BLT3014 Examples

BSc (Hons) in Civil Engineering

BLT3014 Hydraulics 3C

Worked Examples

1

Uniform flow in open channels Chezy example

A rectangular channel 3.5m wide has a bed slope So of 0.005 (1 in 200) and a Chezy ‘C’ value of 50. Determine the flowrate in the channel when the depth is 1.2m

1.2m

3.5m

A = 3.5 x 1.2 = 4.2m2 , P = 1.2 + 3.5 + 1.2 = 5.9m R = A = 4.2 = 0.712m P 5.9

= 12.53m3/sec

Uniform flow in open channels Manning example

A trapezoidal channel is to be designed to convey 15m3/sec of water at a maximum velocity of 2m/sec so as to minimise erosion of the channel bed. The channel is to be cut through natural ground, which will give a Manning ‘n’ value of 0.025. The base width of the channel will be 2m and the sides will slope at 1 vertical to 2 horizontal. Determine the depth of flow y and the required bed slope So for the channel.

y

2.0m 2y

A = Q = 15.0 = 7.5m2 = ( 2 + 2 + 4y ) x y , 2y2 + 2y - 7.5 = 0 , y = 1.5m V 2.0 2.0

P = = 8.71m

R = A = 7.5 = 0.861m P 8.71

15 = 7.5 x 0.8612/3 x So

1/2 , So = ( 15 x 0.025 )2 = 0.0031 = 1 in 323 0.025 7.5 x 0.8612/3

Specific energy example

2

Water flows at a rate of 8m3/sec in a 2.5m wide rectangular channel at a depth of 1.5m. A bed hump is located in the channel which causes the water surface to drop by 0.1m. Determine the height of the hump. What height of hump would cause critical depth flow over the hump?

Specific energy linev1

2/2g v22/2g

1.5m (1.4 – h) 1.4m

h

1 2

Specific Energy at any point (relative to the bed at that point) (pg 7)

Assuming no energy loss across the bed hump =

y2 + h = 1.4m , v1 = Q = 8 = 2.133 m/sec A1 2.5 x 1.5

Therefore = hence v2 = 2.552 m/sec

v2 = Q 2.552 = 8 thus y2 = 1.254m and h = 1.4 – 1.254 = 0.146m A2 2.5 x y2

For critical flow over the hump:

q = flow per unit = 8/2.5 = 3.2m3/sec/m, = = 1.014m (pg 9)

vc = Q = 8 = 3.156 m/sec, ( = = 3.154 m/sec) Ac 2.5 x 1.014

= , h = 0.208m

Broad crested weir example

3

A broad crested weir (bcw) 1.5m high is located in a 50m wide river. Measurements taken during a flood event showed the upstream water level to be 1.2m above the crest of the weir. Determine the approximate flood flow rate in the river.

Specific energy line v1

2/2g vc2/2g

2.7m yc

1.5 m

1 2

Providing weir is not drowned (water level downstream artificially raised) then critical depth flow will occur at some point on the weir.

Q = A x v = yc x B x vc , now (pg 9)

= yc1.5 x B x 9.810.5

now yc = 2 E2 (pg 9) (Note E2 = Ec) 3

Therefore Q = = 1.705 x B x E21.5 (bcw equation)

We don’t know Q and therefore cannot determine vc and E2. As a first guess assume v1 is small (hence v1

2/2g is small) and therefore E1 is approximately equal to y1. Assuming no energy loss across the weir then:

E1 = E2 + h , thus 2.7 = E2 + 1.5 hence E2 = 1.2m

Therefore Q = 1.705 x 50 x 1.21.5 = 112.1m3/sec

v1 = Q = 112.1 = 0.830 m/sec, v12 = 0.83 2 = 0.035m, E1 = 2.735m & E2 = 1.235m

A1 2.7x50 2g 19.62

Q = 1.705 x 50 x 1.2351.5 = 117.0 m3/sec

(Iterating one step further gives v1 = 0.848 m/sec and Q = 117.4 m3/sec)

Venturi-flume Example

4

A venturi-flume, situated in a long rectangular channel of width 1.0m, has a throat width of 0.4m. A sluice gate is situated some distance downstream of the flume which can be used to adjust the water level in the channel. Determine:

a) The discharge, when the sluice is lowered causing a depth just upstream of the flume of 0.55m and a depth at the throat of 0.47m (i.e. flume is drowned).

b) The depth upstream of the flume, when the sluice is raised and hence the downstream water level is lowered, to create critical depth flow at the throat for the same discharge.

c) The upstream depth if a 0.2m bed hump is introduced at the throat. Flow at the throat remaining critical for the same discharge.

1.0m 1 Flow 2 0.4m 3

Plan view of flume

Specific energy line

1 2 3

Longitudinal sectional view of flumea)

B1 = 1.0m, B2 = 0.4m, y1 = 0.55m, y2 = 0.47m

Assuming no energy loss through flume, then E1 = E2 = E3

, thus

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0.55 + 0.169 Q2 = 0.47 + 1.442 Q2, 0.08 = 1.273 Q2

= 0.251 m3/sec

b)

q = Q (flow rate per unit width) B

q1 = q3 = 0.251 = 0.251 m3/sec, q2 = 0.251 = 0.628 m3/sec1.0 0.4

y2 = = 0.343m

E2 = Ec = 3 y2 = 1.5 x 0.343 = 0.515m 2

By iteration y1 = 0.502m

c)Specific energy line

y1 yc

0.2m1 2

From part b) Ec for this flowrate = 0.515m and y2 = yc

E1 = E2 + 0.2 = 0.515 + 0.2 = 0.715m

By iteration y1 = 0.708mHydraulic Jump Example 1

Water is passed along a 4.5m wide rectangular channel at a flow rate of 17.6 m3/sec. A sluice gate is situated at the downstream end of the channel. The opening under the sluice gate is 0.76m. Just downstream of the sluice a hydraulic jump forms. Assuming no energy losses across the sluice and that the coefficient of contraction CC for flow under the sluice is 0.6 determine:

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1. The depth of flow just downstream of the sluice (i.e just upstream of the hydraulic jump).

2. The depth of water just upstream of the sluice gate.3. The depth of water just downstream of the hydraulic jump.4. The energy and power lost in the hydraulic jump.5. The resultant thrust on the sluice gate.

Specific energy line

y1

y3

y2

1 2 3

1) y2 = 0.6 x 0.76 = 0.456m

2) E1 = E2 therefore

= 4.206 by iteration y1 = 4.160m

(pg 16)

v2 = 17.6 = 8.577 m/sec 4.5 x 0.456

= = 4.055

= 2.397m

v3 = 17.6 = 1.632 m/sec 4.5 x 2.397

4) E = E2 - E3 = -

7

E = - = 4.206 - 2.533 = 1.673m

Power loss P = gQE = 9810 x 17.6 x 1.673 = 288854 watts (288.9 Kw)

5)

M1

M2

Resultant force on gate P = (M1 - M2) g (pg 12)

,

P=

x

9810

P = (40.628 - 15.856) x 9810 = 243013N

Hydraulic Jump Example 2Water flows along a 3m wide rectangular channel at rate of 6.0m3/sec. A broad crested weir is located in the channel which causes a hydraulic jump to form in the channel just downstream of the weir. The depth of flow just downstream of the jump is 1.2m. Find the height of the weir h above the base of the channel assuming no energy loss across the weir.

Specific energy line

8

y1 yc 1.2m y3

h1 2 3 4

v4 = Q = 6 = 1.667 m/sec A4 3 x 1.2

= = 0.486 (subcritical)

= 0.420m

v3 = Q = 6 = 4.762 m/sec A3 3 x 0.42

= = 2.346

= = 1.576m, E3 = E2 + h , therefore h = E3 – E2

y2 = yc as critical depth flow occurs over weir.

Flow per unit width q = Q = 6 = 2.0 m3/sec/m b 3

y2 = = 0.742m and v2 = = 2.697 m/sec

E2 = Ec = 3 y2 = 1.5 x 0.742 = 1.113m 2

h = 1.576 – 1.113 = 0.463mGradually Varied Flow Worked example – Depth calculated from distanceA rectangular channel 1.5m wide has a uniform slope of 1 in 1600 and a uniform flow depth of 0.75m when the discharge is 0.56m3/s. A sluice is lowered increasing the depth just upstream of it to 1.0m. How far upstream of the sluice will the depth be 0.83m?

9

For uniform flow, width = 1.5m, depth = 0.75m, So = 1 1600

R = A = 1.5 x 0.75 = 0.375m P 2(0.75) + 1.5

vo = Q = 0.56 = 0.498 m/sec A 1.5 x 0.75

Chezy, , 0.498 = C

Hence C = 32.529m½ sec

For an accurate solution a series of small steps are chosen between the depth at the sluice (1.0m) and the required depth (0.83m). (Note the backwater curve becomes asymptotic near the normal depth therefore for calculation purposes the curve is deemed to finish 50mm above this depth). The variation of Chezy C with depth is considered.

C = 1 R1/6 (n is the Manning number) n

From the above, at uniform flow: c = 32.529 & R = 0.375m

Hence n = 0.375 1/6 = 0.026 32.529

Therefore in general C = 1 R1/6 or C2 = 1479 R1/3

0.026

Referring to the table overleaf. Depth intervals are chosen from 1.0m to 0.83m as shown in column 1 (the smaller the interval the greater the accuracy)

Columns 1 to 9 are compiled at each section

Column 10 lists the mean values of v 2 at each section

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C2 R Columns 12 and 13 list E and x respectively (x is the distance between sections)

Column 14 summates x to give the required distance

Gradually varied flow example

Work back upstream from the sluice using depth increments of 50mm.

At the sluice depth y = 1.0m

Breadth of channel b = 1.5m, therefore Area A = 1.0 x 1.5 = 1.5m2

Wetted perimeter P = (1.0 x 2) + 1.5 = 3.5m

Hydraulic radius R = A = 1.5 = 0.429m P 3.5

C2 = 1479 R1/3 = 1479 x 0.4290.333 = 1115

Velocity v = Q = 0.56 = 0.373m/sec A 1.5

Kinetic energy = V 2 = 0.373 2 = 0.0071m 2g 19.62

Specific Energy E = y + v 2 = 1.0 + 0.0071 = 1.0071m2g

v 2 = 0.373 2 = 2.909 x 10-4

C2R 1115 x 0.429

At the depth y = 0.95m

Breadth of channel b = 1.5m, therefore Area A = 0.95 x 1.5 = 1.425m2

Wetted perimeter P = (0.95 x 2) + 1.5 = 3.4m

Hydraulic radius R = A = 1.425 = 0.419m P 3.4

C2 = 1479 R1/3 = 1479 x 0.4190.333 = 1107

Velocity v = Q = 0.56 = 0.393m/sec A 1.425

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Kinetic energy = V 2 = 0.393 2 = 0.0079m 2g 19.62

Specific Energy E = y + v 2 = 0.95 + 0.0079 = 0.9579m2g

v 2 = 0.393 2 = 3.330 x 10-4

C2R 1107 x 0.419

Work out the mean value of sf (slope of TEL) between the two depths

= 3.120 x 10-4

= 3.13x10-4

For a quick approximate solution (ignoring the variation of Chezy C with depth), proceed as follows;

C for uniform flow = 32.529 m1/2/sec (as previously calculated)

Depth at sluice = 1.0m and depth at point considered = 0.83m

Take intermediate point where depth = 1.83 = 0.915m 2

12

At this point R = 1.5 x 0.915 = 0.412m 1.5 + 2(0.915)

v = Q = 0.56 = 0.408 m/sec A 1.5 x 0.915

For non uniform flow and hence

Sf = v 2 = 0.408 2 = 0.000382 C2 R 32.5292 x 0.412

Now the surface slope dy = So - Sf

dx 1 – Fr2

And Fr2 = v 2 = 0.408 2 = 0.019 g y 9.81 x 0.915

And bed slope So = 1 = 0.000625 1600

Therefore dy = 0.000625 - 0.000382 = 0.000248 dx 1 - 0.019

Therefore dx = dy = 1.0 - 0.83 = 685.5m 0.000248 0.000248

Compares with 692m from step method.

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y(m)

A(m2)

P(m)

R(m)

C2

(m/s2)V

(m/s)v2/2g(m)

E(m)

v2/C2Rx 10-4

(v2/C2R)m

x 10-4 (so -v2/C2R)m

x 10-4 E(m)

x(m)

x=x(m)

1.0 1.50 3.5 0.429 1115 0.373 0.0071 1.0071 2.909

3.120 3.130 0.0492 157.2 0

0.95 1.425 3.4 0.419 1107 0.393 0.0079 0.9579 3.330

3.583 2.667 0.0491 184.1 157.2

0.90 1.35 3.3 0.409 1098 0.415 0.0088 0.9088 3.835

341.3

0.85

0.83

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y(m)

A(m2)

P(m)

R(m)

C2

(m/s2)V

(m/s)v2/2g(m)

E(m)

v2/c2Rx 10-

(v2/C2R)m

x 10-4 (So -v2/C2R)m

x 10-4 E(m)

x(m)

x=x(m)

1.0 1.50 3.5 0.429 1115 0.373 0.0071 1.0071 2.909

3.120 3.130 0.0492 157.2 0

0.95 1.425 3.4 0.419 1107 0.393 0.0079 0.9579 3.330

3.583 2.667 0.0491 184.1 157.2

0.90 1.35 3.3 0.409 1098 0.415 0.0088 0.9088 3.835

4.143 2.107 0.0490 232.6 341.3

0.85 1.275 3.2 0.398 1088 0.439 0.0098 0.8598 4.451

4.596 1.654 0.0195 117.9 573.9

0.83 1.245 3.16 0.394 1084 0.450 0.0103 0.8403 4.741

691.8

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Pump characteristics example1

A scale model of a pump has been tested and the following results obtained:

Impeller diameter (D) 100mm

Power input (P) 2.5Kw

Speed (N) 1200 r.p.m.

Flow (Q) 10 litres/sec

Head (H) 15m

Efficiency (E) 60%

Determine the speed and size of a geometrically similar pump which would deliver a flow rate of 25 litres/sec against a 22m resistance head. Determine the power input required by this pump.

Using the above coefficients, where m = scale model & p = prototype:

,

Hence NP DP3 = 0.025 x 1200 x 0.1 3 = 3.0

0.01

And NP = 3.0 - equation (1) DP

3

Hence NP2 DP

2 = 22 x 1200 2 x 0.1 2 = 21120 - equation (2)

15

16

Substitute for Np from equation (1) into equation (2)

Therefore NP = 3.0 = 1005 r.p.m 0.1443

Hence PP = 2.5 x 1005 3 x 0.144 5 = 9.092 Kw 12003 x 0.15

(Check Input power = Output power = g Q H = 9810 x 0.025 x 22 = 8.993 Kw) Efficiency E 0.6

Pump characteristics example2

A centrifugal pump has been tested at 600 rev/min and found to deliver 42 litres/sec at a head of 36m when operating at a maximum efficiency of 65%. The pump is to be installed in a pipe system with a static head lift of 25m and a frictional head loss of 4m when the flow rate is 20 litres/sec. The pump is to operate at maximum efficiency in this pipe system. Determine the speed at which the pump should be operated and the discharge and head difference across the pump.

Using the above coefficients, where t = test and o = operational.

and Ho = 36 x No2 = 1.0 x 10-4 No

2 - equation (1)

17

360000

and Qo = 7.0 x 10-5 No - equation (2)

For the system hf = K Q2 , hf = 4.0m when Q = 20 litres/sec

4.0 = K x 0.022 , K = 4.0 = 10,000 0.022

Static head hs = 25m

Ho for system = hs + hf = hs + KQ2 = 25 + 10,000 Qo2

Subs for Ho and Qo from equations (1) and (2)

1.0 x 10-4 No2 = 25 + 10,000 x (7.0 x 10-5 x No)2

1.0 x 10-4 No2 = 25 + 4.9 x 10-5 No

2

1.0 x 10-4 No2 - 4.9 x 10-5 No

2 = 25

5.1 x 10-5 No2 = 25

Ho = 1.0 x 10-4 No2 = 1.0 x 10-4 x 7002 = 49m

Qo = 7.0 x 10-5 No = 7.0 x 10-5 x 700 = 0.049 m3/sec

PUMP MATCHING EXAMPLES

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1)

Water from a land drainage scheme is to be lifted from a proposed pumping station to a nearby river. Four centrifugal pumps connected in parallel are to be installed in the station. Each pump has the characteristics given below:

Q (litres/s) 0 30 60 90 120

H (m) 15 14 12.5 10 6

E (%) 0 54 66 60 50

P (Kw) 9 9.8 10.5 11 11.2

The pumps are set up such that in low flows 1 or 2 pumps operate and in high flows all four pumps operate.

The pumping main is 450mm in diameter, 300m in length (allowing for minor losses) and has a roughness height Ks of 1.5mm.

The main has a free outfall into the river 7.5m above the start level for one pump, 7m above the start level for two pumps and 6m above the start level for four pumps.

Determine:

(a) The quantities delivered at low and high flows.(b) The power consumed in each case.(c) The efficiencies in each case.

(2)

Two of the above pumps are set up in series. They discharge into a 225m diameter main, 500m in length with a Ks value of 1.5mm. Determine the quantity delivered, power consumed and pump efficiency when the static head is 12m.

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20

21

22

23

24

25

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PUMP MATCHING TABLE - 4 PUMPS IN PARALLEL Flow (l/s) 0 30 60 90 120 180 240 360 480 H for one pump alone (m)

15 14 12.5 10 6

H for two pumps in parallel (m)

15 14 12.5 10 6

H for four pumps in parallel (m)

15 14 12.5 10 6

hf for one pump alone (m)

0 0.05 0.13 0.3 0.52

hf for two pumps in parallel (m)

0 0.13 0.52 1.20 2.10

hf for four pumps in parallel (m)

0 0.52 2.10 4.70 8.33

HT for one pump alone (m)

7.5 7.55 7.63 7.80 8.02

HT for two pumps in parallel (m)

7.0 7.13 7.52 8.20 9.10

HT for four pumps in parallel (m)

6.0 6.52 8.10 10.70 14.33

Power for one pump alone KW)

9.0 9.8 10.5 11.0 11.2

Efficiency for one pump (%)

0 54 66 60 50

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Pumps in parallel graphs

0

2

4

6

8

10

12

14

16

0 100 200 300 400 500 600

Flow (litres/sec)

Hea

d (

m) Single pump

Two pumps in parallel

Four pumps in parallel

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0

2

4

6

8

10

12

14

16

0 100 200 300 400 500 600

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PUMP MATCHING TABLE FOR TWO PUMPS IN SERIES

Flow (l/s)

0 30 60 90 120

Head for one pump alone (m)

15 14 12.5 10 6

Head for two pumps in series (m)

30 28 25 20 12

hf for two pumps in series (m)

0 2.2 8.6 19.6 35.5

HT for two pumps in series (m)

12 14.2 20.6 31.6 47.5

Power for two pumps in series (KW)

18 19.6 21 22 22.4

Efficiency for two pumps in series (%)

0 54 66 60 50

30

Pumps in series graphs

0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140

Flow (litres/sec)

Hea

d (

m)

Single pump

Two pumps in series

31

Pumps in series graphs

0

5

10

15

20

25

30

35

40

45

50

0 20 40 60 80 100 120 140

Flow (litres/sec)

Hea

d (

m)

32

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PJ3/J4 Tutorial Questions

1. Broad crested weir tutorialA broad crested weir 1.2m high is located in a 20m wide river. Measurements taken during a flood event showed the rate of flow in the river to be 32m3/sec. Assuming critical depth flow over the weir, determine the approximate depth of water just upstream of the weir. (2.13m)

2. Venturi-flume tutorialA long rectangular channel of width 6m is fitted with a venturi flume of throat width 3.0m. Some distance downstream of the flume a sluice gate is lowered resulting in flow depths of 3.5m upstream of the flume and 3.0m at the throat of the flume. Determine the flow rate in the channel. ( 31.2m3/sec)

Determine the depths just upstream and downstream of the flume, when the sluice is raised and hence the downstream water level is lowered, to create critical depth flow at the throat for the same discharge. (3.25m, 0.72m)

3. Venturi-flume tutorialA venturi flume is 1.2m wide at the entrance and 0.6m wide at the throat. Developing your work from the specific energy equation and neglecting energy losses, calculate:

a) the discharge when the depth upstream and downstream of the flume is 0.6m, and the depth at the throat of the flume is 0.56m;

b) the depth of flow at entrance to the flume if the downstream water level is reduced to create critical flow in the throat for the same discharge;

c) the depth of flow at entrance which would result if a hump 0.3m high is now fitted in the throat, the flow remaining critical in the throat.

4. Hydraulic Jump Tutorial 1Water is passed along a 2m wide rectangular channel at a flow rate of 5m3/sec. A sluice gate is situated at the downstream end of the channel. Just downstream of the sluice a hydraulic jump forms which has a downstream depth of 1.6m. Assume no energy losses across the sluice and that the coefficient of contraction CC for flow under the sluice is 0.6. Determine the height of the opening under the sluice gate and the depth of water just upstream of the sluice. (0.666m, 2.332m)

5. Hydraulic Jump Tutorial 2Water flows along a 1.2m wide rectangular channel at a flow rate of 1.4 m3/sec. A venturi-flume is situated in the channel which has a throat width of 0.6m. A hydraulic jump forms in the channel just downstream of the venturi-flume. Assuming no energy losses across the flume, determine the depth of water just downstream of the venturi-flume and just downstream of the hydraulic jump. (0.268, 0.892m)

6. Pump characteristics tutorial

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A centrifugal pump has the following characteristics when operating at 800 revs/min.

Discharge (litres/sec) 0 10 20 30Head (m) 36 32 24 8Efficiency % 0 54 62 54

The pump is to be installed in a pipe system with a static head lift of 20m and a frictional head loss of 6m when the flowrate is 18 litres/sec. The pump is to operate at maximum efficiency in this pipe system. Determine the speed at which the pump should be operated and the discharge and head difference across the pump.

7. Gradually varied flow tutotrialA long rectangular channel 2m wide with a Manning n value of 0.014 conveys water at a flowrate of 6m3/s. The flow in the channel is steady and uniform at a depth of 1.8m. A sluice gate with coefficient of contractions 0.6 is then lowered into the flow such that the gate opening is 0.5m and the flowrate remains the same.

a) Determine the bed slope of the channel.

b) Sketch and justify the form of the water profile.

c) Determine the distance from the sluice gate to the point where a hydraulic jump occurs, using one finite difference step of integration.

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