Critical flow

1. Calculate the critical depth in a trapezoidal channel with a base width of 10m and slide slopes of 1:2 when the discharge is 340 m3s-1.

[3.792m]2. Water flows in a 5.0m wide rectangular channel at a uniform depth of 2.0m. If the

channel has a bed slope of 1 x 10-4 and a Manning’s n of 0.02, determine the discharge, mean velocity and Froude number. What is the critical depth for this flow? What bed slope would be required for the flow to run uniformly at the critical depth?

[5.364m3s-1, 0.536 ms-1, 0.121, 0.450m, 6.32 x 10-3]

3. Show that the critical depth in a circular culvert running part full is given by the relationship:

Q2

g D5=(θ−sinθ )3

512sin( θ2 )where is the angle subtended at the centre by the free surface and D is the diameter of the culvert.

4. Show that for flow in a ‘V’ shaped open channel with equal side slopes of 1:s (vertical: horizontal), the critical depth and velocity are given by:

yc=[2Q2

gs2 ]1 /5

=4 E /5 & V c=¿¿

5. Show that the critical depth in an open channel with a parabolic cross section is given by:

yc=3 E4

What is the wetted perimeter of a parabolic cross section when expressed in terms of the channel centre line?

[~T+8h2/3T]

1. A = (b+sy)yP = b +2y(1+s2)0.5 R = A/PT = b + 2sy

E= y+ Q2

2 A2g →

dEdy

=1− Q2

g A3dAdy

=0

but dAdy

=b+4 y

Hence, 1=Q 2

g A3(b+4 y )= Q2

g (by+2 y2 )3(b+4 y)

g (by+s y2 )3

Q 2 =(b+4 y )

g (10 y+2 y2 )3

3402=(10+4 y )

y=√ 3√ 3402g(10+4 y )−10 y

2

Hence y = 3.793m (normally you would not do this many iterations).

2.

Q=1nA R2/3S1 /2= 1

0.02(5 x 2 )(5 x 25+4 )

23 (1 x10−4 )0.5=5.364m3/s

V = Q / A = 5.36 m/s

yc=3√ Q2

B2 g=0.4896m

Fr= V√gy

=5.36√2 g

=0.121

For critical slope, y =yc.

Q=1nA R2/3S1 /2

S=( nQ

A R23 )2

=6.317 x10−5

3. For critical depthQ2Tg A3

=1 or Q2

g= A3

T

For critical depth A=d2

8(θ−sinθ )

Considering the triangle below, sin(π−θ2 )=x /( d

2)

Hence 2 x=dsin(π−θ2 )

But sin(π−θ2 )=sinπ cos( θ2 )−cosπ sin( θ2 )

Hence, sin(π−θ2 )=sin( θ2 )

Thus, 2 x=T=d sin( θ2 )Hence Q2

g= A3

T=(d28 (θ−sinθ ))

3

/d sin( θ2 )

/2

/2

x

h

h-d/2

d/2

d/2

ThereforeQ2

g d5=

(θ−sinθ )3

sin( θ2 )4. T = 2sy A=0.5Ty = sy2 P=2 y √1+s2 R= A

P= sh2√1+s2

For critical depthQ2Tg A3

=1

Q2= gs3 y6

2 sy= gs2 y5

2

yc=( 2Q2

g s2 )1 /5

And E= y+ Q2

2 g A2= y+ g s2 y5

2 gs2 y 4= y+ y

4=5 y4

yc=4 E5

And E= y+V 2

2 g

V c2

2g=E c− yc=E−4 E

5= E5

V=¿¿

5.

Let y = ax2 where a is a constant.

Therefore, dA=2( ya )1 /2

dy

h

x

y

1

s

Hence, A=∫0

h

2( ya )1/2

dy=[ 2a12 23 y3 /2]0

h

=4h

32

3a1/2=4 h3 ( ha )

1 /2

But at the free surface h=a(Ta )2

or T=2( ha )1/2

Therefore, A=4h3 ( ha )

1 /2

=2hT3

E=h+ Q2

A22 g=h+ gA

2>¿=h+ A2T

=h+ h3=4 h3

¿

Rapidly varied flow tutorial questions

1. 240m3s-1 of water flows along a rectangular channel which is 10m wide at its narrowest. If the minimum depth is 4.2m, prove that depth changes downstream will affect those upstream?

2. Water flows under a wide sluice gate into a 4m wide rectangular channel. A hydraulic jump starts to form where the local depth is 40cm. calculate the conjugate depth and the power dissipated in the jump when the discharge is 8 m3s-1.

[23.7kW]

3. Water flows in a rectangular channel at a discharge of 1.6 m3s-1 per m width and a speed of 4 ms-1. To what height must a weir structure back-up the water to cause a hydraulic jump to form in the channel?

[0.96m]

4. A stilling basin of width 80m has to take a maximum flow of 2210m3/s from a gated spillway. The upstream head corresponding to this flow is 5m and the velocity of approach to the spillway is 2m/s. The floor of the stilling basin is 50m below the spillway crest level and it can be assumed that 25% of the total energy above the stilling basin floor is lost in the fall. Estimate the depth and supercritical velocity at the entrance to the stilling basin and determine the type of jump and the sequent depth of flow. If the tailwater depth (t) downstream of the stilling basin is given by t = 0.38Q1/2, where Q is the flow, determine whether t is greater than the sequent depth of flow. What effect will this have?

[0.981m; 28.16m/s; 12.115m]

1. q = 240 / 10 = 24m.

yc=3√ q2

g= 3√ 24 x24g

3.9m

Since min depth (4.2) > critical depth , the flow is subcritical hence disturbances will travel upstream.

2. V 1=QA

= 84 x 0.4

=5m / s

Fr1=V 1

√g y1= 5

√0.4 g=2.52

y2y1

=12 (√1+8Fr12−1)=12 (√1+8 x 2.52❑2 −1 )

Hence y2 = 1.24m

Power lost = weight of work / sec x total head loss across jump

¿ϱQg [( y1+ V 12

2 g )−( y2+ V 22

2g )]¿1000 x8 x 9.801[(0.4+ 252 g )−(1.24+ 8

2/ (1.24 x4 )2

2 g )]= 23.7kW

3. q = V1 y1, hence y1 = 1.6 / 4 = 0.4mE = 1.22mEmin = 3yc/2 = 1.5(q2/g)1/3=0.959E = Emin + zz =0.257m

Fr1=1.6

√0.4 g=2.02

Using hydraulic jump equation gives y2 = 0.96m

4

q = Q / b = 2210 / 80 = 27.625 m3/s / m

The total head above the stilling basin = 50+ho+uo2

2 g=50+5+ 2

2

2g=55.20394m

Assuming 25% energy loss = 0.25 x 55.20394 = 13.801mHence, E1 = 55.204 – 13.801 = 41.403m

At section 1:E1=h1+u12

2 g=h1+

q12

2gh12

41.403=h1+27.625❑

2

2gh12

41.403=h1+38.908h12

Try h1 = 1.0m, RHS = 39.908 h1 = 0.95m, RHS = 44.061 h1 = 0.98m, RHS = 41.492 h1 = 0.981m, RHS = 41.411 (close enough).

u1 = q/h1 = 27.625/0.981 = 28.180 m/s

Fr1 = 28.180 / (9.807 x 0.981)^0.5 = 9.079

Hence, h2h1

=12 [√1+8Fr12−1 ]=12 [√1+8 x9.079❑

2 −1 ]=12.349mh2 = 12.329 x 0.981 = 12.115m

tailwater depth, t = 0.38Q0.5 = 0.38 x 22100.5 = 17.864m

since t > h2 the tailwater will drown out the jump causing it to occur on the spillway. This is a safe design but not efficient since little energy will be dissipated. To overcome this problem either build a sloping apron or provide a drop in the channel bed.

Channel transitions

1. A 10m wide rectangular channel discharges 100 m3s-1. At a particular location the width is reduced locally to 5m by means of a smooth transition. Plot the depth versus specific energy relations for q = 10 m3s-1 /m and q = 20 m3s-1 /m in the range of 0 < h < 6m. What is the depth of flow at the throat for upstream depths of 5.5m and 1.0m? evaluate the critical depths in the channel and at the throat and compare them with given by the graphs.

2. Water flows at a velocity of 3ms-1 and at a depth of 3.0m in an open channel of rectangular cross section. Find the downstream depth and the change in water level produced by (a) smooth upwards step of 0.3m, and (b) a smooth downward step of 0.3m in the channel bed. Also find (c) the maximum allowable size of upward step for the upstream flow to be as specified.

[(a) 2.495m, -0.205m; (b) 3.402m, 0.102m; (c) 0.427m]

3. A venture flume installed in a horizontal rectangular channel 700mm wide has a uniform throat width of 280mm. When water flows through the channel at 0.14 m 3/S, the depth at a section upstream of the flume is 430mm. Neglecting friction and given that the discharge passes through the critical depth at the throat, determine the depth of flow at the throat, the depth at a section just downstream of the flume where the width is again 700mm, and the force exerted on the stream in passing through the flume. (Assume = 1000 kg/m3).

[294mm, 74.6mm, -305.3N]

4*. To investigate the problem of flow around bridge piers use the concept of specific energy and critical depth and assume that the conditions are as shown in Figure 1 with the critical depth occurring in between the piers. Show that under chocking conditions, i.e. h2 = hc, the various hydraulic parameters are related by the equation:

σ 2=27 Fr1

2

{β3 (2+F112 )3}Where E2 = E1 = downstream specific energy, with being some coefficient to allow for energy loss, Fr = Froude number and = (B-b)/B.

If B = 20m, Q = 80m3/s and the upstream depth h1 = 2.0m determine the maximum permissible pier width b for subcritical conditions to remain in the channel if the energy loss by the flow past the bridge pier is 10% of the specific energy.

[3.2m]

1. b = 10m & 5mQ = 100 m3/sq1 = 100 / 10 = 10 and q2 = 20

………

E1=h1+u12

2 g2.

E1=h1+u12

2 g=3+ 3❑

2

2g=3.4589m

(a) upward step of 0.3mE2 = E1 – 0.3 = 3.1589m

E2=h2+q22

2 gh22

3.1589=h2+4.1297h22

Solving by trail and error gives h2 = 2.495m. Hence, water level is 0.3 + 2.495 = 2.795m above the original bed level. Therefore, change in water level = 3 – 2.795m = 0.205m (downwards).

(b) downwards step of 0.3mE2 = E1 + 0.3 = 3.7589mRepeat of the above gives h2 = 3.402m. Water level is 3.402 – 0.3 = 3.102m above the bed. Hence, change in water level = 3 – 3.102m = -0.102m (upwards).

(c) max step size

hc=3√ q2g = 3√ 92g =2.0214m

Emin = 1.5 hc = 3.0321mZ = E1 - Emin = 3.4589 – 3.0321 = 0.427m

3.

4.E1=h1+

u12

2 g E2=h2+

u22

2 g βE1=E2

β [h1+ u12

2 g ]=E2

β h1[1+ u12

2g h1 ]=E2

β [1+ Fr12

2 ]= E2h1

β [2+Fr12 ]=2 E2h1

Choking conditions at throat give hc = h2 = 2/3Emin = 2/3E2

β [2+Fr12 ]= 2h132h2=3

h2h1

h2h1

=β [2+Fr12 ]

3

Continuity gives Q1 = Qc

q1B = qc (B-b)

q1=( B−bB )qc=σ qc

u1h1 = uchc

u12h1

2 = 2uc2hc

2 = 2ghc3

(since uc2 = ghc (Fr = 1))

u12

g h1=σ2( hc

h1 )3

=σ2( h2h1 )3

Hence, Fr12

σ2=( hch1 )

3

=β3 [2+Fr12 ]

3

27

Gradually varied flow

1. The rate of change of depth of a steady flow along an open channel of rectangular cross section can be expressed as:

dhdx

=(So−S f )/(1+ ugdudh )

Where So is the channel bed slope, Sf is the frictional slope, U is the mean velocity and h is water the depth. In a wide horizontal rectangular channel of constant6 width 10m the depth of flow decreases from 0.9m at x = 0 to 0.6m at x = L. if the discharge in the channel is 5m3s-1 and Sf = 0.01U2/2gh, determine the distance L between the two depths. Check the Froude number at both positions and the sign of dh/dx.

[972.7m, 0.187, 0.343, negative]

2. Water flows down a trapezoidal channel having a constant longitudinal bed slope of 0.0005 and a Manning roughness coefficient of 0.024. The channel cross section has a base width of 15m and slide slopes of 1:1. At a certain point in the channel the depth of flow is found to be decreasing at a rate of 1 in 10,000 and at this point the depth is 5m. Working with the gradually varied flow equation estimate the discharge in the channel.

[230 m3s-1]

1. dhdx

=(So−S f )/ (1−Fr2 )

Fr2=u2/ ghfor a rectangular channel q = uh. Hence, dq = udh + hdu = 0. Hence,u/h = -du/dh or Fr2=−u/h (du /dh) which gives:

dhdx

=(So−S f )/(1+ ugdudh )

dx=[1+ ugdudh ](−1S f )dh

q = uh and du/dh = -u/h = -q/h2

dx=[1+ qgh (−q

h2 )](−2gh20.01q2 )dh

∫0

L

dx=[−200 g h4

4q2+200h]

h1

h2

L=−200 g4 x 0.52

(0.64−0.94 )+200 (0.6−0.9 )=972.677 m

2. At point of interest:A = (15 + 1x5)x5 = 100m2

P = 15 + 2x5 (1+12)0.5 = 29.1421mR = A / P = 3.4315 mT = 15 + 2x1x5 = 25m

dhdx

=So−Q2n2/ ( A2 R4 /3 )

1−Fr2=

So−Q2n2 /( A2R4 /3 )1−Q2T /g A3

dh/dx = -1/10000 and So = 0.0005. Hence,

−0.0001=0.0005−Q20.0242/ (10023.43254 /3 )1−Q225/g1003

−0.0001=0.0005−Q2x 1.11287 x10−8

1−Q2 x2.5492 x10−6

-0.0001 + 2.5492Q2x10−6=0.0005−1.11287Q2 x10−8

Q=229.6 m3s-1