Chapter 5- Open Channel Non-uniform Flow

g

v

2

2

1

y1

SoL

Figure 4.6. Non-uniform gradually varied flow

Uniform flow is found only in artificial channels of constant shape, slope, although under these conditions the flow for some distances may be non-uniform, as shown in Figure 4.1. However, with natural stream the slope of the bed and the shape and size of the cross-section usually vary to such an extent that true uniform flow is rare. Hence, the application of Manning equation for uniform flow can be applied to non-uniform flow with accuracy dependent on the length of reach L taken. In order to apply these equations at all, the streams must be divided into several reaches within which the conditions are approximately the same.

E.G.L

. hL=SL

g

v

2

2

2

y2

EGL Slope= S

Channel bed, slope = So

∆x

1 2

Chapter V

STEADY NON-UNIFORM FLOW OR VARIED (S≠ SO)FLOW IN OPEN CHANNELS

∆𝑥 =𝐸2 − 𝐸1

𝑆𝑜 − 𝑆

From the above figure, energy equation between section 1-2

𝑉12

2𝑔+ 𝑦1 + 𝑆𝑜 ∆𝑥 =

𝑉22

2𝑔+ 𝑦2 + 𝑆 ∆𝑥

𝑆𝑜 ∆𝑥 − 𝑆 ∆𝑥 = 𝑉2

2

2𝑔+ 𝑦2 −

𝑉12

2𝑔+ 𝑦1

∆𝑥 𝑆𝑜 − 𝑆 = 𝐸2 − 𝐸1

where :

𝑆 =𝑛2𝑉𝑚

2

𝑅𝑚

43

𝑉𝑚 = 𝑉2 + 𝑉1

2 (𝑚𝑒𝑎𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝑎𝑛𝑑 2)

𝑅𝑚 = 𝑅2 + 𝑅1

2 (𝑚𝑒𝑎𝑛 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑅𝑎𝑑𝑖𝑢𝑠)

𝑅1 = 𝐴1

𝑃1

𝑅2 = 𝐴2

𝑃2

Slope can be determined by Manning’s Equation

There are two types of non uniform flow depending upon the change of depth of flow over the length of the channel. If the depth of flow in a channel changes a gradually over a length of the channel, the flow is said to be Gradually Varied Flow (GVF). If depth of flow changes abruptly over a small length of the channel, the flow is said to be a local non-uniform phenomenon or Rapidly Varied Flow (RVF). Gradually varied flow can occur with either subcritical or supercritical flow, but the transition from one condition to the other is ordinarily abrupt, as between D and E in Figure 4.1. Other cases of local non-uniform flow occur at the entrance and exit of a channel, at channel at changes in cross sections, at bends and at on obstruction such as dams, weirs or bridge piers. See Figure 4.7 for steady non-uniform flow in a channel.

Depth of flow for non-uniform flow conditions varies with longitudinal distance. It occurs upstream and downstream control sections.

Rapid varied flow of occurs on the following condition: 1. Occurrence of hydraulic jump 2. Flow entering a steep channel from lake or a reservoir 3. Flow close to a free out fall from a channel 4. Flow in a vicinity of an obstruction such as bridge pier or sluice gate

Gradual varied flow occurs on the following condition:

1. Backwater created by a dam place in a river 2. Drawdown of a water surface as flow approaches a falls

RVF GVF RVF

RVF GVF

GVF

RVF

GVF

Figure 4.7. Steady Non-uniform flow in a channel.

WATER-SURFACE PROFILES IN GRADUALLY VARIED FLOW

Water surface profiles are classified two different ways: according to the slope of the channel (mild, steep, critical, horizontal, or adverse) and according to the actual depth of flow in relation to the critical and normal depths (zone 1, 2, or 3). The first letter of the type of slope (M, S, C, H or A) in combination with 1, 2, or 3 defines the type of surface profile. If the slope is so small that the normal depth (uniform flow depth) is greater than critical depth for the given discharge, then the slope of the channel is mild, and the water surface profile is given an M classification. Similarly, if the channel slope is so steep that a normal depth less than critical is produced, then the channel is steep, and the water surface profile is given an S designation. If the slope’s normal depth equals its critical depth, then we have a critical slope, denoted by C. Horizontal and adverse slopes, denoted by H and A, respectively, are special categories because normal depth does not exist for them. An adverse slope is characterized by a slope upward in the flow direction. The 1, 2, and 3 designations of water surface profiles indicate if the actual flow depth is greater than both normal and critical depths (zone 1), between the normal; and critical depths (zone 2) or less than both normal and critical depths (zone 3). The basic shape of the various possible profiles are shown in the Table 4.4.

Table 4.4. Types of Varied Flow

Problem: 1. A rectangular canal is 2.0m wide and carries 2.4m3/s of water. The bed slope is 0.0009 and

the channel roughness n=0.012. At a certain section the depth is 0.90m and at another section the depth is 1.20m. a. Determine which depth is downstream b. Determine the distance between sections with the given depths using one reach. c. Determine the distance between the sections with the given depths using three

reaches.

2. Water discharges from under a sluice gate into a horizontal channel at a rate of 1m3/s per meter of width as shown in the figure. What is the classification of the water surface profile? Quantitatively evaluate the profile downstream of the gate and determine whether or not it will extend all the way to the abrupt drop 80m downstream. Make the simplifying assumption that the resistance factor f is equal to 0.02 and that the hydraulic radius R is equal to the depth y.

3. A rectangular concrete channel 4.50m wide is carrying water. At an upstream point, the

depth of water is 1.50m and a downstream point 300m away, the depth of flow is 1.17m. If the channel bed slope is 0.0010, find the theoretical flow rate. Use n=0.013.

1. A rectangular canal is 2.0m wide and carries 2.4 m3/s of water. The bed slope is 0.0009 and the channel roughness n=0.012. At a certain section the depth is 0.90m and at another section the depth is 1.20m. a. Determine which depth is downstream. b. Determine the distance between sections with the given depths using one reach. c. Determine the distance between the sections with the given depths using three reaches. GIVEN: Q = 2.4 m3/s; So = 0.0009; n = 0.012

0.82

1.20

0.90

yc= 0.523

2m

SOLUTION:

a.)

@ Critical flow

channel ofh meter widtper /sm2.12

4.2 3b

Qq

smVc /276.2528.081.9

2056.1528.02 mbYA cc

mYbP cc 056.3528.0222

mRc

c

P

A

c 346.0056.3

056.1

0009.0003071.0346.0

276.2012.03/4

22

3/4

22

o

c

cc S

R

VnS

Flow is subcritical. Actual slope is mild.

mg

qYc 528.0

9.81

2.13

2

3

2

@ Uniform flow

ooo YbYA 2

ooo YYbP 222

o

o

Po

A

oY

YR o

22

2

2/13/21SRA

nQ oo

2/1

3/2

0009.022

22

012.0

14.2

o

oo

Y

YY

3/2

3/5

148.0

o

o

Y

Y

Let 3/2

3/5

1 o

o

Y

YM

By Trial & error:

Assume oY M

1.0 0.630

0.82 0.482

oY =0.82m < 0.90m

Since Y > oY > cY and the slope is mild, the depth 1.20m is downstream of depth 0.90m. Type of

profile is 1M .

Δx

1.20

0.90

b.) using one reach

@ Section 1:

mY 90.01

2

11 80.190.02 mbYA

mYbP 80.390.0222 11

mRP

A474.0

80.3

80.11

1

1

smA

QV /333.1

80.1

4.2

1

1

Then

mg

E 991.02

333.190.0

2

1

@ Section 2:

mY 20.12

2

22 40.220.12 mbYA

mYbP 40.420.1222 212

mRP

A545.0

40.4

40.22

2

2

smA

QV /1

40.2

4.2

2

2

Then

mg

E 251.12

0.120.1

2

2

Mean Velocity

sm

VVVm /167.1

2

1333.1

2

21

Mean Hydraulic Radius

m

RRRm 5095.0

2

545.0474.0

2

21

Slope

0004813.05095.0

167.1012.03/4

22

3/4

22

m

m

R

VnS

,Therefore

mx 96.6200009.0000481.0

251.1991.0

Δx2 Δx2 Δx1

1.10 1.0

Δx

1.20

0.90

c.) using three reaches

@ Section 3:

mY 0.13

2

33 0.20.12 mbYA

mYbP 0.40.1222 33

mRP

A50.0

0.4

0.23

3

3

smA

QV /20.1

0.2

4.2

3

3

Then

mg

E 073.12

20.10.1

2

3

@ Section 4:

mY 10.14

2

44 20.210.12 mbYA

mYbP 20.410.1222 44

mRP

A524.0

20.4

20.24

4

4

smA

QV /091.1

20.2

4.2

4

4

Then

mg

E 161.12

091.11.1

2

4

Mean Velocity

sm

VVVm /267.1

2

311

sm

VVVm /145.1

2

432

sm

VVVm /045.1

2

243

Mean Hydraulic Radius

m

RRRm 487.0

2

311

mRR

Rm 512.02

432

mRR

Rm 534.02

243

Slope

000603.0487.0

267.1012.03/4

22

3/4

1

2

1

2

1 m

m

R

VnS

00046.0512.0

145.1012.03/4

22

3/4

2

2

2

2

2 m

m

R

VnS

000363.0534.0

045.1012.03/4

22

3/4

3

2

3

2

3 m

m

R

VnS

,Thus

mSS

EEx

o

09.2761

31

1

mSS

EEx

o

73.1971

432

mSS

EEx

o

52.1681

243

,Therefore

52.16873.19709.276 x

∆𝒙 = 𝟔𝟒𝟐. 𝟑𝟒 𝒎

80 m

10cm

2. Water discharges from under a sluice gate into a horizontal channel at a rate of 1m3/s per meter of

width as shown in the figure. What is the classification of the water surface profile? Quantities

evaluate the profile downstream of the gate and determine whether or not it will extend all the way

to the abrupt drop 80m downstream. Make the simplifying assumption that the resistance factor f is

equal to 0.02 and that the hydraulic radius R is equal to the depth y.

GIVEN:

q=1m3/s per meter width

f=0.02

So=0

R=y

Ys=0.10m (depth of the flow from sluice gate)

SOLUTION:

Critical depth

mYmY

g

qY

sc

c

10.0467.0

9.81

13

2

3

2

(With horizontal bed slope, the water surface profile is classified as type H3, see table 4.4)

Using direct step method

of SS

EEx

21

Where: 3/4

22

22.2 m

mf

R

VnS (Manning Equation English Unit)

m

mf

gR

fVS

8

2

(Darcy-Weisbach Equation)

2

21 VVVm

2

21 RRRm

g

VyE

2

2

111

Sample Computation:

Velocity, @ y=0.10m

smy

qV /10

10.0

1

Using change in depth mooy 4.

mooy 4.

smy

qV /14.7

14.0

1

mx

gx

752.15

0156.0

2

5110014.01.0

Section No.

Depth, y (m)

Velocity @ section, V

(m/s)

Mean Velocity in reach,

Vm

V12

Mean Hydraulics

Radius, 𝑅𝑚

𝑆𝑓 =𝑓𝑉𝑚

2

8𝑔𝑅𝑚

∆𝑥 =

𝑦1 − 𝑦2 +(𝑉1

2−𝑉22)

2𝑔

𝑆𝑓 − 𝑆𝑜

Distance

from gate (m)

1 0.1 10 100 0

8.57 73.4 0.12 0.156 15.7

2 0.14 7.14 51 15.7

6.35 40.3 0.16 0.064 15.3

3 0.18 5.56 30.9 31.0

5.05 25.5 0.2 0.032 15.1

4 0.22 4.54 20.6 46.1

4.195 17.6 0.24 0.019 13.4

5 0.26 3.85 14.8 59.5

3.59 12.9 0.28 0.012 12.4

6 0.3 3.33 11.1 71.9

3.135 9.8 0.32 0.008 10.9

7 0.34 2.94 8.6 82.8

∴ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑓𝑖𝑙𝑒 𝑒𝑥𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑎𝑏𝑟𝑢𝑝𝑡 𝑑𝑟𝑜𝑝 80 𝑚 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚

0

10

20

30

40

50

60

0 20 40 60 80 100 120x (m)

y (cm)

4.50m

y 2

1

y1 = 1.5 m

y2 = 1.17 m

L

3. A rectangular concrete channel 4.50m wide is carrying water. At an upstream point, the

depth of water is 1.50m downstream point 300m away, the depth of flow is 1.17m. if

the channel bed slope is 0.0010, find the theoretical flow rate. Use n=0.013.

Solution:

𝐿 = 𝑉2

2

2𝑔+𝑦2 −

𝑉12

2𝑔+𝑦1

𝑆𝑜− 𝑆

𝑉2

2

2𝑔+ 𝑦2 −

𝑉12

2𝑔+ 𝑦1 = 𝑆𝑜𝐿 − 𝑆𝐿

𝑉2

2

2𝑔−

𝑉12

2𝑔+ 𝑆𝐿 = 𝑦1− 𝑦2 + 𝑆𝑜𝐿 ..Eq.1

@section 1

𝐴1 = 4.5 𝑥 1.5 = 6.75 𝑚2

𝑅1 = 𝐴1 𝑃1 = 6.75 4.5 + (2𝑥1.5) = 0.90𝑚

𝑉1 = 𝑄 𝐴1 = 𝑄/6.75 = 0.148𝑄

𝑉1

2

2𝑔= 0.00112𝑄2

@section 2

𝐴2 = 4.5 𝑥 1.17 = 5.265 𝑚2

𝑅2 = 𝐴2 𝑃2 = 5.265 4.5 + (2𝑥1.17) = 0.77𝑚

𝑉2 = 𝑄 𝐴2 = 𝑄/6.75 = 0.19𝑄 𝑉2

2

2𝑔= 0.0018𝑄2

𝑅𝑚 =𝑅1+𝑅2

2=

0.9+0.77

2= 0.835 𝑚

𝑉𝑚 =𝑉1+𝑉2

2=

0.148𝑄+0.19𝑄

2= 0.169𝑄

𝑆 = 𝑛𝑉𝑚

𝑅𝑚2 3

2=

0.013𝑥0.169𝑄

0.8352 3 2

= 0.00000614𝑄2

From Eq.1

0.0018𝑄2 − 0.00112𝑄2 + 0.00000614𝑄2 𝑥300 = 1.5 − 1.17 + 0.001(300)

∴ 𝑸 = 𝟏𝟓. 𝟖𝟏 𝒎𝟑/𝒔

HYDRAULIC JUMP

A hydraulic jump is a transition flow from supercritical to subcritical flow.

Hydraulic jump is one means of reducing the velocity of flow. It may also be used to separate lighter

solids from heavier ones.

y2>yc

V1 y1 < yc

HYDRAULIC JUMP IN A RECTANGULAR CHANNEL

Consider a freebody of water containing hydraulic jump

F2

F1

W

P1 = γ y1 P2 = γ y2 N

y2

y1 V1

y2

y1 Q

V2

Considering the Impulse-Momentum Equation

𝛴𝐹 = (𝜌𝑄𝑉)𝑜𝑢𝑡 − (𝜌𝑄𝑉)𝑖𝑛

𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛

𝐹1 − 𝐹2 − 𝐹𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

where: Ef = neglected (if distance between sections is relatively small)

𝐹1 = 1

2𝑃1𝑦1𝑏 =

1

2 γ𝑦1𝑦1𝑏 =

1

2 γ𝑦1

2𝑏

𝐹2 = 1

2𝑃2𝑦2𝑏 =

1

2 γ𝑦2𝑦2𝑏 =

1

2 γ𝑦2

2𝑏

Then, 1

2 γ𝑦1

2𝑏 −1

2 γ𝑦2

2𝑏 − 0 = γ

𝑔 𝐴2𝑉2 𝑉2 −

γ

𝑔 𝐴1𝑉1 𝑉1

1

2 𝑔𝑏 𝑦1

2 − 𝑦22 = 𝐴2𝑉2

2 − 𝐴1𝑉12

1

2 𝑔𝑏 𝑦1

2 − 𝑦22 = ( 𝑏𝑦2 )𝑉2

2 − ( 𝑏𝑦1 )𝑉12

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑦2𝑉2

2 − 𝑦1𝑉12

From continuity equation

𝑄1 = 𝑄2

𝐴1𝑉1 = 𝐴1𝑉1

𝑏𝑦1𝑉1 = 𝑏𝑦2𝑉2

𝑽𝟐 = 𝒚𝟏𝑽𝟏

𝒚𝟐

Substitute values

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑦2

𝑦12𝑉1

2

𝑦22 − 𝑦1𝑉1

2

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑉1

2𝑦1 𝑦1

𝑦2− 1

1

2 𝑔 𝑦1

2 − 𝑦22 =

𝑉12𝑦1

𝑦2 𝑦1 − 𝑦2

1

2 𝑔 𝑦1 − 𝑦2 𝑦1 + 𝑦2 =

𝑉12𝑦1

𝑦2 𝑦1 − 𝑦2

1

2 𝑔 𝑦1 + 𝑦2 =

𝑉12𝑦1

𝑦2

𝑽𝟏𝟐 =

𝟏

𝟐 𝒈

𝒚𝟐

𝒚𝟏 𝒚𝟏 + 𝒚𝟐

But

𝑉1 = 𝑄

𝐴=

𝑏𝑞

𝑏𝑦1=

𝑞

𝑦1

𝑞2

𝑦12 =

1

2 𝑔

𝑦2

𝑦1 𝑦1 + 𝑦2

𝒒𝟐 = 𝟏

𝟐 𝒈𝒚𝟏𝒚𝟐 𝒚𝟏 + 𝒚𝟐

ENERGY LOST AND POWER LOST IN A JUMP

Energy Equation 1 – 2

𝑃1

𝛾+ 𝑧1 +

𝑉12

2𝑔=

𝑃2

𝛾+ 𝑧2 +

𝑉22

2𝑔+ ℎ𝐿

𝑦1 + 𝑉1

2

2𝑔= 𝑦2 +

𝑉22

2𝑔 + ℎ𝐿

𝐸1 = 𝐸2 + ℎ𝐿

𝒉𝑳 = 𝑬𝟏 − 𝑬𝟐 energy head lost

Power Lost: 𝑷 = 𝜸𝑸𝒉𝑳

Depth of Hydraulic Jump

Solve for y2: consider the equation:

𝑞2 = 1

2 𝑔𝑦1𝑦2 𝑦1 + 𝑦2

𝑦2 𝑦1 + 𝑦2 = 2𝑞2

𝑔𝑦1

𝑦22 + 𝑦1𝑦2 =

2𝑞2

𝑔𝑦1

𝑦22 + 𝑦1𝑦2 +

1

2𝑦1

2=

2𝑞2

𝑔𝑦1 +

1

2𝑦1

2

𝑦2 + 1

2𝑦1

2 =

2𝑞2

𝑔𝑦1 +

1

4𝑦1

2

𝑦2 + 1

2𝑦1

2 =

1

4𝑦1

2

8𝑞2

𝑔𝑦13 + 1

Extract the square root:

𝑦2 + 1

2𝑦1 =

1

2𝑦1

8𝑞2

𝑔𝑦13 + 1

𝑦2 = − 1

2𝑦1 +

1

2𝑦1

8𝑞2

𝑔𝑦13 + 1

𝑦2 = 1

2𝑦1 −1 +

8𝑞2

𝑔𝑦13 + 1

But 𝑞2

𝑔𝑦13 =

𝑄2

𝑏2

𝑔𝑦13 =

𝐴2𝑉2

𝑏2

𝑔𝑦13 =

𝑏2𝑦12𝑉1

2

𝑏2

𝑔𝑦13 =

𝑉12

𝑔𝑦1= 𝑁𝐹

2

Hence, 𝒚𝟐 = 𝟏

𝟐𝒚𝟏 −𝟏 + 𝟖𝑵𝑭𝟏

𝟐 + 𝟏 𝑁𝐹1> 1

Likewise,

𝒚𝟏 = 𝟏

𝟐𝒚𝟐 −𝟏 + 𝟖𝑵𝑭𝟐

𝟐 + 𝟏 𝑁𝐹2< 1

2

y2

y1

1

yc2

y2

y1

yc1

HYDRAULIC JUMP IN A NON-RECTANGULAR SECTION

section thru 1 - 1 section thru 2 - 2

Impulse-Momentum Equation:

𝛴𝐹𝑥 = (𝑝𝑄𝑉)𝑜𝑢𝑡 − (𝑝𝑄𝑉)𝑖𝑛

𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

𝛾𝐴1𝑦𝑐1− 𝛾𝐴2𝑦𝑐2

− 0 = 𝛾

𝑔 𝐴2𝑉2𝑉2 − 𝐴1𝑉1𝑉1

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴2𝑉22 − 𝐴1𝑉1

2

Continuity Equation:

𝑄1 = 𝑄2

𝐴1𝑉1 = 𝐴1𝑉1

𝑉2 = 𝐴1𝑉1

𝐴2

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴2𝐴1

2𝑉12

𝐴22 − 𝐴1𝑉1

2

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴1𝑉12

𝐴1

𝐴2− 1

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴1𝑉12

𝐴1−𝐴2

𝐴2

𝑽𝟏𝟐 =

𝒈𝑨𝟐

𝑨𝟏 𝑨𝟏𝒚𝒄𝟏

−𝑨𝟐𝒚𝒄𝟐

𝑨𝟏− 𝑨𝟐 or 𝑽𝟏

𝟐 = 𝒈𝑨𝟐

𝑨𝟏 𝑨𝟐𝒚𝒄𝟐

−𝑨𝟏𝒚𝒄𝟏

𝑨𝟐− 𝑨𝟏

*Another solution

𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛

𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

γ𝐴1𝑦𝑐1 − γ𝐴2𝑦𝑐2 =γ

𝑔 𝐴2𝑉2

2 − 𝐴1𝑉12

𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =1

𝑔 𝐴2

𝑄2

𝐴22 − 𝐴1

𝑄2

𝐴12

𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =𝑄2

𝑔

1

𝐴2−

1

𝐴1

𝑸𝟐

𝒈=

𝑨𝟏𝒚𝒄𝟏−𝑨𝟐𝒚𝒄𝟐

𝟏

𝑨𝟐−

𝟏

𝑨𝟏

1.2 m

2.0 m

1.2 m

1. Water flows in a rectangular channel with a width of 4.0 m at a uniform depth of 1.2 m. Adjustment is made downstream to raise water level to 2.0 m. consequently causing hydraulic jump. a. Calculate the discharge in the canal. b. Determine the power lost in a jump.

Solution:

a.) 𝑞2 = 1

2 𝑔𝑦1𝑦2 𝑦1 + 𝑦2

𝑞2 = 1

2 (9.81)(1.2)(2) 1.2 + 2

𝑞 = 6.138 𝑚3 𝑠 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑛𝑎𝑙 𝑄 = 𝑞𝑏 = 6.138 4

𝑸 = 𝟐𝟒. 𝟓𝟓 𝒎𝟑 𝒔 b.) Power Lost, 𝑃 = 𝛾𝑄ℎ𝐿 ℎ𝐿 = 𝐸1 − 𝐸2

where 𝐸1 = 𝑦1 + 𝑉1

2

2𝑔= 1.2 +

24.55

1.2 (4)

2

2𝑔= 2.533 𝑚

𝐸2 = 𝑦2 + 𝑉2

2

2𝑔= 2 +

24 .55

2 (4)

2

2𝑔= 2.480 𝑚

ℎ𝐿 = 2.533 − 2.480 = 0.053 𝑚

Thus, 𝑃 = 𝛾𝑄ℎ𝐿 = (9.81)(24.55)(0.053) 𝑷 = 𝟏𝟐. 𝟕𝟔𝟒 𝒌𝑾

2. A hydraulic jump occurs in a 5 m wide rectangular canal carrying 6 m3/s on a slope of 0.005. the depth

after the jump is 1.4m.

a.) Calculate the depth before the jump.

b.) Calculate the power lost in a jump.

y2 = 1.4 m

y1 =?

Given: b= 5 m S= 0.005

Q= 6 m3/s yafter= 1.4 m

Solution:

a.) 𝐴2 = 𝑏𝑦2 = 5 1.4 = 7 𝑚2

𝑉2 = 𝑄

𝐴2=

6

7= 0.857 𝑚/𝑠

𝑁𝐹2=

𝑉2

𝑔𝑦2=

0.857

9.81 x 1.4= 0.23 < 1

There is a hydraulic jump that occurs. And the depth before the jump is

𝑦1 = 1

2𝑦2 −1 + 8𝑁𝐹2

2 + 1 =1

2(1.4) −1 + 8(0.23)2 + 1

∴ 𝒚𝟏 = 𝟎. 𝟏𝟑𝟔 𝒎

b.) 𝑉1 =𝑄

𝐴1=

𝑄

𝑏𝑦1=

6

5 x 0.136= 8.82 𝑚/𝑠𝑒𝑐

𝐸1 = 𝑦1 + 𝑉1

2

2𝑔= 0.136 +

8.82 2

2𝑔= 4.1 𝑚

𝐸2 = 𝑦2 + 𝑉2

2

2𝑔= 1.4 +

0.857 2

2𝑔= 1.44 𝑚

Therefore,

ℎ𝐿 = 𝐸1 − 𝐸2

ℎ𝐿 = 4.1 − 1.44 = 2.66 𝑚

Thus,

𝑃 = 𝛾𝑄ℎ𝐿

𝑃 = 9.81(6)(2.66)

𝑷 = 𝟏𝟓𝟔. 𝟓𝟕 𝒌𝑾

3. A rectangular canal has a width of 4.0m and carries water at the rate of 12m3/s. its bed slope is 0.0003 and roughness is 0.02. To control the flow, a sluice gate is provided at the entrance to the canal.

a. Determine whether a hydraulic jump would occur when the sluice gate is adjusted so that minimum depth after the gate is 0.40 m.

b. If a hydraulic jump would occur in letter (a), how far from the sluice gate will it occur?

y1=depth req. to

cause a jump

yo

ys=0.4

0m

∆𝑥

Given: Q = 2m3/s; b = 4m; n = 0.02 ; So = 0.0003 Solution:

𝑞 =𝑄

𝑏=

12

4= 3 𝑚3 𝑠 𝑝𝑒𝑟 𝑚 𝑤𝑖𝑑𝑡ℎ

@critical flow

𝑦𝑐 = 𝑞2

𝑔

1 3

= 32

9.81

1 3

= 0.972 𝑚

𝐴𝑐 = 𝑏𝑦𝑐 = 4 0.972 = 3.887 𝑚2 𝑃𝑐 = 𝑏 + 2𝑦𝑐 = 4 + 2 0.972 = 5.944 𝑚

𝑅𝑐 =𝐴𝑐

𝑃𝑐= 0.654 𝑚

𝑉𝑐 =𝑄

𝐴𝑐 𝑜𝑟 𝑉𝑐 = 𝑔𝑦𝑐 = 3.087 𝑚 𝑠

𝑆𝑐 =𝑛2𝑉𝑐

2

𝑅𝑐4 3 =

0.00223.0872

0.654 4 3 = 0.006715 > So = 0.0003 ∴ 𝑠𝑙𝑜𝑝𝑒 𝑖𝑠 𝑚𝑖𝑙𝑑.

@Normal depth 𝐴𝑜 = 𝑏𝑦𝑜 = 4𝑦𝑜 𝑃𝑜 = 𝑏 + 2𝑦𝑜 = 4 + 2𝑦𝑜 = 2(2 + 𝑦𝑜)

𝑅𝑜 =𝐴𝑜

𝑃𝑜=

4𝑦𝑜

2(2+𝑦𝑜 )=

2𝑦𝑜

(2+𝑦𝑜 )

𝑄𝑜 =1

𝑛𝐴𝑜𝑅𝑜

2

3𝑆𝑜

1

2

12 =1

0.024𝑦𝑜

2𝑦𝑜

(2+𝑦𝑜 )

2

30.0003

1

2

2.812 =𝑦𝑜

5 3

(2+𝑦𝑜 )2 3 let 𝑀 =𝑦𝑜

5 3

(2+𝑦𝑜 )2 3

Trial and error: Assume y M 1.0 0.481 3.053 2.182 ∴ 𝑦𝑜 = 3.053 𝑚

yc

yo

Depth required to cause a jump:

𝑦1 =1

2𝑦𝑜 −1 + 8𝑁𝐹𝑜

2 + 1

where 𝑉𝑜 =𝑄

𝐴𝑜=

12

𝑏𝑦𝑜=

12

4(3.053)= 0.983 𝑚/𝑠𝑒𝑐

𝑁𝐹𝑜2 =

𝑉𝑜2

𝑔𝑦𝑜=

0.983 2

9.81 𝑥 3.053= 0.032

𝑦1 =1

2

3

053 −1 + 8 0.032 2 + 1 = 0.184 𝑚 < 𝑦𝑠 = 0.40𝑚

∴ "𝑻𝒉𝒆𝒓𝒆′𝒔 𝒏𝒐 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒋𝒖𝒎𝒑 𝒐𝒄𝒄𝒖𝒓𝒔"

𝑦𝑜 > 𝑦𝑐 > 𝑦 𝑠𝑢𝑝𝑒𝑟𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑓𝑙𝑜𝑤

y1

yo=1.60

m ys=0.50m

Δx

4. A rectangular channel has a width of 5m, so=0.0009 and n=0.012. its uniform flow depth is 1.60m. if a sluice gate is adjusted such that a min. depth immediately downstream of the gate is 0.50m.

a. Determine whether a hydraulic jump would occur, and if it occurs b. how far downstream will it occur c. type of profile

Solution: a.) @ normal depth, yo=1.60 m 𝐴𝑜 = 𝑏𝑦𝑜 = 5 1.60 = 8 𝑚2 𝑃𝑜 = 𝑏 + 2𝑦𝑜 = 4 + 2 1.60 = 8.20𝑚

𝑅𝑜 =𝐴𝑜

𝑃𝑜=

8

8.20= 0.976 𝑚

𝑉𝑜 =1

𝑛𝑅𝑜

2 3 𝑆1 2 =1

0.012(0.976)2 3 (0.0009)1 2 = 2.46 𝑚/𝑠𝑒𝑐

𝑁𝐹𝑜2 =

𝑉𝑜2

𝑔𝑦𝑜=

2.46 2

9.81 𝑥 1.60= 0.385

Depth required to cause a jump

𝑦1 =1

2𝑦𝑜 −1 + 8𝑁𝐹𝑜

2 + 1 =1

2 1.60 −1 + 8 0.385 2 + 1 = 0.816 𝑚

𝑦1 = 0.816 𝑚 > 𝑦𝑠 = 0.5 ∴ 𝒋𝒖𝒎𝒑 𝒘𝒊𝒍𝒍 𝒐𝒄𝒄𝒖𝒓 @ critical flow 𝑄 = 8 2.46 = 19.68 𝑚3/𝑠 𝑞 = 19.68 5 = 3.936 𝑚3/𝑠 𝑝𝑒𝑟 𝑚 𝑤𝑖𝑑𝑡ℎ

𝑦𝑐 = 𝑞2

𝑔

1 3

= 3.9362

9.81

1 3

= 1.16 𝑚 < 𝑦𝑜 = 1.60 𝑚 ∴ 𝒔𝒍𝒐𝒑𝒆 𝒊𝒔 𝒎𝒊𝒍𝒅

b.) distance between ys=0.50 m to y1=0.816 m using one reach

∆𝑥 =𝐸𝑠−𝐸1

𝑆−𝑆𝑜

Mean velocities: 𝑉𝑠 =𝑄

𝐴𝑠=

19.68

5(0.5)= 7.872 𝑚/𝑠

𝑉1 =𝑄

𝐴1=

19.68

5(0.816)= 4.824 𝑚/𝑠

𝑉𝑠 =1

2 𝑉𝑠 + 𝑉1 = 6.348 𝑚/𝑠

Mean hydraulic radius: 𝑅𝑠 =𝐴𝑠

𝑃𝑠=

5(0.5)

5+2(0.5)= 0.417 𝑚

𝑅1 =𝐴1

𝑃1=

5(0.816)

5+2(0.816)= 0.615 𝑚

𝑅𝑚 =1

2 𝑅𝑠 + 𝑅1 = 0.516 𝑚

Then, 𝑆 =𝑛2𝑉𝑚

2

𝑅𝑚4 3 =

(0.012)2(6.348)2

(0.516)4 3 = 0.014

𝐸𝑠 = 𝑦𝑠 + 𝑉𝑠

2

2𝑔= 0.5 +

(7.872)2

2𝑔= 3.658 𝑚

𝐸1 = 𝑦1 + 𝑉1

2

2𝑔= 0.816 +

(4.824)2

2𝑔= 2.002 𝑚

Thus, ∆𝒙 =𝑬𝒔−𝑬𝟏

𝑺−𝑺𝒐=

𝟑.𝟔𝟓𝟖−𝟐.𝟎𝟎𝟐

𝟎.𝟎𝟏𝟒−𝟎.𝟎𝟎𝟎𝟗= 𝟏𝟐𝟔. 𝟒𝟏𝟐 𝒎

c.) Type of profile

Since 𝑦𝑁 > 𝑦𝑐 > 𝑦 ,

∴ 𝑻𝒉𝒆 𝒑𝒓𝒐𝒇𝒊𝒍𝒆 𝒊𝒔 𝑴𝟑

5. Examine the flow conditions in a very long 10ft wide open rectangular channel of rubble

masonry with n=0.017 when the flow rate is 400cfs. The channel slope is 0.020 and an ogee weir

5ft high with Cw=3.80 are located at the downstream end of the channel.

yn

yn

Solutions:

Normal depth of flow in the channel,

𝑄 =1.48

𝑛𝐴𝑅2/3𝑆1/2

400 =1.49

0.017x10𝑦𝑛

10𝑦𝑛

10+2𝑦𝑛

2

3(0.020)

1

2

By trial and error:

𝑦𝑛 = 2.36 𝑓𝑡.

Critical depth: 𝑦𝑐 = 𝑄2

𝐵2𝑔

1

3=

4002

10232.2

1

3= 3.67 𝑓𝑡

Since yn < yc , the flow is supercritical. The head required on the weir to discharge:

𝑄 = 𝐶𝑤𝐿𝐻3

2

400 = 3.80x10 ℎ +

400

5+ℎ x10 2

64.4

3

2

By trial and error, ℎ = 4.53 𝑓𝑡 depth of water upstream weir is 9.53 which is greater than yc.

The flow @ this point is subcritical & hydraulic jump must occur upstream. The depth y2 after

the jump is:

𝑦2 = −2.36

2+

2.362

4+

2 400

23.6

2x2.36

32.2

1/2

= 5.42 𝑓𝑡

The distance from the weir to the jump

𝑦𝐴 = 5.42 𝑓𝑡 𝑉𝐴 = 400/54 = 7.39𝑓𝑡/𝑠𝑒𝑐 𝑉𝐴2 2𝑔 = 0.85𝑓𝑡

𝑦𝐵 = 9.53 𝑓𝑡 𝑉𝐵 = 400/97.4 = 4.20𝑓𝑡/𝑠𝑒𝑐 𝑉𝐵2 2𝑔 = 0.27𝑓𝑡

𝑉𝑎𝑣𝑒 = (VA + VB )/2 = 5.74 ft/sec

𝑅𝑎𝑣𝑒 = 2.95 ft

𝑆 = 0.017x5.79 2 (1.486x2.490)4 3 = 0.00104

𝑥 = 5.42+0.85−9.53−0.27

0.00104−0.02= 186.18 ft

5 ft

4.53 ft

5.40 ft 2.36 ft

EGL

𝑉2

2𝑔= 0.85 𝑓𝑡 𝑉2

2𝑔= 0.26 𝑓𝑡 𝑉2

2𝑔= 4.45 𝑓𝑡

186.18 ft

2

y2 = 1.80m

y1 = 1.20m

1

4m 4m

𝐴23 𝐴22

𝐴13

𝐴12

1.80

1.20 𝐴11

𝐴21

6. A hydraulic jump occurs in a trapezoidal section with bottom width of 4m and side slope of 1:2. The

depth before the jump is 1.20m and after the jump is 1.80m.

a.) Calculate the flow rate in the canal.

b.) Calculate the power lost.

Solution:

𝐴1 = 𝐴𝑛 + 2𝐴12

𝑥

𝑦=

1

2

𝑥1 = 1

2𝑦1 = 0.60 𝑚

𝐴1 = 4 1.20 + 2(1

2 × 0.60 + 1.20)

𝐴1 = 5.52 𝑚2

𝐴2 = 𝐴21+ 2𝐴22

𝑥

𝑦=

1

2

𝑥2 = 1

2𝑦2 = 0.90 𝑚

𝐴2 = 4 1.80 + 2(1

2 × 0.90 × 1.80)

𝐴2 = 8.82 𝑚2

𝐴1𝑦𝑐1= 𝐴11𝑦𝑐1 1

+ 2𝐴12𝑦𝑐1 2

𝐴1𝑦𝑐1= 1.20𝑥4 0.60 +

2(1

2𝑥 0.60 𝑥 1.2)(0.40)

𝐴1𝑦𝑐1= 2.88 + 0.288

𝐴1𝑦𝑐1= 3.168 𝑚2

𝐴2𝑦𝑐2= 𝐴21𝑦𝑐2 1

+ 2𝐴22𝑦𝑐2 2

𝐴2𝑦𝑐2= 4 1.8𝑥 0.90 + 2

1

2𝑥0.90𝑥1.80 𝑥0.60)

𝐴2𝑦𝑐2= 7.452 𝑚2

𝑉12 =

9.81(8.82)

5.52

7.452−3.168

8.82−5.52 = 20.349

𝑉1 = 4.511 𝑚/𝑠

a.) Flow rate in canal

𝑄1 = 𝐴1𝑉1 = 5.52 4.511

𝑸𝟏 = 𝟐𝟒. 𝟗𝒎𝟑/𝒔

b.) Power Lost, 𝑃 = 𝛾𝑄ℎ𝐿

ℎ𝐿 = 𝐸1 − 𝐸2

where 𝐸1 = 𝑦1 +𝑉1

2

2𝑔= 1.2 +

4.5112

2𝑔= 2.237𝑚

𝐸2 = 𝑦2 +𝑉2

2

2𝑔= 1.8 +

2.8232

2𝑔= 2.206𝑚

Then, ℎ𝐿 = 2.237 − 2.206 = 0.031𝑚

Therefore,

𝑃 = 𝛾𝑄ℎ𝐿

𝑃 = 9.81(24.9)(0.031)

𝑷 = 𝟕. 𝟓𝟕𝟐 𝒌𝑾