1 Δ S0 S( m) m (b) An A profile is ontained within the reah (whih has an adverse slope with < 1 ) E E1 Evaluate Q using Δ x S 0 S ( m) Q Q Q E E 1 1 ga ( ) Q Q Q ga 9.81 (.5 1.) 1 1 m ( 1+ ) ( ) 1.15 m 4/ ( m) 10 / ( bm) Qn b + S( m) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

15 4/ Q 0.01 ( ) 10/ 1 ( ) [ ( )] ΔxS S E E 0 m Q 5 Q Q 50( Q ) whih redues to Q 0.10 Solving, Q 15.5 m /s Q 15.5 g m b.5 The profile is an S urve Write energ eqn. aross the gate to ompute the disharge: q E E or g1 g 1 1 q g( 1 ) ( ) m /s Q bq m /s q 10.6 q / g 1.9 / m Compute 0 using Chez-Manning relation: Qn S (4 ) / 0.81 (4 / + 0 ) Solving, m. Sine 0 >, mild hannel onditions exist. Upstream of the gate there will be an M 1 profile, with an M downstream of the gate, terminating in a hdrauli jump to normal flow onditions. Compute the depth upstream of the jump, onjugate to 0 : 0 q g0 Fr / 1.9 /( ) ( 0 ) ( ) 0 j 1+ 8Fr m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

16 Use the step method to ompute water surfae and energ grade line: q E g 981. Qn( 4+ ) S ( ) 10/ ( 4) 4/ ( 4 + ) 4/ 4 10/ 10/ ( 4+ ) / / M 1 urve upstream of gate E m S( m ) Δx x (m) (m) (m) (m) (m) M urve downstream of gate E m S( m ) Δx x (m) (m) (m) (m) (m) Hdrauli jump ours ~5 m from the gate. (a) Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

17 Compute normal depth using Chez-Manning eqn: Qn S (.66 ) (.66 + ) 5/ 0 / 0 Solving, 1.65 m 0 Compute ritial depth: q Q/ b 15.8/ m /s q / g 4.0 / m Upstream of slope hange the hannel is steep, and downstream, the hannel is mild. The hdrauli jump will terminate at normal depth. (a) Find depth onjugate to 0 : q g0 Fr / 4.0 /( ) ( 0 ) ( ) 0 j 1+ 8Fr m (b) Step method: q E g / Qn( b+ ) ( ) S ( ) 10/ 10/ 10/ ( b) ( 66. ) 4 4/ 10/ (.66 + ) 4/ E m S( m ) Δx x (m) (m) (m) (m) (m) L j () An M urve exists downstream of the hange in hannel slope, terminating in a hdrauli jump, approximatel 50 m from the slope hange Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

18 10.68 This is a design analsis problem. Some extensive alulations are required. (a) Find the depth of flow immediatel upstream of the jump (all it loation 1): Q V 1.67 m/s, q 0.95 m /s. A Fr 0.467, 1 ( ) 0.46 m Compute the loss aross the jump, and subsequentl the dissipated power: hj E1 E m W γ Qh watt, or 71 kw j (b) The length of the apron is the distane from the toe to downstream of the jump. First, ompute the distane from toe to 1 using a single inrement of length: E P toe m m E m, m, ( ) 0.17 m, m Am, m, Sm ,.46 4/.46 / 0.5 E1 Etoe x1 xtoe 59. m S S m ( ) The length of the jump is six times the downstream depth, or m. Hene the required length of the apron is L m, or approximatel 65 m. Normal design would require additional length as a safet fator. Use the varied flow funtion to ompute the water surfae profile /10 qn m S / q m g 9.81 Therefore the estuar has a mild slope and an M 1 urve exists upstream of the outlet. For a wide retangular hannel (Ex ), J.5, N., M, and N/J 1.. Use Eq to ompute x: Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

19 x u F( u,.) + F( v,.5) , 70 (,.) 0.04 (,.5) [ u F u F v ] The results of the alulations are as follows: (m) u F(u,.) v F(v,.5) x (m) x 69,800 (m) , ,700 10, ,500 0, ,500 0, ,700 55,100 This is a design analsis problem, and requires extensive alulations that would best be performed on a spreadsheet. Let loation A be 400 m upstream of loation B. Conditions at B are known. The energ equation between A and B is predited using total energ H + z+α V / g: 10.7 * HA HB+ hl H L B+ ( SA+ SB) In the equation, H B and S B are known, and H A and S A are unknown. A trial and error solution is required. With the table of given data, loation A is assoiated with x 0, and loation B with x 400 m. The following parameters are omputed at loation B, where B.0 m: Loation A B (m ) P B R B K B (Eq ) (m) (m) Side hannel Main hannel ,10 Now α B, V B, H B, and S B an be omputed: (60 + 7) 11 18,10 αb 1.48 (Eq ) + (11+ 18,10) 60 7 Q 80 VB 0.84 m/s A B S B H B (Eq ) (11+ 18,10) m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

20 With L 400 m, the energ equation between A and B is predited: * 400 H ( S ) S The total energ at loation A is A A A VA VA HA A + za + αa A αa g g The trial and error solution proeeds b assuming values of A, omputing the * orresponding α A, V A, S A, H A, and H A until the latter two are in lose agreement. For A. m, the hdrauli parameters at loation A are provided in this table: Loation A A (m ) P A (m) R A (m) K A (Eq ) Side hannel Main hannel ,000 The orresponding values of H A and H A * are α A (17 + 4) , (Eq ) + ( ,000) 17 4 V A S A Q m/s A A (Eq ) ( ,000) 0.54 H A m 9.81 * A H m Hene the depth at the upstream loation is A. m Cengage Learning. All Rights Reserved. Ma not be sanned, opied or dupliated, or posted to a publil aessible website, in whole or in part.

2. The Energy Principle in Open Channel Flows

2. The Energy Principle in Open Channel Flows . The Energy Priniple in Open Channel Flows. Basi Energy Equation In the one-dimensional analysis of steady open-hannel flow, the energy equation in the form of Bernoulli equation is used. Aording to this