cive 530 - lecture chow 04
 

 

CIVE 530 - OPEN-CHANNEL HYDRAULICS

LECTURE 4: CRITICAL FLOW

4.1  CRITICAL FLOW


  • Critical flow occurs when the following conditions are met:

    • The specific energy E is a minimum for a given discharge Q.

    • Discharge Q is a maximum for a given specific energy E.

    • The specific force F is a minimum for a given discharge Q.

    • Discharge Q is a maximum for a given specific force F.

    • The velocity head hv is equal to 1/2 of the hydraulic depth D (D= A/T).

    • The Froude number is equal to 1 (F = 1).

    • The mean velocity is equal to the relative celerity of small surface perturbations [v = (gD)1/2]


    Emergency spillway at Turner reservoir, San Diego County.

  • The flow can be subcritical or supercritical.

  • In subcritical flow, the secondary surface perturbation can travel upstream [v < |(gD)1/2|]

  • Subcritical flow is controlled from downstream.

  • In supercritical flow, the secondary surface perturbation cannot travel upstream [v > |(gD)1/2|]

  • Supercritical flow is controlled from upstream.

  • In critical flow, the secondary surface perturbation remains stationary [v = |(gD)1/2|]

  • Critical state (critical depth) can occur:

    • Along the channel: critical channel (critical depth along the channel).

    • At a cross section: critical cross section, in either gradually or rapidly varied flow

  • The slope of the channel that sustains critical depth is called "critical slope."

  • The uniform flow equation in terms of slope, friction coefficient, and Froude number (for a hydraulically wide channel) is:

    So = f' F2

  • f' is 1/8 of the Darcy-Weisbach friction coefficient.

  • From now on, we will refer to f' as f.

    So = f F2

  • For F = 1:

    So = f = Sc

  • Thus, the friction coefficient f is equal to the critical slope.

  • In other words, f is the critical slope.

  • The slope is proportional to the square of the Froude number; the factor of proportionality is f.

  • In subcritical flow: So < Sc   ⇒ F2 < 1.

  • In supercritical flow: So > Sc   ⇒ F2 > 1.

    So = Sc F2


    Taymi Canal, near Posope Alto, Chiclayo, Peru.


    Fig. 2-2 (Chow)

Example 4-2. Compute the critical depth and velocity of the trapezoidal channel shown above, carrying 400 cfs.

Solution.- The following proportion holds:    z / 1 = x / y

From which: x = zy

T = b + 2x = b + 2zy

A= (1/2) (b + T) y = (1/2) (b + b + 2zy) y = (b + zy) y

D = A/T = [(b + zy) y ] / (b + 2 zy)

V = Q/A

V = (gD)1/2

Q/A = (gD)1/2

Q/(g)1/2 = A(D)1/2 = A(A/T)1/2

Q2/g = A3/T

(Q2/g) T - A3 = 0

[Q2/g] (b + 2zyc) - [(b + zyc) yc ]3 = 0

Divide by 8:

[Q2/(2g)] (b/4 + zyc/2) - [(b/2 + zyc/2) yc ]3 = 0

For Q = 400 cfs, g = 32.2 ft/s2, b = 20 ft, z = 2:

2484 (5 + yc) - [(10 + yc) yc ]3 = 0

Solve by trial and error: yc = 2.15 ft.

Ac = (b + zyc) yc = 52.2 ft2

Vc = Q/Ac = 7.66 fps.

The critical flow equation is:

f(yc) = [Q2/g] (b + 2zyc) - [(b + zyc) yc ]3 = 0

Changing variable to x for simplicity:

x = yc

f(x) = [Q2/g] (b + 2zx) - [(b + zx) x ]3 = 0

 

To solve this equation by trial and error, the following simple algorithm is suggested:


  1. Assume x = 0.

  2. Assume Δx = 1.

  3. Increment x:    x = x + Δx

  4. Calculate f(x)

  5. Stop when Δx is small enough (Calculate output variables Ac and Vc).

  6. If f(x) > 0, return to step 3

  7. If f(x) < 0, set Δx = 0.1 Δx

  8. Set x = x - 9 Δx and return to step 4

Example: onlinechannel02.php


4.5  CONTROL OF FLOW


USGS stream-gaging station at Campo Creek, near Campo, San Diego County, California.


  • A channel control means the establishment of a definite relationship between the stage and the discharge.

  • Channel control is achieved at a control section.

  • The control section is a suitable site for a gaging station and to develop the discharge rating curve.

  • The Parshall flume forces critical flow at the throat, so that a stable control section can be obtained.


    Parshall flume at Cucuchucho constructed wetland, Michoacan, Mexico.


    Fig. 4-5 (Chow)

Example 4-4. Using the theory of critical flow, derive an equation for the discharge over a broad-crested weir.


Detail of 8000-ft weir, Boeraserie Conservancy, Guyana.
Solution.-

Consider the section on the weir crest where critical flow occurs.

The discharge per unit of width is:     q = Vc yc

q = (gyc)1/2 yc

q = (g)1/2 yc3/2

yc = (2/3) He      (He = specific energy head at the critical section)

q = (g)1/2 [(2/3) He ]3/2

q = (g)1/2 (2/3)3/2 He 3/2

q = C He 3/2

C = (2/3)3/2 (g)1/2

C = 3.09 in U.S. customary units.

C = 1.7 in SI units.

This is a theoretical equation, since the critical section is usually difficult to locate with certainty.

For practical purposes, the equation is written as:

q = C H3/2

where H is the elevation of the upstream water surface above the weir crest.

This assumes that the approach velocity, at a section sufficiently far from the weir, is Va ≈ 0.

Therefore: H ≈ He

Example: onlinechannel14.php

Crossing the El Cora ford on a motorcycle

Go to Chapter 5.

 
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