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1 ( ) Vρ C st s T t 0 wc Ti s T s Q s (8) K T ( s) Q ( s) + Ti ( s) (0) τs+ τs+ V ρ K and τ wc w T (s)g (s)q (s) + G (s)t(s) i G and G are transfer functions and independent of the inputs, Q and T i. Note G (process) has gain K and time constant τ. G (disturbance) has gain and time constant τ. gain G(s0). Both are first order processes. If there is no change in inlet temperature (T i 0), then T i (s) 0. System can be forced by a change in either T i or Q (see Example 4.).

2 Transfer Function Between Q and T Suppose T i is constant at the steady-state value. Then, () Ti t Ti Ti t 0 Ti s 0. Then we can substitute into (0) and rearrange to get the desired TF: T s K Q s τ s+ ()

3 Transfer Function Between and T T : Suppose that Q is constant at its steady-state value: i Qt Q Q t 0 Q s 0 Thus, rearranging T T s τ s+ i ( s) (3) 3

4 Comments:. The TFs in () and (3) show the individual effects of Q and on T. What about simultaneous changes in both Q and? T i T i Answer: See (0). The same TFs are valid for simultaneous changes. Note that (0) shows that the effects of changes in both Q and T i are additive. This always occurs for linear, dynamic models (like TFs) because the Principle of Superposition is valid.. The TF model enables us to determine the output response to any change in an input. 3. Use deviation variables to eliminate initial conditions for TF models. 4

5 Example: Stirred Tank Heater K 0.05 τ T Q No change in T i s + Q(0) 000cal/sec. Then, step change in Q(t):500 cal/sec 500 Q s T s+ s s(s+ ) What is T (t)? t / τ 5 T () t 5[ e ] T() s s( τ s+ ) T t e t / 5[ ] 5

6 Solve Example 4. and Solve Example 4. if you have any question ask me! 6

7 Properties of Transfer Function Models. Steady-State Gain The steady-state of a TF can be used to calculate the steadystate change in an output due to a steady-state change in the input. For example, suppose we know two steady states for an input, u, and an output, y. Then we can calculate the steadystate gain, K, from: y y K u u (4-38) For a linear system, K is a constant. But for a nonlinear u, y. system, K will depend on the operating condition 7

8 Calculation of K from the TF Model: If a TF model has a steady-state gain, then: K s 0 lim G s (4) This important result is a consequence of the Final Value Theorem Note: Some TF models do not have a steady-state gain (e.g., integrating process in Ch. 5) 8

9 . Order of a TF Model Consider a general n-th order, linear ODE: n n m d y dy dy d u an + a n n + Ka n + a0y b m + dt m dt dt dt m d u du bm + K+ b m + b0u dt dt (4-39) Take L, assuming the initial conditions are all zero. Rearranging gives the TF: G s Y s U s m i 0 n i 0 bs i as i i i (4-40) 9

10 Definition: The order of the TF is defined to be the order of the denominator polynomial. Note: The order of the TF is equal to the order of the ODE. Physical Realizability: For any physical system, n min (4-38). Otherwise, the system response to a step input will be an impulse. This can t happen. Example: ay 0 b du + bu 0 and step change in u (4-4) dt n > m0 Impulse response to step change. Physically impossible. 0

11 nd order process General nd order ODE: d y dt a + b dy dt + y Ku Laplace Transform: roots Y ( s) G( s) U ( s) as s, b ± b a b 4a > b 4a < [ as + bs +] Y ( s) KU ( s) K + bs + 4a : real roots : imaginary roots

12 . Examples 3s + 4s + b a > e s. -bt 3s + 4s+ (3s+ )( s+ ) 3( s+ )( s+ ) 3 t 3 t transforms to e, e ( real roots) (no oscillation) s b + s + 4a 4 < L ω 3 3 sinωt s + s+ ( s j)( s+ 0.5 j) ( s + b) +ω + s+ 3 (s+ 0.5) cos, sin 0.5t 0.5t transforms to e t e t (oscillation)

13 3. Additive Property Suppose that an output is influenced by two inputs and that the transfer functions are known: Y( s) and Y s G s G s U s U s Then the response to changes in both U and Ucan be written as: + Y s G s U s G s U s The graphical representation (or block diagram) is: U (s) U (s) G (s) G (s) Y(s) 3

14 4. Multiplicative Property Suppose that, Y s U s G( s) and U s U3 s G3( s) Then, Y s G s U s and U s G s U s Substitute, Or, Y s 3 3 Y s G s G s U s U3 s G s G s U s G s G s Y s 4

15 Example : Place sensor for temperature downstream from heated tank (transport lag) Tank: Sensor: T(s) K G U(s) + τ s G T s(s) T(s) Overall transfer function: Ts U Ts T T U K e + τ G -θs s G K Distance L for plug flow, Dead time θ V fluid velocity, KK e + τ s θs L V τ is very small (neglect) 5

16 Solve Example 4.3 and Solve Example 4.4 if you have any question ask me! 6

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