NON UNIFORM FLOW IN OPEN CHANNEL

LEARNING OUTCOME

-determine and calculate critical depth, critical slope and specific energy

Introduction

The energy grade line, water surface and channel bottom are not parallel; that is,

Sf ≠ Sw ≠ So

where;

Sf = slope energy grade line

Sw= slope of the water surface

So= slope of the channel bed

Where;z is the elevation of the channel’s centerline y the water depth

H = z + y + av²/2g

Where H = Total of Energy

z = potential head (distance from the datum to channel bed)

y = Depth of flow

av²/2g = velocity head (energy of velocity)

av²/2g

yE

H

z

Ө

EGL

Free surface

Channel BedDatum

=

Non Uniform Flow schematic

If we neglect the energy loss, the energy equation combined with continuity can be written as

E1 = E2

y1 + Q²/2gA1² = y2 + Q²/2gA2² + ∆zWhere

y = depthQ = dischargeA = Cross-sectional area of flow∆z= z2 –z = change in bottom elevation from

cross section 1 to 2

v²/2g

y1Q

Q y2 = ?

∆z

EGL

Transition with bottom step

Specific Energy Equation in Open Channel;-(Specific energy = energy of water per unit weight (water) at cross section which is measured from channel bed)

Total of Energy = height + depth of flow + energy of velocity

H = z + y + av²/2g

if bed of channel given as datum, then

E = y + av²/2gwhere a= Coriolly coefficient (non uniform velocity correction coefficient)

Normally in an open channel, velocity head is av²/2g where 'a' = 1.0 --> 1.36

From 'v = Q/A', therefore;

E = y + aQ²/2gA²

Considered square channel (prismatic and straight)

E = y + q²/2gy²where, q = flowrate per unit width (m³/s/m)

q = Q/b , a = 1.0 and A = by

Energy, E, depth of flow, y and flowrate, q may be written/defined in 2 conditions as below;-

E and y if q is constant q and y if E is constant

E and y if q is constantE = y + q²/2gy²

y³ – Ey² + q²/2g = 0

Specific Energy Diagram (E-y)

y

oE = y + q²/2gy²

DB

A

Cyc

Emin

y

yv²/2g (head of velocity)

y > yc

y < yc

45º

Specific Energy Diagrams

*Note q is constant.

Emin

y

yc

y=E

EEo

y1

y2

At C, specific energy is minimum and normal depth at this point is 'critical depth', yc

If y > yc ; v < vc ==> Subcritical flow (steady)

y < yc ; v > vc

==>Supercritical flow (turbulant)

Differentiation of

E = y + aQ²/2gA²

dE/dy = 1 - a (Q²/2g)(2/A³)dA/dy

dE/dy = 1 – a(Q²/gA³).T

= 1 – (av²/gA).T

= 1 – av²/gD

At critical point, E is minimum i.e. dE/dy = 0

therefore;-

v²/gD = 1 ; (froude, Fr = 1)

v²/2g = D/2 or v/√(gD)= 1

T

dy

dA = T.dydA/dy = T

For a rectangular channel,

Hydraulic Depth, D = A /T = by/b = y

Therefore, at critical condition ==>>

Fr = 1; y = yc , v = vc

vc /√(g yc ) = 1

vc²/2g = yc/2

From the schematic diagram;- (E = min, y = yc )

E = y + q²/2gy

dE/dy = 1 – q²/gyc³ = 0

q² = gyc³

yc = ³√(q²/g)(c)

q² = gyc³ but q = vy = vcyc

vc ²yc² = gyc³

vc ² = gyc

vc =√(gyc) or vc

²/2g = ½ yc

Emin = yc + q²/2gyc²

= yc +(g yc³) / (gyc²)

= yc + yc√2

Emin = 1.5yc or yc =(2/3)Emin

The point of minimum E is found by setting dE/dy equal to zero, and solving for y. The result is yc = 2E/3, which is called the critical depth yc. The corresponding velocity V is called the critical velocity Vc. The critical depth divides the energy curve into two branches. On the upper branch, y increases with E, while on the lower branch y decreases with E.

ii) q and y if E is constant

E = y + q²/2gy²q² = 2gy² (E - y)

E constanty

y1

yc

y2

q qmax q

E

yc=2/3E

q – y curve

At critical point, dq/dy = 0Differentiate

q² = 2gy² (E - y) = 2g(Ey² – y³)

2qdq/dy = 2g(E 2yc – 3yc² = 02ycE = 3yc²

E = 1.5yc

yc = 2/3E

qmax² = 2gy²(E – y) = 2gy²(1.5yc – yc)= gyc³

qmax = √(gyc³)

NoteSubcritical and supercritical flow s normally depend on the channel slope, S. Therefore, for the supercritical flow, value of S is high.*Critical Slope = slope at critical depth.

Critical DepthCritical flow criterias (square/rectangular channel)

Fr = 1.0 'E' is minimum for 'q' constant

Emin = 1.5yc

yc = ³√(q²/g)

'q' is maximum at E constant

yc = 2/3Emin

qmax = √(gyc³) Velocity head (vc²/2g) is one-half of critical depth, yc

vc²/2g = yc /2 Critical velocity (vc);

vc = √(gyc)

In general, critical flow will occur when Fr = 1.0, it will expressed as below;-

dE/dy = 1 – (Q²/gA³). T = 0

Q² T/gA³ = 1 (for all channels)

Froude Number, F

q2/gyc3 = 1

Then, vc2/gyc = 1 at critical conditions

So, at critical conditions, the Froude number =1!

cgyv is known as the Froude Number, Fcgyv

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.

Flow characteristics of flow in rectangular channels

Critical Depth in non-rectangular channels

Critical conditions for channels of various shape

Example 1

Water flows at 15ft/s in a rectangular channel at a depth of 2ft. Find the critical depth for

a) this specific energy

b) this rate of discharge

Example 2

Water flows down at a wide rectangular channel of concrete (n = 0.014) laid on a slope of 2.4mm/m. Find the depth and rate of flow for critical conditions.

Example 3

A rectangular channel carries 200cfs. Find the critical depth and the critical velocity for

a) A width of 12 ft and

b) A width of 9 ft.

c) What slope will produce the critical velocity in (a) if n = 0.020?

Example 4

A rectangular channel, 30 ft wide, carries 270cfs when flowing at 3.00 ft deep. What is the specific energy? Is the flow subcritical or super critical?

Example 5

A trapezoidal channel has a bottom width of 20ft, side slopes 1:1, and flows at a depth of 3.00 ft. For n= 0.015, and a discharge of 360 cfs, calculate

a) the normal slope,

b) the critical slope and critical depth for 360 cfs

A wide and straight river flows with 3.5m³/s/m flow rate.

a) What is the value of the critical depth?b) If normal depth is 4.6m, calculate the Froude

number for this flow rate. (Type of flow: sub critical or supercritical).

c) Calculate the critical slope if Manning’s Coefficient is 0.035.

Example 6

Example 7

A trapezoidal channel with side slopes of 2 horizontal to 1 vertical is to carry a flow of 590cfs. For a bottom width of 12 ft, calculate the

a) critical depth,

b) critical velocity.

Example 8

A rectangular channel with 3m width flows water at12m³/s flow rate when Froude number is 0.8. Determine the depths (y1 and y2) for the same flow rate and specific energy.

Control Section

Control Section may be defined as ;

A section where a certain relationship can be established between flowrate and water levelIt also controls the flow so that it can prevent the changes of flow types from happening (critical flow, subcritical & supercritical)Gauge station – to get flow rating curve which represents the 'flowrate' vs 'depth' relationship for the channel.

CONTROL POINT

point where depth of steady flow can be determined

due to grade change, dam, weir, etc. Examples;

The change of slope from mild to steep Free drop Entrance point from reservoir to steep

channel Outlet point from steep channel to reservoir Flow over weir

Presence of Broad Crested Weir

A rectangular channel with width b (constant along the channel) flows with q m³/s/m. Assume this channel's slope is 0 degree (flat) and no roughness coefficient (subcritical flow).

E1

y1

q²/2gy2²

EGL

q²/2gy1²

y2

E2

∆z

q

Broad Crested WeirA

B

Curve of Depth,y (Channel Bed) versus Specific Energy,E for

the Presence of Broad Crested Weir

45º

y

EE2

A

A'

E1

C

B

yc

y2

y1

∆Z

q (constant)

E = y + q²/2gy²

E1 = E2 + ∆Znote:- ∆Z = height of Broad crested weir

y1 + v1²/2g = y2 + v2²/2g + ∆Z

or

E2 = E1 - ∆Z

From that figure, depth of water flow become lesser from point A to point B,

Specific Energy at point A, E1 > E2 (at point B)If y2 = yc ; E2 = Emin ;

therefore ∆z = ∆zc (critical flow and this broad crested weir represent as control point)

If the weir increase more than before, specific energy will be decreased and water depth, y2 become lower until one point (point C). Specific energy, E2 become minimum and y2 turn to yc. At this point, ∆z = ∆zc, flow is critical and weir known as control point.

If the height of weir increase greater than (∆z > ∆zc), E-y curve for same q can not be used because minimum point for this curve is achieved and E2 < Emin. Therefore, for E2 and water depth above weir is constant, yc so ∆z move to right side from point C. At this point, E1 not enough for same q.

E1’ = Emin + ∆z

In this condition, total flow at point A can not flow over weir but maintain at the back of the weir. This condition called ‘choke’ and water depth at the upstream is increase. The depth at the upstream called backwater situation. At the downstream of the weir supercritical flow will be happened.

Examples of Weirs

Change of channel’s width(Narrowing channel)

The similar concept like weir will be applied in this topic but the relationship between q-y is used because the width of channel will be changed and q also.

For same Q;Q = q1b1 = q2b2

q1 = Q/b1 q2 = Q/b2

b2 < b1 , therefore; q2 > q1

Plan

Side plan

At critical point, b2 become minimum and q is maximum at same specific energy.

When the channel’s become smaller, E1 do not enough to support q so E1 need to increase for achieve suitable specific energy, E’ for critical depth yc’. Therefore, the depth at upstream, y1 increase for E’.

y1’ + q1²/2gy1²’ = E’

y1’+ Q²/2gb1²’y12’ = E’ where E’

=1.5yc

and

yc ‘ = ³√(q’²/g) = ³√(Q²/b’2g)

Therefore,

if E’ > E1, b’ < bmin

and specific energy at upstream =

E’ = 1.5y’c If control situation happened, the structure

will be controlled flow at upstream and this structure called venturi flume.

Example of flume

Example 9

The water flows uniformly at 16.5 m3/s in a rectangular channel with 3.0 m width and 1.8 m depth. If one part of the channel was narrowing, calculate the maximum width of narrowing that can obtain critical water depth.

COURSE OUTCOME (CO1)

-Analyse and evaluate the non uniform transitional flow