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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