Hydraulics - Tutorials and Solutions

MEMORIAL UNIVERSITY OF NEWFOUNDLAND FACULTY OF ENGINEERING AND APPLIED SCIENCE

Hydraulics Engineering 6713 Problems & Solutions Dr. Leonard Lye Professor of Civil Engineering

1

TUTORIAL 1 TURBULENT PIPE FLOW 1. In a chemical processing plant it is desired to deliver benzene at 50°C (rel. density 0.86) to point B with a pressure of 550 KN/m2. A pump is located at point A 21 m below point B, and the 2 points are connected by 240 m of plastic pipe having an inside diameter of 50 mm. If the volume flow rate is 110 liters/min, calculate the required pressure at the outlet of the pump. How much would the required pressure change if welded steel pipes are used instead of plastic pipes? (µ = 0.00042 N-s/m2). B

EL = 0 m PA=?

EL = 21 m PB = 550 KN/m2 L = 240 m D = 50 mm µ = 0.00042 Ns/m2 ρ = 0.86 * 1000 = 800 kg/m3

A

Q  110L / min  0.001833 m3 / s Q 0.001833 v   0.934 m / s A 0.05 2  4 vD 860  0.934  0.05 Re    95624  9.56  10 4  0.00042 From Moody diagram for smooth pipes, ƒ = 0.018 Therefore h f  0.018 

240 0.934 2   3.84 m 0.05 2g

Applying Bernoulli between A & B:

PA PB   21  h f g g PA 550  10 3   21  3.84  90.03 m g 860  9.81 PA  860  9.81 90.03  759.6 KN / m2 If welded steel pipes are used, ε = 0.000046 m Therefore



D



0.000046  0.00091 0.05

From Moody diagram, with Re = 9.56 104 and

 D

 0.00091 ,

240 0.934 2   4.70 m 0.05 2g PA  860  9.81 90.89  766.8 KN / m2

ƒ = 0.022 and h f  0.022 

2

2. What diameter of cast iron pipe would be required to ensure that a discharge of 0.20 m3/s would not cause a head loss in excess of 0.01 m/ 100 m of pipe length? Assume water temperature of 20°C.

Q  0.2 m3 / s ,

hf L



0.01 ,   0.000244 m 100

From the Darcy-Weisbach equation,

L v2 D 2g 100 0.2 2 4f 0.01  f    5 2 D  D 2  D  12.1026    2 g  4  f 0.01  0.3305 5 D 5 --------- (1) D  33.05 f hf  f

Re 

vD



0.000244 D

D







0.2 D 252878   2 6 D D 1.007  10  4

--------- (2)

--------- (3)

Assume ƒ = 0.02 (arbitrary value) D  0.92 m From (1) From (2) Re  2.74  105 From (3)



D

 0.00027

From Moody diagram, new ƒ = 0.017 D  0.89 m From (1) From (2) Re  2.84  105 From (3)



D

 0.000274

From Moody diagram, new ƒ = 0.017 (closest one can read) Therefore D  0.89 m

3

3. A 2000 m long commercial steel pipeline of 200 mm diameter conveys water at 20°C between two reservoirs, as shown in the Figure below. The difference in water level between the reservoirs is maintained at 50 m. Determine the discharge through the pipeline. Neglect the minor losses. (1) hf = 50m (2)

  0.045mm

L = 2000 m

D  200mm

Applying Bernoulli between (1) and (2):

0  0  50  0  0  0  h f Therefore,

L v2 2000 v 2  f  D 2g 0.2 2 g 0.0981  fv 2 0.0981 --- (1) v2  f vD v  0.2 Re    198610v --- (2)  1.007  10 6  0.045   0.000225 D 200

50  f

Assume flow is in fully rough zone, with From (1), v  2.557 m/s From (2), Re  5.08  10 5

 D

 0.000225 , f  0.015

From Moody diagram, new f  0.016 From (1), v  2.476 m/s From (2), Re  4.92  10 5 From Moody diagram, new f  0.016 Therefore Q  2.476   

0.2 2  0.078 m3/s 4

4. Compare answers for Q1, Q2, and Q3 using explicit equations. 5. Use the Hazen-Williams and Manning’s equations to solve Q1, Q2, and Q3. 6. Use Flowmaster to solve Q1, Q2, and Q3. 4

TUTORIAL 2 SIMPLE PIPE PROBLEMS AND MINOR LOSSES 1. A 6-km-long, new cast-iron pipeline carries 320 litres/s of water at 30°C. The pipe diameter is 30 cm. Compare the head loss calculated from (a) the Hazen-Williams formula, (b) the Manning formula, and (c) the Darcy-Weisbach formula.

L  6000 m, Q  0.32 m3/s, D  300 mm,   0.804  10 6 m2/s,   0.000244 m, C HW  130 , Manning’s n  0.011 a) Hazen-Williams formula

Q  0.278  C HW  D 2.63  S 0.54

0.32  0.278  130  0.32.63  S 0.54 S 0.54  0.21 hf S  0.0556  6000 Therefore h f  333.6 m b) Manning formula

D    4

2

1

3 S 2 R S v n n Q 0.32 v   4.527 m/s A 0.3 2  4 2

3

1

2

1

0.1778  S 2 4.527  0.011 h S  0.0784  f 6000 Therefore h f  470.4 m c) Darcy-Weisbach formula

L v2 6000 4.527 2  f  20890.65 f D 2g 0.3 2 g  4.527  0.3  0.00081 , Re   1.7  10 6 6 D 0.804  10 From Moody diagram, f  0.019 Therefore h f  0.019  20890.65  396.9 m hf  f

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2. Two reservoirs 1200 m apart are connected by a 50 cm smooth concrete pipe. If the two reservoirs have an elevation difference of 5 m, determine the discharge in the pipe by (a) the Hazen-Williams formula, (b) the Manning formula, and (c) the Darcy-Weisbach formula.

5m

L = 1200 m D = 0.5 m a. Hazen-Williams formula: Q  0.278  C HW  D 2.63  S 0.54 Q  0.278  140  0.5 2.63  (5 / 1200) 0.54 Q = 0.326 m3/s

b. Manning formula:

D  Q  VA  4

2

3

S

n

1

2



D 2 4

0.125 3  51200 2  0.52 Q  2

1

0.011

4

Q = 0.288 m3/s

c. Darcy-Weisbach formula:



0.9mm  0.0018 D 500 h f  5 ,   1.007  10 6 m2/s (assume 20°C water) 

Using Swamee and Jain’s explicit equation for Q,

  9.81  0.5  5  0.0018 1.784  1.007  10 6 2 Q =  0.965  0.5  ln   3.7 1200 9.81  0.5  5  0.5 1200  Q =  0.034489  ln 0.0004865  0.000025133 Q = 0.261 m3/s

6

     

3. An old pipe 2 m in diameter has a roughness of ε=30mm. A 12-mm-thick lining would reduce the roughness to ε=1mm. How much in annual pumping costs would be saved per kilometer of pipe for water at 20°C with discharge of 6 m3/s? The pumps and motors are 80% efficient, and power costs 4 cents per kilowatt-hour.

6  1.91 m/s 22  4 1000 1.912 h  f    92.96 f m/km Head loss of old pipe = f 2 2g  30 vD 1.91  2   0.015 , Re    3.8  10 6 6 D 2000  1.007  10 Old pipe: D = 2 m, ε=30mm, v 

Q  A

From Moody diagram, f  0.044 Therefore, h f  92.96  0.044  4.09 m/km New pipe D = 2000 – 24 = 1.976 m, ε = 1 mm, v 

Q  A

6  1.957 m/s 1.976 2  4

1000 1.957 2 hf  f    98.739 f m/km 1.976 2g  1 vD 1.957  1.976   0.000506 , Re    3.84  10 6 D 1976  1.007  10 6 From Moody diagram, f  0.017 Therefore, h f  98.739  0.0165  1.679 m/km Saving in head = 4.09 – 1.679 = 2.411 m/km

Qh Energy = Power x Time =

e

 365  24 

9.81  6  2.411  365  24  1554.212 KW 0.8

Therefore annual savings = 1554.2 x $ 0.04 = $ 62168/km

7

4. What size commercial steel pipe is needed to convey 200 L/s of water at 20°C 5 km with a head drop of 4 m? The line connects two reservoirs, has a reentrant entrance, a submerged outlet, four standard elbows, and a globe valve. (1) Reentrant entrance

4m

Standard elbows

(2) Globe valve

Q = 0.2 m3/s Submerged outlet

Applying Bernoulli between 1 and 2:

0  0  4  0  0  0  he  4helbows  hGV  ho  h f v2 v2 v2 v2 L v2 4  ke  k elbows  k GV  ko f 2g 2g 2g 2g D 2g k e  1.0 , k elbows  0.9 , k GV  10.0 , k o  1.0 Therefore,

v2  5000  1.0  4  0.9  10.0  1.0  f  2g  D  v2  f  4 15.6  5000  2g  D Q 0.2  4 0.2546 v   2 D   D2 D2  4 4

Therefore,

0.2546 2  f  78.48  15.6  5000  ---------(1) 4 D D   0.000046  ---------(2) D D vD 0.2546 256830 Re    6 --------- (3)  D D  1.007  10 1.01121 6.4821 , from (1) f  0 . 02 78 . 48   Assume D4 D5 By trial and error D  0.62 m  5 From (2),  0.000074 , Re  4.14  10  f  0.0145 D 1.01121 4.69953  From (1), 78.48   D  0.583 m D4 D5  From (2),  0.000079 , Re  4.4  105 D 8

New f  0.0145 , therefore D  0.583 m Note: Equivalent length method would have been easier

5. What is the equivalent length of 50 mm diameter pipe, f=0.022, for (a) a re-entrant pipe entrance, (b) a sudden expansion from 50 mm to 100 mm diameter, (c) a globe valve and a standard tee?

f  0.022 , D  50 mm kD Le  f a.

k  1.0 for re-entrant pipe entrance. 0.05 Therefore, Le  1   2.27 m 0.022

b. Sudden expansion from 50 mm to 100 mm. 2

  D 2    50  2  1    1   k  1      0.5625   D2     100   0.5625  0.05 Therefore Le   1.278 m 0.022 c.

2

k  10.0 for globe valve, k  1.8 for standard tee. Therefore  K  11.8 0.05 Le  11.8   26.82 m 0.022

6. Solve Q1 and Q1 using Flowmaster.

9

TUTORIAL 3 EGL, HGL, AND PIPES IN SERIES 1. Sketch the energy grade line and the hydraulic grade line for the compound pipe shown below. Consider all the losses and the change in velocity and pressure heads. EGL HGL

2. Two sections of cast-iron pipe connected in series bring water from a reservoir and discharge it into air at a location 100 m below the water surface elevation in the reservoir through a globe valve. The first pipe section is 400 mm diameter and is 1000 m long, and the second pipe section is 200 mm diameter and 1200 m long. If the water temperature is 10°C, and square connections are used, determine the discharge. Sketch the EGL and HGL. (A)

he 2

v1 2g D1 = 0.4 m L1 = 1000 m

hc

2

v2 2g

D2 = 0.2 m L2 = 1200 m

Energy between (A) and (B): 2

he  h f 1  hc  h f 2

v  hv  2  100 2g

2

2

v 1000 v1 he  0.5 1 , h f 1  f1   2g 0.4 2 g 2

v hc  0.33 2 2g

k c  0.33 (assumption) 2

hf 2  f2  2

hv  k v

1200 v2  0.2 2 g 2

v2 v  10 2 2g 2g

10

100 m

(B)

Therefore, 2

100  (1  10  f 2

2

v 1200  1000 v  0.33) 2   f1  0.5  1 0.2 2g  0.4  2g

From continuity,

A1V1  A2V2  0.42 v1   0.22 v2 4 4 Substituting for v1 , we get:

1962 11.36  156.25 f1  6000 f 2 vD v  0.4 Re1  1 1  1  7.63  10 4 v2 6 1 1.31  10 v D v  0.2 Re 2  2 2  2  1.53  10 5 v2 6 2 1.31  10

v2  2

1 D1

2 D2

 0.00065  f1  0.0178  0.0013



f 2  0.0205

Solving for v 2 ,

1962 11.36  156.25(0.0178)  6000(0.0205) v2  3.78 m/s v1  0.25  3.78  0.94 m/s Re1  7.63  10 4  3.78  2.88  105  f1  0.019 Re 2  1.53  105  3.78  5.78  105  f 2  0.021 v2  2

Use new values of f 1 and f 2 to calculate v 2 ,

1962 11.36  156.25(0.019)  6000(0.021) v2  3.739 m/s v1  0.25  3.739  0.935 m/s v2  2

Re1  7.63  10 4  3.739  2.85  105  f1  0.019 Re 2  1.53  105  3.739  5.72  105  f 2  0.021 Therefore, Q  v1 A1  v2 A2  3.739 

  0.2 2 4

 0.117 m3/s  0.12m3/s

11

3. Two new cast-iron pipes in series connect two reservoirs. Both pipe are 300 m long and have diameters of 0.6 m and 0.4 m, respectively. The elevation of water surface in reservoir A is 80 m. The discharge of 10°C water from reservoir A to reservoir B is 0.5 m3/s. Find the elevation of the surface of reservoir B. Assume a sudden contraction at the junction and a square-edge entrance. 80 m H

Q = 0.5 m3/s L1 = 300 m D1 = 0.6 m

L2 = 300 m D2 = 0.4 m

0.5  1.77 m/s 0.6 2  4 0.5 v2   3.98 m/s 0.4 2  4 vD 1.77  0.6 Re1  1 1   8.08  10 5  1.31  10 6 v D 3.98  0.4 Re 2  2 2   1.22  10 6  1.31  10 6 v1 

1

Q  A1

 0.00043 ,

2

 0.00065 D1 D2 f1  0.017 , f 2  0.018 , from Moody diagram Bernoulli between two reservoirs, 2

H  0.5

2

2

2

2

v1 v v v 300 v1  0.017    0.24 2  0.018  2  2 2g 0.6 2 g 2g 2g 2g

1.77 2 300 1.77 2 3.98 2 300 3.98 2 3.98 2  0.017    0.24  0.018    2g 0.6 2g 2g 0.4 2g 2g H  0.0798  1.357  0.1938  10.899  0.807 H  13.337 m H  0.5

Therefore, the surface elevation of reservoir B = 80-13.337 = 66.66 m

12

4. Pipeline AB connects two reservoirs. The difference in elevation between the two reservoirs is 10 m. The pipeline consists of an upstream section, D1 = 0.75 m and L1 = 1500 m, and a downstream section, D2 = 0.5 m and L2 = 1000 m. The pipes are cast-iron and are connected endto-end with a sudden reduction of area. Assume the water temperature at 10°C. Compute the discharge capacity using the graphical approach. See figure of previous problem except different lengths and diameters. Bernoulli between two reservoirs, 2

H  0.5

2

2

2

From continuity,

v1 A1  v2 A2 0.75 2 0.5 2 v1     v2    4 4 v2  2.25v1 Substituting for v 2 in (1)

H  0.35  101.94 f1  516.05 f 2 v1

H  0.35  101.94 f1  516.05 f 2 

2

Q 2 (4) 2

  0.75 

2 2

H  5.1234Q 2 0.35  101.94 f1  516.05 f 2 

1

D1

2

v1 v v 1500 v1 1000 v2  f1    0.24 2  f 2    2 --- (1) 2g 0.75 2 g 2g 0.5 2 g 2 g

 0.00035 ,

2

D2

 0.00052

Assume Q = 0.5 m3/s:

0.5  1.132m / s 0.75 2  4 v1 D1 1.132  0.75 Re1    6.48  10 5 6  1.31  10 v2 D2 2.546  0.5 Re 2    9.71  10 5 6  1.31  10

Therefore v1 

Therefore H = 14.304 m

v2  2.546m / s

f1  0.015 f 2  0.018

too high

3

Assume Q = 0.4 m /s

v1  0.9054m / s Re1  5.18  10 5 Re 2  7.78  10 5

v2  2.037m / s f1  0.016 f 2  0.0175

Therefore H = 9.03 m too low 13

H (m) vs. Q (m3/s)

From the graph, Q = 0.42 m3/s for H = 10 m

14

TUTORIAL 4 HYDRAULICS 6713 BRANCHING PIPES AND PIPE NETWORKS 1. A two-loop pipe network has node designations as shown below. Inflows of 0.4 m3/s and 0.45 m3/s enter points A and B, respectively. Equal withdrawals are made at points C, D, and F. The pipe characteristics are as follows: Pipe AB BC AF BE CD DE EF

Length (m) 500 400 650 750 700 550 900

A

Diameter (m) 0.4 0.5 0.5 0.35 0.4 0.5 0.6

B

C

Loop 1

F

Loop 2

E

0.4 m3/s

Friction factor ƒ 0.017 0.016 0.014 0.015 0.013 0.016 0.015

D

0.45 m3/s L = 500, D = 0.4, f = 0.017

A L = 650, D = 0.5, f = 0.014

B 1 +

0.28333 m3/s

L = 400, D = 0.5, f = 0.016

C

L = 750, D = 0.35, f = 0.015

2 +

L = 700, D = 0.4, f = 0.013

0.28333 m3/s F 0.28333 m /s 3

L = 900, D = 0.6, f = 0.015

E

L = 550, D = 0.5, f = 0.016

15

D

Trial 1 Loop 1

2

Pipe AB BE EF AF

R 68.619 177.067 14.352 24.072

BC CD DE EB

16.93 73.462 23.279 177.067

Pipe AB BE EF AF

R 68.619 177.067 14.352 24.072

BC CD DE EB

16.93 73.462 23.279 177.067

Pipe AB BE EF AF

R 68.619 177.067 14.352 24.072

BC CD DE EB

16.93 73.462 23.279 177.067

Q 0.4 0.4 0.28333 0

hL 10.979 28.331 1.152 0 40.462 ΔQ = 0.198 0.45 3.428 0.16667 2.041 -0.1166 -0.317 -0.202 -7.225 -1.439 ΔQ = -0.0123

2hL/Q 54.895 141.655 8.13186 0 204.6819 0.198 15.237 24.492 5.435 71.535 116.699 -0.0123

new Q 0.202 0.202 0.08533 -0.198

Q 0.202 0.1897 0.08533 -0.198

hL 2.7999 6.3719 0.1045 -0.9437 8.3326 ΔQ = 0.07796 0.4623 3.6183 0.17897 2.353 -0.1043 -0.2535 -0.1117 -2.2108 3.507 ΔQ = 0.0406

2hL/Q 27.72178 67.1787 2.449314 9.532323 106.8821 0.077961 15.65347 26.29491 4.858183 39.57043 86.377 0.040601

Q 0.12404 0.15234 0.00737 -0.2759

2hL/Q new Q 17.02306 0.084586 53.9488 0.112886 0.211669 -0.03208 13.28584 -0.31541 84.46936 0.039454 14.27878 0.40115 20.32984 0.11782 6.749034 -0.16551 39.97821 -0.13344 81.33586 0.02055

0.4623 0.17897 -0.10436 -0.1897

Trial 2 Loop 1

2

new Q 0.124039 0.111739 0.007369 -0.27596

0.421699 0.138369 -0.14496 -0.15234

Trial 3 Loop 1

2

hL 1.05577 4.10928 0.00078 -1.8331 3.33265 ΔQ = 0.03945 0.4217 3.01068 0.13837 1.40652 -0.1449 -0.4891 -0.1128 -2.2565 1.67146 ΔQ = 0.02055 16

Trial 4 Loop 1

2

Pipe AB BE EF AF

R 68.619 177.067 14.352 24.072

BC CD DE EB

16.93 73.462 23.279 177.067

Pipe AB BE EF AF

R 68.619 177.067 14.352 24.072

BC CD DE EB

16.93 73.462 23.279 177.067

Q 0.08459 0.13344 -0.0320 -0.3154

hL 0.491 3.1529 -0.0147 -2.3948 1.23433 ΔQ = 0.01646 0.40115 2.7244 0.11782 1.01977 -0.1655 -0.6376 -0.1169 -2.4230 0.68344 ΔQ = 0.00854

2hL/Q new Q 11.60894 0.068126 47.2557 0.116976 0.920823 -0.04854 15.18531 -0.33187 74.97077 0.016464 13.58295 0.39261 17.31064 0.10928 7.705758 -0.17405 41.42657 -0.12552 80.02592 0.00854

Q 0.06813 0.12552 -0.0485 -0.3318

2hL/Q 9.350066 44.45092 1.39349 15.97752 71.172 0.005946 13.29375 16.05582 8.103419 42.34373 79.79672 0.003135

Trial 5 Loop 1

2

hL 0.31851 2.78974 -0.0338 -2.6512 0.4232 ΔQ = 0.00595 0.39261 2.60963 0.10928 0.87729 -0.1740 -0.7052 -0.1195 -2.5315 0.2502 ΔQ = 0.003135

To three decimal places, the final flow values are: AB = 0.062 m3/s BC = 0.390 m3/s CD = 0.106 m3/s DE = -0.177 m3/s EF = -0.054 m3/s AF = -0.338 m3/s EB = -0.123 m3/s

17

new Q 0.062184 0.119574 -0.05449 -0.33782

0.389475 0.106145 -0.17719 -0.12271

2. Determine the flow into and out of each reservoir in the Figure below if the connecting pipes are made of the same material with ε = 0.05 mm and water temperature at 20°C. The pipe characteristics are as follows: Reservoir A B C

Elevation (m) 100 80 70

Pipe a b c

Length (m) 3000 4000 5000

Diameter (m) 0.8 1.2 0.6

A a

B

c

b

C

J

Try both the iterative method and the graphical method. Plotting first few points from table: 90

85

80

P

75

70 -2

-1

0

1

ΣQ

P≈ 81.4 m (Assume fully turbulent flows, correct for ƒ when near correct answer)

18

2

100

P

A

80 a

B

c

b J Assume fully turbulent flow

Initial estimates: 𝜀𝑎 0.05 = 800 = 6.3 × 10−5 , 𝑓𝑎 = 0.011, 𝑅𝑎 = 8.3212 𝐷 𝑎

𝜀𝑏 𝐷𝑏

= 1200 = 4.2 × 10−5 , 𝑓𝑏 = 0.010 , 𝑅𝑏 = 1.3282

0.05

𝜀𝑐 𝐷𝑐

=

0.05 600

= 8.3 × 10−5 , 𝑓𝑐 = 0.0115, 𝑅𝑐 = 61.099

Trial 1 (assume P = 90 m) 𝑧𝐴 − 𝑃 = 𝑅𝑎 𝑄𝑎2 = 100 − 90 = 10 𝑃 − 𝑧𝐵 = 𝑅𝑏 𝑄𝑏2 = 90 − 80 = 10 𝑃 − 𝑧𝐶 = 𝑅𝑐 𝑄𝑐2 = 90 − 70 = 20 So, 10 𝑄𝑎 = 8.3212 10 𝑄𝑏 = 1.3283 20 𝑄𝑐 = 61.099

1

1

1

2

2

2

= 1.096 = 2.744 (−) = 0.572 (−)

𝑄 = −2.219

Continue trial with a smaller P. e.g. Trial 4, P = 81.5 m 𝑄𝑎 = 1.491 𝑄𝑏 = 1.063 (−) 𝑄𝑐 = 0.434 (−)

19

70 C

𝑄 = −0.0055

Now check Re and get new 𝑓s. 𝑅𝑒𝑎 = 2.25 × 106 𝑅𝑒𝑏 = 9.97 × 105 𝑅𝑒𝑐 = 8.56 × 105

new 𝑓𝑎 = 0.0121 𝑓𝑏 = 0.0125 𝑓𝑐 = 0.0131

 𝑅𝑎 = 9.1533 𝑅𝑏 = 1.6603 𝑅𝑐 = 69.60

With P = 81.5 m, 𝑄 = 0.0646 Try P = 81.7 m 𝑄 = −0.0079 Try P = 81.68 m 𝑄𝑎 = 1.4147 𝑄𝑏 = 1.0059 (−) 𝑄𝑐 = 0.4097 (−) 𝑄 = −0.00084 (𝑔𝑜𝑜𝑑 𝑒𝑛𝑜𝑢𝑔𝑕) Should check Re and get new 𝑓s. But it should be very similar to last values.

3. Four pipes are connected in parallel. Their characteristics are as follows: Pipe No. 1 2 3 4

Diameter (m) 0.15 0.30 0.45 0.60

Length (m) 3000 3000 3000 3000

Roughness (mm) 0.06 0.06 0.09 0.09

Determine the discharge through each pipe if the total flow is 1.4 m3/s. Assume that the pipe flow is fully turbulent. Total flow = 1.4 m3/s

20

Four pipes in parallel.

[1] Q = 1.4 m3/s

A

[2]

B

Q = 1.4 m3/s

[3] [4]

Assume fully turbulent flows, 𝜀1 𝐷1

= 0.0004, 𝑓1 = 0.016, 𝑅1 = 52228.42

𝜀2 𝐷2

= 0.0002, 𝑓2 = 0.014 , 𝑅2 = 1428.12

𝜀3 𝐷3

= 0.0002, 𝑓3 = 0.016, 𝑅3 = 214.93

𝜀4 𝐷4

= 0.00015, 𝑓4 = 0.016, 𝑅4 = 51.00

𝑄=

𝑄𝑖 = 1.4 𝑚3 /𝑠 1

𝑕𝐴𝐵 𝑄1 = 𝑅1

2

𝑕𝐴𝐵 ; 𝑄2 = 𝑅2

1

2

𝑕𝐴𝐵 ; 𝑄3 = 𝑅3

1

2

𝑕𝐴𝐵 ; 𝑄4 = 𝑅4

Or 𝑄2

𝑕𝐴𝐵 =

2

1 1

𝑅1 =

2

+

1 1

𝑅2

1 52228.42

2

1

+

1 1

𝑅3

2

+

1 1

𝑅4

2

1 + 2 1428.12

1.42 1

1 + 2 214.93

1

1 + 2 51.00

= 34.29 𝑚 Therefore, 𝑄1 = 0.026

𝑚3 𝑚3 𝑚3 , 𝑄2 = 0.155 , 𝑄3 = 0.399 , 𝑄4 = 0.82 𝑚3 /𝑠 𝑠 𝑠 𝑠 21

2 1

2

1

2

4. Two reservoirs have a difference in elevation of 6 m and are connected by a pipeline which consists of a single 600 mm diameter pipe 3000 m long, feeding a junction from which 2 pipes, each 300 mm diameter and 3000 m long, lead in parallel to the lower reservoir. If ƒ = 0.04, calculate the flow rate between reservoirs.

6m

D2 = 300 mm L2 = 3000m D1 = 600 mm L1 = 3000m f = 0.04

Q1  Q2 Q 3

8 f1 L1Q1

2



 2 gD15 8 f 2 L2 Q2



 2 gD2 5

Q3

------(1)

8 f 2 L2 Q2

2

Q2

Q1

2

6

 2 gD2 5 8 f 3 L3Q3

------(2)

2

------(3a)

 2 gD3 5

Since f 2  f 3 , L2  L3 , D2  D3 , so Q2  Q3 From (2)

127.51Q1  4080.34Q2  6 2

2

------(3b)

------(4)

From (1) and (3b), Q1  2Q2 Substitute into (4):

127.51 4Q2  4080.34Q2  6 2

2

Therefore Q2  0.0013071 , Q2  Q3  0.0362 m3/s, and Q1  2Q2  0.0723 m3/s 2

5. Solve Q1, Q2, Q3, and Q4 using WaterCad.

22

TUTORIAL 5 HYDRAULICS 6713 UNIFORM FLOWS IN OPEN CHANNELS 1. The cross-section of a canal is as shown below. The canal slope is 1/4000.

2m

1:3

1:3

3m a. Determine the discharge if Chezy’s C is 60 m½/s.

Q  AC RS 1 A  3  3  2  2  32  18m 2 2 P  3  2 4  36  15.65m A R   1.15m P Therefore,

Q  18  60  1.15  0.00025  18.31

m3 s

b. Determine the discharge if Manning’s n is 0.025. 2

AR 3 S Q n

1

2



18  1.15

2

 0.00025 0.025 3

1

2

 12.50

m3 s

c. What value of C corresponds to n=0.025? 1

1

1

R 6 1.15 6 m 2 C   40.9 n 0.025 s d. What value of n corresponds to C = 60 m½/s? 1

1

R 6 1.15 n  60 60

6

 0.017

2. A trapezoidal canal has a bottom width of 5 m, side slopes of 1:2 and a slope of 0.0004. Manning’s n is 0.014. The depth is 2 m. Determine the discharge. 2

AR 3 S Q n

1

2

23

A  5  5  2  2  2  18m 2 P  5  2 4  16  13.944m R  1.29m Therefore, 2

18  1.29 3  0.0004 Q 0.014

1

2

 30.49

m3 s

3. Calculate for the same canal as in Problem 2 the water depth when the discharge is 75 m3/s. Answer must be accurate to the nearest cm. 2

AR 3 S Q n

1

2

Since A and R are functions of depth, rearranging Mannings eqn.

Qn S

1

2

 AR 3 , where the LHS is known = 52.50

2

A  5  5  4y 

1 10  4 y y  5 y  2 y 2 2

P  5  2 y2  4y2  5  2y 5 10  4 y R 5  2y 5

m3 Since Q > 30.49 (from Q2), y must be > 2 m. s y 3 4 3.1 3.2 3.11

A 33 52 34.72 36.48 34.89

P 18.42 22.89 18.86 19.31 18.91

R 1.792 2.272 1.841 1.889 1.846

R⅔ 1.475 1.728 1.502 1.528 1.505

A R⅔ 48.68 89.87 52.15 55.75 52.50

Ans: y = 3.11 m 4. A reinforced concrete aqueduct of rectangular cross-section is to be designed to carry 10 m3/s with a velocity of 2 m/s. Determine the water depth and the width of the cross-section so that the required slope of the aqueduct is minimized.

y b 24

Q  VA 10  2 A , A  5m 2 For minimum slope, P must be minimized for a given cross section.

5  2y y dP  5  2 20 dy y

P

2y2  5 Therefore,

y  1.582m b  3.162m 5. Design a trapezoidal cross-section canal with an area of 60 m2, a hydraulic radius of 2 m, and side slopes of 1:3.

y

1:3

A  60m 2 R  2m

1:3

B

Side slope = 1:3

R

A , therefore P

P  30m  B  2 y 2  9 y 2  B  2 y 10 ------------------- (1) 1 A  60  ( B  B  2  3 y) y  By  3 y 2 ------------------- (2) 2 From (1), multiplying by y, we get: By  6.325 y 2  30 y ------------------- (3) (3) – (2) gives:

3.325 y 2  30 y  60  0 30  900  4  3.325  60 30  10.1  2  3.325 6.650 y  6.053m (not feasible) or 2.992m

y

Therefore, B  30  6.325  2.992  11.08m

6. Solve Q1, Q2, and Q3 using Flowmaster. 25

TUTORIAL 6 HYDRAULICS 6713 ENERGY CONCEPTS IN OPEN CHANNEL FLOW 1. Water is flowing in a rectangular channel at a velocity of 3 m/s and a depth of 2.5 m. Determine the changes in water surface elevation for the following alterations in the channel bottom: a. An increase (upward step) of 20 cm, neglecting losses. (1)

TEL

(2)

y1

y2 h = 0.2 m

m s y1  2.5m

v3

Check whether flow is sub or supercritical.

Fr1 

v gy



3 9.81  2.5

 0.61 , subcritical (water level will drop or encounter a hump)

E1  h  E2

q2 q2 y1   0.2  y 2  2 2 2 gy1 2 gy 2

2.5 

(3  2.5) 2 (3  2.5) 2  0 . 2  y  2 2 2  9.81  2.5 2 2  9.81  y 2

2.959  0.2  y 2  2.759  y 2 

5.734 2 2 y2

2.867 3 2 f ( y 2 )  y 2  2.759 y 2  2.867  0 2 or y2

Using Newton’s Method:

y 2,n  y 2, 0 

f ( y2 ) 2 f ' ( y 2 )  3 y 2  5.518 y 2 , where f ' ( y2 )

26

y2,0 2.0 2.175 2.128 2.123 Therefore, y2 = 2.123 m

f(y2) -0.169 0.104 0.0096 0.0005

f’(y2) 0.964 2.190 1.843 1.807

y2,n 2.175 2.128 2.123 2.123

b. The maximum increase allowable for the specified upstream flow conditions to remain unchanged, neglecting losses.

yc 

3

3 q2  1.79m , Ec   1.79  2.684m 2 g

Therefore,

max h  E1  Ec  2.959  2.684  0.274m c. A “well-designed” decrease (downward step) of 20 cm.

E1  0.2  E2

2.959  0.2  y 2 

7.5 2 2 2 gy 2

2.867 3 2 f ( y 2 )  y 2  3.159 y 2  2.867  0 2 , or y2

3.159  y 2 

f ' ( y 2 )  3 y 2  6.318 y 2 2

y2,0

f(y2)

f’(y2)

y2,n

2.7

-0.479

4.811

2.8

2.8

0.052

5.830

2.79

2.79

-0.0053

5.725

2.791

2.791

0.0004

5.735

2.791

Therefore, y2 = 2.791 m

27

2. Water is flowing in a rectangular channel whose width is 5 m. The depth of flow is 2 m and the discharge is 25 m3 /s. Determine the changes in depth for the following alterations in the channel width: a. An increase of 50 cm, neglecting losses.

q1

5m

5.5 m

q2

(1) (2)

Q  25

m3 b1  5m s ,

Therefore,

q1  5m 3 / s / m , y1  2m Check:

Fr 

q

 0.56 , subcritical therefore with decrease in q, depth increases

gy 3

Neglecting energy losses, E1  E2

E1  y1 

q2 

2

q2 2 gy 1

2

 2

q 25  2.319m  y 2  2 2 2 gy 2 2g  2

25  4.545m 3 / s / m 5.5

Therefore,

4.545 2 3 2 f ( y 2 )  y 2  2.319 y 2  1.053  0 2 , or 2 gy 2

2.319  y 2 

f ' ( y 2 )  3 y 2  4.638 y 2 2

y2,0

f(y2)

f’(y2)

y2,n

2.0

-0.223

2.724

2.08

2.08

0.019

3.332

2.074

2.074

-0.0009

3.2852

2.074

y 2  2.074m

28

b. A decrease of 25 cm, assuming a “well-designed” transition. A decrease in width means an increase in depth.

E1  E2

2.319  y 2 

25 / 4.752 2 gy 2

2

f ( y 2 )  y 2  2.319 y 2  1.419 3

2

f ' ( y 2 )  3 y 2  4.638 y 2 2

y2,0

f(y2)

f’(y2)

y2,n

2.0

0.143

2.724

1.948

1.948

0.0112

2.3493

1.943

1.943

-0.00049

2.3141

1.943

y 2  1.943m c. The maximum decrease allowable for the specified upstream flow conditions to remain unchanged, neglecting losses. For no change in upstream flow condition, y c 

2 E  1.546m 3

Maximum q at section 2 should be,

qm  2 g  1.546 2 (0.773)  6.02m 3 / s / m Therefore,

bc 

Q 25   4.152m q 6.02

Therefore, Maximum decrease

 5  4.152  0.848m

29

3. A lake discharges into a steep channel. At the channel entrance the lake level is 2.5 m above the channel bottom. Neglecting losses, find the discharge for the following geometries:

E1

2.5 m

yc

(2) (1) a. Rectangular section, b = 4 m.

b  4m Rectangular channel. At the channel entrance, depth =yc Assuming no losses , E1  Ec (since v =0) 1 Therefore, y c 

2 2 Ec   2.5  1.667m 3 3

At critical flow, q 

gyc  9.81 1.667 3  6.74m 3 / s / m 3

3 Therefore, Q  6.74  4  26.96m / s

b. Trapezoidal section, b = 3 m, side slopes = 1:2.5. B yc

1: 2.5

3m 2

v Q2 Ec  y c  c  y c  2g 2 gA 2 At critical depth,

Q2B 1 gA 3

1 2.5  2.5yc 2 A 2  yc  Therefore, Ec  y c  2B 23  2.5  2.5 y c  3 yc 

2.5  y c 

3 y c  2.5 y c 6  10 y c

2

15  25 yc  6 yc  10 yc  3 yc  2.5 yc 2

2

30

12.5 yc  16 yc  15  0 2

16  256  4  15  12.5 16  31.72   1.909m 25 25 A  3  1.909  2.5  1.909 2  14.83m 2 B  6  10 1.909  25.09m

Therefore, y c 

Q

gA3 9.81  14.833 m3   35.71 B 25.09 s

31

TUTORIAL 7 HYDRAULICS 6713 MOMENTUM CONCEPTS IN OPEN CHANNEL FLOW 1. A 3 m wide rectangular channel carries 15 m3/s of water at 0.6 m depth before entering a hydraulic jump. Compute the downstream water depth and the critical depth.

q

15  5m 3 / s / m 3

Critical depth, y c 

v1 

q 2 3 52   1.60m g 9.81

q 5   8.33m / s y1 0.6

Fr1 

y2 

3

8.33 g  0.6

 3.43 , supercritical





y1  0.6 2 1  8  3.432  1  4.38m  1  8Fr1  1   2 2

2. A long rectangular channel 3 m wide carries a discharge of 15 m3 /s. The channel slope is 0.004 and the Manning’s roughness coefficient is 0.01. At a certain point in the channel where the flow reaches the normal depth, a. Determine the state of the flow. Is it supercritical or subcritical?

15  5m 3 / s / m 3 From question 1, yc  1.60m q

2

1

AR 3 S 2 Q  From Manning’s eqn, n A y1b b  3m A1  y1b , R1  1  P1 2 y1  b ,  3 y1  1 Therefore, 15  0.01 3 y1  2 y  3   1  Solving for y1, y1  1.08m , Fr1 

23

q gy 3

0.0041 2  1.42

Since y1  y c , flow is supercritical.

32

b. If a hydraulic jump takes place at this depth, what is the sequent depth at the jump?





y1  1.08 2 1  8  1.42 2  1  1.698m  1  8Fr1  1   2 2

y2 

c. Estimate the energy head loss through the jump. 3  y 2  y1  Head loss E 

y1 y 2



0.623 4  1.7  1.08

 0.032m

Or P  QE  9810  15  0.032  4709W

3. A spillway, as shown, has a flow of 3 m3 /s per meter of width occurring over it. What depth y2 will exist downstream of the hydraulic jump? Assume there is no energy loss over the spillway.

5m y2 y1 (0)

(1)

For no losses:

y0 

q2 q2  y  1 2 2 2 gy 0 2 gy1

32 32 5  y1  2 2g52 2 gy1 y1  0.312m

Fr1 

q gy1

3

3



g  0.312 3 y

1 Therefore, y 2  2

 5.49

 1  8  5.49  1  2.27m 2

33

(2)

TUTORIAL 8 HYDRAULICS 6713 DESIGN AND ANALYSIS OF CULVERTS 1. A rectangular concrete conduit is to be used as a culvert on a slope of 0.02. The culvert is 15 m long and has a cross-section of 2.13 m x 2.13 m. If the tail water elevation is 1.8 m above the crown at the outlet, determine the head water elevation necessary to pass a 10 m3/s discharge. Assume a square-edged entrance (Ke = 0.5).

hL

HW

TW 1.8 m

Q = 10 m3/s SoL

2.13 m S = 0.02

“outlet control”

L  15m n  0.013

K e  0.5 R

2.132 0.5325m 4  2.13

A  2.132  4.5369m 2

hL  TW  S o  L  HW hL  HW  TW  S o  L hL  HW  3.93  0.02  15 hL  HW  3.63

34

2.13 m 2.13 m

  Q2 n2 L hL   K e  4  2 g  1  2 R 3   2 gA   0.013 2  15 10 2  hL   0.5   2 g  1 4  2 g  4.5369 2 0.5325 3   hL  0.5  0.1152  1  0.24762 hL  0.39995m Therefore,

HW  3.63  0.39995  4.03m

HWelevation  4.03  0.3  4.33m [Using Culvert Master, HW elevation = 4.327m] (see attached printout)

2. A culvert is 11 m long and has upstream and downstream inverts of 263.4 and 263.1 meters, respectively. The downstream tailwater is below the downstream pipe invert. a. For a square-edged entrance and Manning’s n of 0.013, what is the minimum diameter for a concrete circular culvert (in mm) required to pass 1.4 m3/s under a roadway with a maximum allowable headwater elevation of 265.2 m? b. What is the headwater elevation for the selected culvert?

265.2 m h 265.4 m D

Q = 1.4 m3/s

263.1 m TW

For the given condition  Inlet control Use orifice equation,

Q  Cd A 2 gh

35

D D  h   265.2  263.4    1.8  2 2  Assume Cd  0.62 1.4  0.62 

D 2

D   2 g 1.8   4 2 

Squaring both sides and simplifying,

D  1.96  4.6532 D 4 1.8   2 

-----------(1)

By trial, D  0.736m Closest size available is probably 0.75m, Therefore use D  0.75m With D  0.75m , from (1)

0.75   1.96  4.6532  0.75 4  HW   2   HW  1.7062m Or HWelevation  265.106m Using Culvert Master, D  0.75m , HWelvation  265.131m Reason for the difference is that Culvert Master uses a slightly different form of equation for inlet control (if Cd = 0.61, we get the same answer as Culvert Master).

3. Twin 1220 by 910 mm box culverts (n = 0.013, 90° and 15° wingwall flares entrance) carry 8.5 m3/s along a 31 m length of pipe constructed at a 1.0 percent slope. The tailwater depth is 0.61 m. a. What is the headwater depth? b. Are the culverts under inlet or outlet control conditions? Best to use Culvert Master:

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a. Headwater depth = 2.634 m (remember to subtract the upstream invert elevation from the headwater elevation). b. Culvert under inlet control. [See attached printout from Culvert Master]

4. A 12.2 m long 920 by 570 mm concrete arch pipe (n =0.013, groove-end with headwall entrance) constructed at a 0.8 percent slope carries 1.84 m3/s. a. If there is a constant tailwater depth of 0.3 m, what is the headwater depth for both inlet and outlet control conditions? b. Is the culvert flowing under inlet or outlet control conditions? c. What would be the result if the tailwater was 0.5 m deeper? Use Culvert Master: a. Inlet control headwater depth is 2.34 m, outlet control headwater depth is 2.13 m. b. Culvert is flowing under inlet control. c. If TW is 0.5 m deeper, we get outlet control and headwater depth is 2.36 m. [See printouts from Culvert Master]

5. Twin culverts are proposed to discharge 6.5 m3/s. The culverts will be 36.6 m long and have inverts of 20.1 and 19.8 m. The design engineer analyzed the following three culvert systems. Which of the following proposed culverts will result in the highest headwater elevation? The lowest? Tailwater elevation is below the downstream invert. [Hint: Use Culvert-Master to either solve the problems or use it to check your solutions]. a. 1200 mm circular concrete pipes (n = 0.013, square-edged entrance); b. 1200 x 910 mm concrete box culverts (n = 0.013, 90° and 15° wingwall flares entrance); c. 1630 x 1120 mm steel and aluminum arches (n = 0.025 and Ke= 0.5). Use Culvert Master: a. 1200 mm circular concrete pipes (n = 0.013, Ke = 0.5), HW elevation = 21.92 m b. 1220 x 910 mm concrete box culverts (n = 0.013, 90° and 15° wingwall flares entrance), HW elevation = 21.94 m. c. 1630 x 1120 mm steel and aluminum arches (n= 0.025, Ke = 0.5), HW elevation = 21.70 m. d. Therefore, box culverts have the highest headwater; the arches have the lowest. (See attached printouts from Culvert Master).

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TUTORIAL 9 HYDRAULICS 6713 PUMPS 1. From the manufacturer’s data, a pump of 254 mm impeller diameter has a capacity of 76 L/s at a head of 18.6 m when operating at a speed of 900 rpm. It is desired that the capacity be about 95 L/s at the same efficiency. Determine the adjusted speed of the pump and the corresponding head.

D1  254mm , Q1  76l / s , H 1  18.6m , N1  900rpm , Q2  95l / s Therefore,

H  Q2 N  1.25  2   2  Q1 N1  H 1 

1 2

N 2  1.25  900  1125rpm

H 2  1.25 2  18.6  29.06m

2. The following performance curves were obtained from a test on a 216 mm double entry centrifugal pump moving water at a constant speed of 1350 rpm: Q (m3/min) H (m) η

0 12.2 0

0.454 12.8 0.26

0.905 13.1 0.46

1.36 13.4 0.59

1.81 13.4 0.70

2.27 13.1 0.78

2.72 12.2 0.78

3.8 9.0 0.74

Plot H vs. Q and η vs. Q. If the pump operates in a system whose demand curve is given by H = 5+ Q2, find the operating point of the pump and the power required. In the demand curve, Q is given in m3/min.

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22 20 18 16

H (m)

14 operating point

12 10

Pump Curve

8

System Curve

6 4 2 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Q (m3/s)

100

80

η %

60

40

Efficiency

20

0 0

0.5

1

1.5

2

2.5

Q (m3/s)

m3 H  12.2m ,   0.78 min , QH 2.72 1 Input power =  9810   12.2   6.96 KW  60 0.78 At operating point, Q  2.72

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3

3.5

4

3. With reference to the pump data in Problem 2, if the pump is run at 1200 rpm, find the discharge, head, and required power.

N1  1350rpm , N 2  1200rpm 1

1

N 2 1200 Q2  H 2  2  P2  3      0.89    N1 1350 Q1  H 1   P1  Therefore,

Q2  0.89  2.72  2.42

m3 min

H 2  0.89 2  12.2  9.66m P2  0.893  6.96  4.91KW

4. Water is pumped between two reservoirs in a pipeline with the following characteristics: D = 300 mm, L = 70 m, ƒ = 0.025, ΣK = 2.5. The radial flow pump characteristic curve is approximated by the formula: Hp = 22.9 + 10.7Q – 111Q2 Where Hp is in meters and Q is in m3/s. Determine the discharge Q and pump head H for the following situations: a. Total static head = 15 m, one pump placed in operation; b. Total static head = 15 m, with two identical pumps operating in parallel; c. Total static head = 25 m. System Curve: 2 0.81  fLQ 2  K  Q  H p  Hs     g  D 5 D4 

H p  Hs 

0.81  0.025  70  Q 2 2.5  Q 2     9.81  0.35 0.34 

H p  H s  84.95Q 2 Pump Curve:

H p  22.9  10.7Q  111Q 2 a.

H s  15m

(one pump) 40

Operating point when:

15  84.95Q 2  22.9  10.7Q  111Q 2 Or,

195.95Q 2  10.7Q  7.9  0 QD 

10.7  10.7 2  4  195.95  7.9 10.7  79.41 m3   0.23 2  195.95 391.9 s

Therefore, operating head,

H o  15  84.95  0.232  19.49m b. For two pumps in parallel: 2

Q Q H p  22.9  10.7   111   22.9  5.35Q  27.75Q 2 2 2 Equating this to the system curve,

15  85Qo  22.9  5.35Qo  27.75Q0 2

2

112.8Qo  5.35Qo  7.9  0 2

Qo  0.29

m3 s

Therefore,

H o  15  85  0.29 2  22.2m c. Since the static head is greater than the single pump shut off head (ie. 25 > 22.9), it is necessary to operate with two pumps in series. The combined pump curve is:





H  2 22.9  10.7Q  111Q 2  45.8  21.4Q  222Q 2 The system demand curve is changed since Hs = 25m. It becomes:

H o  25  85Q 2 Equating the pump curve and system curve, we get:

25  85Qo  45.8  21.4Qo  222Qo 2

2

307Qo  21.4Qo  20.8  0 2

Qo  0.3

m3 s

Therefore,

H o  25  85  0.32  32.7m

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5. A pumping system is to deliver 28.3 L/s of water at 15°C. The suction line is 152 mm in diameter in a 91 m long cast iron pipe. The suction inlet is 6 m above the reservoir level. The atmospheric pressure of 101 kPa exists over the reservoir. The required NPSH of the pump is 2 m. Determine whether the system will have a cavitation problem. (Vapour pressure at 15°C is 16.8 kPa, kinematic viscosity of water is 1.14 x 10-6 m2/s).

Patm  Pv

NPSH 



 H s  hL

Q 0.0283 m   1.56 A   s 2  0.152 4 vD 1.56  0.152 Re    2.08  10 5 6  1.14  10  0.00024   0.0016 D 0.152

v

From Moody Diagram, f  0.023

hf 

fL v 2 0.023  91 1.56 2    1.708m D 2g 0.152 2g

Minor losses, (assume 1 exit and 1 bend)

K  0.5  0.9  1.4 Therefore,

v 2 1.4  1.56 2 hm  1.4    0.174m 2g 2g And,

NPSH 

101  1.68  6  1.708  0.174  2.24m 9.81

Since NPSH > 2m (required), there is no cavitation problem.

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