CEE 370 Fall Laboratory #3 Open Channel Flow

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CEE 370 Fall 2015

Laboratory #3 Open Channel Flow

Objective:

The objective of this experiment is to measure the flow of fluid through open channels using a V-notch weir and a hydraulic jump.

Introduction:

Flow of water in a conduit is of two types: open-channel flow and pipe flow. These types of flows are similar but differ in one important aspect: open channel flow must have a free surface which is exposed to atmospheric pressure whereas pipe flow has no free surface since it if flowing full. Open channel flows are driven by gravity alone, and the pressure gradient at the atmospheric interface is negligible. Thus the basic force balance in an open channel is between gravity and friction.

Figure 1. A sketch of open channel flow

Reynold’s number of open channel flow:

As in pipe flow, three states of flow are possible, laminar, turbulent and transitional. The effects of viscosity relative to inertial forces of the flow determine the state of flow. The relative effects of velocity to inertia can be represented by the Reynold’s number which is a dimensionless number derived through dimensional analysis of the flow. The Reynold’s number for open channel flow is given by:

𝑅𝑅 =^𝑣𝑅_𝜈^ℎ Eq. 1

Where, Re is the Reynold’s number; v is the mean flow velocity (in m/s); Rh is a

characteristic length (the characteristic length for open channel flow is the hydraulic radius) (in m), and ν is the kinematic viscosity of the fluid (m²/s).

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Rh is calculated using the cross sectional area (A) divided by the wetted perimeter (P) where b is the width of the channel, and y is the depth of water in the channel (Eqs. 2 and 3).

𝑅_ℎ = ^𝐴_𝑃 Eq. 2

𝑃 = 𝑏 + 2𝑦 Eq. 3

For a full pipe, hydraulic radius is calculated as (Eq. 4 and Fig. 2):

𝑅_ℎ = ¹₄𝐷_ℎ Eq. 4

Figure 2. A graphic presentation of a wetted perimeter

V-notch weir in open channel flow:

Weirs are overflow structures built across open flow channels to measure the discharge (or flowrate) (Fig. 3). Some general terms pertaining to weirs are:

• Notch: the opening which water flows through

• Crest: the edge which water flows over

• Nappe: the overflowing sheet of water

• Length: the “width” of the weir notch

Figure 1. A sketch of V-notch weir in open channel

The basic principle is that discharge is directly related to the water depth above the bottom of the V; this distance is called head (h). The V-notch design causes small changes in discharge

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to have a large change in depth allowing more accurate head measurement than with a rectangular weir.

To calculate the flow rate of an open channel flow, the Bernoulli equation (Eq. 4) for a 0 surface pressure is used (Eq. 5). Q is fluid discharge at any given point, which is the product of fluid velocity (V) and cross-sectional area (A). The cross-sectional area can be calculated as the product of the width of the free surface and the constant head (Eq. 5).

Bernoulli’s equation: ^𝑃1

𝜌𝜌+_2𝜌^𝑣2+ 𝑧₁ = ℎ₁ Eq. 4

Where ^𝑃1

𝜌𝜌 is the pressure head, ^𝑣2

2𝜌 is the velocity head, and z1 is the velocity or potential head.

ℎ1 =^𝑉_2𝜌² Eq. 5

where h₁ is the head associated with velocity at a point in the weir; and g is gravitational constant.

𝐴 = 𝑏ℎ Eq. 6

where b is the width of free surface.

𝑏 = 2(ℎ2− ℎ1) tan^𝛼₂ Eq. 7

where h2 is head associated with the total height of the weir; and α is the angle of the V-notch weir.

The values of b and α are used to calculate the cross-sectional area.

However, the flowrate changes when hydraulic head changes (Eqs. 8 and 9);

𝑑𝑑 = 𝑉𝑏 𝑑ℎ

and ∫ 𝑑𝑑 = ∫��2𝑔ℎ1�[2(ℎ₂ − ℎ₁) tan^𝛼₂] 𝑑ℎ₁ Eq. 8

h2

h₁ b

α

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Figure 4. A sketch of discharge calculation

The result of integration is represented by Eq. 8:

𝑑 = 𝐶𝐷�𝑔 tan �^𝛼₂� ℎ2 5

2 𝑎𝑎𝑑 𝐶𝐷 =^8√2₁₅ Eq. 9

where C_D is discharge coefficient. The theoretical value represented by Equation 9 precludes losses, upstream kinetic energy, and jet contraction. Neat, handwritten calculations are appropriate, with inclusion of a control volume sketch. State all assumptions.

Hydraulic jump:

In an open channel flow, a supercritical flow can change rapidly to a subcritical flow by passing through a hydraulic jump (Figure 5). The dimensionless number which indicates the relative effects of gravity and inertial forces on the state of flow is the Froude number. It is also derived through dimensional analysis and is defined as (Eq. 10):

Fr =_√𝜌𝑔^𝑉 Eq. 10

Where v is the average flow velocity, g is the gravitational constant, and y is the channel depth.

Figure 2. A sketch of hydraulic jump in open channel flow

The Froude number defines three regimes of flow in open channels: critical, subcritical, and supercritical flow (Table 1).

Table 1. The three regimes in an open channel

Fr Type of flow Criteria

1 Critical flow Y = yc; V= Vc

<1 Sub-critical flow Y>y_c, V<V_c

>1 Super-critical flow (fast flow-high energy state) Y<yc, V>Vc

The critical depth (yc in m) and critical velocity (Vc in m/s) are calculated as follows (Eqs.

10-13):

Page 5 of 10 𝐹𝐹𝑐 = _�𝜌𝑔^𝑉𝑐

𝑐= 1 Eq. 10

Where Fr_c is the critical Froude number.

where 𝑉𝑐 = _𝐴^𝑄

𝑐= �^𝐴𝑐_𝑏^𝜌 Eq. 11

and 𝐴𝑐 = ( ^𝑏𝑄_𝜌²)¹³ Eq. 12

and 𝑦𝑐 = (_𝑏^𝑄₂²_𝜌)¹³ Eq. 13

Figure 6. A picture and schematic of a sluice gate for a hydraulic jump

For this experiment, the discharge per unit width of a rectangular channel is calculated as (Eq.

14):

𝑞 =^𝑄_𝑏 = 𝑉𝑦 Eq. 14

where q is the discharge per unit width of a rectangular channel.

To get the sluice gate force, 𝐹_𝑒𝑒𝑒 is calculated as (Eq. 15):

𝐹𝑒𝑒𝑒 = 𝜌𝑏𝑞²�_𝑔¹

2−_𝑔¹

1� + 𝜌𝑔^𝑏₂(𝑦₂²− 𝑦₁²) Eq. 15

Page 6 of 10 Examples of Froude number is given in Table 1.

Table 1. Description of Froude number

Fr Jump Dissipation

1 Impossible

1-1.7 Standing wave (undular jump) about 4 y₂ long <5 %

1.7-2.5 Weak jump 5-15%

2.5-4.5 Unstable, oscillating 15-45%

4.5-9 Stable, well-balanced, steady jump 45-70%

>9 Rough jump-intermittant and strong 70-85%

For Today’s Experiment:

Apparatus:

The apparatus consists of the following equipment, manufactured by G. Cussons, Limited, of Great Britain, and maintained in Marston 10:

• Hydraulic bench (P. 6100)

• Hook gauge and scale (P. 6107)

• Rectangular open channel (P. 6245)

• 90 degree (α) V-notch weir

• Stop watch and scale Procedure:

Part 1: V-Notch Weir

1. Note the water temperature in the hydraulic bench.

2. The cross sectional area of the drainage canal approaching the weir should be noted.

3. Measure the height of the bottom of the “V” to the bottom of the weir, as well as the cross sectional area that will be used to measure the volume of flow. (Referring to the low and high flow regions in the stepped measuring tank of the hydraulic bench).

4. Establish a low recirculating flow in the bench by turning the pump on and maintaining a low setting on the rheostat, with the measuring tank drain open. The discharge should be sufficient to create a free jet through the V-notch weir. Observe the progression of flow through the drainage canal, V-notch weir, and into the measuring tank and sump.

5. Record the water surface elevation in the drainage canal using the hook gauge

6. Be ready with a stopwatch. When measuring the flow rate, select a volume in the lower step of the tank (as well as the high step if chosen to do so). Turn the knob at the sump horizontal to stop drainage and allow for flow to accumulate to the selected volume. The timer should start at the time the drainage of the sump is stopped. There should be a total of 3 replicates for this flow.

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7. Increase the discharge for two more values. Make replicate measurements for each new flow, so that you'll have a total of 3 replicated flows for 3 different flows (giving a total of 9 data points) for the V-notch weir experiment.

8. Turn off the pump and open the measuring tank drain.

Part 2: Hydraulic Jump

1. Measure the width of the open channel and the opening under the sluice gate.

2. Carefully turn on the pump and establish a subcritical depth y1 upstream of the sluice gate (without flooding the lab), a supercritical jet of depth y2 downstream of the sluice gate, a hydraulic jump, and a downstream subcritical depth y₃.

3. Note the progression of flow from the inlet section of the flume, through the sluice gate, through the exit gate, and into the hydraulic bench measuring tank. Measure the discharge into the tank with a stopwatch and sight glass. Use a ruler to measure y₁, y₂, and y3.

4. Measure the downstream locations of the sluice gate, start of the hydraulic jump and end of the hydraulic jump.

5. Turn off the pump and open the measuring tank drain when you're through.

Data Analysis:

Part 1: V-notch weir

Estimate observed values for CD and the open channel Reynolds number (defined in class, based on upstream section) for the 3 discharges of the V-notch weir.

Part 2: Hydraulic Jump

Compute the approach, jet, and exit Froude numbers Fr1, Fr2, and Fr3 for the jump.

Compute the power dissipated across the hydraulic jump and the magnitude of the sluice gate force F_EXT. Compute the discharge per unit width q.

Results and Discussion:

• Why would you expect CD observed to be less than the theoretical Eq. (1b)? Does CD

vary with Reynolds number?

• Verify the sub or supercriticality of the three open channel flow regions. Where are the controls located for y1, y2, and y3?

• What fraction of the approach energy is dissipated across the jump? What is the ratio y₂ to the sluice gate opening?

• Compare the observed y2 and y3 values with predicted values and comment.

Basic contents of lab report:

• Introduction

• Materials and methods

• Results and discussion

• Conclusions

• References

• Appendix

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Open Channel Flow Laboratory – Part 1

Lab Group: _________ Name: _______________________ Date: _________

V-Notch Weir

Temp (C): ___________ Approach width, cm: _________

Height of the bottom of the “V” to the bottom of the weir, cm: ___________

Area high Q section cm²: ________ Area low Q, cm²: ____________

Low Flow

Replicate ∆𝜉 (cm) ∆𝑡 (𝑠) Q (cm³/s) h₁ (cm) h₂ (cm) C_D

Mid Flow

Replicate ∆𝜉 (cm) ∆𝑡 (𝑠) Q (cm³/s) h1 (cm) h2 (cm) CD

High Flow

Replicate ∆𝜉 (cm) ∆𝑡 (𝑠) Q (cm³/s) h₁ (cm) h₂ (cm) C_D

Open Channel Flow Laboratory – Part 2

Lab Group: ______ Name: _______________________ Date: ___________

Hydraulic Jump and Sluice Gate

Temp (C): ___________

Channel width, cm: ___________ Sluice gate opening, cm: ___________

Channel Depth, Y1, cm: ___________ Distance from flow source, X1, cm: ___________

Channel Depth, Y₂, cm: ___________ Distance from flow source, X₂, cm: ___________

Channel Depth, Y₃, cm: ___________ Distance from flow source, X₃, cm: ___________

Page 9 of 10 Jump start, XSTART, cm: ___________

Jump end XEND, cm: ___________

X_GATE, cm: ___________

∆𝜉 (cm): ___________

∆𝑡 (𝑠): ___________

Q, cm³/s: ___________

Example:

Water flows in a wide channel at q = 10 m³/(s-m) and y1 = 8 m. If the flow undergoes a hydraulic jump, compute

a) Y2

b) V2

c) Fr₂ d) Y₃ e) V3

f) Fr3

g) h_f

h) % dissipation of energy 1. Find the upstream velocity, V₁

s m m

m s m y

V q 1.25 /

8 / 10 ³

1

1 ⋅ =

=

=

2. The upstream Froude number, Fr1 is 14 . 0 ) 8 )(

/ 8 . 9 (

/ 25 . 1

2 1

1

1= = =

m s m

s m gy

Fr V (This is a subcritical flow)

3. Is the hydraulic jump possible?

a. Find the depth y2

b. Find the velocity after the gate

s m m

m s m y

y

V V 11.9 /

84 . 0

8 / 25 . 1

2 1 1

2 = = × =

c. The supercritical Froude number is

15 . 4 ) / 8 . 9 )(

84 . 0 (

/ 9 . 11

2 2

2

2 = = =

s m m

s m gy

Fr V

d. The depth, y3, is 𝑦₂³− �𝑉₁²

2𝑔 + 𝑦¹� 𝑦₂²+𝑉₁²𝑦₁²

2𝑔 = 0; 𝑦² = 0.84 𝑚

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2 / 1 3 2

2 / 1 2 2 2

3 1 [1 8(4.15) ]

84 . 0 ] 2 8 1 [ 2 1

+ +

−

=

= +

+

−

= m

Fr y y

y

m

y₃ =4.53

e. The downstream velocity, V3, is

s m m

m s

m y

y

V V 2.2 /

53 . 4

) 84 . 0 )(

/ 9 . 11 (

3 2 2

3 = = =

f. The downstream Froude number

33 . 0 ) 53 . 4 )(

/ 8 . 9 (

/ 2 . 2

2 3

3

3 = = =

m s

m s m gy

Fr V

g. Dissipation head loss across the jump is m m m

m m

y y

y

h_f y 3.3

) 84 . 0 )(

53 . 4 ( 4

) 84 . 0 53 . 4 ( 4

)

( ³

3 2

3 2

3− = − =

=

h. % Dissipation

) 100 (

%

E

loss = h

s m m

s m m

g y V

E 8 . 07

) / 8 . 9 ( 2

) / 9 . 11 84 (

.

2 0

2 2

2 2

2

= + = + =

% 41 ) 100 07 (

. 8

) 3 . 3

% = ( =

m

loss m