Tutorial answers

If the ashes already exists, different scenarios squeeze out be tried using he model before tests being carried by on the real governing body. Alternatively if the system does non exist, the model undersurface be ingestiond to help decide on the final picture of a system. Often there are constraints on the target that make to be investigated e. G. Constraints on cost, space, etc. Modification to systems once they exist can be expensive and so it is master(prenominal) to try and get design of systems right-first- snip and this is where modeling and simulation can be useful engineering tools.Example reassure public lecture check offs weekly under second basetion Why is modeling important ?. 2 fit nones 3. Components of system Inputs Outputs States Environment Tank Valve Pipes Inlet Flows of A and B Liquid level Level change in storage tank Upstream of inlet to tank and downstream of subject valve 4. essay notes 5. See notes 6. Bookwork (as coursework 1) TUTORIAL AN SWERS 2 El . A proportional relationship for a component is here considered to be an unchanging relationship (and is often referred to in modeling terms as a constitutive or personal relationship).These are the natural physical laws which the individual components of the system obey e. G. For an electrical system, the relationship between potency and current and in the special eccentric of an ideal resistance Ohms Law FRR. I E. Kerchiefs Current Law algebraical summation of exclusively currents leading into a junction of a interlock is zero. Kerchiefs Voltage Law algebraic summation of all potential differences playing virtually a loop of a circuit is zero. Examples see section 2. 2. Of lecture notes. E. The impedance of an element is its voltagecurrent ratio.Multiple picking galvanizing Systems El(b) E(C) Problems Electrical Systems El . 3. 3 0, Ohms Law is obeyed since the resistance is unvarying as the voltage varies Q. 40 A E. IV E. (a) 4. 4 0 (b) 16 (c) 4. 4 E. (a) 0. 68 V (b) 0. 47 V (c) 0. 34 V E. (a) 0. 2 V, 205 ma (b) 1. 03 V, 52 ma (c) 1. 16 V, 193 ma E. = 30/84 v = 0. 357*12 volts = 4. Volts E. VI = 366/191 = 1. 92 V, TUTORIAL ANSWERS 3 MI . (I) spring f = xx= k(XSL -xx) where x is the displacement (or extension) and k is the rest constant called the spring constant with units of force/displacement e. . N/m. damp f = BE k(FL vi) where v is the stop number and B is the proportionality constant called a viscous friction coefficient or constant. Its dimension is force/velocity e. G. NSA/m. MM. See lecture notes. The force eternal sleep law demands that (a the acceleration). Analogies This is analogous to Kerchiefs voltage law, particularly if one treats the inertia acceleration as an equivalent force. scar In evidence to model a mechanical system, the usual put is to number a free bole diagram around each inertia (mass) component.One will then end up with a specify of simultaneous differential equations, the solution of whic h dictates the dynamics and constitutes the system model. In the suit of clothes where there are no mass components, then ensure a force balance at selected points in the system. That is the net force acting on any point must be zero, I. E. Multiple Choice mechanic Systems MI . G) MM. (a,b) Problems Mechanical Systems MI. O. AN MM. 0. 05 arms Question/ Variable (NSA/m) 812 834 biology I 2 1217 24/19= 1 . 263 4 15/8 60/47 = 1 . 276 Questions Thermal and bland Systems TFH .A system is said to be in equilibrium when its behavior is steady I. E its output and inputs are unchanging. For the liquid level system with an inlet and outlet flow, this corresponds to the inlet and outlet flows being the same. Multiple Choice Thermal and Fluid Systems TFH. (b) TFH. (c) Problems Thermal and Fluid Systems TFH . The flow cannot be assumed to be stratified as the proportionality constant is not inner as the flow increases through the pipe I. E. I/R resistance (Pa. s/mm) 3 TUTORIAL ANSWERS 4 IQ . Methods that can be used to observe the gradient of the straight line at t=2 sec are (I) plot a graph and determine where t=2 or (it) differentiate x(t) with respect to t and substitute t=2. Q. (a) Q. See lecture notes Q. Completing the carry over gives Electrical Component Equation Mechanical Component Rotational Component Inductor Inertia Rotating inertia ohmic resistance Damper Capacitor Spring Torsions spring Q. (a) Using free body diagram on the mass-damper system of Fig. 5. 1, the mass and ampere can be considered to be in parallel.Force balance gives where Hence (b) Similarly for the spring-damper system of Fig. 5. 2 Force balance gives where , (c) For the rotational pulley of Fig. 5. 3, a torque balance is required Torque balance givesand where, , Q. (a) For a resistor and capacitor in series of Fig. 6. 1 Apply Kerchiefs Voltage Law gives (b) For a resistor and inductor in series of Fig. 6. 2 where , , (c) For a 5 resistors and a capacitor system of Fig. 6. 3, observe tha t this is nearly the same system as shown in Tutorial Sheet 2 Problem E and with the addition of the opacity.Hence where with and Hence as in Qua. Q. Q. Material balance on tank rate of change of mass floodwater = mass flow in mass flow out assume constant density 0 and area Given , A = 7 mm, R = 0. 14 her/mm and Sin = 100 mm/her substituting gives Tutorial Answers 5, 6 IQ Bookwork straight from notes Ask in a tutorial if stuck and/or use MENTAL to generate solutions and image against you work. E. G. For 1st of these t=alliances(O,2,100) ext=subs(x,t) fugue(l reset Q throughout Q assume a model of the form Steady-state is 0. 6. Initial cheer is -1. spring up is inclined as 1. . 63% of trick up is wedded by which implies x(t) has this value at about t=O. 25 sec and therefore T=O. 25, Steady-state is 30. Initial value is 2. Rise is given as 28. 63% of progress is given by 0. 63*28=17. 64 which implies x(t)=19. 64. X(t) has this value at about t=5 sec and therefore T=5, k=30. Steady-state is 50. Initial value is 20. Rise is given as 30. 63% of wage increase is given by 0. 63*30=18. 9 which implies x(t)=38. 9. X(t) has this value at about t=50 sec and therefore T=50, k=50. Q Maximum current is at t=O and given as V/R Therefore R=V/I = 5/0. 004 = mashes. metre constant is given by ARC, so T=ms implies that C=. 005/1250 = 4 micro. Q Parameters give a time constant of cosec so after 30 sec aircraft at 95% of steady-state land speed. 1 MPH is the same as mutterer pH or (1609/3600)m/s MPH is the same as mom/s Steady-state is given as f/B. Therefore min f required is BIB scale by (11. 95) to be precise. Q Model is Bad/dot +xx=f or (B/k) DXL/dot +x =f/k Desired time constant is about 0. Sec, therefore (B/k)=O. 8 so k=NON/m Steady-state displacement is given as (1 /k)f = 0. 04, and therefore f=AN is required. Tutorial Answers 7 1 .Find the Lovelace transform of the pursual signals Students should use MENTAL to check their functional here, e. G. Ray the comm and 2. rehearse partial derivative fractions, a lookup table and inverse Lovelace to find the underlying signals with the following transforms. Students should use MAT to check their working, e. G. 3. What is the final value for signals with the following transforms? Use the Pit provided note that (I) there is no final value if the signal is divergent which is the circumstance for 5th (obvious from negative sign) and (it) for convergent signals, the final value must be zero if there is no integrator.Hence only 2nd and 6th discombobulate a non-zero values which must be 4 and 0. 5 respectively. 4. Which of the following transforms has the fastest settling time? What are the settling generation to deep down 5% of steady-state? Time constants are negative inverses of poles. One can watch time to 5% error as approximately three times slowest time constant (exact for 1st order but no strict generalization when many poles due to uncertainty about partial fractions). Time constant is the negative inverse of the pole. So pole at -0. 25 gives T=4, etc. 5. Sketch the poles and zeros of the following transforms on an fancy diagram.By marking the LAP and RAP clearly, hence determine which represent still and un immutable behavior. Students should use MENTAL to check their working for his, for example, doing 4th as follows will produce a fugue with poles marked in Y and zeros in o Systems are stable if and only if all the poles are in the LAP the origin is counted as being in the LAP. The positions of the zeros do not affect stability. Tutorial Answers 8 1 . The inverse Lovelace transform of a transfer function is called the impulse response function. If a system has an impulse response function given by g(t) t(l-sin(t)).Compute its transfer function, G(s). 2. Use Lovelace methods to solve the following ODE equations. 3. Give examples of type O, type 1 and type 2 systems. Has does this affect the expected behavior? Bookwork 4. Which of the following transforms for 1st order ODES has the highest gain? What are the gains? What are the time constants? Determine and brief the mensuration responses for each of these. Gains are 4, 3, 1. 5 and 1. 125 respectively. Time constants are 4, 0. 2, 1. 25, 0. 5 respectively. As these are 1st order, sketching step response follows same procedures as tutorial 5,6.Tutorial Answers 9, 10 1 . Bookwork read some control schoolbook books to broaden your views on the uses and potential of control. 2. This is also straight from the notes but your reasonableness will also be improved by some wider reading. Dont clean stick to your main discipline, but look at examples from chemical, aerospace, automotive, medical, electrical, biological, etc. 3. truthful application of the Pit. 2nd set has an integrator and hence the head start is know to be zero. Otherwise, use formula. establish this with MENTAL, I. E. Plot is seen to settle at 0. 52 4. The 1st part is taken direct from the lecture slides so not repeat ed here. The closed-loop time constant and rise time are Time constant +AKA), closed-loop game = AKA/(I+AKA), where A=4/5, -r=o. 2 Hence 0. 2/(1 +K/5)O. 8 which gives K 4+3. K or 0. K4 or K5. Confirm this using MENTAL, ii. Use G=TFH(4,1 %% plot in a figure It is clear that the time closed loop pole polynomial is (s+ 1 +AKA/T) and hence the pole is in the LAP for all positive K which implies closed-loop stability. discourse of large K is bookwork read some text books. 5.This brain is intentional to make a student think and experiment. To meet specifications, the closed-loop is given as Clearly the steady-state gain is unity as expected so the offset requirement is met. The closed-loop poles are determined from the roots of the denominator and we want the poles to be to the left wing of -2. 5 e. (s+2. 5) is equivalent to (0. As+1). Both roots can be placed at 2. 5 if In the future students will actualize that lower values of K will give a gradual pole and higher values of K wil l give rise to oscillation. 6. Standard question.Form closed-loop transfer function and find characteristic polynomial for all 3 cases. You will need to do the partial fractions for all 3 and sketch, but you can use MENTAL to check your answers. E. G form the three closed-loop transfer functions and then type feedback(GO,GO,GO) to see all 3 together. N.B. 63 is seem 2 content. Clearly Just proportional is fastest, but gives a large offset. GIG is smooth (2 real poles) ND no offset. But poles are well spaced so this is conservative. 63 has interchangeable response time to GIG (same slowest time constant), but has obscure poles and thus oscillation.Conclusion, PI is best Typical exam type question outline answer a) Let the internal temperature be given as T degrees. The rate of shake up supplied is given as The heat exit OHIO(T+50) Hence the temperature is given by In steady-state we desire T=20 which implies that b) If the external temperature drops by 10 degrees, then the model becomes which implies the new steady-state temperature will be 6 degrees The time constant is clearly 1000 sec. Students should sketch a graph showing the temperature moving from 20 to 6 with the sequester time constant. ) If the heat input from the passengers is increased, the model becomes In the case, the change in temperature is negligible which suggests that for this scenario the key factor is the external temperature and heaters rather than any heat coming from the passengers. D) Clearly the open-loop choice of voltage does not maintain the temperature properly in general and so some control is needed. It is known that the constitute steady-state can only be achieved in the presence of uncertainty if intrinsic action is included.The steady-state error too change in desired temperature is given by because K(O) is infinite, irrespective of changes in the gain of G or disturbances such(prenominal) as changes in external temperature Students should first put the equations for the model and underlying control law into Lovelace transforms about the steady-state Hence The closed-loop transfer function is given as Students should validate that the time constants are reasonable and that the closed- loop is stable The time constants are given from the roots of the closed-loop denominator. Students should note that these are resembling to the original time constant and thus satisfactory.