ENGR202 2017 Q A1Past paperOld spec ENGR2022:125 marks30 min

ENGR202 Summer 2017 Q A1[VALID] official answers

The relationship between the pressure X(s)X(s) and the flow rate control input U(s)U(s) in an oil fractionator is described by the following Transfer Function model:

X(s)=b1s+a1 U(s)(1a)X(s) = \frac{b_1}{s + a_1}\,U(s) \qquad \text{(1a)}

Using standard notation, generalised first and second order differential equations are:

τdxdt+x=Ku(t)andd2xdt2+2ζωndxdt+ωn2x=Ku(t)(1b)\tau\frac{dx}{dt} + x = Ku(t) \quad \text{and} \quad \frac{d^2x}{dt^2} + 2\zeta\omega_n\frac{dx}{dt} + \omega_n^2 x = Ku(t) \qquad \text{(1b)}

Formulas you may need
  • Generalised forms: τdxdt+x=Ku(t)\tau\dfrac{dx}{dt} + x = Ku(t) and d2xdt2+2ζωndxdt+ωn2x=Ku(t)\dfrac{d^2x}{dt^2} + 2\zeta\omega_n\dfrac{dx}{dt} + \omega_n^2 x = Ku(t) (given in the question)
  • Frequency response: M=∣G(s)∣s=jωM = |G(s)|_{s = j\omega}, ϕ=Arg(G(s))∣s=jω\phi = \mathrm{Arg}(G(s))|_{s = j\omega} (given in the question)
  • Laplace with zero initial conditions: dxdt→sX(s)\dfrac{dx}{dt} \to sX(s), d2xdt2→s2X(s)\dfrac{d^2x}{dt^2} \to s^2X(s) (learn this)
  • Steady state gain: set s=0s = 0 (learn this)
  • Pole placement: match the closed-loop characteristic equation to s2+2ζωns+ωn2s^2 + 2\zeta\omega_n s + \omega_n^2 (learn this)
  1. (a)
    Showing your working and indicating any assumptions made, develop the Transfer Functions forms of the generalized first and second order differential equations (1b), expressed in terms of τ\tau, KK, ωn\omega_n and ζ\zeta.
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  2. (b)
    Assume a1=5a_1 = 5 and b1=2b_1 = 2 for this part of the question. Determine the pole and stability condition of the oil fractionator model (1a). Determine the time constant and steady state gain. For a flow rate u(t>0)=5u(t > 0) = 5, what is the steady state pressure.
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  3. (c)
    The steady state response of the oil fractionator model (1a) for u(t)=sin⁡(ωt)u(t) = \sin(\omega t) when t>0t > 0, is Msin⁡(ωt+ϕ)M\sin(\omega t + \phi). Determine relationships for MM and ϕ\phi as a function of frequency ω\omega and the model parameters a1a_1 and b1b_1. Note that for a Transfer Function G(s)G(s), then: M=∣G(s)∣s=jωM = |G(s)|_{s = j\omega} and ϕ=Arg(G(s))∣s=jω\phi = \mathrm{Arg}(G(s))|_{s = j\omega}
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  4. (d)
    An automatic control system is used to regulate the pressure in the oil fractionator model. The closed-loop Transfer Function is shown below, where D(s)D(s) is the set point and (k1,k2)(k_1, k_2) are gains, X(s)=k1b1s+k2b1s2+(a1+k1b1)s+k2b1 D(s)X(s) = \frac{k_1 b_1 s + k_2 b_1}{s^2 + (a_1 + k_1 b_1)s + k_2 b_1}\,D(s) What is the stready state gain of the closed-loop Transfer Function? Comment on the significance of your answer. Assuming that a1=5a_1 = 5 and b1=2b_1 = 2, design a control system (i.e. determine values for the control gains) such that the closed-loop system has a damping ratio ζ=1.2\zeta = 1.2 and natural frequency ωn=0.1\omega_n = 0.1.
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