Fox J.A. Transient Flow In Pipes, Open Channels And Sewers
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96
Boundarv conditions
Dividing
top
and bottom
by
(N/N-)2
gives:
Wr: sign(T)
lrtr.l
(N/N-)2
+
(QlQ.)'
Wn:sign(ry)
J1+%
where
v:
HlN2fromgHlN2Dz
andolNfrom
QlNDt.
Also
lch.4
J
Again, dividing
top and
bottom
by
(N/N-)2
gives:
wr:sign(I)
J-ffi
where L:
TlNz
from
TpN'Dt.WrandWTcan
then
be
plotted
against
0
where
0
is
given
by
o:arctan(#)
This
is the same
as 0
:
arctan(<D/{D*).
4.7.1 The use
of the Suter
method
Figs. 4.5
and
4.6
contain
the
same
information
as is
provided in
Fig.
4.4. The
reduction
in complication
is impressive
and
clearly
these
two
graphs
can
be stored
in
a computer
as a
pair
of
arrays.
If
each
quadrant is allotted
15
points for
the
description
of
the shape
of
the
graph within the
quadrant,
quite an
accurate
definition of it
will be
possible. Thus
two arrays
containing
60
real
variable
values
each
will
be sufficient
to
define
the
characteristic
curves
of
the
punp.
As
these
points
will be located
at 6o
intervals,
linear
interpolation
between
adjacent
points will
give
sufficient
accuracy
for any
engineering
Purpose.
There are two
cases
to
consider
when
performing a four-quadrant
pump analysis:
these
are start-up
from
zero
speed and shut-down,
or
pump trip,
from
steady
operational
speed. Of
course,
it is
possible
to
start up
from
zero speed,
to
run
up
to
sieady
operaiional
speed
and
after a
period
at
this speed
to
perform
a
pump trip.
Start up from zero speed
requires
a little
thought
as the speed
will be initially
zeto
and
so
will
the flow.
Then
0
will
be undefined.
However,
when the
pump
motor
is started,
the zero-speed motor
torque
is available
to accelerate
the
pump
impeller
but
the
pump
will not
deliver
flow
until it
has
reached a specific
minimum
speed,
i.e. until
it
develops
a
no-flow
head equal
to
the
head on the
pump.
Sec.
4.71
The Suter
presentation
Fig.4.5.
97
WH
1.0
0.8
0.6
o.4
o.2
0
-o.2
-0.4
-0.6
-0.8
-1.0
WT
1.0
0,8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
Fig.4.6.
98
Boundary conditions
The
pump
torque
can be obtained by
determining
the
value of
conditions at
start
up and
then calculating
the torque
as follows:
lch.
4
from the
r,
:
r*w
r,tlw,,1
[
(#)'
.
(#)']
Simiiarlv
H
r
:
H*w
n,tlw
u,lt
(#)'
.
(&)'l
From the
torque equation
for the
pump
set and
motor:
T^- Tr:
I{2
To obtain
the
value of
I-, it
is necessary
to have an
array of T*
values for the range
of motor
speeds
from
zero
to
operational
speed
with a constant speed
increment.
This increment
should
be small
if the operational
steady final
speed
is to
be
accurately
predicted.
Given
the
value of N,
then T-
can be
obtained
from
this array
by interpolation.
So
and so
Nz:Nt-
60
A(r-
-
Tr)
At the speed
N, the
value of
T-can be obtained
from
the motor torque
array;
Trcan
be found
from
the
W.curve
using the
value
of 0
appropriate
to the speed
Nt
and flow
Qr.
The
value
of N2
can
then be
readily calculated.
The
value
of the
head
upstream
and downstream
of
the
pump
togetherwith
the new
flow through
the
pump
can
then
be found as follows.
The
positive characteristic
invariant
for the
upstream side of
the
pump is:
(hr".n- h^)+ft(r",.n-
un)
*#:o
where dr",,
is the
diameter
of the
suction
pipe;
scn
denotes suction.
The negative
characteristic
invariant
for the downstream
side of
the
pump
is:
2Ir
The
Suter
presentation
99
-
(hro"r-
/r")
*
3(r"o",
-
us)
*
t#tr:
O
where
d6.,
is
the
diameter
of the
delivery
pipe;
del denotes
delivery.
Adding
these
two expiJssions
gives
-
(ft"o.,
-
ft"".r,)
in
terms
of
D.o.,
and
u"".,,'
Now
hro"r-
h"'"n
is
I{,
the head
generated
by
the
pump;so
this equation
can
be
rewritten
in
the
form
Sec.
4.71
where
and
":**foo.,
-H+
nQ+
F:0
This
equation
together
with
theWrversus
0
graph
and
the
definition
of
0,
i'e'
0
:
arctan(NQ.lN"Q\
gives
the
solution
for
the
new
point of
operation
of
the
pump'
Because
N is
know-n
(it
is
N,
calculated
as
shown
above)
but
Q
has
not
yet been
found,
the
process
of
solution
has
to
be iterative.
First
setNto
N2
and
Qto
Q.t'
Next
evaluate
0.
Interpolate
in
thewr,
afiay
for
the
new
value
of
wr, evaluate
H and,
using
the
above
equation,
calculite
the
value
of
Q.
Another
value of
0
can
now
be
obtined
using
this
new
value of
Q
and
the
process iterated
to
give new
values
of
I/,
I
and
e.This
piedictor<orrector
process cin
be
repeated
until
successive
values of
0
have
no
significant
difference.
If ihe
sequence
includes
the
recalculation
of
Nr,.using
the new
n"lu",
of
the
pump
torque,
there
will
be only
a small
degree
of
approxima-
tion
in the
entire
ptoiett.
For
this
technique
to
work
a iet
of
four-quadrant
pump
characteristics
is
required.
More
and
morspurnp
manufacturers
are supplying
them
and
they
can
be
emptoyed
as
illustrated
above
but
this
is
only
true of
larger
pumps.
It
is
very iiffrcutt
to
oUtain
four-quadrant
pump characteristics
for
smaller
pumps.
A
suggeition
of
how
to
handle
this
problem
is to
use
existing
four-quadrant
pump
characteristics
for
different
pumps
which
have
different
specific
speeds'
It
may
be
found
acceptable
to
interpolate
bitween
these
graphs
using
the
pump's
specific
speed
as
the
basis
of
the
interpblation.
Next
the
values obtained
from
the
manufacturer's
pump characteristic
curve
may
be
plotted onto
the
same
interpolated
diagram
and,
if
the
results
from
the
characteristiC
curve
match
those
obtained
from
the
interpola-
tion,
it seems
reasonable
to
assume
that
the
interpolated
result
will apply
throughout
the
entire
four-quadrant
range.
To
assist
users
of
this
type
of
technique
to build
up
a
library
of
four-quadrant
pu*p
characteristics,
Table
4'1
is
supplied
(based on
Donsicy's
[13]
paper
which
givei
the
necessary
data).
This
gives
the Suter
diagramsof
three
pu-pr
*tti.tr,
between
them,
cover
much
of
the
range
of
specific
speeds'
The
100
Boundary conditions
Table
4.1
lch.4
0
(rad)
Nr:35
N.: 147
N': 261
wH
wH
wH
wr
w7:
wr
0.
0.168
0.318
0.4u
0.588
0.69s
0.785
0.876
0.983
1.t07
1.249
I.406
1.57L
I.736
1.893
2.034
2.L59
2.266
2.3s6
2.447
2.554
2.678
2.820
2.976
3.L42
3.307
3.463
3.60s
3.730
3.836
3.927
4.018
4.124
4.249
4.39L
4.547
4.712
-0.728
-
0.548
-
0.639
-
0.394
-
0.445
+
0.095
-0.L79
+0.400
+
0.398
+ 0.545
+ 0.576
+ 0.&4
+0.707
+0.707
+
0.806
+
0.745
+
0.904
+
0.772
+0.992
+
0.785
+ 1.069
+
0.771.
+ 1.120
+
0.725
+ 1.136
+ 0.663
+I.I29
+
0.608
+1.102
+ 0.585
+ 1.107
+ 0.587
+ 1.039
+ 0.606
+ 1.010
+ 0.661
+0.997
+0.721
+0.979
+
0.777
+0.947
+ 0.831
+ 0.939
+
0.885
+ 0.901
+
0.926
+
0.876
+
0.940
+ 0.831
+
0.927
+0.789
+ 0.887
+0.754
+ 0.828
+0.727
+ 0.743
+
0.710
+
0.654
+
0.709
+ 0.565
+0.7tL
+ 0.480
+0.721
+ 0.376
+ 0.740
+ 0.263
+0.764
-
0.155
+
0.788
-
0.379
+ 0.801
-
0.600
+0.794
-
0.819
-
r.249
-
1.249
-
1.048
-
0.951
-
0.789
-
0.651
-0.529
-0.297
+
0.186
+0.477
+ 0.555
+ 0.630
+ 0.707
+ 0.707
+0.791
+ 0.807
+
0.881
+ 0.807
+
0.984
+
0.853
+ L.094
+ 0.939
+
L.216
+ I.07L
+ 1.400
+1.217
+ 1.405
+
1.240
+ 1.479
+
t.244
+ 1.505
+1.274
+
1.536
+
1.308
+ 1.573
+ 1.381
+ L.624
+ 1.M2
+ L.674
+ 1.535
+
I.703
+ 1.594
+L.725
+ 1.650
+
1.700
+ 1.658
+ L"620
+ 1.580
+ 1.473
+ 1.450
+ I.241
+ 1.235
+ 0.996
+ 1.018
+0.785
+ 0.815
+ 0.&4
+
0.622
+ 0.528
+0.428
+0.624
+ 0.000
+ 0.335
-0.414
+0.204
-
0.564
-
0.310
-
0.709
-
0.502
-
0.843
-0.669
-
1.030
-
0.819
-
r.225
-
0.707
-
0.748
-
0.935
-
0.776
-
0.828
-
0.736
-
0.632
-
0.559
-
0.n6
+
0.144
+
0.468
+ 0.550
+0.707
+ 0.707
+ 1.403
+ 0.861
+ 1.043
+
0.861
+ 1.187
+ 0.951
+ 1.348
+
r.L02
+ 1.506
+ I.275
+t.652
+ 1.400
+ t.784
+ I.520
+ 1.864
+
L.627
+ 1.891
+ I.713
+ L.873
+ L.741
+ 1.803
+
I.761
+ 1.809
+ 1.660
+ 1.689
+ 1.596
+ 1.576
+ L.477
+
1.470
+
I.342
+ 1.350
+ 1.201
+ 1.040
+
0.818
+ 0.887
+
0.646
+
0.839
+0.644
+
0.785
+ 0.710
+
0.680
+ 0.610
+
0.510
+ 0.326
+
0.255
-
0.274
-
0.407
-
0.570
-
0.645
-0.763
-
0.829
-
0.938
-
1.013
-
1.082
-
r.228
-
1.240
-
1.480
-
r.526
Sec.
4.81
Pumps
equipped
with
non-return
valves
pump can
be
tested
after
installation
and
results
obtained
from
it
can
be
plotted
back
onto'the
graphs. Hence
a library
of
graphs
for
pumps of
different
specific
speeds
can
be
develop"i.
tt
should
be remembered
that
pumps of
the same
specific
speed
do
not
necessarily
have
identical
characteristics'
4.8
PUMPS
EQUTPPED
WITH
NON-RETURN
VALVES
pumps
equipped
with
a non-return
valve cannot
operate
in
the
second
or
third
quadrant'U"ii,rt.
the
flow
can never
reverse.
They are
very unlikely
ever
to
get into
the
fourth
quadrant
as
to
do so
they
would
have
to be
driven
backwards.
Essentially
they
operaie
in the
first
quadrant
only.
Thus
only
the
TL,
DI
and
P1 zones
are
of
of the
ZL
and the
DI
zones.
From
the
head
versus
flow
curve
the Stepanov
equation
can
be
derived-
This
equarion
is f/
:
AN2
+ BNQ
-
CQ'
and
is
derived
as follows.
If the
assumption
is
-ud.
that
the
pump is
a
radial
flow
type,
the
head
Il is
given
by
kP?
k"D'
O*:U-DfCOtY
where
u: nDN/60.
D is
the
impeller
diameter
and
N is the
speed
in revolutions
per
minure.
u, is the
velocity
of
flowand
is equal
to
QlA"where
Q
is
the
flow
and.4"
is
the
area
of
exit.
v2:
u?
+ u?,
and
101
2g
2g
H:u*u
g
lo2
Boundary conditions
ur: DfCOSeCT
where
y
is the exit
angle
of
the blades.
So
H: ANz
+ BNQ
-
CQ'
where
2(u-
urcotl)u
k1.t!coseczy
k"[u?
+
(u-
urcotl)2]
'-
2g
-
u
-
,g
After
reduction
this
gives
lch.4
and
^
(ki+k,)cosec2l
-:
zgAJ
-
'
are
not available.
The
fi
zone
can
also
be solved
using
the
values of
A, B
and
C and
the
manipulation
following
later.
The Pl
zone
can
readily
be solved.
Write the two
characteristic
equations
for
the
dr
lengths
adjacent
to
the
PumP:
h",.n-
h**f,{r",.,
-
un)
*
T:
O
and
Sec.
4.8]
Pumps
equipped
with
non-return
valves
-
(hro.r-
hr)+
1(r.o.,
-
u")
+
F""
:
g
8
"--'
I
103
by addition
an
expression
for
hro.r-
hr,.nis
obtained
and
this
equals
the
expresslon
for
the
head
across
the
PumP:
H:
ANz
+ BNQ
_
CQ,
where
H:
h"o"r-
hrr.n
This
gives an equation
in
which
Q
and
Dp
appear
as
variableS.
uP,.,,
and
up6.1
can,
of
aour*,
be expr-esSed
as
QlAr.nand
QlAa.r
where
Ar.,,
and,4o",
are
the
areas
Of
the
suction
pipe and
the
delivery
pipe of
the
pump
respectively.
Therefore
the
equation
is
a
quadratic
in
e
andso
is
easily
solved.
It
is
vitally
important
to
choose
the correct
sign
in solving
the
quadratic
equation:
Q:
-B+J(0'-asr)
2a
where a,
p
and
y
are
the
constants
in
the equation
resulting
from
the
reduction
of
the
pump
equation'and
the
two
characteristic
equations.
Having
solved
the
flow
and
it"uO
it is
possible
to obtain
the
water
power
wQH,
and
this
divided
by
the efficiency
E
will
give the
power supplied
to
the
impeller
by the
motor.
To
do
this
it
is necessary
to
havl
un
"quution
for-E.
The
output
power of
the
pump
is
given
by
P:
wQ(AN2
+ BNQ
-
CQ')
'
The
input
power can
be
written
as
P,:
weV
*
Po
T are
constants.
Then
104
Boundarv conditions
ANZQ+
BNQZ-CQ'
RN'Q+SNO2+fN,
lch.
4
A, B and Ccan
be
readily determined
from the steady state,Flversus
Q
characteristic
curve
obtainable
from the
pump
manufacturer.
R, S
and Tcan be obtained
from the
efficiency
versus
p
characteristic
that is usually
supplied
with
the
I{
versus
O
curve.
Then using current
values for
N and
Q
the
value of E can be
readily calculated
within
the
pump
subroutine.
This
will
permit the
value
of
the shaft
power wQHIE to be
determined
and this,
in turn,
will
permit
the
speed
at the end of the
next
Ar to be
found. To do
this,
the technique
given
later
can be used using
the shaft
power
in
the
place
of
the
variable
Pwr. Thus
the solution
of the
Plzone is simple
and a subroutine
in a computer
program to solve
a network
containing
a
pump
or
pumps for the PL
zone
is
easy to
write. This
same
piece
of
programming
can
be used
to solve the
Dl
zone
because, in
this case,
the only
difference
between
the two cases is
that in
the DL
zone the head across
the
pump is negative.
The solution
can handle
a negative
head
with no
difficulty.
Clearly
the
productwQH which is the rate
at which
power is being
transferred to the
water by
the
pump
is negative
when FIis negative
and so
the
pump
must be absorbing
energy
from
the water flowing through
it.
This is the
definition
of
dynamometer
action.
The
nature of
the flow
in this condition
must
be
clearly
differentiated
from
that
when in
the 11
(turbining)
mode.
When the
pump
is being
driven
by its
motor
and
is
generating
a negative head,
it is absorbing
energy
from
the
flow
and
destroying
it.
When the
pump
is not being driven
by
its motor, the
pump
can
absorb
energy
from
the flow
and
be accelerated
by it or,
perhaps, not slow
down so
rapidly-as it
would
do if
the
water were not
driving it. Remember
that
a
turbining
pump
is being
driven
by the
flow through
it but
it is also
destroying some
energy
by
the action
of friction
between
the fluid and
the surfaces
inside
the
pump,
by the losses
occurring in bearings,
windage in
the driving
motor and
losses caused
by
boundary
layer
separations
in the
flow
within the
pump.
4.E.1 Forwardturbining
When
a
pump
is
turbining,
the
head across
it must
be negative;
this assumes
that
normal
pumping,
gives
a
positive head. For
normal
pumping
the delivery
head
must
be larger than the
suction
head,
i.e. h*r- h,.nmust
be
positive.
For
turbining,
this
head must
be
negative.
The head equation
can
now
be
written
for the
turbining
case:
-
tf
-Y
+k-'A +
k..*
I
'
^R28
'*"28
This
is
a statement
that
the
energy
of
the flow
per
unit
weight available
to drive
the
flow through the
pump
is
partly used
to
give
energy
to the
pump
itself
(the
u*ulg
term),
partly
used
to overcome
the
frictional resistance
of the
liquid
with the moving
impeller
blades
(the
k*
ullZg
term)
and
partly
used to supply
the losses that occur
as
it leaves
the impeller
and enters
the
volute mixing with the
slower
moving liquid
in
the
volute
(the
k"u!2"
term).
The energy
given
to the
pump
then is
partly
lost in
Sec.
4.81
Pumps
equipped
with
non-return
valves
overcoming
the
friction
losses
that
the
impeller
experiences
in
windage
and
in
bearing
losies.
If there
is
any
left,
the
pump
will speed
up
but,
if
there
is a
deficit,
the
pump
initl rlo*
down.
From
these
considerations
the turbining
pump
case
can
be
solved.
The
derivations
of
A,
B
and C
in the Stepanoff
pump equation
can
be
used
to
obtain
the
values
of
k"and
the other
variables
in
the turbining
equation
above,
and
from
these
the
value
of
u*
ulg
can
be
determined'
Un:
il
-
Df COIY
'
Dr:
DICOSOC]
I
u?-u3,+7
.
Substituting
these
results
into
the
equation
gives
105
u*u
tt
krulcosety
-:
-tla-
g29
(k, + k")u?cosec1
2g
-
k"(u'-Zuurcoty
So
DnU
rr
-:
-
r7--
I
)o
then
Y
:
-
H
-
cQ'
-*
t(flz
-znuQ'*r]
from
the
derivation
of
the Stepanov
equation
earlier
and
the
above
equations
k":2-*(fr)'"
and
coty_
60g8
A.
-
(k"-l)tD
'
The
values
of
A, B
and
Cmust
have
already
been
obtained
when
writing
the
portion
of
the
subroutine
dealing
with
the
Pl
and
DL
zones so
they
are
available.
Using
the
above
three
equations
the
value
of
u*ulg
can
therefore
be calculated-
As
stated
106
Boundary conditions
lch.
4
before,
this
value
minus
the losses
due to bearing
losses, disc friction
and
windage
losses in
the
motor is available
to accelerate
the impeller
and motor.
If P6 is
the
power
consumed
by
the
pump
at
zero flow at steady
running speed
N* then these
bearing,
disc friction
and windage losses can be
represented
by Pg(N/N*)3.
Thus
the
power
Pwr
avallable to
accelerate the impeller
will
be
wQu*ulg
-
P'(N/N*)3. So,
when the
pump
has been
tripped and is turbining,
Tf,)
:
pwr
andthis result
is due to
the
fact
that
the
pump
is
tripped;
so
fmoto,
:
0.
If Pwr is
positive
and Q is also
positive
then the
torque
I being
applied to the
pump will be causing it
to accelerate as it
is
positive. If this torque
is negative
then the
pump will decelerate
and after
pump
trip
it
is certain
that
the
pump
must eventually
decelerate
and come
to rest unless
there are
unusual
circumstances
such
as may occur
when
a
booster
pump
is inserted
in a
pipeline
for
which the
upstream reservoir
level is
higher than the
downstream
level.
In
such
cases
when
pump trip
occurs
it is usual to
apply
a brake to the
pump
to bring
it
to rest and
a
bypass
pipe
around the
pump
equipped
with a non-return
valve is fitted.
This
permits
the flow
under the nafural
head
(without
the
pump) to occur
but
prevents recirculation
when
the
pump
is in use.
As before
(see pp.
69:70)
T= I{l
where O is the
angular
velocity
and
/ is the
moment of inertia of
the impeller
plus
that
of the
motor
armature
plus
that of the
water
contained
in the impeller. So
Pwr
(2r160)NtI
and
N,:N'.W
Thus
N2can
be calculated
and may
be larger or smaller
than
N2 according
to the
relative
values of
P6(N/N*)3
and
wQvplq.
This means that the
pump,
when it
is
turbining,
may
speed
up
and then slow
down in
a
complex
fashion
according
to the
pressure
transients
travelling
back and
forward in
the
pipe,
so changing
the
value of
,FIon
the
pump.
There
can
be many
ways in which the
network may react
to
pump trip
and these will
be decided
by such
factors as the
lengths of
the
pipes,
the relative
levels
of the reservoirs
and the
magnitudes
of
the
wave
speeds
in the
various pipes making
up the network.
The
techniques
by
which the P1.,
Dl and
Tl zones may be simulated
all from
the
Sec.
4.81
Pumps
equipped
with
non-return
valves
107
manufacturer's
steady
state
characteristic
curves
have
been
outlined
above'
It
is
now
necessary
to
indicateihe
methods
of
analysing
the
situations
that
are
most
usually
of
interest
to
the
analYst.
4.E.2
Start-uP
of
the
PumP
pump
start-up
is
difficult
to
simulate
if
the
motor
characteristic
is
not
available'
very
frequently
pumps
are
driven
by
large
motors
which
have
star-delta
starters
or
timed
starters.
A
star-delta
starter
connects
the
three-phase
supply
across
the
windings
of
the
motor
initially
in star
connection
and
after
iO"tuy
changes
this
connection
into
delta
connection.
Effectively
this
places
a
large
voltage
across
the
windings
initially
and
then,
when
the
current
starts
to
rise,
ptaJes
a
lower
voltage
across
the
windings
and
supplies
sufficient
current
to
produc"
tttt
required
steady
state
power'
Timed
starters
work
by
om"riog
additionairesistance
in
th-e
supply
circuit
initially
which
gets
hot
and
radiates
heat.
This
limits
the
current
through
the
motor
windings.
As-
!h.e
motor
speeds
up,
the
windings
start
to
generate
a
blck
electromotive
force
which
opposes
the
current.
The
resiitance
wtriitr
was
inserted
into
the
the
circuit
initially
,uono*uereo.,ceibywithdrawing:fi
:H;t;"r:li:t#:ft
i*:ffi:'1,#:.::T:::
rer
generated
by
the
motor
equals
that
e
subroutines
to
describe
these
processe.s
iary
to
do
so'
When
a
pump
starts
up,
rt
ve up
the
suction
pipe and
a
positive
wave
uvwu
rrrv
vv4r!
-^J
r^r---
-
e
r
r
^,erio-ds
thewavesreachthe
endsof
these
pipes
and
return
iJrfi"
pumgt
Jo,
it
ttt"
putttp
has-reached
its
tull
speed
in
this
time'
the
situation
will
be
like
a
sudden
valve
.iornr.
and
the
pressure
rise
generated
by
the
pump
will
be
the
same
as
it
would
have
been
if
the
puilrp
had-reach:g
tlt
full
speed
in
zero
time
-
equivalent
to
an
instantaneous
valve
closure'
It
would
therefore
seem
logical
to
analysis
a
pump
$f-np
as
if
were
an
instantaneous
start-up'
In
other
n"Jrdr,
start
thi
pump
at
its
full
steady
speed'
that
it
is
possible
for
the
pump
to
generatt
solution.
Of
course,
if
it
is
known
thatthe
I
low
starting
torque
and
the
pump
has
a
relatively
large
moment
oI
lnerlra
,r"""rrrry
6
wriie
a subroutine
which
will
accurately
describe
the
interact
pumpandthemotorduringitsstart-upP!a;e.Themethodofdoingthit
outlined
above.
li
*"V
be
ttought
*orth
*hile
to
run
the
start-up
on
until
it
reaches
'pqt1T:T"':*::;x;i*r':Jff
3"ffi
,Tff
il#"ll:l',H;f
il
jil:l'::J;
:urate than
the
results
of
steady
network
ons
of
accuracy
set
in
most
steady
state
rmulae
com-only
used
for
friction
in
such
a
is
stitl
regarded
as
the
best
available
description
of
friction.
4.8.3
PumP
triP
The
method
of
aialysis
of
pump
trip
is
entirely
based
on
the
methods
outlined
above'
When
the
power
supply
to
the
*oio,
is
inteirupted
for
whatever
reason'
the
pump
eitherentersthe
n
zoteorstaysinthe
plzone.ItcannotentertheDl
zonebecause
108
Boundary
conditions
lch.4
in that
zone the
pump
must
be
driven
by the
motor.
Next the head
across the
pump
at
the 6r
increment under
examination must
be
assessed; if it is
positive,
the
pump
is in
the Pl
zone, if it is negative,
the
pump
is in the
T1 zone. If
the
pump
is in the
Plzone,
the
value
of
the
variable
Pwr must be
obtained
from
wQHIE
and,
if it is in the Z\
zone,
then Pwr must
be assigned
the
value
of.wQ(u*ulg)
-
Po(N/N*)3.
The Nrvalue
can be
calculated
and
the
process
repeated
until the
entire
time
of the simulation has
been
covered.
4.9
THE RISING MAIN
The
rising
main is,
probably,
the commonest
case of the use
of apump.
It
arises
in the
situation in which
a liquid has to
be
pumped
from a low level
reservoir
to
a
higher
level reservoir via
a
pipeline.
This circumstance
covers a
wide
range
of cases; from
the small
sewage
pumping
station with
a rising
main of a
few miles length to sections
of an
oil
pipeline
1000
miles long.
Consider
a
small sewage
scheme. In Fig. 4.7
the layout
of the system is
given.
Fig.4.7.
There is
an
inflow
to the suction
well
Qi.
The well
has
a
plan
arcaof
A and
the height
of the
pump's
switch-on
electrode
is I/""
and the
height
of the switch-off
electrode is
Hoff.
I.et
the
height
of the
water
surface relative
to
the well's
base at time r
be h.
Then
dh
-
Ai*
Qi=
QP
where
Oo
is the flow out of the suction well via
the
pump.
Assume
that the initial
level
of the
sewage
surface is
set at the level
of
the
pump
switch-on
electrode
F/o,, and the initial
speed of the
pump
is zero. This
electrode will
be
set
at the highest level that can be accommodated
in the
suction
well
with a free-
board for
safety. Then, as there
is
an inflow but
the
pump
is not running,
the level will
rise above
the f/o" level. When this happens,
the
switch-on
electrode
will
cause the
pump
to
switch on and
to start
to reduce
the level. Even
when
the level falls below the
F/," level
but is still
above
the F/"*r level
the
pump
must
stay on until
the level falls
below
the
switch off, i.e. I/o6level. Then the
pump
must
switch
offand
stay off until
the level
rises above the 1/o,,level again. This
sequence
of
events dictates
the way in
Sec.4.10l
Pumps
in series
which
the
computer
program
must
be
written
and
how the
pump
must
be
switched
off
and
on.
The sequence
of events
is
as follows.
Stage
I
h)
H.,nand
N:
0.0;
then set
N
:
N*.
Stage
2
h)
H.,o,
hlHonand
N:
N* then N:
N*.
Stage
3
hlHonand
N=N*;then
N:0-0.
Stage
4
h,
H.,rr,
hlHonand
N:0'0i
then
N:0.0.
Using
the
value
of N
so
determined,
carry
out a
complete
step
of
the
analysis,
calculatiig
heads
and
velocities
at
all nodes.
The
new
level in
the
suction
well
can
also
be determined;
so,
to
perform
the
next step
of
the analysis,
it
is first
necessary
to
re-
examine
the
electrode
situation.
The
fact
that a
pump
is small
and
the
scheme
into
which
it
is fitted
is
also small
does
not
diminish
tle
nicessity
of
examining
the
possibility of
D1
and
ZL
operation-
In
the
case
of a
'rising'
main
in
which
there
is
little or
no static
head
and
the
pump
head
is
required
to
drive
the
flow
against
friction
the
neglect
of
the_turbine
effect
will
produce
a completely
erroneous
analysis.
In such
a case the
pipeline
is
likely
to
be
iong. Such
pipilines
usually
are
employed
where
the
flow
has
to
be delivered
over
sigriificant
dirtutt""r
across
flat
ground.
The
inertial
head
of
such
a long
column
of
fi{uid
is considerable;
so,
when
the
pump
is
switched
off,
the
head
on
the
delivery
side
of
the
pump
will
drop
and
eventually
the suction
head
will
become
larger
than
the
head
on
the
deliverysid.,
ro
causing
the
pump to turbine.
If this
effect
is
ignored,
the
pump will have
been
assumed
to
have stopped
when
the
power
was switched
off
but,
of
.ourr",
it
will
not
have
done
so;
it
witl still
be
passing
fluid.
Although
the
delivery
head
will have
dropped
below
the suction
head,
the
flow
will
continue
and
so
the
head
will
not
drop
to-ttre
low
value
that
it
would
have
dropped
to
if
the
assumption
that
the
pump had
stopped
were true.
As a consequence,
wfren
the
flow
in
the
iipeline
has
re-duced
its
velocity
to zero,the
head
will
be
comparable
with
that
on
the
suction
side
and
not
be
very
far under
atmospheric
pressure.
The
reverse
flow
that
then
occurs
in the
delivery
pipe towards
the
pump will
be
accompanied
by
a
pressure
rise
which
will
be
a
great deal
smaller
than
that
that
would
have
been
lxpected
if
turbiningwere
negJcted.
If turbining
had not
been
taken
into
account,
it
is
irobable
that a
nity
large
head
would be
predicted consequent
upon-reflux
valve
closure.
If the
scheme-weti
to be
built,
the
analyst
would
be embarrassed
to
find
very
little
trace of
such
a
high
pressure swing.
Large heads
have
been
predicted
consequent
to
pump trip
for
such
'flat'
schemes
and
these
do
not
occur
if
the
pump
behavis
as
a
turbio".
itr.
reader
is strongly
warned of
the
need
to
consider
this
situation
most
carefullY.
4.10
PUMPS
IN
SERTES
pumps
are operated
in series
so
as
to
produce a
larger
head
than
a single
pump
co,uld
prouid". Th.V
operate
by
supplying
the
pump downstrearn
of
themselves
with
a
flow
"t
"
trigtr"r
headthan
th;t
af
whictrthey
received
it.
In a system
of
pumps
which
are
joineiin
series
by
pipes
which
in
total
are
shorter
than
a
Ax
length
it
will
be
109
Boundary
conditions
[ch.
4
reasonable
to assume that the
flow through
each
pump
is
the same
as through all
others
but the
head
across the
group
is the
sum of
the heads
generated
by each
pump.
For
series operation
the
pumps
do not
have
to be identical. Thus the relationship
between
H
and
Q
for a
group
of n
pumps
arranged in series is
H-
)
r"""*
)
uo"ooo-
2
,oe,
p:l,n p:l,n
p:l,n
in
which the
subscript
p denotes a
value
for an individual
pump.
The
summation
process
must include all the
pumps
in the
group.
The
value
of
Qpmust
be
the same
for
all the
pumps
in series.
Let
subscript e denote a
value
for the single
pump
equivalent to the
pump group.
Then
A.N3
:
ArN?
+
A2N3
+ 1l.N?+ . . .,
B"N.
:
BrNr
+ B2N2+ BrNr+ . . .,
C": Cr+ C2+
Cs*
.. . .
Although
it is
possible
to calculate
the constants
of such
an equivalent
pump
which
will
make it operate at the same head
when delivering the
same
flow
-
as illustrated
above
-
and
it is also
possible to calculate
the
moment
of
inertia
of such.an
equivalent pump,
all the
pumps
will
not run
down in speed in the same
period
of time
as
one
another,
upon
pump
trip,
unless they are all identical. This means that the
calculation
of
the
equivalent
pump's
run down
in speed
will
not be representative of
the pump group's
behaviour.
Except
when
the
pumps
are identical the following
calculation
should
therefore be seen
as an approximation to reality. The running
speed
of the
equivalent
pump
must
be
selected arbitrarily
unless all the pumps are
running
at the same speed.
The
speed
of
the
fastest
pump
in
the
group
is a suitable
choice.
Consider the energy of
the rotating masses in the
pumps:
Sec.4.10l
Pumps
in
series
Then
the equivalent
pump's
head
versus
flow
equation
will be
H":
A"NZ
+ B"N"Q"-
C.Q3
where
2
o,*3
A.:_t*_,
)
u"""
B.:ft_,
C.:)ar.
P:l'n
It
will be
obvious
that
the
power
given to
the
water
will
be
the
sum
of
the
powers
given
by
the
individual
pumps
and
this
will
equal
the
power of
the
equivalent
pump.
So
E"wQ"H":
)
nr*]rtto
P:l'n
and,
as
p"
equals
the
value
of the
flow
Qo
for each
individual
PumP,
the
value
of
Q"
on
th" rigit-iranO-side
can
be cancelleO
wittr the
Qrvalue
on
the
left-hand-side.
w
cancels
throughout
also;
so
111
110
i,"n=:
Z l,pr,
P:l,n'
where
f,)o denotes
the
running
speed of an individual
pump
in
radians per
second.
1-1
ir.ruz:
o4,,i,o*3
where
No
denotes
the
running speed
of an individual
pump
in revolutions per
second,
l.e.
L12
Boundary conditions
lch.
4
2
u,'o
r: - P:l,n
ue
H.
From
these results
the equivalent
pump
can
be specified
and
this
can
be used in the
mathematical model
instead of
the
group
of
pumps.
4.I1 PUMPS IN PARALLEL
Pumps
are operated
in
parallel, usually,
to
provide
a
larger flow than can
be
provided
by a
single
pump. They operate
under a
common head
and the flow that
they
provide
is
equal to
the sum of
the
flows
of all the
pumps.
It is
usual to use identical
pumps
running
at the same
speed
when operating
in
parallel because, if dissimilar
pumps
are
used there is a
large risk
of
instability in operation.
If
pumps
are used,
without non-
return
(or
reflux)
valves,
there
is the
possibility that a
pump
which
generates a
higher
head than
the
others
in a
paralleled
group
could
force liquid
back through
the lower
head
pumps
and
they
would then
operate in
dynamometer mode
and
destroy energy.
Even if they
were equipped
with
non-return
valves, they
would deliver
no
flow
and the
high head
pump which
was
operating
normally
would
be
delivering
a flow
far
lower
than
that
which
had
been
planned
for the
whole
group.
Pumps should
therefore
have
identical
H and
Q
and
E and
Q
curves.
(E
is the
symbol
for
efficiency.) Defining
p
as
the
flow through
the entire
group,
then, if
all the
pumps are
identical,
the
flow
through
each
must be
Qln
where
n is the number of
pumps in the
group:
Then
Hr: AJ.'l'+
BJ'JQI-
CrQS
and this
head
must
equal
the
head of the
equivalent
pump.
Oo
is equalto
Qln
where
Q
is
the flow through
the entire
pump group.
4.I2
SURGE SUPPRESSION
OF
TRANSIENTS GEI\IERATED
BY
PUMP
TRIP
There
is a large
number
of methods of
suppressing
transients but
not all
are relevant
to the
pump
trip
circumstance.
Two methods
are
particularly
appropriate
in this
situation
and
these
are
(1)
increasing the
moment
of inertia
of the
rotating
parts
of the
pump
and
motor
by
fitting
a
flywheel
to
the
pump
and
(2)
fitting air
vessels or surge
tanks to the
pipe
just
downstream of the
pump.
4.12.1
Increasing the
moment
of
inertia of
the rotating
parts
By increasing
the
moment
of
inertia of the
rotary
parts
of the
pump
and
motor,
the
run-down
in
speed
of
the
rotating
pafts
upon
pump
trip
can be
greatly
extended.
If it
is
possible
to
extend
the
pump speed run-down
to
a longer time than the
pipe period,
Sec.
4.12]
Surge
suppression
of
transients
generated
by
pump
trip
113
a
reduction
in the
magnitude
of
the
pipe-line
transient
will
occur.
Whether
this
reduction
will be
suffiJient
will be
dependent
upon
the
reduction
achieved
in
the
deceleration
in
the
flow.
If
it
is
not
zufficient,
it
may
be
possible to
increase
the
moment
of
inertia
of
the
rotating
parts still
further
and
so
to
produce
the
required
reduction.
By
performing
"
n.t*b..
of
simulations
using
progressively
larger
inertia
values
u
gruptr such
as
Fig.
a.8
can
be
constructed
in
which
the
magnitude
of
the
Fig.4.8.
largest
transient
is
plotted
against
the
moment
of
inertia
of
the
flywheel'
From
this
dialgram
the
size
oi
ttt"
flyw:heel
required
can
be
chosen'
It
may
happen
that
the
miriimum
pressure
generated
by
the
transient
is
critical
rather
than
the
maximum
pressure,
in
which
case
it
may
be
necessary
to
plot
a
diagram
of
minimum
pressure
against
moment
of
inertia-
The
approximately
horizontal
section
of
the
first
portion of
this
graph
is due
to
the
fact
that
up to
a
certiinvalue
of
the
flywheel
inertia
the
pump
run-down
will still
occur
in a
time
shorter
than
the
pipe
peiio
d2Llc
and
so
will
have
no
effect
upon
the
magnitude
of the
transient.
It
ijonlywhen
the
inertia
is large
enough
for
the
pump
run-down
to
be
longer
than
the
pipiperiod
that
it
has
any
effect
upon
the
size
of
the
transient.
This
tech'nique
can
U-eiatisfactory
for short
pipelines
in-which
the
pipe
period is
relativety
smatt.
For
long
pipelines
the
flywheels
have
to
be
so
large
that
ihey
become
impiacticable.
When
a
ilywheel
is
very
large,
it
becomes
difficult
to
prouid"
adequatl
bearings
for
it and
the
start-up
torque
of
the
motor
requires
to
be
large
so
as
to
get the
pum; started
in a
sensible
time.
For
these
reasons
it
is suggested
tha-t
the
flywlieel
-eihod
ir
not
suitable
for
pipelines
longer
than
3
k-
4.12.2
Air
vessels
and
surge
tanks
Air
vessels and
surge
tanks
are
devices
which
are suitable
for
longer
pipelines'-Air
vessels
are
suitableior
long
pipelines
but
may
not
be suitable
for
petrochemicals
or
hydrocarbons
with
low
nasf,
pointr
whereas
surge
tanks
are
better
suited
to
medium
length
pipelines
which
are
iairty
flat
so
that
the
surge
tank
does
not
have
to
be
e*c"ersiu"iy
high.
The
necessary
height
of
the
surge
tank
can
only
be
obtained
by
carrying
out
;fu[
computer
simulation
of
the
entire
scheme.
A first
guess
for
the
tr
E
E
Lr4
Boundary conditions
lch.
4
height of a simple
surge
tank can
be
obtained,
however,
by taking
the surface
height
of the water
in the suction
well and adding
the
no-flow
head of the
pump. This
will
give
the height
of
the
surge
tank which
will not be
overtopped.
The
real height
will be
rather less
in
fact but this
gives
a first
approximation
which will allow
the designer
to
establish initial
values
of
height, diameter,
amplitude
and
period
of
the oscillation.
Generally
air
vessels are a
very effective
way of
providing
excellent
surge
suppression
but they
are
quite
expensive
and
need
to be designed
with considerable
care.
They
have
to resist
high internal
super-atmospheric
pressures
and
also
low sub-
atmospheric
pressures; so they
will experience
tensile stressing
and
compressive
stressing
with a consequent
risk
of
buckling
and therefore
they
must be internally
braced. They
must be equipped
with an
air
supply
to
replace air removed
from
the
tank by the
solution
of
air that
occurs
when water
is exposed
to the
high and
low
pressure processes that occur
as
the
water enters
and
then leaves
the tank.
Usually
this
supply
comes
frorn
an air
compressor.
As the
risk involved
in
losing all
the
air
in
an air vessel
is unacceptable,
the air
compressor
supplies
a compressed
air
bottle,
which then
supplies
the
tank.
This
arrangement
provides
some
protection
against
the
possibility
of
failure
of
the compressor,
its motor
or
the
electricity
supply.
Complete
air loss from
an air
vessel
involves
the
generation of
the full
undamped
transient
in
the
pipeline, all
protection being
lost.
Air
vessels
are frequently
used
in conjunction
with
pumps
but not invariably
so.
Analysts feel
the need
to develop
two
subroutines,
one
a
pump-air vessel complex
in
which
the
pump
delivers
straight
to the air
vessel and
the
other
being
that
for
an air
vessel
only
which
can
be used
anywhere
in
the network.
4.13
THE
ANAITYSIS OF
AIR VESSELS
An
air
vessel contains
air
which is available
to be compressed
and so
provides a force
to
decelerate
the
liquid
in the
pipeline.
The
forces
generated
are much smaller
than
those created
by
pressure transients
and so
do
not decelerate
the liquid
as rapidly.
The
oscillatory
motions
so
created
are called
mass
oscillations
and
have much
smaller
magnitudes
than
do transients
but
they have
much
longer
periods.
In
Fig.
4.9 the
cross-sections
of
'two
common
types of air-vessels
are
depicted.
plan
is
circu
lar
side
elevation
is circular
(a)
Fig.
4.9
-Two
common
types
of
(b)
air
vessel:
(a)
vertical cylinder;
(b)
horizontal
cylinder.
Sec.
4.13l
The
analvsis
of air
vessels
Both types of
air
vessel can
be solved
by the same
technique.
The
only
significant
difference is that
the area
of
the free surface
in the
vessel does not
vary
with depth
in
one
case but
does
varv in the
other.
4.13.f The
vertical cylinder
First, consider
a
vertical cylinder
(Fig.
a.10).
Let the
number of Ax
values
in
the
pipes
xR
N.,Nz
s
Y
x
(a)
(b)
Fig.
4.10
-
(a)
The
vertical
cylinder
and
(b)
its associated
x
versus r
diagram.
in the network
be
held
in an
array, netno,
which
will have two
bounds,
the
first
to
hold the
pipe
number
and
the second
to
hold the
number of
nodes
in each
pipe
length. Then,
if
I is the
number
of the
upstream
pipe and
i
* 1
is the
number
of
the
downstream
pipe,
it
is
possible to
write
the
continuity
equation
in a typical
computer
expression
as
.,{(f)ul(i,
neto(i))
-
A(i
+ 1)u1(i
+ 1, L)
:
Q,t
In
explanation
of this equation,
arcay Aholds
the cross-sectional
areas
of the
pipes in
the n-etwork and
the
array
u1
holds the
velocities
at the
Ax nodes
at
time
t
and
a uZ
alray is used
to
hold the
velocities at the
6x nodes
at
time
t + Lt.
Q,
is the
flow entering
the
air
vessel at time
r and
115
lo'
ah
er:
Ar
at
.
The
continuity
equation
of
the
air
vessel
will
therefore
be
A(i)
ul(i,netno(i))
-
A(i+
1)
ur(i + 1, 1)
:
Or*
ll