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

10.8 Differentiation as a linear operator 383
and has a constant value independent of the operating point:
δq i
=k p
δv in
(10.3)
Therelationshipbetweenpressureatthebottomofthetankandthewaterheight
isalsolinear.
Since p = ρgh we have dp dp
= ρg. Because is a constant wecan write
dh dh
δp
δh = dp
dh = ρg
δp= ρgδh (10.4)
Thechangeincontrolvoltage, δv in
,isgivenas0.4.Wealsoknowthatk p
= 0.03.
Therefore, using Equation (10.3),we have
δq i
= 0.03 ×0.4 = 0.012
Now if time is allowed for the system to reach equilibrium with this increased
inputflow,then δq o
= δq i
.Inotherwords,theoutputflowincreasesbythesame
amount as the input flow and the water height once again stabilizes to a fixed
value. Therefore,
δq o
= δq i
=0.012
UsingEquation (10.2),wefind
δp =14000 δq o
=14000 ×0.012 =168
UsingEquation (10.4),weget
δh = δp
ρg = 168
998 ×9.81 = 0.0172
Therefore the new water height is 0.25 +0.0172 = 0.267 m to three significant
figures. The new water flowrate is0.35 +0.012 = 0.362 m 3 s −1 .
Torecap,alltheelementsofthefluidsystemwerelinearapartfromthevalve.
Byobtainingalinearmodelforthevalve,validforvaluesclosetotheoperating
point, it was possible to calculate the effect of changing the control voltage to
themotor.Itisimportanttostressthatthelinearmodelforthevalveisonlygood
forsmallchangesaroundtheoperatingpoint.Inthiscasetheincreaseincontrol
voltage was approximately 3%. The model would not have been very good for
predictingtheeffectofa50%increaseincontrolvoltage.Linearmodelsofnonlinear
systems are particularly useful when several components are non-linear,
astheyaremucheasiertoanalyse.Weexaminetheseconceptsinmoredetailin
Chapter 18.
EXERCISES10.8
1 Differentiate the following functions:
(a) y = 4x 3 −5x 2
(b) y = 3sin(5t) +2e 4t
(c) y = sin(4t) +3cos(2t) −t
(d) y = tan(3z)
(e) y =2e 3t +17−4sin(2t)
(f) y= 1
t 3 + cos5t
2