analysis of water systems examples (1)

8/3/2019 Analysis of Water Systems Examples (1)

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Sea water ( = 1025 kg/m 3 ) flows through a rectangular culvert ( 1.6m wide x 2.4

high). At a certain section, the culvert describes a horizontal circular arc of mean

radius 8m. It is found that a pressure difference of 56 N/m 2 exists between the inner

and outer walls of the culvert. Assuming that the sea water flows through the culvert

under free vortex conditions, calculate the flow rate.

v = r

c

i.e. greater velocity at inner radius

Applying Bernoulli between O and I :-

g

po

+

g2

v2

o + z o =g

p I

+

g2

v2

I + z I

g

pp Io

=g2

c2

2

o

2

I r

1

r

1

=

22

2

8.8

1

2.7

1

2

c

1025

56= 0.0032c2

c0032.0x1025

56= 4.132m 2 /s

Now v =r

cand v =

r

c( v is mean celocity)

v = 8132.4

= 0.517 m/s

Q = v x A = 0.517 x 1.6 x 2.4 = 1.984 m 3 /s

Forced Vortex Example

A closed vertical cylinder, 450mm dia & 700mm high containing some water, isrotated about its vertical axis at a speed of 270 r.p.m. When stationary, the water

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inside the cylinder fills the cylinder to a depth of 640mm, the remainder being air at

atmospheric pressure. Calculate the force on the end plates of the cylinder.

= 270 rpm = 2.60

270= 28.274 rad/s

Water volume =4

d 2x h = 0.64x

4

0.45x 2= 0.102m 3

(this volume remains constant, since the cylinder is closed)

When rotating the water profile becomes the paraboloid CDE, now

Vol of water = vol of cylinder vol of paraboloid

i.e. 0.102 =4

45.0x 2x 0.07 -

2

1x x.

2

or x z

5.940 x 10 3 = 2or x z __________ 1

* Note: Vol of paraboloid = vol. of circumscribing cylinder.

Now, z = cg2

r22

+

(Equation 5 in notes p.6)

Using the origin of the parabola as datum, i.e. D (datum) when r = 0, z = 0 c = 0

And when r = ro , then z = z 1

z 1 =9.81x2

rx28.274

g2

r2

0

22

o

2

=

= 40.745 ro 2 ______ 2

Substituting in 1 :-

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5.940 x 10 3 = ro 2 (40.745 ro 2 )

or = 0.140m

z1 = 0.492m

To determine the force on the base and lid, first consider the pressure distribution

(Equation 6 in notes p.6):-

z-c2g

r

g

p 22 +=

As before, c = 0

g

p

= g2

r22

- z _______3

z at lid = 0.492m

The water is in contact with the lid between r = ro and r = r1 ,

Consider annulus, r wide, force on annulus, F = p . 2r r. So to find theforce on the lid, integrate between r = ro and r = r1

i.e. F

L

=

1rr

ror

==

P . 2

r

r.

but P = g

Z-

g2

r22

( From equation 3)

= g

0.492-

g2

r22

LF = 2 g1r

ro rr0.492-g2

r22

= 616381r

ro [ ]0.492r-r745.40 3 r

= 61638 [ ] 225.0 110.024 0.246r-r186.10

LF = 933N

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Now consider the base,

z = 0.208m (-ve since below datum D)

and F B = 2

g

1r

ro

(-0.208)-

g2

r22

rr

=61638 [ ] 226.0024 r104.0r186.10 +

BF = 1934 N

Alternatively the force on the base can be found by adding the weight of the water to

the force on the lid:-

F B = 933 + (4

d2 x h x x g)

= 933 +4

45.0x.2

x 0.64 x 10 3 x 9.81

= 933 + 999

= 1932 N

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Compound Vortex example

A vertical cylindrical drum (int. dia = 500mm) contains water. A set of paddles (dia =

300mm) rotate concentrically with the axis of the drum at 150 rev/min and produce a

compound vortex in the water.

Assuming that all the water in the central core rotates as a forced vortex, determine:-

i the water velocity at a) 60mm radius

b) 200mm radius

ii the depth of the total depression

iii the pressure head at 200 mm radius relative to the centre of the vortex.

=60

150x 2 = 15.708 rad/s

i a) Forced Vortex, where v = r

v 60 = 0.06 x 15.708

= 0.943m/s

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i b) Common radius forced vortex to free vortex = 150 mm

ie) r =c

0.150 =708.15

c

c = 0.353 m 2 /s

200mm radius i.e. free vortex v 200 =r

c

=2.0

353.0

v 200 = 1.767m/s

ii Total depression = y 1 + y 2 =g

r22

=81.9

0150.0.708.15 22= 0.567m

iii 200mm radius free vortex

ie)g

p

= (H-Z) - 2

2

2gr

c____ p/o

on surfaceg

p

= o

Z = H - 22

2gr

c

= 0.567 -2

2

0.2.9.81.2

353.0

= 0.408m

Pressure head relative to centre = 0.408m of water

( P = g h h =g

p

= pressure head)

Flow under varying head Example 1

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A vertical cylindrical tank, 0.6 in diameter and 1.5m high, has an orifice 25mm

diameter in the bottom. The discharge coefficient is 0.61. If the tank is originally full

of water, what time is required to lower the level by 0.9m?

Let tank area = A ; vs = surface velocity; orifice area = a.o ; vo = outlet velocity

Q = A vs = ao vo

Q = Cd ao gh2 - for an orifice

Cd ao gh2 = A .

dt

dh

dt = dhh2gac

A 1/2-

od

Integrating:-

T = )h(h.2gac

A2 2/12

1/2

1

od

=)6.0(1.5x

)62.194

0.025x(61.0

)4

20.6x(22/11/2

2

Time = 192 seconds

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Flow under varying head Example 2

A cylindrical tank; 0.9m diameter, is emptied through a 50mm diameter pipe, 3.6m

long ( both ends being sharp) for which = 0.04.

Find the time taken for the head over the outlet to fall from 2.4 m to 1.2m

1

Bernoulli between 1 and 2:-

z1 +g

p1

+

g2

v2

1 = z 2 + g

p 2

+

g

v

2

2

2 +2gd

vL2

2

+ KLg2

v2

h =

+

+ 5.0

d

L1

g2

v22 KL = 0.5 Entry loss

h =

+dL5.1

g2v2

2

v =

d

L5.1

gh2

+

v 2 = n h 2/1 where n =d

L5.1

g2

+

Q = v1 A = v2 a where a = pipe area

- Adt

dh= n h 2/1 a

dt =n.a

Ah 2/1 dh

t =

na

A2 2/12

2/1

1 hh

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t =

2/1

d

L5.1

g2a

A2

+

( )2/122/1

1 hh

=

( ))2.1(2.4x

0.05

3.6x04.05.1

62.19

4

0.05

0.9x2 2/11/22/1

2

2/4

+

= 0.454x2.117x00196.0

272.1

Time = 139 seconds

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Surge Tank - Example

A 6m diameter surge tank is located at the downstream end of a 1m diameter water

pipeline. The pipeline is 3km long with = 0.02 and a velocity of flow = 3m/s.

Determine the difference in water level between the upstream reservoir and the surge

tank, and the period of oscillation of the water level in the surge tank, when the flowbecomes unsteady.

Using a time interval of 10 seconds and a finite difference integration method,

calculate the water level in the surge tank and the velocity in the pipeline at a time of

40 sec after load rejection.

Level difference between reservoir

& surge tank =2gd

VL 2

= =1.0x9.81x2

3.0x3000x02.0 2

27.523m

Period of oscillation, T = 2ag

AL(pg 9 in notes)

= 29.81x1

3000x62

2

= 659.3 sec

Equations appropriate are:-

dv = -

dg2

vvLz

l

gdt (eqn 5 pg 10 in notes)

dz = v aA

dt(eqn 6 pg 10 in notes)

Equation 6 becomes:-

dz = v2

2

6

110 = 0.278v

andgd2

L=

1.0.81x9.2

3000x02.0= 3.058

Equation 5 becomes

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dv = - [ ]10vv058.3z3000

81.9

dv = - 0.033 [ ]vv058.3z

Initially z o = - 27.523 and v o = 3.0 m/s i.e. at t = 0 secs

dz = 0.278 x 3 = 0.834m;

dv = 0.033 [-27.523 + 3.058 x 3.0 x 3.0] = 0

At t = 10secs, z 10 = -27.523 + 0.834 = -26.689m

v10 = 3.0 0.0 = 3.0m/s

dz = 0.278 x 3 = 0.834m

dv = -0.033 [ -26.689 + 3.058 x 3 x 3] = - 0.028 m/s

At t = 20secs; z 20 = -26.689 + 0.834 = -25.855m

v 20 = 3 0.028 = 2.972m/s

dz = 0.278 x 2.972 = 0.826 m/s

dv = -0.033 [-25.855 + 3.058 x 2.972 x 2.972]= - 0.038 m/s

At t = 30secs, z 30 = - 25.855 + 0.826 = -25.029m

v 30 = 2.972 0.038 = 2.934m/s

dz = 0.278 x 2.934 = 0.816m

dv = - 0.033 [-25.029 + 3.058 x 2.934 x 2.934] = - 0.043 m/s

At t = 40secs, z 40 = - 25.029 + 0.816 = -24.213m

v 40 = 2.934 0.043 = 2.891m/s

etc

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Open channel surge waves: Positive upstream surge example

A rectangular concrete channel, 3m wide, slope = 1in 900 and Mannings roughness

coefficient n = 0.012 conveys a flowrate of 15 m 3 /s. The downstream flowrate is

suddenly reduced to 9m 3 /s by the partial closure of a sluice gate.

Determine the initial depth of flow and the celerity of the positive surge wave.

Assuming uniform flow conditions exist initially then:

Q =n

AR 3/2 S o 2/1 Mannings equation

15 =

2/13/2

1

11

900

1

2y3

y3.

0.012

y3

+

Trial and error gives, y 1 = 2.0m

v1 =1

1

A

Q=

2.315 = 2.5m/s

v 2 =2

2

A

Q=

2y0.3

0.9=

2y

3

Continuity gives:-

A1 (v 1 + c ) = A 2 ( v 2 + c )

3.0 x 2 .0 (2.5 + c) = 3.0 x y 2

+ cy32

c =2

2

2 y

Now, c =( )

1

2/1

1

122 vy

yy

2

gy

+

( ) 5.20.20.2y

2y9.81

22

2/1

22

2

+=y

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Trial and error gives:- y 2 = 2.66m = depth of flow

c = 266.2

2

= 3.03m/s = wave celerity

If the gate is rapidly closed completely, determine the increase in the depth of the

surge wave.

v1 = 2.5m/s y 1 = 2.0 v 2 = 0 y 2 = ?

Continuity:-

A1 (v1 + c) = A 2 (v 2 + c)

3.0 x 2.0 (2.5 + c) = 3.0 x y 2 (0 + c)

c =2y

5

2

c =( )

1

2/1

1

122 vy

yy

2

gy

+

( ) 5.20.20.2y

2y9.81

2y5

2/1

22

2

+=

Trial and error gives y 2 = 3.26m

Increase in depth = 3.26 2.66 = 0.6m

Hydraulic Modelling Example

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A fixed bed model is to be constructed of a 15km length of river (direct length =

10km) for investigation of flood wave flow. At normal winter discharge of 270m 3 /s

it is known that the average depth and width are 4m and 45m respectively. The

Mannings roughness coefficient n, is estimated to be 0.035. The maximum flood

discharge is 910m 3 /s.

If the laboratory is 22m long, calculate :-

i suitable scales for the model

ii scales for horizontal velocity and acceleration in the downstream

direction

iii the maximum discharge required from the circulating pump

Kinematic viscosity of water = 1.15 x 10 6 m 2 /s.

i Max horizontal scale = 1 :22

10000= 1 : 454.5

say 1 : 455

Say distortion of 1 : 5 vertical scale = 1.91

Time scale horizontal

p

m

T

T

= p

m

X

X

2/1

m

p

Y

Y

p

m

T

T=455

1

2/1

1

91

= 1:47.7

ii Horizontal

Velocity scale =m

p

p

m

p

m

T

T

X

X

V

V=

=

1

47.7

455

1= 1 : 9.54

Acc n scale =m

p

m

p

p

m

m

p

p

m

T

T

T

T

X

X

T

T

V

V

T

V ==

=455

1

T

T

X

X2

m

p

p

m =

(47.7) 2 = 1 : 5

iiip

m

Q

Q=

2/3

p

m

p

m

Y

Y

X

X

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p

m

Q

Q =

2/3

91

1

455

1

= 2.5 . 10 6 m 3 /s

-6

m 10x2.5x910Q = = 2.275 x 10 3 m 3 /s = 2.275 /sec

b)

6/1

p

m

p

m

Y

Y

n

n

= e 2/1

2/1

m

p

p

m

6/1

p

m

p

m X

X.

Y

Y

Y

Y

n

n

=

2/16/1

m 455x91

1

91

10.035n

= = 0.037

Check Re m > 800 y = =91

40.044m B =

455

45= 0.099m

R =P

A=

0.0992(0.044)

0.099x0.044

+= 0.023m

p

m

V

V=

2/1

p

m

Y

Y

=

2/1

91

1

= 0.105

Vp =p

p

A

Q=

45.4

270= 1.5m/s

mV = 1.5. 0.105 = 0.158m/s

mRe =

RV=

6-10.1.15

0.023.158.0= 3160 > 800Turbulent & ok

pRe =

RV

45.4

270

A

QVp == = 1.5m/s

610.15.1

396.3.5.1= pR = 5.44.2

45.4

+=

P

A= 3.396

= 4.430 . 10 6+ Turbulent

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analysis of water systems examples (1)

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