082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

614 Chapter 20 Ordinary differential equations II
Finally,anoutputequationisneeded.Inthiscaseitistrivialastheoutputvariable
isthe same as one of the statevariables. So,
ω m
= ( 0 1 )( )
i a
ω m
Engineeringapplication20.3
State-spacemodelforacoupled-tankssystem
Derive a state-space model for the coupled tank system shown in Figure 20.9. The
tankshavecross-sectionalareasA 1
andA 2
,valveresistancesR 1
andR 2
,fluidheights
h 1
and h 2
, input flows q 1
and q 2
. Additionally, there is a flow, q o
, out of tank 2.
Assumethatthevalvescanbemodelledaslinearelementsandletthedensityofthe
fluid inthe tanks be ρ. Letq i
be the intermediate flowbetween the two tanks.
q 1
q 2
h 1
q i
qo
h 2
Tank 1, R 1 Tank 2, R 2 Figure20.9
area A 1 area A 2 Coupled tanks.
Solution
A convenient choice of state variables is the height of the fluid in each of the tanks,
although a perfectly acceptable choice would be the volume of fluid in each of the
tanks.For tank 1,conservation of mass gives
q 1
−q i
=A 1
dh 1
dt
Fortheresistanceelement,R 1
,thepressuredifferenceacrossthevalveisequaltothe
productoftheflowthroughthevalveandthevalveresistance.Thiscanbethoughtof
asafluidequivalentofOhm’slaw.Notethatatmosphericpressurehasbeenignored
as itisthe same on both sides of the valve. So,
ρgh 1
−ρgh 2
=q i
R 1
q i
= ρg (h
R 1
−h 2
) (20.13)
1
Combining these two equations gives
dh 1
= − ρg h
dt R 1
A 1
+ ρg h
1
R 1
A 2
+ 1 q
1
A 1
(20.14)
1
For tank 2,
q i
+q 2
−q o
=A 2
dh 2
dt
ρgh 2
=R 2
q o
(20.15)