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

382 Chapter 10 Differentiation
p (N m –2 )
2448
0.35
q o (m 3 s –1 )
Figure10.19
Relationshipbetween pressureacross
valve (p)and flowthrough valve (q o ).
Alinearmodelforthevalveisoneinwhichtherelationshipbetweenpand
q o
isapproximated by the straightlinewhich forms a tangent tothe curve atthe
operatingpoint.Theoperatingpointisthepointaroundwhichthemodelisvalid.
Itisclearthatifthestraightlineapproximationisusedforpointsthatarealarge
distance from the operating point, then the linear model will not be very accurate.
However, for small changes around the operating point the approximation
isreasonablyaccurate.Clearlyadifferentoperatingpointwillrequireadifferent
linearapproximation.Inordertoobtainthegradientofthislineitisnecessaryto
differentiate the function relating valve pressure tovalve flow. So,
p = 20000q 2 o
dp
dq o
= 40000q 0
At the operating pointq o
= 0.350. Therefore,
∣
dp ∣∣∣qo
= 40000 ×0.350 = 14000
dq o =0.350
This value is the gradient of the tangent to the curve at the operating point.
Smallchangesaroundanoperatingpointareusuallyindicatedbythenotation δ.
Therefore,
δp
≈ dp ∣ ∣∣∣qo
= 14000
δq o
dq o =0.350
δp = 14000δq o
(10.2)
Note that equality has been assumed for the purposes of the linear model. It is
easy to relate a change in pump flow to a change in control voltage because the
relationship islinear and so a linear approximation isnotrequired. So,
q i
=k p
v in
Differentiating thisequation w.r.t. v in
yields
dq i
dv in
=k p
Inthis case
δq i
δv in
= dq i
dv in
=k p