FluidandThermdSystems
( | BE
J GTIVES
T h ef o c u so f t h i s c h a p t e ri s o n m o d e l i n fgl u i d a n d t h e r m asl y s t e m sa, n d t h e m a i n
addressed
topicsare
" F l u i d( l i q u i da n d p n e u m a t i ce)l e m e n t s - i n e r t a n ccea, p a c i t a n c e
r e, s i s r a n c a
en
,o
f l u i de n e r g ys o u r c e s .
'n Thermalcapacitance
and resistance
elements.
al modelingof fluid systemsand formulation
" Mathematic
of the naturaland
forcedresponses.
', Mathematical
modelingof thermalsystemforceoresponse.
U
s
e
o
f
M
A
T
L
A Bi@
n s y m b o r iac n d n u m e r i c acra r c u r a t i o nasn, d i n e v a r u a t i oonf
"
fluidsystemseigenvalues
and eigenvectors.
y o d e la n d p l o tt h e t i m e r e s p o n soef
t o g r a p h i c a l lm
" ' A p p l i c a t i oonf s i m u l i n k @
f o r c e df l u i ds y s t e m w
s i t h l i n e a ra n d n o n l i n e apr r o p e r t i e s .
INTR(lDUCTION
This chapter is dedicatedto modeling the dynamics of fluid (liquid
and pneumatic)
systemsas well as thermal systems.These systemsare modeled
using elements
similar to those for mechanical and electrical systems:inertance, capacitance,
and
resistance.
Systemmodelsarederivedwhen thesecomponentsarecoupledin various
systems.In caseonly inertance and capacitanceproperti", ur" p."r"nt
and no resistive lossesoccur,the naturalresponseof fluid systemsis studied.In
many applications, the inertia properties of liquids and gasescan be neglected,and
the resulting
mathematicalmodels are basedon only the capacitive and resistive properties
used
to formulate the forced responseof fluid and thermal first-order systerns.
The use of
MATLAB@and Simulink@in solving for the natural and forced responses
of fluid and
thermalsystemsis illustrated by severalsolved examples.
152
5 FluidandThermal
Svstems
CHAPTER
MODELING
S-XLIOUID
SYSTEMS
In liquid (or hydraulic) systems,the medium of energy transmissionis a liquid. We
review a few basic liquid laws first, such as Bernoulli's law and the law of mass conservation.Next, we discussthe basic liquid elementsof inertance,capacitance,and
resistance,together with the sourcesgeneratingliquid system motion. You will learn
liquid systems
how to formulate the natural (free) responseof inertance-capacitance
as well as the forced responseof liquid systemscontaining capacitanceand resistance
components.
The notion of flow rate is utilized in liquid and pneumatic systemswith different
meanings.Inliquid systems,thevolumeflowrate is employed,which is denotedhere
by qu; whereaspneumatic systemsuse the massflow rate, denotedby 4.. The two
amounts are connectedby meansof the mass density:
Q,=
LV
N:
L,m
LxA = v A ) Q . = E
=
A,
AY =
Q
QQ,
*
(s.1)
where V is volume, m is mass,v is the fluid velocity, x is distanceffavelled by fluid, and
A is areaperpendicularon flow direction.
LawandtheLawof MassConservation
$"3.'gBernoulli's
An important instrument in modeling liquid dynamics is Bernoulli's law. For a
conservative liquid (with no energy losses or gains) flowing in a pipe and when
the liquid is consideredincompressible(and therefore its mass density is constant),
Bernoulli's law statesthat
pvT
Pzr
z
l- QBhz = Pt
pv?
+j
z
l- QBht
(s.2)
which is based on Figure 5.1, and where p is the static pressure, ft is the vertical
distance (also named head) from a referenceline, and v is the liquid velocity at the
center point of a cross-section.When lossesare accountedfor (of a viscous nature)
and energy is input into the system (such as by pumps or hydraulic actuators),
Eq. (5.2) changesto
u,* + * psh,= prt +.
t sw- e7hr
e7hr
(5.3)
where w is the specific work produced by a hydraulic source (it is energy per unit
mass, being measuredin N-m-kg-t in SI) and hy is the lost head (it is due mainly
to viscous friction). Equations (5.2) and (5.3) are the pressure-form equations of
Bernoulli's law.
The law of volume/massconservation statesthat, if there is no accumulation or
loss of liquid between points I and2 of Figure 5.1, then the volume flow rate will
not chanse:
Qr : Q,z
(s.4)
Modeling 153
5.1 LiquidSystems
5.I
TIGURE
Plane.
Pipein a Vertical
cross-Section
Variable
through
Traveling
column
Liquid
on Eq.(5.1)
or,based
vrA, = vrA,
(s.s)
Fora pipeof lengthl, thelost headis calculatedas
lf
,,2
7r" = - : 7 X ^
zP
cl.
(s.6)
where/is the Moody friction factor and dn is the hydraulic diameter, which is calculatedas
(s.1)
a" n4 4=
E
with P, being the wetted perimeter of the pipe internal cross-section.For laminar
flow (which is defined shortly), the Moody friction factor is determinedas
(s.8)
r"-6R4e
The Reynolds number, Re, is the ratio of the inertia to viscous friction effects
associatedwith the relative motion between a solid and a fluid' The mathematical
expressionof the Reynolds number is
avd,
R":-p.=f
vd,
(s.e)
154
CHAPTER
5 FluidandThermal
Svsrems
where dn is the hydraulic diameter (for a spheremoving in a fluid, dnis thediameter;
similarly, for fluid flowing in a pipe, dn is the pipe innir diameter), p is rtre
dynamic
viscosity, and v is the kinematic viscosity.Laminarflow has a smooth, linear character and is defined by generally paraltel streamlines. rt is charactenzed by large
viscosity effects and small inertia properties and occurs for low Reynolds numbers,
for instance,the flow in a pipe is consideredlaminar when Re < 2000. For very small
Reynolds number, such as Re < 1, the flow is known as creep or Stokesflow. On the
contrary, turbulentflow has a nonlinear characterand is defined by a chaotic motion
containing vortices and eddies. In pipelines, turbulent flow occurs for Reynolds
numbers larger than 4000 and where inertia effects are predominant. When 2000 <
Re < 4000, the flow is consideredto be of a transition nature, becauseit combines
laminar and turbulent traits.
Example5.1
A pump is usedto send iiquidverticaliy
througha pipe of circularcross-section
with
innerdiameterd = 0.02 m. Assumethe pipe,whoseheightis / = 6 m, is openat itsend
opposite
to the pump;alsoassumetheflowis laminarandthe incompressible
liquidhasa
p = 0.00001N-s/m2.Knownalsois
massdensityp = 1000kglm3and dynamicviscosity
the inputvolumeflowrategvi= 0.0001m3/s.Calculate
the specificworkof the pump w
thatis necessary
to sendliquidto thetopof the pipeby considering
thefrictionlosses.
The
pressure
at the pumpintakeand at the pipe'sfreeend is atmospheric.
Solution
Figure5.2 showsschematically
the pumpandverticalpipesystem,
wherepoint0 is at the
intake(input)to the pump,point1 is at the outtake(output)of the pump,and point2 is at
theendof theverticalpipesegment.
Application
of Bernoulli's
law betweenpoints0 and 1, whichare assumedat the same
hpioht
lp:elc tn
, Qv2, = P o t
?r+
^.,2
Pro
^z + p w
2
(s.10)
gives
The lawof massconservation
qr,: v6A : vtA
(s.1 )
whichassumedthatthe pumpintakeand outtakeareasare identical;
the resultis
ro= vr:
4e,t
;i
6.12)
Becausevo= vr, Eq.(5.10)simplifies
to
P1: Pn-t Pw
(s.13)
155
SystemModeling
s
5.1 Liquid
Pipe cross-section
Pipe segment
5.2
TIGURE
PipeSegment.
PumpwithVertical
il
1
lawis nowappliedbetweenpoints1 and 2:
Bernoulli's
pv:
)
p,*
;
pv,,
+ pgl : p, +
;
- p7h.r or pgl : Qw - QBhr
(5.14)
Thefrictionheadis found,usingEqs.(5"6)
sinceyz = y1becauseof massconservation.
through(5.9),as
,
'1
o
321tlv,
(s.1s)
pgd'
II
I
r
I
that the hydraulicdiameteris equalto the actualinner
whichtakesinto consideration
c o r ko f
p i p ed i a m e t edr . c o m b i n i n g
E q s .( 5 . 1 2 )(, 5 . 1 3 )(, 5 . 1 4 )a, n d ( 5 . 1 5 )t,h e s p e c i f i w
t h eo u m oi s
, , 1281t"1q,,
W=gt -r
,
Ttpd."
h
l/
workis w = 58'8m2ls2'
valueofthepumpspecific
Thenumerical
.*"
W :,
15 l6\
I
: Liquid
Elements
Similar to mechanicalor electrical systems,which are formed of elementswith inertia, storagecapacity,losses,and energy input, liquid systemscan be defined by such
elements.Inertanceelementsportray liquid inertia effects, whereascapacitancesand
resistancescharacteizeliquid storageand loss features,respectively.The pressureor
the differenceof level (head) among various componentsof a liquid systemrepresent
sourceelementsthat set the liquid into motion. As a consequenceof the dual manner
of generatingliquid motion, the elements'definitions can be provided by either using
1.56 e["{AP"R"E,R
S Fjuidand Therrrral
Systems
[ t r l cP l - c s s t l lor 'll' t h c ] t c a c i{l r i u l t l ; i i o r r[ o o t l r c r i r i ] l ( r . u ] { s
o X ' i n t c r . e s t . ' i ' hscu S s c r . i i r t 1 i r .
lr
l ) r o s s t l l l ro)l ' i i ( l i r r i t c a t l ) a c e o l t l l t r t r t idchs cI c t I c l ' 1 ( s r l l n r i i i rl i l:r I i c ; u i r l i)n I l r c l o r l l g w i i r g .
Inertansp
n h t : i t r c t ' l t t r r ( ct ; t r l l l t l i l i c sl h c i n e r " t i l cl t r l c c l r r r r
iirluirl svstcn)ti rrrtlis pis-tie6Nllr.iy
l l l l l l 0 l ' 1 ; t l il l{l t o r l l lc t t l l t i t l i [ :s; t t c l ll l s
P i p c s .l l o r - l l r r r r i n l l l - f t r o v r ' ai lr]utl" l r . j r ]gst rp r - c s s a l . c .
I h c i l l c ll l l l t c c ( w l l i e t r irs ' - i c l t o t e rht ly / ) l s t i c f i l r c t {
l l s r t r r cp r t r s s u r cr j i t r { c ' r c r l r]cr x e c * i s ; r r . !
1 o f r ' ( i r i r ! c {l l. ru n i { e l l i u i r r ci n { h c r . u t co l ' c h a r l g co l
l i r c v q l r r r p r 1, l . w * l t r - . :
l\rr
{,.,,{lj
Ao
(ttll
Altrlt
dq,
{ 5 .i 7 )
tlt
'I'irc'5i
r n r i l o l ' / r . r ,i s N s ' l r r . ( o l . l i 1 ,u l r . ) .{'hc
hclrrl n'trlilcr-iillr.',r.1;1r;(-"g
rfc/iniIiorr is
s i l l r i l l r ' t al lrl a to t r ' I r l (. 5 .t r 7 ) :
,
_ A/r
i1,
l\fi
tlt1,
Alrtli
tlt1,
( - 1 I. f i )
tlt
'i'ht.
').
r (ol.sr
S I u n i { o 1 ' / 7 .i7s ,l i g l t { ]
irr
[ ] y { l r l , i l l gi n { o l r c c o r r r r[[l l r { u s I l n t r c
ltn.r-:ss u l r :c { i l ' { i : l - c , ti-s' cc ' l l r r c e . i c. rrIlrl rr I l l c
{ r c u cci l i l . { c l . c l r e c ; r s
A/,
1tt'l,lr
r...l{)1
i { n o l l , o w :{ ;h l r {t l l r I w ( ) i i r c l . t u l j r cr t r c f i n i t i o n s
o X . r, tr 1 s(.- 5 ]. 7 ) l l l r i t (r - 5 . l g )l n . ct . t ] u [ r ] ( X
il
ft,,-li,1lr.r,
(.5.10)
'n-lrc:
hircIic orcrgy trilir crl'r-cs'.rrris
r' ircrrllrcc ls r,,cr.ysi'rirar. I o t l l c
iiiili:{ie
c r c r g y , i ' i l r c c h u r i c u l l r r r {c l c c . t r - i c u l
s y s { c ' 1 s .l l,e r i r c t i r r s
,. ,: io ,,lrlLl(l)lr
:
._,u,4,!,,,,:
),,,,,,:
( - 5 . 2) 1
T i r e' r c s s u r e - t r c f i r i l i r ' ' 1 ' E q( '- 5 . 2 r )
i s k c p t r L e r i - ' ( r v h .sr Is eu l l i {i s . r * r r d - r i l ) "
usthc
hcrttlclclirlctlellelllyresilhsirl a clu.nti{y
ul-,,,,
in'r., ac[u'lilva' crerlry(the i'tcrcs{ecl
t'eatlcr
ririghfwftltt1.ocl-reck
tlle unitsoi' 7,,,= ,./.(Al,/4,)r:i1.
A n r i c r o c h a n nuesl e d r n a m i c r o l l u i d i c
a p p l i c a t i o rnr a si h e s h a p ea n d r i i n r e n s i o n s
i n d i c a t erdn f - i g u r 5
e 3 c a l c u l a tteh e p r e s s u r e - d e f i n e d
i n e r l a n coef l h e l i q r , r rt h
d a riii o w s
t n r o u g ht h i sc h a n n esr e g m e n tc,o n s i d e r i n g
t h a , k n o r , r an r e t h e e n d w i d , r s w r a t l d
w),
t h e t h i c k n e shs, t h e r - ' n g t h a n dt h e
1
l i q u i dm a s sd e n s i t p
y . o b t a i nt h e i n e r t a n cteo r i h e
partrcula
d re s i g nw i l h w 1: * . , .
Modeling 157
5.1 LiquidSystems
5.3
FIGURE
View;
PipewithLaminarFlow:(a)Three-Dimensional
Tapered
Rectangular
Cross-Section
(b)SideView.
Solution
interms
ofxas
canbeexpressed
5.3(b)
ThewidihwofFigure
Wz- wt
(s.22)
w = w ( x ) : w , *' t+ x
parallelepiped
thickness
h,andlen$hdxisfirst
ofwidthr,v,
ofanelementary
Theinertance
bysumming
allsimilar
elementary
iscalculated
afteruvard
Thetotalinertance
determined.
(whichmeansintegration
prisms
secondlawof
overthetaperedpipelength/). Newton's
theexternal
forceas
of lenghdx byexpressing
is usedforthe prismatic
element
motion
difference:
areatimespressure
,dv
dm:; - A(x)Vt(x)- p(x + dx)l = A(x)d(x)
(5.23)
dt
whereit has beenconsideredthat p(x) = p(x-t dx)+ dp\). Knowingthat in = Aldv(t)/
of the elementas
the inertance
dfl,Eq.(5.23)enablesexpressing
.-
dp(x)
--'t'P
urt
- -
a'
d*i
dv(t\
A@)'zlg
dt
dv(t\
oA(x)dx-6'
Aa'49
dt
pdx
pdx
A(x)
w(x)h
(s.24)
inertance
of
the elementary
MathToolboxrM
is usedto integrate
Symbolic
MATLAB@'s
code:
Eq.(5.24)usingthefollowing
> > s y m sw l w 2 h I x r h o
)) w: w1+(w2-wl)/l*x;
)) r'n:limit(int(1/w,x),x,1,' lefl' )-lt'mit(int(1/w,x),
x , 0 , ' r " ig h t ' ) :
) ) i n e r t a n c e: r h o / h * in
158
Systems
5 FluidandThermal
CHAPTER
sequencereturns:
The lastMATLAB@
commandin the previous
wc
al
T,,,=
=-Xln;:
Ia,,.o
n\w2-
/5 r5\
fvl
wt)
betweenthe limitsof 0 and / in twosteps:
It can be seenthatthe variablex is integrated
as
thenthe definiteintegralis evaluated
integral
of 7lw(x)is calculated,
firstthe indefinite
integral;
on attempting
between
the upperand lowerlimitsof the indefinite
the dlfference
t o d i r e c t lcya l c u l a tteh e d e f i n i t ien t e g r abl y m e a n so f t h e c o m m a n di n t ( 1 / w , x , 0 ' I ) ,
cannotdeterminewhethert,v(x)is between
is returned,as MATLAB@
an errormessage
the limits0 and /.
prismbecomes
of
withan inertance
a parallelepiped
Forw1= w2= w,thetrapezoidal
.olo1
(s.26)
f - '
-t.P
hw
A
co
@m m a nIdi m it ( i n er t a n c e , w 1 , w 2 , ' 1e f t ' ) eA T L A B
w h i c hi so b t a i n euds i n g t hM
the limitof /i.,whenwr reachesw2fromthe left"Thefull MATLAB@
thiscommandcalculates
(5.26),whereA is the innercrosswebsite.
Equation
codecan be foundon the companion
pipe.
I
sectional
area,is validfor anyconstantcross-section
Capacitance
The capacitance in the liquid domain reflects the storage capacity by a tank-type
device.The capacitancecan be definedin terms of static pressureas the ratio between
the volume flow rate and the rate of pressurevariation:
dv(t) .
q,(t\
q,ft)dt
-dt
clt
dv
dp(t)
dp(t\
dp(t)
dp
I
" t'P
(5.27)
dt
In the International System (SI) of units system, the pressure-definedliquid
capacitanceis measuredin m5-N-r (or ma-s2-kg-r).In terms of head, the capacitance quantifiesthe volume flow rate necessaryto changethe head rate variation
by one unit:
^
"Lh
q,(t) q,U)dt
=dh(t)
dhu)
dt
dv(t) .
o'
d,
dhu)
dv
dh
(s.28)
Modeling 159
5.1 LiquidSystems
two
Due to the connection between head and static pressure,Eq. (5.19),the
of Eqs. (5.27) and(5.28) are relatedas
capacitances
Ctn : QBCt,p
(s,29)
The SI unit of Or is m2.
the
Similar to springs in mechanicalsystemsand capacitorsin electrical systems,
as
energy
pressure-form
potential
liquid capacitanceis incorporatedinto
I ..
u,,o=lc,.o1Lp1'2
#= |tnntnrr
(5.30)
The hydraulic energy,according to Eq. (5.30), is measuredin N-m'
5.3
Example
vesselsketchedin
cylindrical
cross-section
of the variable
the capacitance
a. Determine
Figure5.4(a).
b . U s et h e r e s u l t o c a l c u l a tteh e c a p a c i t a n coef t h e c o n i c asl e g m e not f F i g u r e5 ' 4 ( b ) '
Knownare the end diametersdt and dz,the heighth, as well as the liquid mass
g.
acceleration
densityp and the gravitational
A1
(a)
5.4
FIGURE
VerticalTank'
(a)Variable
VerticalTank;(b) Conical-Segment
Cross-Section
160
CHAPTER
5 FluidandThermal
Systems
Solution
a. Thepressure-defined
capacitance
canbeexpresseo
as
dv
u,--T-T
dv
dh
dp
dp
(s.31)
dh
Thepressure
at the bottomof the tank is
p:pgh
(5.32)
Thevolumeoccupiedbythe liquidin a variable
cross-section
tankwitha head(height)
of h is calculated
as
V = lAu\dx
J
/5 ??\
0
whereA(x)is thetankvariable
cross-sectional
areaar a distance
x measured
from
thebottom
of thetank.Bytakingderivatives
of p and t/in Eqs.(5.32)and(5.33)with
respect
to h andsubstituting
thesederivatives
intoEq.(5.31),the pressure-defined
and head-defined
capacitances
become
4 io,*,0*
c,,=#, ff =Ti
cn=fri o,.tor
(s.34)
b" The diameterdefiningthe variableareaA(x) is the one givenin Eq. (5.22)with d
insteadof wand h insteadof /. Thetankvolumeis therefore
expressed
as
-rn,*)'o,
=ffta;+dd,+
, : XiQ,*o'
d7)
Calculating
the derivatjve
of l/with respectto h in Eq. (5.35)and substituting
it into
Eqs.(5.34)yields
x(a|+d,d"+d1\
C,.r= --l2W
i Cr.n=
r(ai + dd2+ d3)
I2
(s.36)
Fora cylindrical
tankwith dt = dz = d, Eqs.(5.36)changeto
c,,=#tc,o:#
(5.37)
I
Modeling 161
5.1 LiquidSYstems
tl
ll
I
I
""lF
-----------,>l
+\1
l- "'l=
(a)
5.5
TIGURE
(b) Parallel'
in (a)Series;
Uqrd Swtg. T-ks Connected
il
I
I
In
th
)o
way as
in seriesand in parallel in the same
Liquid capacitancescan be connected
of the
website Chapter 5 gives the derivation
electricalcapacitances'ftt" totp*ion
on Figure 5'5:
siries and parallel capacitances'based
ffiut"nt
l1
'
C,r'
Cu
= CnI Cn
lc,,
[1
--
-,LJ--
1c,'
(s.38)
r)
I
,'d
Resistance
from changesin
or co^nduitsencountersresistance
The liquid motion through pipes
which act as
of valves' or other constrictions'
the conduit direction, ttt-ee*istence
action-isquantified by meansof resistances'vety
energydissipaters.The dissipative
with a valve
5.6 shows the portion of apipeline
similar to electrical ,yr,;;r:Fi;rre
pressuredrop
flow and the net result of it is a
on it. The valve change; ;" a;" of
tt"Tro;?#ucary,
ratio of rhe pressuredrop to rhe
the liquid resistanceis defined as the
volumeflow rate:
Lp
Pt- Pz
Rr,o=-qr=-q:
is N-s-m-s
The SI unit for R7,o
resistanceis defined as
(s.3e)
the liquid
(or kg-s-t-m-a)' In terms of equivalenthead'
n,.n=
*
(s.40)
162
CHAPTER
5 FluidandThermal
Svstems
^
y1
+
Valve
R,
p2
--lQv
FIGURE
5.6
Pipeline
withValveand Pressure
Drop.
Laminar Turbulent
(linear) (nonlinear)
(q, Lp)
6(AP,)
LPn
6eu,,
FIGURE
5.7
LaminarandTurbulent-Regime
Relationships
between
Pressure
DropandVolume
FlowRate.
Rr,r,is measuredin s-m-2in SI units. Again, when the static pressure-head
relationship is considered,the two resistances
ofEqs. (5.39) and (5.40) are connectedas
R,.r=
pph
Lp
q,=;;=eiRln
(5.41)
Both definitions assumea linear relationship between pressure(or head) variation and volume flow rate. While this linearity covers the laminar flow domain, for
turbulent flow, the relationship between these variables becomesnonlinear and it is
acceptedthat pressurevaries with the squareof the volume flow rate:
Lp = kq?,
(\ 4)\
where k is a constant determined experimentally. Figure 5.7 shows the variation of
the pressurein terms of volume flow rate for the laminar and turbulent regimes.
Assuming the pressurevaries nonlinearly with the volume flow rate, this relationship can be linearized using a Taylor series expansion about a given nominal
(or operational) point, such as the one indicated with the letter n in Figure 5.7. The
Taylor seriesexpansioncan keep the first two (linear) terms:
d(Ap) I
a
Lp = Lp, + -r_: |
@,- q,,^\
oQ, Ia,=a,.,,
(s.43)
Modeling 153
5.1 LiquidSystems
As seenin Figure 5.7,
LP -
LP,= b(AP,); Q,- Q,.n= E4u,n
(5'44)
consequence,
representsmall variations in pressure and volume flow rate. As a
Eq. (5.a3)can be written as
d(Ap) |
bq",,
D(Ap,)= :I
oe, ln,=r,.,
(5'45)
point is
Equation(5.45) suggeststhe linearizedresistanceaboutthe nominal
R.,, ,, =
dtAp)
^-
(s.46)
dQ,
of a small
Equation(5.46) indicatesthat the linearizedliquid resistanceis the ratio
rate and
flow
volume
the
of
variation
small
the
to
variation
variationof the pressure
of the
terms
in
variation
pressure
the
of
derivative
partial
can be calculated as the
linear(5.42),
the
Eq.
account
into
taking
By
point.
nominal
volume flow rate at the
izedresistancedefinedin Eq. (5.46)becomes
='(+),,=,,,
=2ke',n
=ryl^=,,.,
R,,,
(s.41)
the liquid
In other words, the linearized liquid resistancefor turbulent flow is twice
(linear)
flow.
resistanceof laminar
5.4
Example
in termsof volume
in a pipewithturbulentliquidflowis expressed
variation
Thepressure
corresponding
resistance
hydraulic
linearized
the
=
q,+
Compare
3q?.
flowrateas Ap
Lp/q, by
R,,r=
as
defined
be
can
which
resistance,
linear
to
the
to this relationship
=
ratioin termsof gu.calculatethis ratiofor q, 0.01 m3/sand
plotting
thetwo-resistance
=
Qv 2 m3lsSolution
as
is obtained
resistance
hydraulic
to Eq.(5.47),the linearized
According
0tApl
RL,, P. = - * = l * 6 4 ,
(s.48)
dq,
as
Rip is calculated
Atthesametime,the resistance
Ln
Rio- - 4,
'-I
| * 3q,
(s.4e)
164
Systems
5 FluidandThermal
CHAPTER
z
1.9
1.8
1a
l.o
a,rtr1 . 5
1.4
t.\t
1.2
1.1
'l
81012
qu1m3/s)
14
to
18
1V
FIGURE
5.8
of VolumeFlowRate.
Ratioas a Function
Resistance
Hydraulic
Linear-to-Nonlinear
of Eqs.(5.48)and (5.49):
ratiocan be usedto comparethe resistances
Thefollowing
'*:
R,.,
l*6q,
n*=
t*^-z-
I
l+3q.,
^
(s.s0)
which indicatesthe linearizedresistanceis almosttwice the resistancedefinedin
volume
(5.50)alsoshowsthatthe ratioincreases
withthe increasing
Eq.(5.49).Equation
flowrate.andthe limitis
=
c^,*u*
,
Ii*r*:
(5.s1)
ratioof Eq.(5.50)dfecp: 1.029for qu: 0.01m3/s
valuesofthe resistance
Thenumerical
I
=
=
5.8 is the plotof cnas a functionof qu.
Figure
and cp 1.857for e, 2 m3ls.
In the companion website Chapter 5, the Hagen-Poiseuille equation is demonstrated, which gives the resistanceof a cylindrical pipe of length I and internal
diameter d for laminar flow as
Lp
R,.o= q,:
l28ul
*^
(s.s2)
Modeling 155
5.t LiquidSYstems
E x a m p l e5 . 5
of a taperedpipehavinga len$h of /and end diameresistance
the hydraulic
Determine
the resislamlnarflowandcalculate
5.9.Consider
Figure
in
indicated
dz,as
tersof dr and
=
N
=
p 0'001 s/m'z'
t a n c en u m e r i c av la l u ef o r d r = O 5 m , d z = O ' 3m , / 1 0 m , a n d
Solution
portion
portionof lengthdxof the taperedpipeis studied,the respective
lf an elementary
isapproximate|yacy|inderofdiameterd,asshowninFigure5.9;therefore,itspressure
as
equation
is givenby the Hagen-Poiseuille
difference
128udx
d(Lp) = --a
7td
/5 <?\
q,
The diameterd is expressedgeometricallyin terms of x as
d = dz-l (dr-
.,t-x
(s.54)
dr)-----I
the limitsof 0 and /with
between
of Eq.(5.54)intoEq.(5.53)and integration
substitution
the inputandthe outputof the pipe:
between
difference
to x yieldsthe pressure
respect
=
P,- Pz= 6P = [a1LP)
dl,+ d:)
rz8pt(di+
q,
(s.ss)
zttd3,d)
of the taperedpipeof Figure5'8 is
thatthe fluid resistance
whichindicates
R,,, =
tzspt(di+ dld,+ dl)
(s.s6)
3nd3,dl
the taperis zero(thetaperedpipe becomesa
when dr = dz, andtherefore
obviously,
equationresistance'
Hagen-Poiseuille
one),Eq.(5.56)reducesto the classical
cylindrical
to
pipe.Thesolution
cylindrical
cross-section
to a constant
Eq.(5.52),whichcorresponds
code
the
ToolboxTM;
Math
Symbollc
MATLAB@
thisproblemhasbeenobtainedusingthe
the
of thisexample,
website.Forthenumericalparameters
is includedin the companion
I
=
ls Rlp 19'72N-s/ms
resistance
hydraulic
systems,
Similar to dampers in mechanical systems and resistors in electrical
pressure
in
expressed
be
can
loss
this
and
energyis lost through liquid resistances,
form as
=iY = Iorn.
u*:!n,,oar
166
Systems
5 FluidandThermal
CHAPTER
Qv
_-____--->
5.9
FIGURE
PiPewithLaminarFlow.
Tapered
.-------->
4
(b)
(a)
.10
F I G U R5E
(b) Parallel
in (a)Series;
Connected
LiquidResistances
power: The pressurets
It should be mentionedthat Eq. (5'57) actually expresses
N-s/ms' and therefore the unit
measuredin N/m', hydraulic resistanceis measuredin
of Uaris N-m/s, which is the unit for power'
(as shown in Figure 5.10(a))or in
Liquid resistancescan be connectedin series
resistancesR," and R12'
pu.utt"i (as illustratedin Figure 5.10(b)),and the equivalent
which are derived in the companion website Chapter 5' are
[^^
ll
= R 1 1* R p
t&
5.6
Example
ll
R,,
(s.s8)
R,,
c a l c u l a t teh ee q u i v a l e nhty d r a u A m l c r o f l u i dci ch a n n esl y s t e mi s s h o w ni n F i g u r e5 . i 1 .
thatthe liquidlossesare produced
betweenpoints1 and2by considering
lic resistance
t h e p o w e rl o s tl n t h e m i c r o a c c o r d i ntgo t h e H a g e n - P o i s e uei lql eu a t i o nA. l s oc a l c u l a t e
=
20 pm (pipediameter)'and
system.Kno*n rr" / = 100 u.m,p = 0'0005 N-s/m2'd
Lp = pr - p2 = 103N/m2.
5.1 LiquidSystems
Modeling 157
FIGURE
5.11
Six-Component
Pipeline
System
withFlowing
Liquid.
Pt ,!r---2<l--
Pz
_;
(b)
FIGURE
5,12
(a)ActualMicrochannel
System
withLiquidResistances;
(b) Equivalent,
On"_n"rlrtu*
Liquid
Svstem.
Solution
Figure
5.12(a)
showsthe microchannel
system
of Figure
5.11withthe corresponorng
hydraulic
resistances,
whichareidentical.
Theaimistoobtain
theequivalent,
one-resrstance
system
of Flgure5.12(b).Toachieve
that,the resistance,
whichis equivalent
to thefour
actual
resistances
inthemiddle
ofthesystem
of Figure
5.12(a),
canbecalculated
bycom_
bining
in parallel
twogroups
of series
connected
resistances,
whichvields
o,,=ffiffi=*,
(s.5e)
Theequivalent
resistance
is formedby connecting
in seriesthe end resistances
of the
original
systemof Figure5.r2(a)to the middleresistance
of Eq.(5.59):
Ru: Rr + Rrl + Rr
(s.60)
Substituting
Eq.(5.59)intoEq.(5.60)yietds
R,"=3R,=3xn8fi=3844
(5.61)
158
5 FluidandThermalSystems
CHAPTER
is foundto be Rr"=
reslstance
the equivalent
valuesof thisexample,
Withthe numerical
through
(5.57)is usedto calculate
the powerdissipated
Equation
3.8Ig7 x 1013N-s/m5.
pipeline
sYstem:
the
r (Ap)'z
U u , =i - i ,
poweris lJar= I'3I x 10 BW
the dissipated
Numerically,
(s.62)
I
EnergY
of HYdraulic
Sources
motion is
For liquicl-levelsystems,as is discussedshortly in this chapter,the liquid
tanks and
as
such
g"n".ut"d throughthe headdifferenceamongvariouscomponents,
piping.Hydraulicactuatorsarecomponentsthatconverthighinputfluidpressure
into kinetic energy at the output.
to generateflow in a liquid netIn many liquid applications,the energy necessa"ry
liquid
work is providedbyp umps,whtchtransfotmthe input electricenergyinto output
of
several
be
can
work, glnerally manifesiedas flow rate or equivalenthead' Pumps
appliconfigurations,suchas centrifugal,axial, rotary,or reciprocatingin regular-scale
cations'aswellasdiaphragm(ormembrane)inmicro.andnano.applications.The
head variacharacteristiccentrifugal pump, a widely used configuration, shows the
sketchedin
as
nonlinear,
tion as a function of the flow rate at the output is generally
F i g u r e5 . 1 3 .I t se q u a t i o ins
h = hr- Koql,
(s.63)
pump prowhere lzr, the geometric head, is the maximum head-type energy a
related to
coefficient
is
a
Kn
and
friction
through
lost
duces when no energy is
a specithat
indicates
curve
characteristic
the
along
point
the energeticlosses.A
as the
that,
and
rate
flow
volume
of
the
given
value
a
fied head /z correspondsto
pump
the
of
head
the
losses),
corresponding
the
(together
with
flow rate increases
.13
F I G U R5E
Pump'
Centrifugal
of a Generic
RateCharacteristic
Head-Flow
Modeling 159
5.1 LiquidSystems
decreases.An equation similar to Eq. (5.63) can be written when using pressure
insteadof head:
p = po- Koaj
(5.64)
wherepo is the maximum attainablepressureat zeto flow rate and Ko is a constant
dependingon the pump construction.
W m.x.sLiquidSystems
Assembling severalof the liquid elementspresentedthus far in this section leads to
the formation of liquid systems.When only inertance and capacitancesare present
in a liquid system and no forcing source is considered,the natural responsecan be
formulated similarly to mechanical and electrical systems.When external action or
energyis applied in any form in a liquid system,the responseis forced. Both liquid
systemresponsesare studied next.
NaturalResponse
We study the natural responseof free losslessliquid systemsthat are describedby
one single variable (single-DOF systems) as well as for systems whose response
needsto be formulated in terms of more than one liquid-system variable (multipleDOF systems).The natural frequenciesand correspondingmodes (eigenvectors)of
multiple-DOF systemscan be calculated analytically or by MATLAB@, as shown in
previouschapters.
LiquidSystems
Conservative
Single-DOF
Considera lumped-parameterliquid systemthat is definedby inertanceI and capacitanceCt. This systempossessesa natural frequency,which can easily be found using
the energymethod, similar to the modality usedto determinethe natural frequencies
of single-DOF mechanicalor electrical systems.The demonstrationof the following
naturalfrequency
I
,[Le,
(s.65)
is given in the companion website Chapter 5. This natural frequency is very similar
to the natural frequency of an electrical system formed of an impedance L and a
capacitanceC, as discussedin Chapter 4. It can be checkedthat o, of Eq. (5.65) is
measuredin s-1, which is identical to rad-s-r, the unit of natural frequency'
5.7
Example
pipesegmentof givenlength/ and innerdiameterd throughwhicha liquid
A lossless
properties
flowsneedsits hydraulicnaturalfrequencyreducedby 20%. What
of known
can
designchanges be madeto achievethatgoal?
l7O
CHAPTER
5 FluidandThermal
Svsrems
Solution
Based
on Eqs.(5.26)and(5.37),
thepressure-defined
liquidinertance
andcapacitance
(where
p hasbeendropped)
thesubscript
are
,
-1
wt-
pL
4pt
A
rd2
(5.66)
n
+pg
By substituting
Eqs.(5.66)intoEq.(5.65),the hydraulic
natural
frequency
of the pipe
segment
becomes
,,=
ts
| j
(5.6i)
As a side note,the particular
naturalfrequencyof Eq. (5.67)is identicalto the natural
frequency
of a simplependulumof length/ underthe actionof gravity.
(5.67)indicates
Equation
that changesin the naturalfrequencycan be operatedby
usinglengthalterations.
Therequirement
is thatthe newnaturalfrequencvbe
,lin- 0.2an:0.8t0,
(s.68)
w h i c h ,b a s e do n E q .( 5 . 6 7 )i,s
E
"n
4/
vt
,+
(5.6e)
A combination
of Eqs.(5,67),(5,68),and (5.69)yieldsthe newtength:
."t
| =
0S+:
1.561
(s.70)
which showsthat an increaseof 56% in the pipe lengthis necessary
to producea
reduction
of 20% in the naturalfrequency.
I
Multiple-DOF
Conservative
Liquid Systems
The natural responseof multiple-DoF conservativeliquid systemsis studiedemploying the analytical approachand MATLAB@. Schematicliquid circuits can be drawn
and analyzed similarly to electrical systems,as shown in the following example. It
should be mentioned that, while relative agreementexists in terms of the symbols
usedfor hydraulic elementssuch as pumps, actuators,or resistances,there is no consensuson the graphical representationof liquid inertancesand capacitances.Due to
the similitude between electrical and hydraulic systems,the symbol used for electrical inductancesis used for hydraulic inertances,whereashydraulic caDacitancesare
symbolized similar to electrical capacitances.
Modeling 17l
5.1 LiquidSystems
AnalyticalApproachTheanalyticalapproachto findingthenaturalfrequencyof
multiple-DOFliquid systemsis similarto the approachusedfor mechanicaland
electricalsystemsin Chapters3 and4 andis illustratedby an example.
5.8
Example
Utilizethe energymethodto derivethe mathematicalmodelfor the liquid system
and deterwhosecircuitis sketchedin Figure5.14. Calculateits naturalfrequencies
a
p
p
r
o
a
c
hc. o n s i d e r
(
e
i
g
e
n
v
e
c
t
o
r
s
)
a
n
a
l
y
t
i
c
a
l
t
h
e
b
y
m i n et h e c o r r e s p o n d i nmgo d e s
=
3
X
1
0
8
m
4
S
2
k
g
1
'
C
t
C
P
=
C
n
=
l n = l B = l r = 2 X 1 0 6k g l m aa n d
Solution
and
Eq.(5.21)
ontheenergy
5.14andbased
inFigure
indicated
flows
thevolume
Using
is
components
tothefourliquid
corresponding
thetotalenergy
Eq.(5.30),
n,=|ru;,1.
W
* +,i
+ lr,,;,i
(5.71)
whichusesthe
velocity,
withtranslatory
volume(notto be confounded
withy indicating
iszero,whichleadsto
itstimederivative
therefore
isconstant,
Thisenergy
samesymbol).
o
vr)+
;,, * [t,,- ";]+ ;,,1r,v,*
*",]=
*rr,[r,,i;,
(s.72)
(5.72)needsto be satisfied
at alltimes,and sincethe volumeflowratescannot
Eouation
Eq.(5.72)is when
bezeroat alltimes,the onlywayof validating
o
*
[t,u'+,,-*r,=
(s.73)
*),,=o
I,,r,-*,,*(+.
to Eq.(5.73)is harmonic:
Thesolution
= Vrsin(tor)
[rt
[v, = Vrsin(to/)
I
5.14
FIGURE
LiquidSystem.
Conservative
Two-Mesh
(s.74)
L72
Svstems
5 FluidandThermal
CHAPTER
s .q u a t i o n( s5 . 7 4 )a r es u b s t i t u t ei nd t oE q s .( 5 . 7 3 )
w h e r eh a n dV z a r ev o l u m ea m p l i t u d e E
cannotbe zeroat
system,as sin(coD
equations
algebraic
and the resultis the following
a l lt i m e s :
- ,'r,,)v,
*r,=o
I(+
o
l-*, . (+. +,-,'r,,)v,=
(s.75)
UJ
ta
fo
in l/r and 7zwhenthe
solutions
system(5.75)hasnontrivial
equations
Thehomogeneous
of the syslemis zero:
delerminant
-'"" -*
l/,,
l=n
| ; 1,.+-,',,,1
(s.16)
w h i c h ,b y a l s ou s i n gt h e s p e c i f i cv a l u e so f t h e i n d u c t a n c easn d c a p a c i t a n c eosf t h e
p r o b l e mr,e s u l t isn t h ec h a r a c t e r i set iqcu a t i o n
t' cl .t o
c; z!-'+l
=o
(s.77)
of the hydraulic
the two naturalfrequencies
The solutionof this equationin o provides
system:
Onl
(s.78)
=
2ILCt
?re on1= 2.5 radlsand
the naturalfrequencies
dataof the problem,
Withthe numerical
,Jn2
= 6.6 radls.
T h ef o l l o w i nagm p l i t u drea t i oi s o b t a i n efdr o mt h ef i r s tE q .( 5 7 5 ) :
v,:
V2
I
l -
(s.1e)
azI,C,
) n dr e q u i r i ntgh a tt h ec o r r e s p o n d ienigg e n v e c tiosru n i t - n o r mt h, e
U s i n go t r ri n E q .( 5 . 7 9 a
f o l l o w i negq u a t i o nasr eo b t a i n e d :
{fi3 t:o= 1
(f),:,.,=+= 1e;
As
V
(s.80)
5.1 LiquidSystems
Modeting 173
Theeigenvector
corresponding
to Eqs.(5.g0)is
{v}*-.-,
'" : iI,'l- {0.8s}
tz.l t0.53
lv
(5.81)
J
Asseenin Eq.(5.81),the twoflowsyr and vzhave
identical
directions
and theamplitude
t{ is largerthan the ampritudevz duringthe
modarmotionat the resonantfrequency
onr.The secondnaturalfrequencyof Eq. (5.78)
is now substituted
intothe amplitude
ratioof Eq. (5.79)and a unit-normeigenvector
is againsought,which resurtsin the
following
equations:
v,, -0.62;./vl+$=1
/%\
v
\%L=,,,= u=
(s.82)
whosecorrespond
ingeigenvector
is
=
{v}^=,,,,
{Yt:}:{;l;i'}
(5.83)
I n s p e c t i oonf E q .( 5 . 8 3 )s h o w st h a t ,d u r i n g
t h e s e c o n dm o d a rm o t i o na t o r , z t, h e t w o
f l o w sh a v eo p p o s i t ed i r e c t i o n s( t h i si s i n d i c a t e d
b y t h e m i n-uvsvs iog"n/ )u
a "n\d t h e a m p r i _
t u d e l / r i s s m a l l e rt h a n t h e a m p l i t u d et / 2 .
I
usingMATLAPto catcutate
NaturatFrequencies,
probtem
theEigenvatue
Similar
to mechanicaland electrical systems,the natural
responseof a murtiple-DoF hydrau_
Iic system can be formulated in vector-matrix
form as an eigenvarueproblem, and
can be employed to solve for eigenvaluesand
eigelnv""ro.r.Th" equation
{Ar!fn@
describingthe free vibrations of a multiple-boF
hydraulic ,yrt". can be written as
l4l{t} + [cJ {v} : {0}
(s.84)
where [4] is the inertance matrix,
[ci] is the capacitancematrix,and {u} is the volumevector.It can be shown by following a
developmentsimilar to the one applied
to mechanicalsystemsin chapter 3, that when
sinusoidalsolutionof the type
{u} =
{ v} sin(ror)is soughtfor Eq. (5.g4),rhe following equationis
obtained:
der([I,]-'
[cJ - llr]) = o
15R5)
yh":: 1 : 0)2arerhe eigenvalues,[1] is the identiry matrix and
[It]_t[Ci= [D7]is the
liquid dynamic matrix.
O n c e[ D 7 ]h a s b e e n d e t e r m i n et h
de
, MATLAB@command
tV, Drl : ej9(Dr ),
which was utilized in chapters 3 and4,returns
the modal matrir % ,rror" columns
are the eigenvectors,and the diagonal matrix
D1, whose diagonal elements are the
eigenvalues.
174
CHAPTER
5 FluidandThermal
Svstems
Example5.9
Calculate
theeigenvalues
andtheeigenvectors
corresponding
to thehydraulic
system
of
Example
5.8usingtheeigenvalue
method
andMATLAB@.
Solution
(5.75)canbearranged
Equations
in thematrix-vectorform
of Eq.(5.84)with
111 -c"
C
= !,,];rct=
r1,r
[t
-
l
I
I
1l
I
(s.86)
c,, 1- c"l
By takingintoconsideration
thatthe inertances
are identicaland the conductances
are
alsoidentical,
the dynamicmatrixis calculated
usingMATLAB@
symboliccalculation:
r r 1 _l l
l D , l = V , j - ' [ c Ju=, l _ ,
;l
(s.87)
As expected,the eigenvalues
returnedby MATLAB@
are the squaresof the naiural
frequencies
of Eqs.(5.78)and the eigenvectors
are
: {-3.!i};
={;l;T}
{v}.=,.,
{v},=,.,
(s.88)
w h i c ha r e e s s e n t i a ltlhy e e i g e n v e c t oor sb t a i n e di n E x a m p l e5 . 8 ( t h e m i n u ss i g n si n
the first eigenvector
of Eq. (5.88)showsthat the two componentsmovein the same
direction),
I
ForcedResponse
of Liquid-Level
Systems
As mentioned previously in this chapter, one modality of generating the forced
responseof liquid systemsis by meansof tanks, where flow connectsin and out with
pipe lines,for instance.Often, in suchliquid systems,known as liquid-levelsystems,
hydraulic calculationscan be performed without considering the inertia effects; as a
consequence,
suchsystemsconsistofonly capacitances
and resistances.
Example
5.10
A liquid-level
systemis formedof a tank of capacitance
with a pipe
0 communicating
segmentequippedwitha valveof resistance
Rr,as sketchedin Figure5.15.
a. Derivea mathematical
modelfor thissystemby connecting
the output(eitherthe flow
raleqoorthe headh) to the inputflow rateqi. Assumethatthe pressure
at the input
and outputportsis zero.
b. Fora unit rampinput,find the solution9,(f)and plotit againsttime considering
that
R r , : 5 x 1 0 4N - s / m sa n d C 1 , r =2 x 1 0 - 6 m 5 / NT
. h es l o p eo f t h e r a m pi n p u ti s Q =
10 a m3/s.
Modeling L75
5.1 LiquidSYstems
I I G U R5E. 1 5
withTankandValve'
System
Liquid-Level
Solution
the fol|owing
and capacltances,
definitionsof resistances
a.lf we usethe Pressure
canbewritten:
equations
= R,.rq.(t)
[nO
dp(t)
I
-,
'' al
I q,lt)- q,,(t\= C,.,,
t"
(s.89)
w h e r e t h e p r e s s u r e i s a t t h e b o t t o m o f t h e t a n k j u s t b e f o r e t h e v athe
|ve.Takingthetime
is constant, fo||owthe resistance
of the firstEq.(5.89)and considering
derivative
results:
ingequation
dp(t)
d'
dq"(t)
= ^x'''-dt
(s.90)
intothe secondEq (5 89)' produces
whlch,substituted
n,.rcrr#*q,,(t)=1iQ)
( s.el)
Equation(5.91)representsthemathematica|mode|ofthe|iquid-Ieve|systemof
Figure5.i5,whereqiistheinputandgoistheoutput.SincetheorderofthedifferentiaI
one'
systemis a first-order
is one,the physical
equation
that the pressurebeforethe
considering
by
obtained
be
can
model
An alternate
tsactuallY
resistance
pft\ = Pgh(t)
ri Q?)
It is simPleto seethat
dp(t)
dt
dh(t)
= Pg
dt
(s.e3)
176
5 Fluidand ThermalSystems
CHAPTER
(5.92)and(5.93)are usedin conjunction
with Eq.(5.89)to obtainthe firstEquations
:
eouation
orderdifferential
lExa
Der
ino tl
R,.o
^ dh(r) .
* h{D =
R,.,C,r;
7f ot?'t
(5.94)
model,wherethe outputis the headh and the
anothermathematical
which represents
qi.
flow
rate
inputisthe
is th
Solt
Bas
theI
qi = Qf(whereQ is a constant),
the
b. Whenthe inputflowrateis of a rampform,namely,
Eq.(5.91)is
solutionto the differential
r
|
/
\l
q , , ( t ) =L r + \ r - . - l ) r l Q
(5.95)
dC h a p t e r 4 .
M A T L A B @ c o m mdasnodl v e , a s i n t r o d u c ei n
w h i c hc a nb e f o u n du s i n g t h e
=
values
of
thisexamfor
the
numerical
value
of
0.1
s
has
a
r
RpA,p
constant,
Thetime
ple.Dueto itsrampnature,the inputflowrategrowsto infinitywhentimegoesto infinity,
Eq.(5.95):
codegenerates
andthisis shownin the plotof Figure5.16.Thefollowing
) ) d s o l v e ( ' R * C * D q o+ Q O: q * t ' , ' q o ( 0 )
= 0')
I
afterdefiningthe symbolicvariables.
Th
wl
x 1 03
th
0.8
o
o
+c 0.6
0.4
0.2
012345678910
Time (sec)
F I G U R5E
.16
of Time.
OutoutFlowRateas a Function
s odeling 177
5 . 1 L i q u i dS y s t e mM
E x a m p l e5 . 1 1
s k e t c h e idn F i g u r e5 . 1 7b y c o n s i d e r D e r i v e t hm
e a t h e m a t i cm
a lo d eol f t h e l i q u i ds y s t e m
pr and the flowrateq2,whereasthe output
ingtheinputsto the systemarethe pressure
istheflow rale qo.
Solution
as wellas on Figure5.17,
and resistance
definitions,
capacitance
Based
on the hydraulic
relationships
can be formulated:
thefollowing
pr(t)-pt(t) ^
o,(t\- p.ft)
R',' , = - *
\Rn --./\rr-q,(t)
q,(tl
L I
"
-
q , ( t )- q ' ( t )
iCn
p,(t)-p"
q,(rl
qlt)+ qr(t)- q,,(t)
dp)\tl
dPr(t)
dt
dt
(s.e6)
p as
ThethirdEq.(5.96)allowsexpressing
(s.e7)
h(t):R,.q,,(t)*P,,
whichcan be usedto obtainqr fromthe fifth Eq.(5.96),
da (t)
q t G ' t= R , r C , r t d r t * q , , ( t l- q z ( t )
(5.98)
s xpressing
C o m b i n i nngo wt h e s e c o n dE q .( 5 . 9 6 )w i t h E q s .( 5 . 9 7 )a n d ( 5 . 9 8 )e n a b l e e
p:
thepressure
dq.,(r)
pzftl = Rt2Rt3Ct2;
+ ( R , t - l R , r ) q , , ( t-) R , t q r ( t )* p "
r 5 qq'\
lr"
V
Pt
Rn
Pz
Pump-nf l G U R5E
.17
withPump,TwoTanks,andThreeValves
Liquid-Level
System
Ps
Rrs Pa
--------->
no
178
Systems
5 FluidandThermal
CHAPTER
ThefirstEq.(5.96)and Eq.(5.99)yield
- p"l (s.100)
- R t 2 R t 3dq,(t)
q"(t)+ n,rqr(t)
c t 2 - (R,, * R,r)
Q i u=)f i [ o ' , u
dt
withEqs.(5.98)and( 5 . 1 0 0 ) t o
thefourthEq.(5.96)is usedin conjunction
Eventually,
equation:
differential
produce
second-order
thefollowing
Rt)Rt.Cr2 ^ ^ldq,,(tl
+ R'.cn]i
+ -ftf
+ R/i)C/r
[(R,,
d'zq^(t) |
Rt2Rr3C^C,r:il+
dqr(t)
. R , , * R i 2+ R i r
*-ffq .(t) =R,c,,:it
+
R / r + R / 2^ , . , , p t ( l )
R- q,( l) + &
(s.101)
Pu
- R-
for qo, is the mathematical
Equation(5.101),whlch can be solvedindependently
I
modelof the liquidsystemof Figure5.i7.
MODELING
SYSTEMS
S-trPNEUMATIC
In pneumatic systems,the motion agent is a gas,most often air. Gasesare compressible, particularly at large velocities; therefore,pneumatic systemsproduce responses
that are slower than liquid systems,but for velocities that are smaller than the sound
velocity, they are nearly incompressible. We discuss some basic gas laws then
introduce the pneumatic elementsand modeling of pneumatic systems.
S"#.TGaslaws
Gas laws describeeither the state or the transformation Qtrocess)between different
statesof a gaseoussubstance.The perfect (or ideal) gas law postulatesthat, for a
given gas statethat is defined by pressurep, volume % and absolute (Kelvin-scale)
temperature0, the following relationship applies for a gas mass of m:
pv= #Re
(s.102)
where R isthe universalgas constant and M is the gas molecular mass.If the
gas constantRru,is used,which is defined as Rr: RlM,the perfect gas law of
Eq. (5.102)becomes
(5.103)
pV : mRrQ
Equations(5.102)and (5.103)allow expressingthe gas massdensityas
'-
fl=-
mPMP
v
R0
Rro
(5.104)
Modeling 179
SYstems
5.2 Pneumatic
Transformationor processgas laws connect two statesusing specific conditions'
Thepolytropic transformatior is defined by the equation
lmY
p = a p" ' = o \ n
)
(s.10s)
wherea is a constantand n is thepolytropic exponent.Severalparticular transformapolytions, relating to actual physical conditions, can be derived from the general
gas
the
tropic transformation, each defined by a specific exponent n. All assume
exchange
heat
rurg12 is constant.The adiabatic transformation, which considersno
betweenthe gas and its sunoundings, has an exponentdefined as
cp
(s.106)
n=;
wherecois the constant-pressurespecific heat andcuis the constant-volumespecific
temheat.Tiespecific heat is defined as the heat (energy) Q necessaryto raise the
peratureof the massunit by one degree:
-f = -
o
(5.107)
mLj
and
Transformationsthat keep the temperature constant are called isothermal,
as
follows:
=
checked
be
easily
can
1.
That
is
n
exponent
the
for such processes,
(5.105)
Eq. (5.t6lZ)showsthatpV = constant,a condition that also resultsfrom Eq.
satisfy
when n = l. Constant-pressure transformations (also called isobarlc)
(5'102)'
This
Eq'
from
results
as
it
a
constant,
is
VlT
ralio
the
the condition that
(5.105).
=
Eq'
in
checked
be
can
0,
as
n
be
to
needs
exponent
meansthe polytropic
p/0 =
Eventually;,inrtint-rotume transformations rcsult in equations of the type
constant,u* ,""n in Eq. (5.102).It can alsobe checkedout that suchprocessesimply
-+ oo, Eq' (5'105)'
n'+ @ becauseV = constantwhen n
Elements
5.*.2Pneumatic
inerThepneumaticelementsare defined similarly to liquid systems,particularly the
the
chapter,
this
to
introduction
tanci and the resistance,but as mentioned in the
The
liquids'
for
operates
rate
that
massflow rate is used instead of the volume flow
pneumaticcapacitanceis discussedin terms of the specific gas transformation.
Ineftance
A column of gas moving in a duct possesseskinetic energy; therefore, its inertance
is definedas
Ir=
Lp
Lp
r,
pA"
P-I
- = - = - I t
A^
(s.108)
180 CHAPTER
5 FluidunOfhurrnut
Sffi
where 4 is the pressure-defined
inertanceof the liquid
at the beginning of this
chapter.The SI unit of 1, is m-r.
fire tcineiic'ei"rgy or-studied
u gu, i,
rr=*Lq'^=t*!o,ql=pr,
(s.109)
wherc Tt is the liquid kinetic energy;
the gas energy,s SI unit is kg2_m_r_s*2.
Capacitance
The pneumatic capacitanceof
a container is defined as
q,(t)
6- _ 4.(t) _ dm(t) = PLt
rrjpG)
"
dP(t)
dpft)
dt;
(5.I 10)
where c7 is the capacitanceof
a liquid. The SI unit for gas capacitance
is m_s2.For a
polytropic process,wherethe
pr"rrur" f, O"n""j as in Eq. (5.105),
ap= n
(s.111)
oLao
which indicatesthe capacitance
of Eq. (5.110)can be written
L
'
=-
dm
dp
Vda
dp
=_:
for constantvolume as
rl
(s.112)
nRrT
where Eq' (5'104) has been
used.It can be seenthat, for
a constant-pressure
trans_
tt:lff:T;,HH jl?j"rvtropic
"""rn;;;i; n = 0,tr'"pn"u.ui"capacitance
capacitanceisequar;;"::1ilX1[i1trJ"#,]:'l#,;#;Jh
depends
ontemperarure;
for anir"*r".,n"ii.iniro..u,ion o;t.t;;;e
ture is constant),
the pneumaticcapacitancei,
"onr,un,
C' ^ = V
R"0
The energy stored by a pneumatic
capacitive element
(5.1l3)
is
u,=*c,p,=|oc,pr=rs,
and its SI unit is N-kg-m-z (or
tempera_
and equal to
(s.I l4)
kg2_m-r-s-2.;.
Resistance
The pneumatic resistanceis defined
in terms of pressurevariation and
R,=*=#,=*o,
massflow rateas
(s.1
15)
Modeling 18t'
Systems
5.2 Pneumatic
5.18
FIGURE
Curveof a Fan'
Characteristic
in m-r-s-l in the
liquid' Gas-resistanceis measured
whereRr is the resistanceof a
resistanceis
pneumatic
it energy dissipatedthrough
InternationalSystem
"
"r,"iir.
= QIJ
at
u* = !n,o'^= | x f, o'n?,
(s.116)
(s1 ftg2-111-r-3-3';'
andthe SI energy unit is N2-s-m-3
EneryY
of Pneumatic
Sources
systemto allow
n"tds to be suppliedto a pneumatic
Similarto liquid system*,"n"'g-f
for which
blowers'
or
are thefi's
The main pneumaticenergysources
operation.
expressed
is
this
of the aii volume;
thedeliveredpressurets a paraboliciulctiol energylosses,suchasthosedue to
includesthe
bv meansof an equationrrru,urro
impellerfriction or shock:
ou=,,(+-tl
(s.1 7)
dependin Figure 5'18; cr arrd czareconstants'
whosecharacteristiccurve is shown
and performance'
ing on att" type of pneumatic source
SYstems
ffi g.x.gPneumatic
sYstems"is
yi:ry"i1::ssed next'
pneumatic
of conservative
Thenaturalresponse
of a pneumaticsystemwith
the forced
followedby an exampf"iifo't'uting
"'pon'"
losses.
ResPonse
Natural
by only inertanceand capacpneumaticsystemswith no losses,thereforedefined
similarlv to
in termsof their-"{T* t"tplllil
itanceproperties,can b;;;il;
system'the
systems'For a single-DoFpneumatic
liquid,mechanical,or "i";;;
t82
Systems
5 FluidandThermal
CHAPTER
naturalfrequencyiscalculatedbymeansofanequationsimilartoEq.(5.65),where
usingEqs.(5.108)
g (gas),h";i;L; usedinsteadof t, for liquid.Also
thesubscript
ls
system
and(5.110),thenaturalfrequencyof a pneumatrc
1
{ic,
=
@rJ
( 5 11 R )
therefore,asingle-DOFpneumaticsystemandasingle-DOFliquidsystemhave
identicalnaturalfreqoencies.ThecompanionwebsiteChapter5includesexamples
ofcalculatingthenaturalfrequenciesorsingle-andmultiple-DoFpneumaticsysto that
and MATLAST' Uot the procedure is identical
tems using the energy ;;,h"i
a
However'
and is not pursued here'
used for the natural responseof liquid systems
at end of this chapter'
iew proUtemsdedicatedto this topic are proposed
ForcedResPonse
Whenpneumaticsourcesareincludedinapneumaticsystem'thedynamicresponse
pneumatic system with resisis forced. We study the forced responseof a two-DOF
tance lossesand negligible inertia'
5.12
Example
AfanisusedtopressurizethecontainerofcapacitanceCgzasinFigure5.19,where
to thetargetvesselandthefan' Derivethe
csl is connected
anothervesselof capacitance
poto the
the outputpressure
systemthatconnects
modelof thispneumatic
mathematical
inputpressurecreatedbythefan,Pi,byalsoconsideringtheductlossesRgrandRg2'
Solution
components:
arewrittenfor the four pneumatic
equations
Thefollowing
R "qr
ptft) - p(t):
Q^,(t)
R-,=
- q^nul.
q''"(t)
p(tl - p-"(t).
-r r' -c", = 9-,u)
q.o(t)
4P,!!)
w
dt
dt
(5' I 19)
intothe fourthEq' (5'119)'
The massflow rateof the secondEq.(5.119)is substituted
p as
the pressure
whichenablesexpressing
- dP'(t)
p(t)=p.ft)*RrzCsz-;-
(5.120)
andthis resultsin
dp(t)=dp"ft)*R,C-.44!
'"82"82
dt
dt
dt'
(s.121)
Modeling 183
5.3 ThermalSYstems
5E
.19
FIGUR
andTwoValves
withFan,TwoContainers,
System
Fneumatic
( 5 . 1 2 0 ) a n(d5 1 2 1 ) '
T h e f i r s t a n d t h e t hEi rqds(. 5 . 1 1 9 ) a r e u s iendc o n l u n c t i o n wEi tqhs .
equation:
differential
whichyieldsthe iollowing
C*
R*,R*rC*,
d2p ,\t)
al
+
+ Rdcdry
+ (Rsr
(R,1csr
: pi(t) (5'122)
* p,,(t)
t l ey s o l v e dI o rp o , a n dE q .( 5 . 1 2 0 )w' h i c hc a n
E q u a t i o$n. I 2 2 ) , w h i c hc a n i n d e p e n d e n b
subsequentlybesolvedforp,formthemathematica|mode|ofthepneumaticsystemof
I
Flgure3. r:r.
MODELING
SYSTEMS
THERMAL
ffi,i$
variousstatesof a
In thermalsystems,the focus is on heatand massexchangeamong
to heatexchange'
mediumor differentmedia.The analysisin this sectionis restricted
in heat and mass
specializing
texts
from
learned
be
can
but more advancednotions
resistance,followed
transfer.We introduce the thermal elementsof capacitanceand
inertia can safely
thermal
Since
systems.
thermal
of
modeling
by the mathematical
To keep notation unitary
bl neglected,thermal systemJehave as first-order systems.
symbol Qrnor simply
the
systems,
pneumatic
and
fluid,
with that used for electrical,
time derivative of
the
as
defined
whichis
iate,
heat
the
q is usedhere to indicate
flow
theheatfl.ow,or thermal energyQ'.
dO(t\
q,r(tl=Sltl=i
(s.r23)
is preferred to the
Another notation used here is 0 for temperature; this symbol
time'
denote
to
svmbolt, which has been reserved
Elements
WS.i$.x Thermal
of interest afe the
As inertia effects are negligible in thermal systems,the elements
thermalcapacitance(involvedwiththermalenefgystoring)andthermalresistance
a lumped(responsiblefor energy losses). These amounts are assumedto be of
SyStemS
electrical
(current)
in
p*unr"t", nature.The role of the electrical chargerate