# What is the dimensional formula of tension

## Systems of linear equations

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1 2 Systems of linear equations Many simple models are based on linear relationships between different quantities. Problems with several variables and linear relationships between these variables lead to linear systems of equations. Even more complicated processes with non-linear relationships between the relevant parameters can be approximated by linear relationships within a validity range that is often sufficient for practice. In this chapter we will get to know linear systems of equations for the description of electrical networks in the direct current circuit and in the alternating current circuit as well as elastic frameworks and analyze their structure. Another important application is pipeline systems, for example for supplying houses or cities with water or gas; this is addressed in the tasks. Large systems of linear equations can also be obtained by numerical discretizations of partial differential equations; these systems of equations have a lot in common with the systems presented here. 2. Electrical networks Electrical networks are among the most elementary building blocks of the modern world. They are essential for the public power supply, but also for the functioning of many devices and machines, including small ones. We will first discuss the simplest case of an electrical network in a direct current circuit, which essentially consists of voltage or current sources and ohmic resistors. In particular, we are not yet looking at electronic components such as capacitors, coils, diodes or transistors. The network shown in Figure 2 should serve as a concrete example. Springer-Verlag GmbH Germany 207 C. Eck et al., Mathematical Modeling, Springer textbook, DOI 0.007 / _2

2 38 2 Systems of linear equations IV R 4 V R U 2 3 R 3 R R R 2 I U II III 2 Fig. 2 .. Electrical network The network essentially consists of edges in the form of electrical lines, nodes, which are connection points of two or more lines. To describe the network we need the following information about the physics of flowing currents: Kirchhoff's law of currents, too. Called Kirchhoff's law: The sum of the currents in each node is zero. This describes the conservation of the electrical charge, electrons can migrate through the network, but they cannot disappear into nodes or be created. Kirchhoff's law of voltages, also called Kirchhoff's second law: The sum of the voltages across each closed conductor loop is zero. From this follows the existence of potentials at the nodes; the voltage applied to a conductor section is given by the difference between the potentials at the end points. Ohm's law: The voltage drop U across the current-carrying resistor R with current I is U = RI. The current is measured in the system of units commonly used in Europe, the so-called SI system (Système International d Unités) in amperes (A), the voltage in volts (V) and the ohmic resistance in ohms (Ω). Here Ω = (V) / (A). One ampere corresponds to the flow of one coulomb per second, one coulomb corresponds to 6 elementary charges. With a few exceptions, we will omit the units in the exercises in the notation in this chapter.

3 To model the network we define: 2. Electrical networks 39 A numbering of the nodes, in Figure 2. from I V, as well as a numbering of the edges, in Figure 2. of 6. These numbering can be specified as desired. Of course, they influence the concrete representation of the model constructed from it and its solution, but not the physical interpretation of this solution. Establishing a positive direction for each edge. This is shown in Figure 2. by an arrow: The definition of the positive direction says nothing about the actual direction of the current, which we cannot yet know. We also introduce variables, these are the currents and voltages along the conductors and the potentials in the nodes. There are two different types of variables: Node variables, namely the potentials x i, i =, ..., m, where m is the number of nodes, here so = 5. These are combined into a potential vector x = (x, ..., x m) of dimension m. Edge variables are, for example, the currents yj, j =, ..., n, where n is the number of edges, here n = 6. These are in a current vector y = (y, ..., yn) der Dimension n summarized. Correspondingly, a vector e = (e, ..., e n) of the stresses can be formed. The voltages can be calculated from the potentials by ei = xu (i) xo (i), i =, ..., n, (2.) where u (i) is the index of the lower node and o (i) is the index of the top knot is. Bottom and top are defined by the direction of the conductor specified above, for example in Figure 2. I is the bottom and II is the top node of the conductor. An essential step now consists in converting the geometry of the network and the known physical laws into relationships for the introduced variables. This is done with the help of suitable matrices. The relationship (2.) between potentials and voltages can be written as e = Bx where the matrix B = {b ij} ni = mj = Rn, m is defined by if j = o (i), b ij = if j = u (i) and 0 otherwise. In the example of Figure 2, is

4 40 2 Systems of linear equations B = The matrix B is called the incidence matrix of the network. It only describes the geometry of the network, specifically the relationships between nodes and edges. In particular, the incidence matrix does not know its application as an electrical network. The same incidence matrix will also be obtained for pipelines with the same geometry through which there is flow. The incidence matrix has exactly once the entry and exactly once the entry in each line, all other entries are 0. It naturally depends on the numbering of the nodes and edges and the specified direction of the edges. The relationship between voltage vector e and current vector y is established by Ohm's law. You have to consider possible voltage sources. Figure 2.2 shows a piece of conductor with a resistor and voltage source. For the potentials x u, x z, x o the following applies with the voltage b j of the voltage source and thus x z = x u R j y j and x o = x z + b j e j = x u x o = R j y j b j. Here, R j y j denotes the voltage drop across the resistor through which current flows, and b j denotes the additional potential difference generated by the voltage source. In vector notation one obtains y = C (e + b), where C R n, n is a diagonal matrix, the entries of which are the conductance values R j, and b R n is the vector of the voltage sources. With the entries of b one has to pay attention to the sign: We have a positive sign if the polarity runs from to + in a positive direction, as in Figure 2.2, and a negative sign if the polarity runs from to + in a positive direction. In our example, / R / R C = 0 0 / R / R / R / R 6 U 0 and b = U

5 2. Electrical networks 4 + R j x u b x j x 0 z Fig To calculate the current Kirchhoff's current law is still missing. This has the form Ay = 0 in vector notation, where A = {a ij} mi = nj = Rm, n is defined by + if i = o (j), a ij = if i = u (j) and 0 otherwise. In our example, A = By comparison with the incidence matrix B, we see A = B. This is no coincidence, a comparison of the definitions of the entries a ij and b ij shows that this applies to every network. Kirchhoff's law of stress has already been built into the model, namely via the existence of the potential vector x. So we have processed all the information we know about the network. If you summarize everything, then the modeling of an electrical network of m consecutively numbered nodes and n consecutively numbered lines with a given direction is given by a potential vector x R m, a voltage vector e R n, which can be calculated using the incidence matrix BR n, m, e = Bx a current vector y R n, to be calculated by y = C (e + b) with the conductance matrix CR n, n and the vector b R n of the voltage sources and

6 42 2 Systems of linear equations according to Kirchhoff's current law B y = 0. The incidence matrix B describes the geometry of the network, the conductance matrix C the material properties and the vector b the external driving forces. To calculate the currents, voltages and potentials in the network, one selects a variable to be determined, for example x, and derives an equation for x by combining all relationships. One obtains or B C (b Bx) = 0 B CBx = B Cb. (2.2) The matrix M = B CB is symmetric if C is symmetric. From x, mx = Bx, CBx with the Euclidean scalar product, one sees that M is positive semidefinite if C is positive semidefinite, positive definite if C is positive definite and B only has the trivial kernel KernB = {0}. Indeed, for most networks, C is positive definite. However, the incidence matrix B has a nontrivial core. It is easy to see that (, ...,) R m is in the kernel of B. This applies to every incidence matrix, since every incidence matrix has exactly one entry + and one entry in every row and all other entries are 0. As a result, the linear system of equations does not have a unique solution. From a physical point of view, the reason is easy to see: The potentials are only unambiguous up to a constant, or only if you define a zero point for the potential. Since B has a nontrivial kernel, it is not clear from the outset that the system of equations is solvable at all. However, the following applies: Theorem 2 .. Let C R n, n be symmetrical and positive definite and B R n, m. Then for M = B CB: (i) KernM = KernB, (ii) The system of equations Mx = B b has a solution for every b R n. Proof. To (i): Obviously, CoreB CoreM applies. For x core M, 0 = x, b CBx = Bx, CBx applies.

7 2. Electrical networks 43 Since C is positive definite, it follows that Bx = 0, i.e. x coreB. To (ii): The equation is solvable if B b core (M) holds. For x core (M) = coreM = coreB, the statement is proven. B b, x = b, bx = 0, In our example KernB = span {(,,,,,)}. One can therefore derive a system of equations with a positive definite matrix by setting one of the potentials. If you set x 5 = 0, then you have to delete the 5th column in the incidence matrix, i.e. set B =, and use the vector x = (x, ..., x 4) R 4 instead of x R 5. For the numerical example R = R 2 = R 3 = R 4 = R 5 = R 6 =, U = 2, U 2 = 4 one has C = IR 6,6 and b = (2,0,4,0,0 , 0), so B CB = BB = and B Cb = B b = The solution to the system of equations is x = 2. From this one can calculate the voltages and currents, 2 0 e = Bx = 3 0 and y = e + b = 0. An obvious question is whether the incidence matrix of a network always satisfies KernB = span {(,, ...,)} (2.3). The answer to this depends on the following geometric property of the network:

8 44 2 Systems of linear equations Definition 2.2. A network is called connected if you can connect two nodes by a path made of edges. Statement (2.3) is equivalent to saying that the network is connected. Specifically, the following applies: Theorem 2.3. For a network with incidence matrix B, the following statements are equivalent: (i) The network is connected. (ii) B cannot be brought into the form () B 0 B = 0 B 2 by rearranging rows and columns with BR n, m, B 2 R n2, m2, n, n 2, m, m 2. ( iii) kernB = span {(,, ...,)}. Equation (2.2) is not the only way to construct a linear system of equations from the physical relationships between voltages and currents in the network. For example, one can view both x and y as variables to be calculated. If you write Ohm's law in the form Ay = e + b with A = diag (r, ..., rn) = C, you get the system or Bx + Ay = b, B y = 0, () (AB yb B 0) (=. (2.4) x 0) This formulation is particularly useful when one of the ohmic resistances is equal to zero and therefore the matrix C can no longer be formed. Networks in the AC circuit We will now extend the model described to AC circuits with additional components. In an alternating current circuit, there is a current that oscillates over time with a specified frequency, for example I (t) = I 0 cos (ωt).

9 a. 2. Electrical networks 45 In the case of an ohmic resistance, the voltage drop is given by U (t) = RI 0 cos (ωt) = U 0 cos (ωt) with U 0 = RI 0. An alternating current circuit with ohmic resistances without additional components can therefore be used how to describe a direct current circuit, if one uses the amplitudes I 0 and U 0 for current strength and voltage instead of the constant current strengths and voltages of the direct current circuit. Additional electrical components add new effects. We consider capacitors here, denoted by the symbol A capacitor can store electrical charges. The amount of charge stored is proportional to the applied voltage. When the voltage changes, a capacitor can therefore absorb or output currents. This is described by the relation I (t) = C U (t), where C is the capacitance of the capacitor. In an alternating current circuit with I (t) = I 0 cos (ωt), U (t) = I 0 Cω sin (ωt) = I 0 Cω cos (ωt π / 2). So there is a phase shift of π / 2 between current and voltage. Coils, denoted by the symbol. A current-carrying coil generates a magnetic field, the strength of which is proportional to the strength of the current. Energy is stored in the magnetic field; this must be taken from the current in the coil when the magnetic field is built up. This results in a voltage drop across the coil that is proportional to the change in current strength, U (t) = LI (t), where L is the inductance of the coil. In an alternating current circuit, U (t) = LI 0 ωsin (ωt) = LωI 0 cos (ωt + π / 2). So there is a phase shift of π / 2 here. The occurring phase shifts make the calculation more complicated here than with the direct current circuit. It is useful to represent current and voltage with complex numbers. Euler's formula e iϕ = cosϕ + isinϕ is used for this. It then applies

10 46 2 Systems of linear equations cos (ωt) = Re (e iωt) and sin (ωt) = Re (ie iωt). If the current intensity is represented as I (t) = Re (I 0 e iωt), then it follows for the voltage drop across the ohmic resistor at the capacitor and the coil U (t) = Re (RI 0 e iωt), (U (t ) = Re i) ωc I 0e iωt U (t) = Re (iωli 0 e iωt). This can be represented by complex impedances R, i ωc and iωl, which take on the role of the real resistances. As with real ohmic resistances, you can also add complex impedances if you have more than one component in the same section of conductor. The total impedance of a conductor with an ohmic resistance of strength R, capacitor of capacitance C and coil of inductance L is therefore R i ωc + iωl. Example: We consider the network shown in Figure 2.3 with m = 5 nodes and n = 6 edges, the nodes and edges are already numbered there, the positive directions are given. The applied voltages U (t) and U 2 (t) are given by U (t) = U 0 cos (ωt) and U 2 (t) = U 02 cos (ωt). It is important here that the frequencies are the same, otherwise there would be no alternating current network with a known frequency. The phase shifts of U and U 2 are also the same here, but it is not difficult to take different phases into account in the model. To model the network, we use a potential vector x C m, a current vector y C n and a voltage vector e C n.

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