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

10.8 Differentiation as a linear operator 381
v in
Tank Valve
Pump/motor
q i
h
p
q o
Figure10.18
Afluidsystem comprising pump, tank
andvalve.
where k p
= pump/motor constant, v in
= control voltage (V), q i
= flow rate into
the tank (m 3 s −1 ). The valve has a non-linear characteristic given by the quadratic
polynomial
p = 20000q 2 o
where p = pressure at the base of the tank (Nm −2 ), q o
= flow out of the tank
(m 3 s −1 ).Thefluidbeingusediswaterwhichhasadensity ρ = 998kg m −3 .Assume
g = 9.81 ms −2 and thatk p
= 0.03 m 3 s −1 V −1 . Carry outthe following:
(a) Calculate theflowrateoutofthetank,q o
,andthecontrol voltage, v in
,when the
systemisinequilibrium and the height ofthe water inthe tank,h, is0.25 m.
(b) Obtainalinearmodelforthesystem,validforsmallchangesaboutawaterheight
of0.25m.Usethismodel tocalculatethenewwaterheightandflowrateoutof
the tank when the control voltage isincreased by 0.4 V.
Solution
(a) The pressure atthe bottomof the tank isgiven by
p = ρgh = 998 ×9.81 ×0.25 = 2448
The flowrate through the valve isgiven by
q 2 o = p
20000
√ √
p 2448
q o
=
20000 = 20000 = 0.350 m3 s −1
Nowifthesystemisinequilibrium,thentheheightofthewaterinthetankmust
have stabilized toaconstant value. Therefore,
and so
q i
=q o
= 0.350
v in
= q i
k p
= 0.350
0.03 = 11.7 V
(b) Before answering this part it is worth examining what is meant by a linear
model. Figure 10.19 shows the valve characteristic together with a linear approximationaroundanoutputflowrateofq
o
= 0.350m 3 s −1 .Thiscorresponds
toawater height of 0.25 m.
➔