Chapter 1 Fluid Flow

7/21/2019 Chapter 1 Fluid Flow

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NM324 Principal and Application of

Marine Machinery

Introduction:

10 hrs from Peilin Zhou

10 hrs from Gerasimos Theotokatos

Exam: End of semester 2

Final mark: 30% from CW + 70% from exam

Objective

To provide an understanding of major marine

machinery components and systems, their working

principles, design concepts and assembly drawings.

Topics to be covered (PLZ Part)

1.Fluid and pipe flow

2.Centrifugal pumps, including matching of

pumps with systems

3.Hydraulic systems

Recommended Reading

1. Mechanics of Fluid, B. S. Massey

2. Marine Auxiliary Machinery, H. D. McGeorge


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Chapter 1: FULID AND PIPE FLOW

Reference book:

Mechanics of fluids, sixth edition, BS Massey

Marine engineering systems consist of largely

pipe systems performing the functionality of

ships. Pipe systems include water, fuel, oil,

gas/steam and cargo.

The chapter concentrates on fluid flow in pipe

systems, including flow resistance and

calculation in different pipe systems, starting

with the basics of Bernoullis equation.

Bernoulli's Equation

.22

22

2

2

2

2

21

2

1

1

1

2

2

2

2

21

2

1

1

1

constzg

ug

pzg

ug

p

gzup

gzup

=++=++

++=++

or .2

2

constzg

u

g

p=++

Physical meanings:

Z Z

pP

u a ua1

2

2

22

1

1

1

12

Datum


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hg

p=

(m) -- pressure head

g

u

2

2

(m) -- velocity (kinetic) head

z (m) -- potential head

Total head = .2

2

constzg

u

g

p=++

Laminar and turbulent flow

Laminar Flow

Definition: individual particles of fluid follow

paths that do not cross those of neighbouring

particles. Therefore there is a velocity

gradient across the flow.

Main features: flow velocity is very low and

viscous force of flow predominates over the

inertia force.

u

y

Laminarflowalongaflatsurface


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Flow resistance:

Newton's law: y

u

=

ordy

du = for one dimension

flow

where: -- shear stress (N/m2)

-- dynamic viscosity (Ns/m2)du/dy -- velocity gradient.

For flow in a pipe:r

u

=

Flow resistance: F = Awhere: A -- total area, for a pipe A = dl

Turbulent Flow

Definition: random fluctuating components

are superimposed on the main flow in a pipe.

Reynolds number: Re =

udud=

u

Laminarflowinapipe


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where: Re-- Reynolds number (non-

dimension)

-- density of the fluid (kg/m3)u -- velocity (m/s)

d -- diameter of the pipe (m)

-- dynamic viscosity (Ns/m2)= /-- kinematic viscosity (m2/s)

Experiments show that:for laminar flow, Re < 2000

for turbulent flow, Re > 3000

transition flow, 2000< Re < 3000

Head loss in a pipe (turbulent flow)

Darcy equation:

g

u

m

lfh

f2

2

=

where:

f -- friction factor / coefficient

l -- length of the pipe

u -- average velocitym -- hydraulic diameter

m = cross-section area / perimeter in

contact with fluid

For a pipe with an internal diameter of d

4//4

2

dddm ==


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then,g

u

d

lfhf

2

4 2=

Friction factor (f)

f is a function of Reynolds number and the

relative roughness (k/d). where k is called

absolute roughness, ie. the mean diameter of

the grains attached on the inner surface of thepipe.

MOODY DIAGRAM

For Re < 2000,

f = 16/Re (64/Re)

For a smooth pipe when 2000< Re < 100000,

f = 0.079 4/1Re -- Blasius's formula.

For the entire range of Re and k/d,

})71.3

(Re9.6{log6.31 11.110

dk

f+=

-- S.E Haaland

formula

Accuracy of the above equation is 90-95 %.


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Example 1, 2, 3 (page 204, Massey's 6th

edition)

Other head losses in pipes

a). Loss at abrupt enlargement

g

uu

hl 2

2

2

2

1 =

b). Loss at abrupt contraction

g

ukhl

2

2

2=

c). Loss at exit

g

uh

l2

2

1=

d). Losses in

u u1

2

d d2 1/ 0 0.2 0.4 0.6 0.8 1.0

k 0.5 0.45 0.38 0.28 0.14 0

u=0u1

2


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pipe fittings

g

ukhl

2

2

=,


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Fluid power

Work = force x distance

= p A l

= p V

Power =work/time

=p V/t

where: Volume flow rate Q = V/t [m3/s],

Also, Q = V/t = (A l)/t = A (l/t) = A u = area x

flow velocity

Question:For Q4 in the tutorial sheet, what is the

minimum power required for a pump to

transfer water from reservoir 2 to reservoir 1 at

a flow rate of 50 m3/h?

Transmission of Hydraulic Power By

Pipeline

Assume:

H (m) - total head supplied to the inlet

hp(m) - head at the outlet end

A


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hf(m) - head loss by friction

Q (m3/s) - volume flow rate (discharge)

Then,

Power supplied at inlet = gHQPower available at outlet = ghpQ

Efficiency of transmission:

H

h

gHQ

Qgh pp==

h

where: H = hp+ hf

hp= H -hf

H

hf=1h

Variation of hp, power and with Q

hfis proportional to the square of flow

velocityi.e. hfu2, or hf= cu2


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when Q = 0, u = 0, thus hf= 0

Power = gQhp= gQ(H - hf)

= gAu(H - cu2)

where: Q = Au, and hf= cu2

when Q (or u) = 0, Power transmitted = 0

Power transmitted , as Q when cu2= H, Power transmitted = 0

Power

Q

Pmax

Let: 0=du

dPower, gives, H - 3cu

2= 0

Q

hp

hf

QmaxQ


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then, 3 hf= H, or hf=3

1H

Max. power transmission occurs when hf= 3

1

H,

or hp=3

2H.

Multip Pipe systems

1. Pipe in series

Total loss:

Entry loss,

Friction loss in pipe 1,

Abrupt enlargement,

Friction loss in pipe 2,

Exit loss.


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g

u

g

u

d

lf

g

uu

g

u

d

lf

g

uH

2)

2

4(

2

)()

2

4(

25.0

2

2

2

22

21

1

22

1 ++

++=

2. Pipes in parallel

Q = QA+ QB

When steady flow is established:(hf)A = (hf)B

orBA

g

u

d

lf

g

u

d

lf )

2

4()

2

4(

22

=

lA = lB

thenB

B

B

A

A

Ad

uf

d

uf

22

=

fA= f(Re, k/d), and

lu=Re


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The trial and error method should be used to

solve the problems involving parallel flow.

Pipe Networks

A

B

CD

Za

Zc

Z

Zd

ZbJ

Qa

Qb

QcQd

Piezometer

J

Assume the directions of flow in pipes to be

toward junction J.

At Junction J:

Qa+ Q

b+ Q

c+ Q

d= 0

gd

Q

d

lf

g

u

d

lfhZZ

a

a

a

a

a

a

a

a

aafJa2

16)(4

2

4)(

42

22

===

where:4

2ad

a

a

aa

Q

A

Qu

==

then, 42

2

2

16 a

a

a d

Q

u =


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Thus, 252

2

2

64aa

a

aaa

Ja Qkgd

QlfZZ ==

where: gd

lf

ka

aa

a2

6452=

Now: 2aaJa QkZZ =

2

bbJb QkZZ =

2

ccJc QkZZ =

2

ddJd QkZZ =

0=+++

dcba QQQQ To solve 9 unknowns with 5 simultaneous

equations, error and trial method should be

used

For each pipe: h = kQ2, and k = h / Q2

then, dQQhkQdQdh 22 ==

Procedure:

1. Assume an estimate ZJ, with the given

conditions.

2. Calculate ka, kb, kc, and kd3. Calculate ha, hb, hc, and hd by ha= Za -

ZJ, etc.

4. Calculatea

aa k

hQ = , etc.

5. If 0=+++dcba

QQQQ , then, the problem is

solved.


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6. If QQQQQdcba

=+++ , then, correct ZJas

follow:

d

d

d

c

c

c

b

b

b

a

a

a

dcba

dhh

Qdh

h

Qdh

h

Qdh

h

Q

QQQQQ

2222+++=

+++=

At J, dh = dha,= dhb,= dhc, =dhd

then, dhQd

aid

Q

i

i==2

1

and,

=

=

d

aid

Q

i

i

Qdh

2

7. Assume a new head at J by ZJ= ZJ+ dh,

and go to step 3 until 0Q .

Example: Four pipes from reservoirs meet at

a point J, viz.:

Pipe Reservoir level

above datum (m)

Pipe

length

(m)

Diameter

(m)

f

a 100 3000 1.5 0.004

b 110 6000 1.00 0.00

7

c 80 3000 1.0 0.00

6


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d 40 10000 2.0 0.00

4

Determine the flow in each pipe, and thepressure at J.

Solution:

40m < ZJ< 110 m, let ZJ= 80 m first. Then,

Pipe k h(m) Q= h k Q/h

a 0.522 100-80=20 6.19 0.31b 13.88 110-80=30 2.16 0.072

c 5.95 80-80=0 0 0

d 13.22 40-80=-40 -3.025 0.075

Q =

5.325Q h/ =

0.457

mhQ

Qdh 27.23457.0/325.52

)/(2

==

=

Make the second estimation:

ZJ= 80 + 23.27 = 103.27 m

Pipe h(m) Q= kh Q/ha -3.27 -25.03 0.765

b 6.73 0.696 0.103

c -23.27 -1.978 0.085

d -63.27 -2.188 0.034

Q

=-5.973 = 0.988


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dh = 2 x (-5.973)/0.988 = -12.09 m

Third estimation

ZJ= 103.27 - 12.09 = 91.18 m

Pipe h(m) Q= h k Q/h

a 8.82 4.11 0.5

b 18.87 1.165 0.062

c -11.18 -1.370 0.122d -51.18 -1.967 0.038

Q=1.93873 = 0.7227

dh = 2 x 1.938/0.7227 = 5.367 m

Take the 4th estimation mdhh 681.22/ ==

ZJ= 91.18 + 2.681 = 93.86 m

then, Q = 0.965

Ans. Qa=4.11; Qb=1.165; Qc=-1.37;

Qd=-1.967 m3/s; and ZJ= 93.86 m

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Chapter 1 Fluid Flow