Tutorial answers

If the scheme already exists, different scenarios can be tried using he feigning before tests being carried out on the real transcription. Alternatively if the system does not exist, the molding can be single- rankd give-up the ghostd to help decide on the final design of a system. a great deal in that location atomic number 18 constraints on the design that need to be investigated e. G. Constraints on cost, space, etc. Modification to systems at one snip they exist can be expensive hence it is important to try and get design of systems right-first-time and this is where moulding and simulation can be useful engineering excessivelyls.Example see lecture notes weekly under fraction Why is modeling important ?. 2 See notes 3. Components of system Inputs Outputs States Environment Tank Valve Pipes Inlet Flows of A and B Liquid level Level change in army tank Upstream of inlet to tank and downstream of outlet valve 4. See notes 5. See notes 6. Bookwork (as coursework 1) TU TORIAL ANSWERS 2 El . A proportional affinity for a component is here considered to be an unchanging relationship (and is often referred to in modeling price as a constitutive or physical relationship).These are the natural physical legal philosophys which the individual components of the system obey e. G. For an electrical system, the relationship between electric potentiality and current and in the special case of an warning resistor Ohms Law FRR. I E. Kerchiefs Current Law algebraic summation of only currents flowing into a junction of a network is zero. Kerchiefs Voltage Law algebraic summation of all electromotive forces acting somewhatwhat 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 Electrical Systems El(b) E(C) Problems Electrical Systems El . 3. 3 0, Ohms Law is obeyed since the resistance is invariant 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) bounciness f = xx= k(XSL -xx) where x is the displacement (or extension) and k is the proportionality constant called the spring constant with units of wedge/displacement e. . N/m. Damper f = BE k(FL vi) where v is the velocity and B is the proportionality constant called a viscous friction coefficient or constant. Its dimension is gouge/velocity e. G. NSA/m. MM. See lecture notes. The force balance law demands that (a the acceleration). Analogies This is analogous to Kerchiefs voltage law, particularly if one treats the inertia acceleration as an alike force. Note In order to model a mechanical system, the usual practice is to form a free body diagram around each inertia ( deal) component.One will then end up with a set of simultaneous differential equations, the solutio n of which dictates the dynamics and constitutes the system model. In the case 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 selection Mechanical Systems MI . G) MM. (a,b) Problems Mechanical Systems MI. O. AN MM. 0. 05 arms Question/ Variable (NSA/m) 812 834 Biota I 2 1217 24/19= 1 . 263 4 15/8 60/47 = 1 . 276 Questions Thermal and smooth 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 laminar 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 .Meth ods that can be used to get 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 table gives Electrical Component Equation Mechanical Component rotational Component Inductor Inertia Rotating inertia Resistor Damper Capacitor Spring Torsions spring Q. (a) utilise 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 thence (b) Similarly for the spring-damper system of Fig. 5. 2 Force balance gives where , (c) For the rotational blocking of Fig. 5. 3, a torque balance is required Torque balance givesand where, , Q. (a) For a resistor and capacitor in serial 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 that this is near ly the same system as shown in Tutorial Sheet 2 Problem E but 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 scene of action Given , A = 7 mm, R = 0. 14 her/mm and Sin = 100 mm/her interchange gives Tutorial Answers 5, 6 IQ Bookwork straight from notes Ask in a tutorial if stuck and/or use MENTAL to generate solutions and disclose a superchargest 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 value is -1. Rise is given as 1. . 63% of rise is given 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 rise 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 i s 50. Initial value is 20. Rise is given as 30. 63% of rise 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 thusly R=V/I = 5/0. 004 = mashes.Time constant is given by ARC, so T=ms implies that C=. 005/ one hundred twenty-five0 = 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 scaled by (11. 95) to be precise. Q Model is magnanimous/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 qualify of the spare-time activity signals Students should use MENTAL to weaken their working here, e. G. Ray the command 2. Use partial fractions, a lookup table and inverse Lovelace to pose 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 stone but note that (I) there is no final value if the signal is divergent which is the case for fifth (obvious from negative sign) and (it) for convergent signals, the final value must be zero if there is no integrator.Hence only when second and 6th have 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 times to within 5% of steady-state? Time constants are negative inverses of poles. One can estimate time to 5% error as approximately threesome times bumper-to-bumper time constant (exact for 1st order but no strict evocation when many poles due to uncertainty about partial fractions). Time constant is the negative in verse of the pole. So pole at -0. 25 gives T=4, etc. 5. Sketch the poles and zeros of the following transforms on an Regard diagram.By scrape the LAP and RAP clearly, hence determine which represent stable and unstable 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 maneuver function is called the impulse rejoinder 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 token O, type 1 and type 2 systems. Has does this affect the expect behavior? Bookwork 4. Which of the following transforms for 1st order ODES h as the highest gain? What are the gains? What are the time constants? Determine and sketch the step 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 text books to broaden your views on the uses and potential of control. 2. This is also straight from the notes but your understanding will also be improved by some wider reading. Dont rightful(prenominal) stick to your main discipline, but look at examples from chemical, aerospace, automotive, medical, electrical, biological, etc. 3. Straightforward application of the Pit. 2nd set has an integrator and hence the offset is known to be zero. Otherwise, use formula. brook this with MENTAL, I. E. Plot is seen to slide down at 0. 52 4. The 1st part is taken direct from the lecture slides so not iterate here. The unope n-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. Discussion of large K is bookwork read some text books. 5.This marvel is designed 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 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 recognize that lower values of K will give a laggard pole and higher values of K will 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. provided poles are well spaced so this is conservative. 63 has similar response time to GIG (same slowest time constant), but has complex 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 commove supplied is given as The heat loss 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 stea dy-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 appropriate time constant. ) If the heat input from the passengers is increased, the model becomes In the case, the change in temperature is measly 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 set uply in general and so some control is needed. It is known that the correct steady-state can only be achieved in the presence of uncertainty if integral 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 as changes in external temperature Students should first put the equations for the model and integral control law into Lovelace trans forms about the steady-state Hence The closed-loop transfer function is given as Students should validate that the time constants are conjectural 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 similar to the sure time constant and thus satisfactory.