caudales ejercicios hidraulica

7/28/2019 caudales ejercicios hidraulica

1/27

1. La distribucin de velocidades en un ro muy ancho de 2.8 m de profundidadvara desde 0.6 m/s muy cerca del fondo hasta 1.9 m/s en la superficie, deacuerdo con la ecuacin:

45.0

8.23.16.0

y

v

Cuales son los coeficientes y cuya inclusin en las ecuaciones de Bernoulliy Momentum respectivamente permite el anlisis unidimensional correcto.

45.0

8.23.16.0

yv

45.08179.06.0 yv

yB

dAv

A

Qv

yBdyByv

8.2

0

45.0

8179.06.0

yB

dyyBv

8.2

0

45.08179.06.0 8.2

8179.06.08.2

0

45.0

dyyv

Resolviendo la integral por Hewlett Packard la integral se tiene:

smv /5.18.2

2.4 smv /5.1

Av

dAv

3

3

yB

dyyB

3

8.2

0

345.0

5.1

8179.06.0

8.25.1

8179.06.0

3

8.2

0

345.0

dyy


102.18.25.1

415.103

sm/102.1

Av

dAv

2

2

yB

dyyB

2

8.2

0

245.0

5.1

8179.06.0

8.25.1

8179.06.02

8.2

0

245.0

dyy


033.18.25.1

51.62

sm/033.1


2/27

2. En una seccin transversal de un puente, las velocidades medidas de flujofueron medidas con los puntos medios de varias subreas, como se

muestra en la figura. Calcule los valores y para la seccin transversal.

Av

dAv

33

Av

Av

3

3

dAAA dAvAvQ

2

16.5

2

4.18m

mmA

smsmmQ /6.5/16.5

32

1

2

2128

2

6.14.1mm

mmA

smsmmQ /6.15/3.112

32

2

2

3168

2

4.26.1mm

mmA

smsmmQ /24/5.11632

3

2

46.178

2

24.2mm

mmA

smsmmQ /16.28/6.16.17

32

4

2

5148

2

5.12mm

mmA

smsmmQ /4.15/1.114

32

2

2

625.5

2

5.17m

mmA

smsmmQ /25.5/125.5 321

245.70 mAA smdAvAvQ /01.94 3

vAQ smvsm

m

sm

A

Qv /33.1/33.1

45.70

/01.942

3


3/27

45.7033.1

25.51141.16.176.1165.1123.16.513

333333

3

3

Av

Av

097.1097.145.7033.1

94.1813

45.7033.1

25.51141.16.176.1165.1123.16.512

222222

2

2

Av

Av

036.1036.145.7033.1

13.1293

3. If a flow of 1.8 m/s is a carried in flume at velocity of 1.2 m/s compute thedimensions of the cross section if it is:

a) Circular with y = 0.72do.b) Rectangular with the width equal to twice the depth.c) Trapezoidal with b = y and side slope 3 horizontal to 4 vertical.

Q = 1.8 m/s

V = 1.2 m/s

22

3

5.15.1/2.1

/8.1mAm

sm

sm

v

QA

vAQ

a) y = 0.72do.

y do

y = 0.72do do

0

00

0

0

5.0

5.072.0

2cos

2

2

2cos

d

dd

d

dy

44.0

2

cos

5.0

22.0

2

cos0

0

d

d 44.0cos

2

1

79.127

053.4180

21.231

rad

rad


4/27

sendA 20

8

1

2

2

2

04776.2

21.231053.4

85.18m

sen

m

sen

Ad

mdmmdmd 57.157.14776.24776.20

2

0

22

0

0

2

1dP mPmP 18.318.357.1053.4

2

1

mRmm

m

P

AR 47.047.0

18.3

5.1 2

2

0

sendT msenmT 416.1

2

21.23157.1

mDmm

m

T

AD 06.106.1

416.1

5.1 2

b) B = 2y

y

2y

yyAybA 222yA

22 25.1 ym

mymm

y 86.086.02

5.1 2

ybP 2yyP 22 mPmmPyP 46.346.386.044

mRmm

mPAR 433.0433.0

46.35.1

2

mDm

m

y

m

b

m

T

AD 87.0

86.02

5.1

2

5.15.1 222


5/27

c) b = y

4 y

3

b = y

22yybA 22yyyA 222 32 yAyyA

5.132 y mym

my 7071.07071.0

3

5.1 2

75.04

3z

2

12 zybP 2

75.017071.027071.0 mmPmP 475.2

mRmm

m

P

AR 6.06.0

475.2

5.1 2

mDm

m

yzy

m

yzb

m

T

AD 85.0

7071.075.027071.0

5.1

2

5.1

2

5.1 222

4. En un canal que conduce un caudal de 9 m/s; existe una transicin desalida, que sirve para unir una seccin rectangular con una trapezoidal,cuyas dimensiones se muestran en la figura.

Indicar cul es la velocidad en la seccin rectangular, considerar que lasprdidas entre la seccin 1 y 2 son solo por transicin, siendo la frmula para

su clculo:

g

vvH

L

2

3.0

2

2

2

1


6/27

g

vyzH

g

vyz

L

22

2

2

22

2

1

11

g

v

g

vv

g

vym

2

3.12

3.02

2.0

2

2

2

2

2

1

2

1

1

yb

Q

A

QvvAQ

11

3

1

37.2

8.3

/9

yy

smv

2

222yzybA 2

2)3.1(5.13.18.5 mmmA

2

2075.10 mA

smm

sm

A

Qv /893.0

075.10

/92

3

2

2

6.19

)893.0(3.1

6.19

)893.0(37.2

3.06.19

37.2

2.02

2

2

1

2

1

1

yy

ym

34.16.19

8.061.5

3.06.19

61.5

2.02

1

2

1

1

yyym


7/27

34.16.19

8.061.5

3.06.19

61.52.0

2

1

2

1

2

1

1

y

y

yym

212

1

2

1

2

1

134.1

6.19

8.061.53.0

286.02.0 y

y

y

yy

2

1

2

13

1

2

134.1

6.19

24.068.1286.02.0 y

yyy

034.10122.00857.0286.02.0 21

2

1

3

1

2

1 yyyy

02003.0013.11

2

1

3

1 yyy

Resolviendo por Hewlett Packard

y1 = 0.63 m

mDb

yb

T

AD 63.063.0

8.3

63.08.3

1

11

1

1

1

smvsmmm

smv /76.3/76.3

63.08.3

/91

3

1

1

1

1

Dg

vFr

151.1

0638.976.3

1Fr Flujo Supercrtico

mmDm

m

yzb

yzyb

T

AD 04.1038.1

7.9

075.10

3.15.128.5

3.15.13.18.5

2

22

22

2

222

2

2

2

2

2

2

Dg

vFr

128.0

04.18.9

893.02

Fr Flujo Subcrtico


8/27

5. Para unas prdidas en el vertedero de 2Nm/N y un caudal unitario de 10m/s, determinar la elevacin del piso para que ocurra un resalto.


9/27

6. Determine the normal discharges in channels having the following sectionsfor y = 1.5 m; n = 0.014, and S o = 1%0

a) A rectangular section 3.2 m wide.b) A triangular section with a bottom angle equal to 90

c) A trapezoidal section with a bottom of 4 m and side slopes 1 : 1.d) A circular section 72 in diameter.e) A parabolic section having an equation y = 0.3 X2.

y = 1.5 mn = 0.014So = 1%0 = 0.001

a) Wide = 3.2 m

1.5 m

3.2 m

2

yzybA 22 8.48.45.12.3 mAmmmA

ybP 2 mPmmmP 2.62.65.122.3

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

smQsmm

mmQ /14.9/14.9001.0

2.6

8.4

014.0

8.4332

13

2

22

b) Angle = 90

X

1.5 m

mxmmxm

x5.15.15.145tan

5.145tan

Z = 1


10/27

2

yzA 222 25.225.2)5.1(1 mAmmA

212 zyP mPmmP 91.591.5115.12 2

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

smQsmm

mmQ /58.4/58.4001.0

91.5

25.2

014.0

25.2332

13

2

22

c) b = 4 m

1 : 1 25.1125.1

VHz

1 1.5 m1.25

4 m

2

zyybA 222 81.881.8)5.1(25.15.14 mAmmmmA

212 zybP

mPmmmP 8.88.8)25.1(15.1242

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

smQsmm

mmQ /91.19/91.19001.0

8.8

81.8

014.0

81.833

2

13

2

22

d) d0 = 72 in

inyinm

inmyinin

05.5905.590254.015.1

)()(

23 in59 in 72 in


11/27

63.0cos236

23

2cos 1

in

in 58.10063.02

58.100

radrad

rad 53.4180

42.259

sendA 20

8

1

22 93.357042.25953.4728

1inseninA

22

2

2

2

)(8.248.24

)12(

193.3570 FtAFt

in

FtinA

Ft

0

2

1dP inPininP 08.16308.1637253.4

2

1

FtAFtni

FtniP

Ft59.1359.13

12

108.163

)(

2

1

0

3

2

49.1 SRn

AQ 2

1

0

3

2

49.1 SP

A

n

AQ

sFtQsFtFt

FtFtQ /64.124/64.124001.0

59.13

59.13

014.0

59.1349.1 332

13

2

22

e) y = 0.3X2

37.137.18.0

5.1 Txx

yTA3

2

xA 3.037.13

222 375.0001.137.13.037.1

3

2mAmA

X > 1

22 1

11

2xxLn

xx

TP


12/27

22 )37.1(137.1

37.1

1)37.1(137.15.0 LnP

mP 718.1

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

smQsmm

mmQ /301.0/301.0001.0

718.1

375.0

014.0

375.0332

13

2

22

7. Using the Manning formula, determine the normal depths in channels havingthe following section when Q = 2 m/s, n = 0.017, S0 = 2%o.

a) A rectangular section 4 m wide.b) A triangular section with a bottom angle equal to 90c) A trapezoidal section with a bottom width of 3.8 m and side slopes 2:1.d) A circular section 1.5 m diameter.e) A parabolic section having an equation y = 0.8 X2.

Q = 2 m/sn = 0.017So = 2%0 = 0.002

a) Wide = 4m

1.5 m

3.2 m

ybA

ybP 2

2

1

0

3

2

SRn

AQ


13/27

2

13

2

3 )002.0(2017.0

/2

yb

ybybsm

044.024

4017.0

4/2

3

2

3

yyysm

3

2

3

)2(

2

)044.0(4

017.0/2

y

yy

m

sm 32

)2(

2193.0

y

yy

Resolviendo por Hewlett Packard:

y = 0.4 m

b) Angle = 90

Z = 1

2

yzA

2

12 zyP

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

2

13

2

2

3

)002.0(12017.0

/2

zy

yzyzsm

2

13

2

2

2

3 )002.0(112

1

017.0

1/2

y

yysm


14/27

2

13

2

2

2

3 )002.0(112

1

017.0

1/2

y

yysm

044.083.2

82.58/23

2

2

23

y

yysm

3

2

2

3

044.05.082.58

/2yy

sm

myyy 175.154.154.1 8 338

c) b = 3.8 m

2 : 12

1

2

V

Hz

1 y2

3.8 m2zyybA

212 zybP

2

1

0

3

2

SRn

AQ 2

1

0

3

2

SP

A

n

AQ

2

13

2

2

22

3

)002.0(12017.0

/2

zyb

yzybyzybsm

044.02128.3

28.3

017.0

28.3/2

3

2

2

22

3

y

yyyysm

044.02128.3

28.3)65.11753.223(/2

3

2

2

2

23

y

yyyysm

Resolviendo por Hewlett Packard:y = 0.372 m


15/27

d0 = 1.5 m

2

0d

y

y d0 = 1.5 m

sendA 20

8

1

01

4

1d

senR

2

1

0

32

SRnAQ 2

1

0

3

2

SPA

nAQ

21

3

2

2

3 002.05.1125.0017.0

5.1125.0/2

sensensm

044.05.1

180

125.0017.0

5.1180

125.0

/2

3

2

2

3

sensen

sm


26.13226.13274.227360360

74.227

75.0

75.0

2cos

y75.075.0

2cos

y

mymy 053.1053.175.075.02

26.132cos


16/27

d) y = 0.8x

yTA 3

2

22

2

832

yTyTR

Si 10 xx

yT

T

yx

44

xTxx

xT

2.32.38.04

2

2

1

0

3

2

SRn

AQ 2

1

0

32

SP

A

n

AQ

2

13

2

22

2

3 )002.0(83

2

017.0

3

2

/2

yT

yTyT

sm

044.08.082.33

8.02.32017.0

8.02.3

3

2

/23

2

222

22

2

3

xxxx

xx

sm

044.012.56.9

38.16

017.0

706.1/2

3

2

42

43

3

xx

xxsm

3

2

2

2

3

3

12.56.9

38.16

044.035.100

/2

x

xx

sm

3

2

2

2

3

12.56.9

38.1645.0

x

xx

Resolviendo por Hewlett Packard:x = 0645

mymyxy 516.0516.0)645.0(8.08.0 22


17/27

8. Un canal trapezoidal revestido de concreto (n = 0.014) de ancho de solera0.5 m, talud 3H:4V y trazado con una pendiente del 1%o, conduce un caudalde 700 Lit/s, en cierto tramo del canal tiene que atravesar una carretera, conuna alcantarilla (n=0.014) de 1.5 m de dimetro.

Para unir la seccin trapezoidal con la circular se construye una transicin conla misma pendiente del canal y de 10 m de longitud. Si las prdidas en latransicin son despreciables, indicar:

a) Cul es la velocidad a la entrada de la alcantarilla?b) Cul es la pendiente con la que se debe trazar la alcantarilla para

conseguir que se establezca el flujo uniforme?

n = 0.014b = 0.5 m3H:4VS0 = 1%o = 0.001Q = 700 Lit /s = 0.7 m/s

d0 = 1.5 m

a)

2

1

0

3

2

SP

A

n

AQ

2

13

2

2 )001.0(12

)(

014.0

)(

zybyyzbyyzb

Q

316.075.0125.0

)75.05.0(

014.0

)75.05.0(/7.0

3

2

2

3

y

yymyymsm


y = 0.67 m

0

3

2

1S

P

A

n

v

001.012

)(

014.0

13

2

2

zyb

yyzbv


18/27

smv /032.1001.075.0167.025.0

67.0)67.075.05.0(

014.0

1 32

2

smv /032.1b) d0 = 1.5 m

2

0d

y

y d0 = 1.5 m

575.0

575.0

2cos

y

575.0

575.067.0

2cos

165.0cos2

1

98.160

360 02.19998.160360

radrad

rad 473.3180

02.199

2

1

0

3

2

SRn

A

Q

21

0

3

2

2

3 5.1125.0017.0

5.1125.0/7.0 S

sensensm

21

0

3

2

2

3

5.1473.3

02.199125.0

017.0

5.102.199473.3125.0/7.0 S

sensensm

21

0

3 273.0857.44/7.0 Ssm

oSSSms

%2.30032.00032.00571.0273.0857.44

7.00

2

0

2

1

0

/3


19/27

9. A channel with the cross section shown in the following figure has aMannings coefficient of 0.040 from station (0) to station (3) and 0.054 fromstation (3) to station (8).

The flow through the channel is 13 m/s, and the water surface is 1.7 m high.

Find the following:

a) Weighted Mannings Coefficientb) Slope of the channel.c) Top channel.d) Wetted Perimeter.e) Flow regime (supercritical or subcritical)

a)

32

1

5.1

Pt

pnn

n

iii

e

P1 =X

P1 0.2

mxx

4.01

5.02.0

mPP 447.0447.04.02.01

22

1


20/27

P2 = 1 m

P3 =

1.5 P3

1

mPmP 8.18.115.13

22

3

P4 = 2 m

P5 =

P5 1

1

mPmP 41.141.111 522

5 P6 = 1 m

P7 =x

0.7 m P7

mxx 7.01

1

7.0

mPmP 99.099.07.07.07

22

7

PTotal = 6.937m

5.15.15.15.15.15.15.1

937.6

054.01054.041.1054.02054.08.104.0104.0447.004.0

en

0567.0937.6

09373.0 32

en


21/27

b)

2

121*2 mA 2

275.0

2

5.1*1mA 2

35.0

2

1*1mA

2

48.27.0*4 mA 2

54.02*2.0 mA 2

604.0

24.0*2.0 mA

2

7245.0

2

7.0*7.0mA

ATotal = 6.735 m

2

1

0

3

2

SP

A

n

AQ

2

1

0

3

2

2

22

3

937.6

735.6

0567.0

735.6/13 S

m

mmsm

%21.10121.00121.011.078.118*98.0

/130

2

0

3

2

1

0 SS

smS

c) mTT 1.77.04.06

d) mPP 647.899.0141.128.11447.0 e)

Dg

vFr

mm

m

T

AD 948.0

1.7

735.6 2

smvsm

m

sm

A

QvvAQ /04.2/04.2

375.6

/132

3

66.0948.0*8.9

/04.2

smFr

166.0Fr Flujo Subcrtico


22/27

10. A stream with the cross section shown in the previous figure has a flowrate of 5 m/s. The stream has a longitudinal slope of 0.002m/m and anatural stony bottom (n = 0.05, stations 0 to 8)

a) Using Mannings equation, what is the water surface elevation of thestream?

b) What is the maximum capacity of the channel?c) How would the capacity of the channel be affected if you were to

pave the center of the channel (n = 0.013) between stations 3 and 5?

Q = 5 m/sn = 0.05S0 = 0.002

a)

Suponiendo un y = 1.5 se tiene:

mP 8.15.1122

1

mP 22

mP 41.111 22

3

mP 14

mP 7071.05.05.0 22

5

mmmmmmPTotal 924.67071.0141.128.1 2

121*2 mmmA

2

275.0

2

5.1*1m

mmA


23/27

2

35.0

2

1*1m

mmA

2

425.0*4 mmmA

2

5125.0

25.0*5.0 mmmA

222222 375.5125.025.075.02 mmmmmmATotal

2

1

0

3

2

SP

A

n

AQ

smQ /0607.4)002.0(

924.6

375.5

05.0

375.52

13

2


mP 8.15.11 221

mP 2

2

mP 41.111 223

mP 1

4

mP 99.07.07.0 225

mP 1

6

mP 447.02.04.0 227

mmmmmmmmP

Total654.8447.0199.0141.128.1

2

121*2 mmmA

2

275.0

2

5.1*1m

mmA

2

35.0

2

1*1m

mmA

2

48.27.0*4 mmmA

2

5

245.02

7.0*7.0m

mmA

2

64.02.0*2 mmmA

2

704.0

2

2.0*4.0m

mmA

22222222 735.604.04.0245.08.25.075.02 mmmmmmmmATotal


24/27

2

1

0

3

2

SP

A

n

AQ

smQ /09.5)002.0(

654.8

735.6

05.0

735.62

13

2

Interpolando los dos datos se tiene:y Q

0607.45.1

09.57.1

mysmQ 68.1/5 3


mP 8.15.1122

1

mP 22

mP 41.111 22

3

mP 14

mP 96.068.068.0 22

5

mP 16

mP 40245.018.036.0 22

7

mmmmmmmmPTotal

572.840245.0196.0141.128.1

2

121*2 mmmA

22 75.0

2

5.1*1m

mmA

2

35.0

2

1*1m

mmA

2

472.268.0*4 mmmA

2

52312.0

2

68.0*68.0m

mmA

2

636.018.0*2 mmmA

2

70324.0

218.*376.0 mmmA

22222222 5936.60324.036.02312.072.25.075.02 mmmmmmmmATotal

2

1

0

3

2

SP

A

n

AQ

smQ /947.4)002.0(

57245.8

5936.6

05.0

5936.62

13

2


25/27

Interpolando Nuevamente se tiene:y Q

9475.468.1

09.57.1

mysmQ 688.1/5 3

Suponiendo un y = 1.688m se tiene:

mP 8.15.1122

1

mP 22

mP 41.111

22

3

mP 14

mP 97.0688.0688.022

5 mP 1

6

mP 42038.0188.0376.0 227

mmmmmmmmP

Total61.842038.0197.0141.128.1

2

121*2 mmmA

2

275.0

2

5.1*1m

mmA

2

35.0

21*1 mmmA

2

4752.2688.0*4 mmmA

2

52366.0

2

688.0*688.0m

mmA


26/27

2

6376.0188.0*2 mmmA

2

70353.0

2

188.0*376.0m

mmA

22222222

65.60353.0376.02366.0752.25.075.02 mmmmmmmmATotal

2

1

0

3

2

SP

A

n

AQ

smsmQ /5/006.5)002.0(61.8

65.6

05.0

65.632

13

2

OK

y = 1.688m

b) y = 2 m se tiene:

mP 8.15.11 221

mP 2

2

mP 41.111 223

mP 1

4

mP 41.111 225

mP 1

6

mP 118.15.05.0 227

mmmmmmmmP

Total738.8118.1141.1141.128.1

2

121*2 mmmA

2

275.0

2

5.1*1m

mmA

2

35.0

2

1*1m

mmA

2

441*4 mmmA

2

55.0

2

1*1m

mmA

2

615.0*2 mmmA

2

725.0

2

5.0*1m

mmA


27/27

22222222 925.015.045.075.02 mmmmmmmmATotal

2

1

0

3

2

SP

A

n

AQ

smQ /21.8)002.0(

738.8

9

05.0

92

13

2

c)

32

1

5.1

Pt

pnn

n

iii

e

3

2

5.15.15.15.15.15.15.1

61.8

97.005.0105.041.105.02013.08.105.0105.04204.005.0

en

ne = 0.0089

2

1

0

3

2

SP

A

n

AQ

smQ /13.28)002.0(

61.8

65.6

05.0

65.6 2132

smQ /13.28

caudales ejercicios hidraulica

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