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4 Pages

### solution hw3

Course: CHE 470, Spring 2009
School: Rice
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Word Count: 689

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1 Problem From table 7.1 in the textbook, the Laplace transform of y(t)=te-t is: L te t = y(s) = 1 (s + 1) 2 (S1.1) Moreover, the Laplace transform of the unit-impulse (t) is: L [ (t) ] = u(s) = 1 (S1.2) Since u(t)=0 all the time but at t=to, assuming that y(t) is in deviation form (i.e. ys=0), the transfer function of the system in exam can be written as follows: G(s) = y(s) 1 = u(s) (s + 1) 2 (S1.3)...

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1 Problem From table 7.1 in the textbook, the Laplace transform of y(t)=te-t is: L te t = y(s) = 1 (s + 1) 2 (S1.1) Moreover, the Laplace transform of the unit-impulse (t) is: L [ (t) ] = u(s) = 1 (S1.2) Since u(t)=0 all the time but at t=to, assuming that y(t) is in deviation form (i.e. ys=0), the transfer function of the system in exam can be written as follows: G(s) = y(s) 1 = u(s) (s + 1) 2 (S1.3) Problem 2 Assuming constant density and reactor volume, the overall material balance can be written as follows: A dh(t) = Fi (t) F = Fi (t) 8h(t)1 2 dt (S2.1) As initial condition for equation (S2.1), we chose the steady state value of the hydrostatic pressure hs given by: 2 h(0) = h s = Fis 64 (S2.2) Linearizing the right hand side of equation (S2.1) about the steady state value (S2.2), one obtains: Fi (t) 8h(t)1 2 [ Fi (t) Fs ] 4h s 1 2 [ h(t) h s ] (S2.3) Defining the deviation variables H(t)=h(t)-hs and Q(t)=Fi(t)-Fs, and considering (S2.3), equation (S2.2) becomes Ah1 2 dH(t) h1 2 s + H(t) = s Q(t) 4 dt 4 subject to the following initial condition (S2.4) H(0) = 0 (S2.5) Therefore, by comparison to the standard form of 1st order systems, the time constant and the gain k are the following: Ah1 2 s = 4 Hence: a) h s = 3ft b) h s = 9ft h1 2 k= s 4 (S2.6) =3 3 4 =9 4 (S2.7) Problem 3 The material balance on the component A is the following: V dc A (t) = F [ c Ai (t) c A (t) ] kc A (t)V dt (S3.1) subject to the initial condition cA(0)=cAss=F/(F+kV)cAi,ss. As the differential equation describing the dynamics of the system is linear, it is already in deviation form when one chooses the deviation variables x(t)=cA(t)-cASS and y(t)=cAi(t)-cAi,SS. Thus, (S3.1) can be rewritten as follows: V dx(t) F + x(t) = y(t) F + kV dt F + kV (S3.2) subject to the initial condition x(0)=0. Therefore, by comparison to standard the form of 1st order systems, the time constant and the gain k of the system are given by: = V F + kV k= F F + kV (S3.3) and the transfer function G(s) can be expressed as follows: x(s) = G(s) = y(s) F F + kV = k V s +1 s +1 F + kV (S3.4) Unit-step change y(s) = 1 s . Therefore x(s) = 1 k 1 = k [s + 1 ] s s s + 1 (S3.5) which yields: x(t) = k(1 e t ) Thus, the sketch to a unit-step change is shown in figure S3.a (S3.6) Unit-impulse change y(s) = 1 . Therefore x(s) = k s +1 (S3.7) which yields: k x(t) = e t Thus, the sketch to a unit-impulse change is shown in figure S3.b Sinusoidal input y(s) = A (s 2 + 2 ) . Therefore (S3.8) x(s) = k A s + 1 s 2 + 2 (S3.9) Repeating what we did in class (see also page 318 of the textbook), one gets: x(t) = kA +1 2 2 sin(t + ) = tan 1 (t) (S3.10) Thus, the sketch to a sinusoidal input is shown in figure S3.c Figure S3 Problem 4 Assuming constant density and reactor volume, the overall material balance can be written as follows: AR dh(t) + h(t) = RFi (t) dt (S4.1) As initial condition for equation (S4.1), we chose the steady state value of the hydrostatic pressure hs given by: h(0) = h s = RFis (S4.2) Since (S4.1) is linear, defining the deviation variables H(t)=h(t)-hs and Q(t)=Fi(t)-Fis (S4.1) can be immediately rewritten in deviation form as follows: AR dH(t) + H(t) = RQ(t) = Ra sin(t) dt (S4.3) subject to the initial condition H(0) = 0 (S4.4) The first order system given by (S4.3)-(S4.4) has time constant and gain respectively given by: = AR k = aR (S4.5) We know from what we did in class that the output of a first order system subject to a sinusoidal input is a sinusoidal wave with the same frequency and with an amplitude ratio given by: ARatio = b = k 22 + 1 = aR A 2 R 22 + 1 (S4.6) Solving (S4.6) for A gives: A= 1 (aR) 2 b 2 R b (S4.7)
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Rice - CHE - 470
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Rice - CHE - 470
Problem 1ySP (s) +k = 1.65 (s + 1)(2s + 1)y(s)The closed loop transfer function can be written as follows:8 y(s) 8 89 (s + 1)(2s + 1) G CL (s) = = =2 = 2 8 ySP (s) 1 + 2s + 3s + 9 2 9s + 1 3s + 1 (s + 1)(2s + 1) Thus, comparing (S1.1) to the standar
Rice - CHE - 470
Problem 1controller valve tank 11 0.2s + 0.4s + 12ySP (s)k C (1 + 3 s)y(s)1Measuring device2Figure 1The characteristic equation for the closed loop in figure 1 can be written as follows:1 + G OL (s) = 1 +2k C (1 + 3 s) 0.2s 2 + 0.4s + 1(S1.1
Rice - CHE - 470
Problem 1 a) The transfer function of this process can be expressed as the product of three first order lag transfer functions. The AR and phase angles of a general 1st order lag are:AR =K +12 2and = tan 1 ()(S1.1)Thus, applying the principle of sup
Rice - CHE - 470
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Rice - CHE - 470
Rice - CHE - 470
Rice - CHE - 470
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Rutgers - PHYSICS - 750:204
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