082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
510 Chapter 18 Taylor polynomials, Taylor series and Maclaurin seriesEngineeringapplication18.1Ad.c.electricalnetworkConsider the d.c. network of Figure 18.2. This network is a linear system. This isbecause the voltage/current characteristic of a resistor is linear provided a certainvoltage isnot exceeded. Recall Ohm’s lawwhich isgiven byV=IRwhereV is the voltage across the resistor,I is the current through the resistor andRistheresistance.Thismakestheanalysisofthenetworkrelativelyeasy.Thevoltagesources,V 1,V 2,V 3,V 4, can be thought of as the inputs to the system. It is possible toanalyse the effect of each of these sources separately, for example the voltage dropacross the resistor,R 5, resulting from the voltage sourceV 1, and then combine theseeffects to obtain the total effect on the system. The voltage drop acrossR 5when allsources are considered would be the sum of each of the voltage drops due to theindividual sourcesV 1,V 2,V 3andV 4.R V 1 2 V 3+ – + ––R 2 R 3 V 4 R 4 R 5++ –V R 6 R 1 7Figure18.2Ad.c. electrical network.Engineeringapplication18.2AgravityfeedwatersupplyConsider the water supply network of Figure 18.3. The network consists of threesourcereservoirsandaseriesofconnectingpipes.Wateristakenfromthenetworkattwopoints,S 1andS 2.Inapracticalnetwork,reservoirsareusuallyseveralkilometresaway from the points at which water is taken from the network and so the effect ofpressure drops along the pipes is significant. For this reason many networks requirepumps toboostthe pressure.The main problem with analysing this network is that it is non-linear. This is becausethe relationship between pressure drop along a pipe and water flow through apipeisnotlinear:adoublingofpressuredoesnotleadtoadoublingofflow.ForthisR 1P 1 P 2R 2 R 3Figure18.3P 3 P 4 P 5 P 6S 1 S 2Awater supply network.
18.2 Linearization using first-order Taylor polynomials 511reason, it is not possible to use the principle of superposition when analysing thenetwork. For example, if the effect of the pressure at S 1for a given flow rate wascalculatedseparatelyforeachoftheinputstothesystem--thereservoirsR 1,R 2,R 3--then the effect of all the reservoirs could not be obtained by adding the individualeffects. It is not possible to obtain a linear model for this system except under veryrestrictive conditions and so the analysis of water networks isvery complicated.Having demonstrated the valueoflinear models itisworthanalysing howand whena non-linear system can be linearized. The first thing to note is that many systems maycontainamixtureoflinearandnon-linearcomponentsandsoitisonlynecessarytolinearizecertainpartsofthesystem.AsystemofthistypehasbeenstudiedinEngineeringapplication 10.6. Thereforelinearization involves deciding which components ofasystemarenon-linear,decidingwhetheritwouldbevalidtolinearizethecomponentsand,ifso, then obtaining linearized models of the components.Consider again Figure 18.1. Imagine it illustrates a component characteristic. Theactual component characteristic is unimportant for the purposes of this discussion. Forinstance, it could be the pressure/flow relationship of a valve or the voltage/currentrelationship of an electronic device. The main factor in deciding whether a valid linearmodel can be obtained is the range of values over which the component is required tooperate. If an operating point Q were chosen and deviations from this operating pointwere small then it is clear from Figure 18.1 that a linear model -- corresponding to thetangent tothe curve atpointQ-- would be an appropriate model.Obtaining a linear model is relatively straightforward. It consists of calculating thefirst-order Taylor polynomial centred around the operating point Q.This isgiven byp 1(x) =y(a) +y ′ (a)(x −a)Then p 1(x)isusedasthelinearizedmodelofthecomponentwithcharacteristicy(x).Itisvalidprovidedthatitisonlyusedforvaluesofxsuchthat |x−a|issufficientlysmall.Asstated, p 1(x)isalsotheequationofthetangenttothecurveatpointQ.Therangeofvalues for which the model is valid depends on the curvature of the characteristic andthe accuracy required.Engineeringapplication18.3PowerdissipationinaresistorThe power dissipated in a resistor varies with the current. Derive a linear model forthispowervariationvalidforanoperatingpointof0.5A.Theresistorhasaresistanceof 10 .SolutionFor a resistorwhereP=I 2 RP = power dissipated (W)I = current (A)R = resistance ().➔
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- Page 517 and 518: 17.2 Trapezium rule 497y{ {y 0 y ny
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510 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Engineeringapplication18.1
Ad.c.electricalnetwork
Consider the d.c. network of Figure 18.2. This network is a linear system. This is
because the voltage/current characteristic of a resistor is linear provided a certain
voltage isnot exceeded. Recall Ohm’s lawwhich isgiven by
V=IR
whereV is the voltage across the resistor,I is the current through the resistor andR
istheresistance.Thismakestheanalysisofthenetworkrelativelyeasy.Thevoltage
sources,V 1
,V 2
,V 3
,V 4
, can be thought of as the inputs to the system. It is possible to
analyse the effect of each of these sources separately, for example the voltage drop
across the resistor,R 5
, resulting from the voltage sourceV 1
, and then combine these
effects to obtain the total effect on the system. The voltage drop acrossR 5
when all
sources are considered would be the sum of each of the voltage drops due to the
individual sourcesV 1
,V 2
,V 3
andV 4
.
R V 1 2 V 3
+ – + –
–
R 2 R 3 V 4 R 4 R 5
+
+ –
V R 6 R 1 7
Figure18.2
Ad.c. electrical network.
Engineeringapplication18.2
Agravityfeedwatersupply
Consider the water supply network of Figure 18.3. The network consists of three
sourcereservoirsandaseriesofconnectingpipes.Wateristakenfromthenetworkat
twopoints,S 1
andS 2
.Inapracticalnetwork,reservoirsareusuallyseveralkilometres
away from the points at which water is taken from the network and so the effect of
pressure drops along the pipes is significant. For this reason many networks require
pumps toboostthe pressure.
The main problem with analysing this network is that it is non-linear. This is because
the relationship between pressure drop along a pipe and water flow through a
pipeisnotlinear:adoublingofpressuredoesnotleadtoadoublingofflow.Forthis
R 1
P 1 P 2
R 2 R 3
Figure18.3
P 3 P 4 P 5 P 6
S 1 S 2
Awater supply network.
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