Dynamic Performance Characteristics of Second-Order Systems: Wind Vanes, Manuais, Projetos, Pesquisas de Meteorologia

Baixe Dynamic Performance Characteristics of Second-Order Systems: Wind Vanes e outras Manuais, Projetos, Pesquisas em PDF para Meteorologia, somente na Docsity! 8 Dynamic Performance Characteristics, Part 2 The first-order model discussed in chap. 6 is inadequate when there is more than one energy storage reservoir in the system to be modeled. If the sensor is linear it can be modeled with a higher-order dynamic performance model. Here the term 'system' refers to a physical device such as a sensor while the equation refers to the corresponding mathematical model. There exists a dual set of terms corresponding to consideration of the physical system or of the mathematical model. For example, an are coefficients of the mathematical model (see eqn. 8.1) but they also represent some physical aspect of the sensor being modeled; thus they can also be called system parameters. 8.1 Generalized Dynamic Performance Models The general dynamic performance model is the linear ordinary differential equation where t = time, the independent variable, x = the dependent variable, an = equation coefficients or system parameters, and x,(t) = input or forcing function. This equation is ordinary because there is only one independent variable. It is linear because the dependent variable and its derivatives occur to the first degree only. This excludes powers, products, and functions such as sin(x). If the system parameters an are constant, the system is time invariant. We can define the differential operator D = d/dt; then eqn. (8.1) can be written as 151 where the right-hand side includes one term for each functionally different form found by examining xf(t) and its derivatives. The constants A, do not depend upon the initial conditions. They are found by substituting eqn. (8.5) into eqn. (8.2). 8.2 Energy Storage Reservoirs Differential equations describe the behavior of physical systems in which a redistri- bution of energy is taking place. In a mechanical system, a mass in motion stores kinetic energy and may store potential energy by virtue of its position in a force field. 152 Meteorological Measurement Systems As before, the solution is x(t) = xT(t) + xs(t), where xT(t) is the transient solution and xs(t) is the steady-state solution. The transient solution has n arbitrary constants which may be numerically evaluated by imposing n initial conditions on eqn. (8.2). The first step in obtaining the transient solution is to calculate the roots of the char- acteristic equation where the operator D has been replaced with a simple algebraic variable, r. The roots of the characteristic equation, r^,r2, ...rn are used to obtain the solution with the following rules: (1) Real roots, unrepeated. For each real unrepeated root r one term of the solu- tion is written as Cert, where C is an arbitrary constant. (2) Heal roots, repeated. For each real root r which appears p times, the solution is written as (C0 + Qf + • • • + Cp_1t p~1)ert. (3) Complex roots, unrepeated. A complex root has the form a ± ib. If the coeffi- cients of eqn. (8.3) are real, which we usually expect, then for each root pair the corresponding solution is eat(c^ cos(irt) + c2 sin(bt)). (4) Complex roots, repeated. For each pair of complex roots which appears m times, the solution is The transient solution is simply the sum of the individual terms. To evaluate the constants GJ, there must be n initial conditions specified. As with the first-order equation, the steady-state solution can be found by the method of undetermined coefficients. It does not work in all conditions but is ade- quate for the present purpose. Given that the input is some function xt(t), repeatedly differentiate x,-(£) until the derivatives go to zero or repeat the functional form of some lower-order derivative. This is also the test for the applicability of the method; if neither of the above conditions prevails, the method of undetermined coefficients rannnt hfi used. Write the stfiaHv-state solution as This solution is plotted in fig. 8-1 for the damping ratio £ = 0, 0.2, 0.4, 0.6, and 0.8, where the amplitude ratio is x(t}/xc. This solution can be generalized for any step function by using the same notation as in eqn. 6.9 where XFS and x/s are the final and initial states of the step function and *c = XFS ~ x/s- Note that, for £ = 0, the solution is a cosine wave. This solution is included for reference; as noted above, it is not realized by any actual system. The various solu- Dynamic Performance Characteristics, Part 2 155 We can evaluate the constants c^ and c2 using the initial conditions: X(0) = ca = xc Then the solution is where the phase shift <b is given by eqn. 8.15. Fig. 8-1 Amplitude ratio, x(f)/xc, for damping various ratios. 156 Meteorological Measurement Systems tions shown in fig. 8-1 are for a device released from some initial position xc and coming to rest at a final position of x(t) — 0. For £ < 1, the system overshoots; that is, it goes from the position x(t} = xc to x(t) = 0 and then past that point to some position x(t) < 0. While some overshoot may be tolerated, excessive overshoot is unaccepta- ble, therefore, sensors are usually designed to have a damping ratio > 0.7. If the damping ratio f > 1, the roots are real and there is no overshoot. The solution is not oscillatory so it would be inappropriate to express the solution in terms of a frequency such as a>n. The transient solution will be of the form But, as noted above, it is inappropriate to express the solution in terms of frequencies, so let us define where r2 > TI since £ > 1. Then and eqn. 8.7 can be written in the form Using the step function previously employed, where x,-(f) = 0 with the initial condi- tions x(0) = xc and Dx(0) = 0, we can obtain the solution Note that if T2^> rlt then that is, the first-order step response. EXAMPLE Consider a step function input with the step from X{S = 0.00 to XFS = —5.00, where a)n — 0.50 and f = 0.30.Then the phase shift is and the undamped period is Tn — 12.5664 while the damped period is Trf = 13.1731. We can explore the solution, using eqn. 8.16, that will be similar to those plotted in fig. 8-1. Note that the cosine term in 8.16 will be zero when the argument of the cosine term, 8.3.2 Ramp Input A ramp input occurs when x,-(£) = at. Following the method of undetermined coeffi- cients, the trial steady-state solution will be xs(t) = k^ + k2t and substitution into eqn. 8.7 yields Iq = — 2a£/con and Ic2 = a. The steady-state solution is therefore where a is the constant slope of the input. A ramp input is shown as the upper line in fig. 8-2 and the complete solution, transient plus steady-state, is plotted as the lower line. In this case, u>n = 0.5, £ — 0.7, and a = 1. Fig. 8-2 Ramp input and output. Dynamic Performance Characteristics, Part 2 157 where / = 1, 2 , . . . . This can be solved for where time and the corresponding solution are listed in the table below. Time t 0.000 3.932 7.225 10.519 13.812 17.105 20.398 23.692 X(f) 0.000 -5.000 -6.773 -5.000 -4.340 -5.000 -5.246 -5.000 Condition Initial X(0) = 0 X(t) = XFS X(t) - local max. X(0 = XFS X(t) = local min. X(t) = XFS X(t) - local max. X(f) = XFS 160 Meteorological Measurement Systems Fig. 8-4 Amplitude ratio versus normalized frequency for c = 0.1, 0.2, 0.4, 0.6, 0.8, 1.0, and 1.2. where a>2n — N/I and £ = (R/2 V)^/N/I. In these equations, N, (aerodynamic torque per unit angle) I (vane moment of inertia), and R (aerodynamic vane lever arm) are assumed to be constants for a given vane since they represent physical parts of the vane. The u>n — f(V) but £ is not a function of V. But the wind speed V is not a constant except in the wind tunnel. In the atmosphere, the wind speed is a variable; therefore a)n is not a constant, and that violates the assumption that the parameters are time-invariant. Then the solu- tions in sect. 8.3, especially the sine input, will not be exactly correct. Normally the wind speed is a fluctuation about the average, so we can use the constant-parameter solutions, above, as an approximation. Then con = 2nV/Xn, where Xn is called the undamped wavelength which is a constant for a particular wind vane. We will use V as the average wind speed. Then we can use wavelength ratio in place of frequency ratio in figs. 8-4 and 8-5. We have used A as the wind input wavelength. The damped natural frequency and the damping ratio of a wind vane can be deter- mined from the step function response (see fig. 8-1) in a wind tunnel. The tunnel flow is held constant at speeds well above the threshold speed of the vane. To perform a step-function test, the initial vane deflection must be < 10° from the tunnel center- line. Beyond 10° deflection the vane will likely be in aerodynamic stall where the lift no longer increases with increasing deflection angle from the wind flow and the vane and this equation can be written as Dynamic Performance Characteristics, Part 2 161 Fig. 8-5 Phase shift as a function of normalized frequency for f = 0.1, 0.2, 0.4, 0.6, 0.8, 1.0, and 1.2. response will be nonlinear and therefore not predictable from eqn. 8.14. The test is usually performed at two speeds: 5 and 10ms"1. Then the dependence on wind speed can be observed. We used a number of terms that have developed in this chapter, such as the time constants Ta, r2, and the damping ratio £. Damped natural frequency, (DJ = &>n\A — £ 2 Damped natural period, Td = l//d = 2n/a)d Undamped natural frequency, con = 2nfn = 2n/Tn Undamped natural period, Tn — 1/fn = 27T/a>n, Now there are some special terms related to wind vanes because of their dependence on the wind speed: Damped natural frequency, a)d = 2nV/^.d Damped natural wavelength, kd = TjV — V/fa Undamped natural frequency, o)n — 2nV/kn Undamped natural wavelength, A,n = TnV = V/fn. Temperature sensors are portrayed as simple first-order sensors with a single energy storage reservoir - the thermal mass of the sensor itself. While it is possible to build a temperature sensor with a single thermal mass, it is difficult. Typically, the raw sensor is enclosed in a protective shield and, frequently, there are shields used to protect the sensor from mechanical damage and from moisture. Unless these protec- tive shields are fused together, which is not usually the case, there are two or more coupled thermal masses and then the sensor is a second-order or possibly even higher-order system. If there are just two thermal masses, the resulting second- 162 Meteorological Measurement Systems order system will have a damping ratio greater than unity and can be represented by eqn. 8.21. The step function solution would be eqn. 8.22. 8.5 Experimental Determination of Dynamic Performance Parameters The dynamic performance parameters, £ and 7^ (or co^ or &>„), can be determined from a step-function test. When £ < 0.6, one can measure the overshoot ratio and the damped period. The overshoot ratio is the ratio of the amplitudes of the two succes- sive deflections of a sensor as it oscillates about the equilibrium position. This can be illustrated as in fig. 8-7. Then £ and o>n can be found from two equations. Equation 8.33 is plotted in fig. 8-6. For example, a step-function test is shown in fig. 8-7 that illustrates some measure- ments which could be used to get the period and overshoots. Use sequential max- imum (minimum) and then minimum (maximum) amplitude. Table 8-1 is an example of the use of experimental data, and corresponds to fig. 8-7. EXAMPLE Determine £ and u>n from the data plotted in fig. 8-7. We can use maximum and minimum to get £. Use points 1 and 3. Xn = 0-5 = —5 and Xn+l = 6.55-5 = 1.55, so \Xn+l/Xn\ = 0.310. Then £ = 0.35. Fig. 8-6 Damping ratio as a function of overshoot for a second-order system with step input.