CIVE 530 - OPEN-CHANNEL HYDRAULICS LECTURE 9: RAPIDLY VARIED FLOW (CHOW CHAPTERS 13 and 14)

 13. CHARACTERISTICS OF RAPIDLY VARIED FLOW

• Rapidly varied flow has very pronounced curvature of the streamlines.

• The change in curvature may become so pronounced that the flow profile is virtually broken, resulting in high turbulence.

• Characteristics of rapidly varied flow:

• The curvature of the flow is so pronounced that the pressure distribution cannot be assumed hydrostatic.

• The rapid variation of the flow regime often takes place in a relatively short reach.

• Unlike gradually varied flow, the boundary friction is comparatively small, and in most cases, insignificant.

• The physical characteristics of the flow are fixed by the (often) rigid boundary geometry.

• The velocity distribution coefficients α and β are usually far greater than unity and cannot be accurately determined.

• Separation zones, eddies, and rollers tend to complicate the flow pattern; flow may be confined between separation zones.

• Theory that assumed hydrostatic pressure distribution is known as Bresse theory, good for uniform and gradually varied flow.

• This theory does not apply for rapidly varied flow.

• A satisfactory general solution for rapidly varied flow does not exist.

• Empirical solutions are used for rapidly varied flow.

 14.1 THE SHARP-CRESTED WEIR

• The sharp-crested weir is a measuring device for open-channel flow and also the simplest form of overflow spillway.

• The profile of the spillway was determined in conformity with the shape of the lower surface of the flow nappe over a sharp-crested weir.

• The shape of the flow nappe over a sharp-crested weir can be interpreted by the principle of the projectile.

 Fig. 14-1 (Chow)

• It is assumed that the horizontal velocity component is constant, and the only force acting on the nappe is gravity.

• In time t, a particle of water in the lower surface of the nappe will travel a horizontal distance x from the face of the weir equal to:

 x = vot cosθ

• In the same time, the particle will travel a vertical distance y equal to:

 y = -vot sinθ + (1/2)gt2 + C'

where C' is the value of y at x = 0.

• C' is equal to the vertical distance between the highest point of the nappe and the elevation of the crest.

• Eliminating t from the above two equations leads to:

 y/H = A(x/H)2 + B(x/H) + C

where

 A = gh/(2vo2cos2θ)

 B = - tanθ

 C = C'/H

• Since the horizontal velocity component is constant, the vertical thickness of the nappe T may be assumed constant.

• This assumption leads to a general equation for the upper surface of the nappe:

 y/H = A(x/H)2 + B(x/H) + C + D

where:

 D = T/H

• Since the equations are quadratic, the nappe surface is theoretically parabolic.

• Equations for the constants in the general nappe equations:

 A = -0.425 + 0.25(hv/H)

 B = 0.411 - 1.603(hv/H) - [1.568(hv/H)2 - 0.892(hv/H) + 0.127]1/2

 C = 0.150 - 0.45(hv/H)

 D = 0.57 - 0.02(10m)2e10m

in which

m = (hv/H) - 0.208

and

hv = velocity head of the approach flow.

• For high weirs, hv is small, and can be neglected.

• Therefore, the constants become:

A= -0.425

B= 0.055

C= 0.150

D= 0.559.

• These equations are not valid for x/H less than 0.5.

• For hv/H > 0.2, additional data for verification are needed.

• This theory applies only if the approach is subcritical.

• A common formula for the discharge over a sharp-crested weir is:

 Q = CLH1.5

• C = discharge coefficient; L = effective length of the weir crest; H = measured head above crest, excluding the velocity head.

• The labyrinth spillway is used to increase the length L in order to increase the discharge. See examples below.

 Valentine Mill Pond, Valentine, Nebraska. Valentine Mill Pond, Valentine, Nebraska. Cloe Lake spillway, Pennsylvania. Melaka, Malaysia. Lower Las Vegas Wash detention basin, Nevada. Black Mountain detention basin, Las Vegas, Nevada. Proposed Lake Brazos Dam, Waco, Texas. Ute Dam, New Mexico.
 East Park Dam, California.

• The Rehbock formula for discharge coefficient is:

 C = 3.27 + 0.4 (H/h)

where h= height of the weir.

• This equation holds up to H/h = 10.

• For H/h greater than 15, the weir becomes a sill, and the discharge is controlled by the critical section immediately upstream from the sill.

• For the sill, the discharge coefficient is:

 C = 5.68[1 + (h/H)]1.5

• Experiments have shown that the coefficient C remains approximately constant for sharp-crested weirs under varying heads if the nappe is aerated.

 14.3 CREST SHAPE OF THE OVERFLOW SPILLWAY

• From 1886 to 1888, Bazin made the first comprehensive laboratory investigation of nappe shapes.

• The use of Bazin's data in design will produce a crest shape that coincides with the lower surface of an aerated nappe over a sharp-crested weir.

• In practice, there exists friction due to roughness on the surface of the spillway.

• Hence, negative pressures cannot be avoided.

• The presence of negative pressures will lead to danger of cavitation damage.

• In spillway design, avoidance of negative pressures is an objective.

 Fig. 14-3 (Chow)

• The Army Corps pf Engineers has developed sharp-crested weir designs, called WES standard spillway shapes (See spillway photos).

• They are represented by the following equation:

 Xn = KHdn-1Y

X and Y are the coordinates of the crest profile, with the origin at the highest point of the crest.

Hd is the design head excluding the velocity head of the approach flow.

K and n are parameters depending on the slope of the upstream face.

 Slope on upstream face K n Vertical 2.000 1.850 3V:1H 1.936 1.836 3V:2H 1.939 1.810 3V:3H 1.873 1.776

• For intermediate values, interpolation is possible.

 14.4 DISCHARGE OF THE OVERFLOW SPILLWAY

• For WES shapes, the formula for discharge of an overflow spillway is:

 Q = CLHe1.5

where He is the total energy head of the crest, including the velocity head in the approach channel.

• Model tests have shown that the effect of the approach velocity is negligible when the height h of the spillway is greater than 1.33Hd, where Hd is the design head excluding the velocity head.

• When h/Hd > 1.33, then He = Hd, and the coefficient of discharge is Cd = 4.03 (U.S. customary units).

• For low spillways, with h/Hd < 1.33, the approach velocity will have an appreciable effect on the discharge and discharge coefficient.
 Fig. 14-4 (Chow)

• Fig. 14-4 relates He/Hd with C/Cd, where Cd is 4.03, and C is the actual discharge coefficient.

• This is valid for vertical-face WES spillways.

• For sloping-face spillways, a correction factor is applied to C.

Example 14-1.- Determine the crest elevation and the shape of an overflow spillway section having a vertical upstream face and crest length L = 250 ft. The design discharge is 75,000 cfs. The upstream water surface at design discharge is at Elev. 1000, and the channel floor is at Elev. 880 ft.

 Fig. 14-5 (Chow)

Solution.-

1. Assume a high overflow spillway, so that h/Hd > 1.33; then, the effect of the approach velocity is negligible: He = Hd, and Cd = 4.03.

2. h + Hd = 1000 - 880 = 120 ft.

3. The discharge equation is:

Q = CdLHe1.5

75000 = 4.03 × 250 × He1.5

He = 17.8 ft.

4. The approach velocity is:

Va = Q/[L × (h + Hd)] = 75000/(250 × 120) = 2.5 fps.

Ha = Va2/(2g) = 0.1 ft.

Hd = He - Ha = 17.8 - 0.1 = 17.7 ft.

7. The height of the dam is:

h = 120 - 17.7 = 102.3 ft.

8. h/Hd = 102.3/17.7 = 5.77 > 1.33.

Assumption in Step 1 was correct.

9. Crest elevation is at:

1000 - 17.7 = 982.3 ft.

10. The crest shape:

Xn = K Hdn-1 Y

Y = Xn/(K Hdn-1)

Y = X1.85/(2 Hd0.85)

Y = X1.85/(2 × 17.70.85)

Y = X1.85/23.

11. For the point of tangency:

Y' = (1.85/23) X0.85

Y' = 0.08 X0.85

Y' = ΔY /ΔX = 1.0/0.6

1.0/0.6 = 0.08 X0.85

X = 35.6 ft.

Y = X1.85/23 = 32.2 ft.

0.5 Hd = 8.8 ft.

0.2 Hd = 3.5 ft.

0.282 Hd = 5.0 ft.

0.175 Hd = 3.1 ft.

 14.5 RATING OF OVERFLOW SPILLWAYS

• The spillway must operate under other heads.

• For lower heads, the pressure on the crest will be above atmospheric, but lower than hydrostatic.

• For higher heads, the pressure will be lower than atmospheric, and it may so low that separation may occur.

• Most experiments indicate that the design head may be safely exceeded by at least 50%.

• Beyond that, harmful cavitation may develop.

• For spillway shapes other than WES, Bradley has developed a universal curve shown in Fig. 14-6.

• This figure shows the relation between He/HD and C/CD.

• HD is the design head including the approach velocity, and CD is the corresponding coefficient of discharge.

• He is any head including the approach velocity, and C is the corresponding coefficient of discharge.

 Fig. 14-6 (Chow)

• To find the coefficient of discharge for the design head for a given shape, use the Buehler method.

• The coefficient of discharge is computed by:

 C = 3.97 (He/HD )0.12

HD is the design head including the approach velocity for a vertical-face spillway.

• The value of HD is obtained from Fig. 14-7, matching a shape.

• If after matching, HD differs on both sides, use the higher value.

• The dashed line indicates HD = 45.

 Fig. 14-7 (Chow)

 RATING OF OVERFLOW SPILLWAYS II

• Pressure on the crest is close to atmospheric for the selected rate.

• The head above the spillway crest, excluding the approach velocity, is the design head Hd.

• The head above the spillway crest, including the approach velocity, is the design head HD.

• The spillway rating is:

• Q = C (2g) 1/2 L H 3/2

• C = discharge coefficient, dimensionless

• The theoretical value of C is based on the broad-crested weir equation:

• Q = (2/3) H L [g (2/3)H]1/2

• C = 2/(3)3/2 = 0.3849

• H = total head, including approach velocity

• For ogee-type spillways:

• Q = CD (2g) 1/2 L HD3/2

• CD is the value of C when H = HD.

• Note that the actual CD ranges from 0.385-0.493, greater than the theoretical value C = 0.385.

• This means that the ogee spillway is very effective in passing the flow.

• Variation of C/CD as a function of H/HD

• This means that the discharge coefficient varies directly with the operating head.

• A spillway crest becomes more efficient at heads that exceed the design head, but this comes at the expense of subatmospheric pressures on the spillway, which leads to cavitation.

EXAMPLE:   SPILLWAY RATING CURVE

• Determine the rating curve for an ogee-type spillway with length = 100 ft. The design head is 30 ft. The spillway crest elevation is at 1,200 ft, and the riverbed elevation is at 1,110 ft. Neglect the approach velocity.

SOLUTION

• HD = 30

• P = 1200 - 1110 = 90

• P/HD = 3

• CD = 0.492

• Q = C (2g)1/2 L H3/2 = 802.12 C H3/2 In the following rating table, calculate C/CD based on H/HD.

 Elevation (ft) H (ft) H/HD C/CD C Q (cfs) 1200 0 0.000 0.00 0.00 0 1205 5 0.167 0.83 0.41 3,677 1210 10 0.333 0.88 0.43 10,907 1215 15 0.500 0.92 0.45 20,970 1220 20 0.667 0.95 0.47 33,720 1225 25 0.833 0.97 0.48 48,127 1230 30 1.000 1.00 0.49 64,582 1235 35 1.166 1.02 0.50 83,044 1249 40 1.333 1.04 0.51 103,490 1245 45 1.500 1.06 0.52 125,910

Check with ONLINE OGEE RATING.

 081027