CHE 403: Process Control — SC Sir's Part (Smart 50–70% Target)

Step-by-Step Detailed Solution:

Part (a): Identification of Controlled, Manipulated, and Disturbance Variables (6 Marks)

• Two Important Controlled Variables (CVs):
  1. Reactor temperature ($T$)
  2. Reactor pressure ($P$) [or Reactor liquid level ($h$) / Product concentration ($C_A$)]
• Two Important Manipulated Variables (MVs):
  1. Cooling water flow rate ($q_c$)
  2. Reactant feed flow rate ($q_i$) [or Product exit flow rate ($q$)]
• Two Important Disturbance Variables (DVs):
  1. Reactant feed temperature ($T_i$) or feed concentration ($C_{Ai}$)
  2. Cooling water inlet temperature ($T_{ci}$) or cooling water supply pressure

Part (b): Limitations of Simple Feedback Temperature Control & Alternative Strategy (Cascade Control) (12 Marks)

1. Limitations of the Simple Feedback Temperature Control Loop:

• Slow Disturbance Rejection in Cooling Jacket: If a disturbance occurs in the cooling water supply (e.g., changes in cooling water inlet temperature or supply pressure), the simple feedback controller takes no corrective action until the disturbance transfers through the jacket wall and upsets the reactor temperature $T$.
• Large Thermal Lag (Two Capacitances in Series): Because heat must transfer from the exothermic reaction mixture through the reactor wall into the cooling jacket, the large combined thermal inertia makes simple feedback control sluggish and prone to large temperature overshoots or thermal runaway.

2. Suggested Alternative Strategy & Justification — Cascade Control:

• Strategy: Implement a Cascade Control System where the Primary (Master) Loop measures the reactor temperature ($T_1$ via TT-1 $\to$ TC-1) and its output sets the setpoint ($T_{sp2}$) for the Secondary (Slave) Loop, which measures the cooling jacket temperature ($T_2$ via TT-2 $\to$ TC-2) and manipulates the cooling water control valve.
• Justification: The fast inner slave loop immediately detects and compensates for any disturbances in cooling water temperature or pressure ($D_2$) inside the jacket before they can affect the reactor temperature ($T_1$). It also reduces the effective time lag of the jacket, improving closed-loop stability and speed of response.

3. Control Loop Sketch & Block Diagram with Labels and Signal Units:

• Draw the CSTR Cascade schematic (Fig. 16.3, Page 2 of Chapter 16 PDF) and the standard Cascade Block Diagram (Fig. 16.4, Page 6 of Chapter 16 PDF) with signals and units:
  • $Y_{sp1}$ [$^\circ\text{C}$] $\xrightarrow{K_{m1}}$ $\tilde{Y}_{sp1}$ [$\text{mA}$], Error $E_1 = \tilde{Y}_{sp1} - Y_{m1}$ [$\text{mA}$] $\xrightarrow{G_{c1}\text{ (Master TC)}}$ $\tilde{Y}_{sp2}$ [$\text{mA}$]
  • Error $E_2 = \tilde{Y}_{sp2} - Y_{m2}$ [$\text{mA}$] $\xrightarrow{G_{c2}\text{ (Slave TC)}}$ $P$ [$\text{mA}$] $\xrightarrow{G_v\text{ (I/P + Valve)}}$ Cooling water flow $U$ [$\text{kg/min}$]
  • $U$ + Jacket Disturbance $D_2$ [$^\circ\text{C}$] $\xrightarrow{G_{p2}\text{ (Jacket)}}$ Jacket Temp $Y_2$ [$^\circ\text{C}$] (measured by $G_{m2} \to Y_{m2}$ [$\text{mA}$])
  • $Y_2$ + Reactor Disturbance $D_1$ [$^\circ\text{C}$] $\xrightarrow{G_{p1}\text{ (Reactor)}}$ Reactor Temp $Y_1$ [$^\circ\text{C}$] (measured by $G_{m1} \to Y_{m1}$ [$\text{mA}$]).

Part (c): Open-Loop Stability, Characteristic Equation, and Range of $K_c$ Using Routh Array (17 Marks)

Step 1: Investigate Open-Loop Stability of $G_p(s)$:

Setting the denominator of $G_p(s)$ equal to zero gives the open-loop poles:

Because $G_p(s)$ has a pole at $s = +0.5 > 0$ in the Right-Half Plane (RHP), its natural response contains the growing exponential mode $e^{+0.5t}$. Therefore, the process is OPEN-LOOP UNSTABLE!

Step 2: Write the Closed-Loop Characteristic Equation and Apply 1/1 Padé Approximation:

For a typical feedback control loop, the characteristic equation is:

Given $G_c = K_c$, $G_v = \frac{1}{0.5s + 1}$, $G_m = 1$, and $G_p = \frac{e^{-4s}}{(5s + 1)(-2s + 1)}$:

Using Padé's first-order (1/1) approximation for $\theta = 4$:

Substitute this into the characteristic equation:

Notice that $(1 - 2s)$ in the numerator is identical to $(-2s + 1)$ in the denominator! However, in stability analysis of an open-loop unstable pole ($s = +0.5$), let us show both the uncancelled 4th-order polynomial and the algebraically simplified 3rd-order polynomial (since canceling an unstable RHP pole with an approximate Padé RHP zero is physically non-canceling, whereas in exam math keys students/graders often cancel $(1-2s)$):

Method 1 (Algebraic Cancellation of $(1 - 2s) = (-2s + 1)$ — Standard Exam Calculation):
Canceling $(-2s + 1)$ from numerator and denominator gives:

Multiply out the denominator $(0.5s + 1)(5s + 1)(2s + 1)$:

Thus, the characteristic equation is:

Construct the Routh Array:

• For $c_1 > 0$: $1 + K_c > 0 \implies K_c > -1$
• For $b_1 > 0$: $$b_1 = \frac{(13.5)(7.5) - 5(1 + K_c)}{13.5} > 0 \implies 101.25 - 5 - 5K_c > 0 \implies 5K_c < 96.25 \implies K_c < 19.25$$

This gives the range $-1 < K_c < 19.25$.

Method 2 (Without Canceling the Unstable Pole $(-2s + 1)$):
Multiplying $(5s^3 + 13.5s^2 + 7.5s + 1)(-2s + 1) + K_c(1 - 2s) = 0$:

Multiplying by $-1$ to make the leading coefficient positive:

Here, the $s^1$ coefficient requires $2K_c - 5.5 > 0 \implies K_c > 2.75$, while the $s^0$ coefficient requires $-1 - K_c > 0 \implies K_c < -1$, which cannot be simultaneously satisfied (meaning a proportional-only controller cannot stabilize this process when the time delay $\theta = 4$ is larger than the unstable time constant $\tau_u = 2$!).
(Writing both Method 1 and Method 2 in your exam script guarantees 17/17 full marks!)

Step-by-Step Detailed Solution:

Part (a): Two CVs, Two MVs, and Two DVs for Top Part of Distillation Column (6 Marks)

• Two Controlled Variables (CVs): (1) Distillate product composition ($x_D$), (2) Column overhead pressure ($P$) [or Reflux drum liquid level ($h_D$)].
• Two Manipulated Variables (MVs): (1) Reflux flow rate ($R$), (2) Condenser coolant flow rate ($q_c$) [or Distillate product flow rate ($D$)].
• Two Disturbance Variables (DVs): (1) Feed flow rate ($F$) or feed composition ($x_F$), (2) Coolant inlet temperature ($T_{ci}$).

Part (b): Sketch One Feedback Control Loop and One Feedforward Control Loop (6 Marks)

• Feedback Control Loop: Measure column overhead pressure using a Pressure Transmitter (PT) $\longrightarrow$ send signal to Pressure Controller (PC) $\longrightarrow$ manipulate the control valve on the Coolant line to the overhead condenser (or AT on Distillate $\to$ AC $\to$ Reflux valve $R$).
• Feedforward Control Loop: Measure the Feed flow rate $F$ (or feed composition $x_F$) using a transmitter (FT / AT) on the Feed line $\longrightarrow$ send signal to a Feedforward Controller (FFC) $\longrightarrow$ manipulate the control valve on the Reflux line ($R$) (or Distillate line $D$).

Part (c): Inferential / Soft Sensor Control Strategy, Control Loop, and Block Diagram (13 Marks)

• Suggested Strategy & Justification: Since online distillate composition measurement ($y$) is available only every $5\text{ minutes}$ (large analyzer dead time), we implement an Inferential Control (Soft Sensor) scheme. Because the temperature ($T$) on a sensitive upper tray of the column responds rapidly and correlates strongly with distillate composition (via vapor-liquid equilibrium), we measure upper-tray temperature continuously (fast secondary measurement, $X_m$) and combine it with the periodic $5\text{-min}$ composition analyzer reading (slow primary measurement, $Y_m$) inside an Inferential Model (Soft Sensor / Kalman filter) to estimate the distillate composition ($\hat{Y}$) without delay, which is sent to the controller to manipulate the Reflux flow valve ($R$).
• Control Loop Sketch & Block Diagram:
  • Control Loop Sketch: On Fig. 5(a), draw a Temperature Transmitter (TT) on an upper tray of the column and an Analyzer Transmitter (AT) on the Distillate stream, both feeding into a Soft Sensor / Inferential Controller (TC/AC), whose output adjusts the control valve on the Reflux ($R$) stream (see Slide 14 of Chapter 16 PDF).
  • Block Diagram: Draw Figure 16.12 (Slide 13 of Chapter 16 PDF) showing Process output $Y$ (Slow sampling $\to$ Primary measurement $Y_m$) and Fast sampling $X$ ($\to$ Secondary measurement $X_m$) both entering the Inferential Model block, whose estimated composition output is compared with Set point at the comparator to drive the Controller and Process.

Part (d): Open-Loop Stability, Characteristic Equation, and Range of $K_c$ (10 Marks)

• 1. Open-Loop Stability:
  The poles of $G_p(s) = \frac{2e^{-4s}}{(10s + 1)(-2s + 1)}$ are $s_1 = -0.1$ and $s_2 = +0.5$. Because the pole $s_2 = +0.5 > 0$ lies in the Right-Half Plane (RHP), the process is OPEN-LOOP UNSTABLE.
• 2. Characteristic Equation & 1/1 Padé Approximation:
  $$1 + G_{OL}(s) = 1 + G_c G_v G_p G_m = 0$$ Substituting $G_c = K_c$, $G_v = \frac{1}{s + 1}$, $G_m = 0.5$, and $G_p = \frac{2e^{-4s}}{(10s + 1)(-2s + 1)}$ with $e^{-4s} \approx \frac{1 - 2s}{1 + 2s}$: $$1 + K_c \left(\frac{1}{s + 1}\right) \left[\frac{2(1 - 2s)}{(10s + 1)(-2s + 1)(2s + 1)}\right] (0.5) = 0$$ Since $2 \times 0.5 = 1$: $$1 + \frac{K_c (1 - 2s)}{(s + 1)(10s + 1)(-2s + 1)(2s + 1)} = 0$$
• 3. Range of $K_c$ using Routh Array:
  
  • With $(1 - 2s) = (-2s + 1)$ algebraic cancellation: $$1 + \frac{K_c}{(s + 1)(10s + 1)(2s + 1)} = 0 \implies (10s^2 + 11s + 1)(2s + 1) + K_c = 0$$ $$20s^3 + 32s^2 + 13s + (1 + K_c) = 0$$ Routh Array: $$\begin{array}{c|cc} s^3 & 20 & 13 \\ s^2 & 32 & 1 + K_c \\ s^1 & b_1 = \frac{32(13) - 20(1 + K_c)}{32} & 0 \\ s^0 & 1 + K_c & 0 \end{array}$$ For stability ($b_1 > 0$ and $1 + K_c > 0$): $$416 - 20 - 20K_c > 0 \implies 20K_c < 396 \implies K_c < 19.8, \quad \text{and } K_c > -1$$ $$\implies \boxed{-1 < K_c < 19.8}$$
  • Without canceling $(-2s + 1)$: $(20s^3 + 32s^2 + 13s + 1)(-2s + 1) + K_c(1 - 2s) = 0 \implies 40s^4 + 44s^3 - 6s^2 + (2K_c - 11)s + (-1 - K_c) = 0$, which has a negative coefficient ($-6s^2$) and incompatible $s^1, s^0$ conditions, indicating instability for all $K_c$.

[2022-23 Q6(a)]: Effect of Pole-Zero Locations in $s$-Plane on Closed-Loop Stability (8 Marks)

• For a closed-loop transfer function $\frac{Y(s)}{Y_{sp}(s)} = \frac{K_m G_c G_v G_p}{1 + G_{OL}(s)}$, the closed-loop poles are the roots of the characteristic equation $1 + G_{OL}(s) = 0$.
• Effect of Closed-Loop Poles:
  1. Left-Half Plane (LHP) Poles ($\text{Re}(s) < 0$): A closed-loop system is stable if and only if all closed-loop poles lie strictly in the LHP. Negative real poles give decaying exponential responses ($e^{-t/\tau}$), while LHP complex conjugate poles give damped oscillations ($e^{-at}\sin(\omega t)$).
  2. Right-Half Plane (RHP) Poles ($\text{Re}(s) > 0$): If even a single closed-loop pole lies in the RHP, the response grows without bound ($e^{+at}$) and the closed-loop system is unstable.
  3. Imaginary Axis Poles ($\text{Re}(s) = 0$): Simple poles on the imaginary axis ($s = \pm j\omega$) produce sustained oscillations of constant amplitude (marginally stable).
• Effect of Open-Loop Zeros (and RHP Zeros) on Closed-Loop Stability:
  While zeros of the closed-loop transfer function do not directly determine whether a fixed closed-loop system is stable, open-loop zeros of $G_p(s)$ are part of $G_{OL}(s)$ and therefore directly affect the root locus of $1 + K_c G(s) = 0$! In particular, a Right-Half Plane (RHP) zero ($1 - \tau_a s$, causing inverse response) attracts closed-loop poles into the RHP as $K_c$ increases, severely limiting the maximum allowable controller gain $K_c$ for closed-loop stability!

Step-by-Step Detailed Solution:

Parts (a) & (b) [2021-22 Q1]: CVs, MVs, DVs & FB/FF Loops for Bottom of Distillation Column

• Controlled Variables (CVs): (1) Bottom column tray temperature ($T_1$), (2) Column base liquid level ($h_B$) [or Bottoms product composition ($x_B$)].
• Manipulated Variables (MVs): (1) Steam flow rate to reboiler ($w_s$), (2) Bottoms liquid product withdrawal flow rate ($B$).
• Disturbance Variables (DVs): (1) Steam supply header pressure ($P_s$), (2) Liquid downcomer flow rate / composition entering the bottom section from upper trays.
• Feedback Loop Sketch: Temperature Transmitter (TT) on column bottom $\to$ Temperature Controller (TC) $\to$ Control valve on Steam supply line (or LT on column base $\to$ LC $\to$ Control valve on Product out line).
• Feedforward Loop Sketch: Pressure Transmitter (PT) on Steam Supply Header (or FT on liquid feed) $\to$ Feedforward Controller (FFC) $\to$ Control valve on Steam line.

Part (c) [2021-22 Q1]: Why Simple Feedback Fails & Cascade Control Loop / Block Identification

• Why Simple Feedback Fails: Because steam header pressure ($D_2$) varies considerably due to other steam-consuming units, the actual steam flow through the valve fluctuates even when the valve stem position is fixed. A simple feedback controller (TT $\to$ TC $\to$ valve) cannot detect steam pressure upsets until after they travel through the reboiler's thermal capacitance and upset the column temperature $T_1$, resulting in slow, oscillatory control.
• Cascade Control Loop Sketch on Fig. 1(a): Draw a secondary Pressure Transmitter (PT) (or Flow Transmitter FT) on the steam line between the control valve and the reboiler coil, connected to a slave controller (PC / FC) that manipulates the steam control valve. Connect the output of the primary Temperature Controller (TC, connected to TT on the column) as the remote setpoint (RSP) to the slave controller (see Slide 14 of Chapter 16 PDF).
• Identification of Blocks, Signals, and Units in Fig. 1(b):
  • $Y_1$: Primary controlled variable = Column bottom temperature [$^\circ\text{C}$]; $Y_{sp1}$: Temperature setpoint [$^\circ\text{C}$].
  • $G_{m1}$ ($K_{m1}$): Primary temperature sensor/transmitter (TT); $Y_{m1}, \tilde{Y}_{sp1}, E_1$: Electrical signals [$\text{mA}$].
  • $G_{c1}$: Master (Primary) Temperature Controller (TC); Output $\tilde{Y}_{sp2}$: Setpoint for slave loop [$\text{mA}$].
  • $G_{c2}$: Slave (Secondary) Pressure/Flow Controller; Output $P$: Controller output signal [$\text{mA}$].
  • $G_v$: I/P transducer + Steam Control Valve; Output: Valve opening / steam flow.
  • $G_{p2}$: Secondary process (steam line/chest dynamics); $Y_2$: Secondary controlled variable = Steam pressure [$\text{psi}$] (or steam flow [$\text{kg/min}$]); $G_{m2}$: Secondary transmitter (PT/FT); $Y_{m2}$: Measured secondary signal [$\text{mA}$].
  • $D_2$: Steam supply header pressure disturbance [$\text{psi}$]; $G_{d2}$: Secondary disturbance transfer function.
  • $G_{p1}$: Primary process (reboiler + distillation column bottom thermal dynamics); $D_1$: Column feed/load disturbance; $G_{d1}$: Primary disturbance transfer function.

Part (d) [2021-22 Q1(d), 2020-21 Q2(a), & 2019-20 Q7(b)]: Derivation of Closed-Loop Transfer Function $\frac{Y_1}{Y_{sp1}}$ and Characteristic Equation

From the Cascade Control block diagram (Fig. 1(b) / Fig. 16.4, setting disturbances $D_1 = D_2 = 0$ for the servo problem):

1. Step 1 — Reduce the Inner (Slave) Loop:
   In the inner loop, the input is $\tilde{Y}_{sp2}$ and the output is $Y_2$: $$Y_2 = G_{c2} G_v G_{p2} E_2 = G_{c2} G_v G_{p2} \left(\tilde{Y}_{sp2} - G_{m2} Y_2\right)$$ $$\implies \left(1 + G_{c2} G_v G_{p2} G_{m2}\right) Y_2 = G_{c2} G_v G_{p2} \tilde{Y}_{sp2}$$ $$\implies G_{\text{inner}}(s) = \frac{Y_2(s)}{\tilde{Y}_{sp2}(s)} = \frac{G_{c2} G_v G_{p2}}{1 + G_{c2} G_v G_{p2} G_{m2}}$$
2. Step 2 — Reduce the Outer (Master) Loop:
   Now the forward path from $\tilde{Y}_{sp1}$ to $Y_1$ is $G_{c1} G_{\text{inner}} G_{p1}$, with feedback $G_{m1}$: $$Y_1 = G_{p1} Y_2 = G_{p1} G_{\text{inner}} G_{c1} \left(K_{m1} Y_{sp1} - G_{m1} Y_1\right)$$ $$\frac{Y_1(s)}{Y_{sp1}(s)} = \frac{K_{m1} G_{c1} G_{\text{inner}} G_{p1}}{1 + G_{c1} G_{\text{inner}} G_{p1} G_{m1}} = \frac{K_{m1} G_{c1} \left(\dfrac{G_{c2} G_v G_{p2}}{1 + G_{c2} G_v G_{p2} G_{m2}}\right) G_{p1}}{1 + G_{c1} \left(\dfrac{G_{c2} G_v G_{p2}}{1 + G_{c2} G_v G_{p2} G_{m2}}\right) G_{p1} G_{m1}}$$ Multiplying numerator and denominator by $\left(1 + G_{c2} G_v G_{p2} G_{m2}\right)$ yields the closed-loop transfer function: $$\frac{Y_1(s)}{Y_{sp1}(s)} = \frac{K_{m1} G_{c1} G_{c2} G_v G_{p2} G_{p1}}{1 + G_{c2} G_v G_{p2} G_{m2} + G_{c1} G_{c2} G_v G_{p2} G_{p1} G_{m1}}$$
3. Step 3 — Characteristic Equation:
   Setting the denominator of the closed-loop transfer function equal to zero gives the characteristic equation: $$1 + G_{c2} G_v G_{p2} G_{m2} + G_{c1} G_{c2} G_v G_{p2} G_{p1} G_{m1} = 0$$

Part (e) [2021-22 Q1(e)]: Range of $K_{c1}$ for Stability

Substitute the given transfer functions into the characteristic equation ($G_{c1} = K_{c1}, \; G_{c2} = 2, \; G_v = \frac{1}{s+1}, \; G_{p2} = \frac{2}{2s+1}, \; G_{m2} = 2, \; G_{p1} = \frac{4e^{-4s}}{(10s+1)(-2s+1)}, \; G_{m1} = 0.5$):

• Inner loop product: $G_{c2} G_v G_{p2} G_{m2} = (2)\left(\frac{1}{s+1}\right)\left(\frac{2}{2s+1}\right)(2) = \frac{8}{(s+1)(2s+1)}$
• Outer loop product (using 1/1 Padé $e^{-4s} \approx \frac{1-2s}{1+2s}$): $$G_{c1} G_{c2} G_v G_{p2} G_{p1} G_{m1} = K_{c1} (2)\left(\frac{1}{s+1}\right)\left(\frac{2}{2s+1}\right)\left[\frac{4(1 - 2s)}{(10s + 1)(-2s + 1)(2s + 1)}\right](0.5)$$ Canceling $(1 - 2s) = (-2s + 1)$ gives: $$= \frac{8 K_{c1}}{(s + 1)(2s + 1)^2 (10s + 1)}$$

Substitute both into the characteristic equation $1 + G_{c2} G_v G_{p2} G_{m2} + G_{c1} G_{c2} G_v G_{p2} G_{p1} G_{m1} = 0$:

Multiply through by $(s + 1)(2s + 1)^2 (10s + 1)$:

Expand each term:

• $(s + 1)(4s^2 + 4s + 1)(10s + 1) = (4s^3 + 8s^2 + 5s + 1)(10s + 1) = 40s^4 + 84s^3 + 58s^2 + 15s + 1$
• $8(20s^2 + 12s + 1) = 160s^2 + 96s + 8$

Summing gives the 4th-order characteristic polynomial:

Construct the Routh Array:

• Calculate $b_1$: $$b_1 = \frac{84(218) - 40(111)}{84} = \frac{18312 - 4440}{84} = \frac{13872}{84} = 165.143 > 0$$
• Calculate $c_1 > 0$: $$c_1 = \frac{b_1(111) - 84(9 + 8K_{c1})}{b_1} > 0 \implies 165.143(111) - 756 - 672 K_{c1} > 0$$ $$18330.86 - 756 > 672 K_{c1} \implies 672 K_{c1} < 17574.86 \implies K_{c1} < 26.15$$
• Calculate $d_1 > 0$: $$9 + 8 K_{c1} > 0 \implies K_{c1} > -\frac{9}{8} = -1.125$$

Thus, the stable range of $K_{c1}$ is $-1.125 < K_{c1} < 26.15$!

Solution for [2020-21 Q2(b)]: Maximum Value of $K_{c1}$ Using Routh Array (Example 16.1, Slide 8 of Chapter 16 PDF)

Given: $G_v = \frac{5}{s+1}, \; G_{p1} = \frac{4}{(4s+1)(2s+1)}, \; G_{p2} = 1, \; G_{m1} = 0.05, \; G_{m2} = 0.2, \; G_{c2} = K_{c2} = 4, \; G_{c1} = K_{c1}$.

• Inner loop product: $$G_{c2} G_v G_{p2} G_{m2} = (4)\left(\frac{5}{s + 1}\right)(1)(0.2) = \frac{4}{s + 1}$$
• Outer loop product: $$G_{c1} G_{c2} G_v G_{p2} G_{p1} G_{m1} = K_{c1} (4)\left(\frac{5}{s + 1}\right)(1)\left[\frac{4}{(4s + 1)(2s + 1)}\right](0.05) = \frac{4 K_{c1}}{(s + 1)(4s + 1)(2s + 1)}$$

Substitute into the characteristic equation $1 + \frac{4}{s + 1} + \frac{4 K_{c1}}{(s + 1)(4s + 1)(2s + 1)} = 0$ and multiply by $(s + 1)(4s + 1)(2s + 1)$:

Construct the Routh Array:

• For $b_1 > 0$: $$46(31) - 8(5 + 4K_{c1}) > 0 \implies 1426 - 40 - 32 K_{c1} > 0 \implies 32 K_{c1} < 1386 \implies K_{c1} < \frac{1386}{32} = 43.3125$$
• For $c_1 > 0$: $5 + 4K_{c1} > 0 \implies K_{c1} > -1.25$.

Therefore, the maximum value of $K_{c1}$ for which the cascade system remains stable is $K_{c1,\max} = 43.31$!

Step-by-Step Detailed Solution:

Part 1: Open-Loop Stability & Pole-Zero Locations in the $s$-Plane (3 Marks)

• Zero of $G(s)$: $-s + 1 = 0 \implies z_1 = +1.0$ (located on the positive real axis in the Right-Half Plane).
• Poles of $G(s)$: $(4s - 1)(2s + 1) = 0 \implies p_1 = +0.25$ (in the Right-Half Plane) and $p_2 = -0.5$ (in the Left-Half Plane).
• Justification: Because the pole $p_1 = +0.25 > 0$ is located in the Right-Half Plane (RHP), the natural response contains the growing exponential mode $e^{+0.25t}$. Therefore, the open-loop process is UNSTABLE.
• s-Plane Plot: Draw the Real vs. Imaginary axes, mark $\times$ at $s = -0.5$ and $\times$ at $s = +0.25$, and mark $\circ$ (zero) at $s = +1.0$.

Part 2: Unit Step Response Sketch of the Open-Loop Process (3 Marks)

For a unit step input $U(s) = \frac{1}{s}$:

Taking the inverse Laplace transform:

At $t = 0$, $y(0) = 0$; as $t \to \infty$, the term $+1.5 e^{+0.25t} \to +\infty$, so $y(t)$ grows exponentially without bound toward $+\infty$!

Part 3: Design a Proportional Controller (Range of $K_c$) for Closed-Loop Stability (6 Marks)

Assuming $G_v = G_m = 1$ and $G_c = K_c$, the closed-loop characteristic equation is:

Multiplying through by $(4s - 1)(2s + 1) = 8s^2 + 2s - 1$:

For a second-order polynomial $a_2 s^2 + a_1 s + a_0 = 0$ with $a_2 = 8 > 0$, the necessary and sufficient Routh stability conditions are that all coefficients must be strictly positive ($a_1 > 0$ and $a_0 > 0$):

1. Coefficient of $s^1 > 0$: $$2 - 2 K_c > 0 \implies 2 K_c < 2 \implies K_c < 1$$
2. Constant term $s^0 > 0$: $$2 K_c - 1 > 0 \implies 2 K_c > 1 \implies K_c > 0.5$$

Combining both inequalities gives the exact range of $K_c$ that stabilizes the closed-loop process:

Step-by-Step Detailed Solution:

Part 1: Derivation of IMC Closed-Loop Relations and Controller Design Equation from Block Diagram

1. Block Diagram Algebra on Fig. 12.6(b):
   In the IMC block diagram:
   • Actual process output: $Y = G P + D$
   • Internal model output: $\tilde{Y} = \tilde{G} P$
   • Feedback signal: $Y - \tilde{Y} = (G - \tilde{G})P + D$
   • Error signal entering $G_c^*$: $E = Y_{sp} - (Y - \tilde{Y})$
   • Controller output: $P = G_c^* E = G_c^*\left[Y_{sp} - (G - \tilde{G})P - D\right]$
   Solving for $P$: $$P = \frac{G_c^*}{1 + G_c^*(G - \tilde{G})}\left(Y_{sp} - D\right)$$ Substituting $P$ into $Y = G P + D$ yields the IMC Closed-Loop Relation (Eq. 12-17): $$Y = \left[\frac{G_c^* G}{1 + G_c^*(G - \tilde{G})}\right] Y_{sp} + \left[\frac{1 - G_c^* \tilde{G}}{1 + G_c^*(G - \tilde{G})}\right] D$$
2. Equivalence with Classical Feedback Controller $G_c$:
   Since $\tilde{Y} = \tilde{G} P$ is fed back positively to $E$, the inner loop around $G_c^*$ and $\tilde{G}$ reduces to the equivalent classical feedback controller (Eq. 12-16): $$G_c(s) = \frac{P(s)}{Y_{sp}(s) - Y(s)} = \frac{G_c^*(s)}{1 - G_c^*(s) \tilde{G}(s)}$$
3. Two-Step IMC Design Equations:
   • Step 1: Factor the process model as $\tilde{G}(s) = \tilde{G}_+(s) \tilde{G}_-(s)$, where $\tilde{G}_+(s)$ contains all time delays and RHP zeros with $\tilde{G}_+(0) = 1$.
   • Step 2: Define the IMC controller as $G_c^*(s) = \frac{1}{\tilde{G}_-(s)} f(s)$, where $f(s) = \frac{1}{(\tau_c s + 1)^r}$ (with $r = 1 \implies f(s) = \frac{1}{\tau_c s + 1}$).
   Substituting $G_c^* = \frac{f}{\tilde{G}_-}$ and $\tilde{G} = \tilde{G}_+ \tilde{G}_-$ into Eq. (12-16) gives the IMC Controller Design Equation: $$G_c(s) = \frac{1}{\tilde{G}_-(s)} \left[\frac{f(s)}{1 - \tilde{G}_+(s) f(s)}\right]$$

Part 2: Complete Derivations for FOPTD Process ($\tilde{G}(s) = \frac{K e^{-\theta s}}{\tau s + 1}$)

Both the PI derivation (using $e^{-\theta s} \approx 1 - \theta s$, for 2023-24 Q4b) and the PID derivation (using 1/1 Padé $e^{-\theta s} \approx \frac{1 - \frac{\theta}{2}s}{1 + \frac{\theta}{2}s}$, for 2022-23 Q7b & 2021-22 Q4b) are written step-by-step in Rank 2 Theory Section 1 (Cases 1 and 2) above.

Step-by-Step Detailed Solution:

Step 1: Derive the General Direct Synthesis (DS) Controller Equation

For a standard unity-feedback loop ($G = G_v G_p G_m$ with $K_m = 1$), the closed-loop transfer function for set-point changes is:

Solving for $G_c(s)$ and replacing the actual process $G(s)$ with the process model $\tilde{G}(s)$ and $\frac{Y}{Y_{sp}}$ with the desired closed-loop transfer function $\left(\frac{Y}{Y_{sp}}\right)_d$:

Since the process contains a time delay $\theta$, a physically realizable desired closed-loop transfer function must also include a time delay $\theta$:

Substituting this into Eq. (1):

Approximating the time-delay term in the denominator using a truncated first-order Taylor series expansion ($e^{-\theta s} \approx 1 - \theta s$):

Step 2: Apply to a Second-Order Plus Time Delay (SOPTD) Process

The standard SOPTD process model is:

Substituting $\tilde{G}(s)$ into Eq. (2):

Expanding the numerator polynomial:

Comparing term-by-term with the ideal parallel PID controller transfer function $G_c(s) = K_c\left(1 + \frac{1}{\tau_I s} + \tau_D s\right)$ yields the DS PID design equations:

Step-by-Step Detailed Solution:

Part A: Solution for 2023-24 Q3(b), 2022-23 Q7(c), & 2020-21 Q4(a) (Smith Predictor Block Diagram)

Step 1: Write the Signal Equations Around Every Block and Summing Junction:

• Process Output ($Y$): $$Y = G P + G_d D \qquad \text{--- (Eq. 1)}$$
• Delay-free Model Output ($\tilde{Y}_1$) and Delayed Model Output ($\tilde{Y}_2$): $$\tilde{Y}_1 = \tilde{G}^* P, \qquad \tilde{Y}_2 = e^{-\tilde{\theta}s} \tilde{Y}_1 = \tilde{G}^* e^{-\tilde{\theta}s} P \qquad \text{--- (Eq. 2)}$$
• Outer Feedback Signal ($Y - \tilde{Y}_2$) and First Comparator Output ($E$): $$E = Y_{sp} - \left(Y - \tilde{Y}_2\right) \qquad \text{--- (Eq. 3)}$$
• Second Comparator Output ($E'$) and Controller Output ($P$): $$E' = E - \tilde{Y}_1 = Y_{sp} - \left(Y - \tilde{Y}_2\right) - \tilde{Y}_1 \qquad \text{--- (Eq. 4)}$$ $$P = G_c E' = G_c \left[Y_{sp} - Y + \tilde{Y}_2 - \tilde{Y}_1\right] \qquad \text{--- (Eq. 5)}$$

Step 2: Substitute $\tilde{Y}_1$ and $\tilde{Y}_2$ from Eq. (2) into Eq. (5) to Solve for $P$:

Group all $P$ terms on the left-hand side:

Step 3: Substitute $P$ from Eq. (6) into Process Output Eq. (1) ($Y = G P + G_d D$):

Multiply both sides by $\left[1 + G_c \tilde{G}^*\left(1 - e^{-\tilde{\theta}s}\right)\right]$ and collect all $Y$ terms on the left:

Therefore, we obtain both required closed-loop relationships:

• 1. Relationship Between $Y$ and Disturbance $D$ (Setting $Y_{sp} = 0$) — [For 2023-24 Q3(b) & 2022-23 Q7(c)]: $$\frac{Y(s)}{D(s)} = \frac{G_d \left[1 + G_c \tilde{G}^*\left(1 - e^{-\tilde{\theta}s}\right)\right]}{1 + G_c \tilde{G}^*\left(1 - e^{-\tilde{\theta}s}\right) + G_c G}$$ (Note: If the model is perfect, i.e., $G = \tilde{G}^* e^{-\tilde{\theta}s}$, the denominator simplifies to $1 + G_c \tilde{G}^*$, completely eliminating the time delay $e^{-\tilde{\theta}s}$ from the characteristic equation!)
• 2. Relationship Between $Y$ and Setpoint $Y_{sp}$ (Setting $D = 0$) — [For 2020-21 Q4(a)]: $$\frac{Y(s)}{Y_{sp}(s)} = \frac{G_c G}{1 + G_c \tilde{G}^*\left(1 - e^{-\tilde{\theta}s}\right) + G_c G}$$ (For a perfect model $G = \tilde{G}^* e^{-\tilde{\theta}s}$, this simplifies to $\frac{Y(s)}{Y_{sp}(s)} = \frac{G_c \tilde{G}^* e^{-\tilde{\theta}s}}{1 + G_c \tilde{G}^*}$).

Part B: Solution for 2019-20 Q8(b) (Standard IMC Block Diagram + Routh Array for $K_c$)

1. Closed-Loop Relation Between $Y$ and $Y_{sp}$ (with $D = 0$):
   From Fig. 8(b): $Y = G P$, $\tilde{Y} = \tilde{G} P$, and $P = G_c^* E = G_c^*\left[Y_{sp} - (Y - \tilde{Y})\right] = G_c^* Y_{sp} - G_c^*(G - \tilde{G})P$.
   Solving for $P = \frac{G_c^*}{1 + G_c^*(G - \tilde{G})}Y_{sp}$ and multiplying by $G$ gives: $$\frac{Y(s)}{Y_{sp}(s)} = \frac{G_c^*(s) G(s)}{1 + G_c^*(s)\left[G(s) - \tilde{G}(s)\right]}$$
2. Finding Maximum Value of $K_c$ Using Routh Array:
   Given $G_c^* = K_c$, $G = \frac{1}{(3s + 1)(s + 1)}$, and $\tilde{G} = \frac{2}{s + 1}$ (notice $\tilde{G} \ne G$, i.e., model mismatch!).
   The closed-loop characteristic equation is: $$1 + G_c^*(s)\left[G(s) - \tilde{G}(s)\right] = 0$$ $$1 + K_c \left[\frac{1}{(3s + 1)(s + 1)} - \frac{2}{s + 1}\right] = 0$$ Multiply through by $(3s + 1)(s + 1)$: $$(3s + 1)(s + 1) + K_c \left[1 - 2(3s + 1)\right] = 0$$ $$(3s^2 + 4s + 1) + K_c (-6s - 1) = 0$$ $$3s^2 + (4 - 6 K_c)s + (1 - K_c) = 0$$ Construct the Routh Array: $$\begin{array}{c|cc} s^2 & 3 & 1 - K_c \\ s^1 & 4 - 6K_c & 0 \\ s^0 & 1 - K_c & 0 \end{array}$$ For all elements in the first column to be positive:
   • Row 2 ($s^1$): $4 - 6 K_c > 0 \implies 6 K_c < 4 \implies K_c < \frac{4}{6} = \frac{2}{3} = 0.667$
   • Row 3 ($s^0$): $1 - K_c > 0 \implies K_c < 1.0$
   Both conditions must hold simultaneously, so the more restrictive bound governs: $$K_c < \frac{2}{3} = 0.667 \quad (\text{i.e., } K_{c,\max} = 0.667)$$

Step-by-Step Detailed Solution:

Let us define the blocks in Figure 3(b) for compactness:

1. Step 1 — Reduce the Inner Positive-Feedback Loop:
   The inner loop has forward block $G_p(s) = \frac{K_p}{\tau s + 1}$ and positive feedback block $G_f(s) = \frac{1}{\tau_c s + 1}$: $$G_{\text{inner}}(s) = \frac{G_p(s)}{1 - G_p(s) G_f(s)} = \frac{\dfrac{K_p}{\tau s + 1}}{1 - \dfrac{K_p}{(\tau s + 1)(\tau_c s + 1)}}$$
2. Step 2 — Write the Outer Loop Equation with $\Gamma^{\text{set}}(s) = 0$:
   With $\Gamma^{\text{set}}(s) = 0$, the error signal is $E(s) = -C(s) = -H(s)\Gamma(s)$.
   The signal leaving the inner loop is $-G_c(s) G_v(s) G_w(s) G_{\text{inner}}(s) H(s) \Gamma(s)$.
   At the output summing junction, this signal adds to $G_F(s) F(s)$ to produce $\Gamma(s)$: $$\Gamma(s) = G_F(s) F(s) - G_c(s) G_v(s) G_w(s) G_{\text{inner}}(s) H(s) \Gamma(s)$$
3. Step 3 — Solve for $\frac{\Gamma(s)}{F(s)}$: $$\frac{\Gamma(s)}{F(s)} = \frac{G_F(s)}{1 + G_c(s) G_v(s) G_w(s) G_{\text{inner}}(s) H(s)}$$ Substituting $G_{\text{inner}}(s) = \frac{G_p}{1 - G_p G_f}$ and multiplying numerator and denominator by $(1 - G_p G_f)$ gives: $$\frac{\Gamma(s)}{F(s)} = \frac{G_F(s)\left[1 - G_p(s) G_f(s)\right]}{1 - G_p(s) G_f(s) + G_c(s) G_v(s) G_w(s) G_p(s) H(s)}$$

Step-by-Step Detailed Solutions for All 4 Exam Cases:

Solution for Case 1 [2023-24 Q4(c) — Evaporator with 3 Control Valves]:

1. Steam Supply Valve (to Evaporator heating coil): Fail-Close (FC / Air-to-Open A/O).
   Justification: In case of power/instrument air failure, steam heating must be shut off immediately to prevent overheating, boiling dry, and overpressurizing the evaporator vessel.
2. Overhead Solvent Vapor Valve (controlled by PC): Fail-Open (FO / Air-to-Close A/C).
   Justification: If air fails, this vapor exit valve must open fully to vent evaporated solvent vapor and prevent dangerous pressure buildup (explosion hazard) inside the evaporator.
3. Bottoms Concentrated Product Valve (controlled by LC/AC): Fail-Close (FC / Air-to-Open A/O).
   Justification: Closing the bottom outlet retains the liquid level above the steam heating tubes (preventing tube burn-out/dry-out) and prevents dumping off-specification or hot concentrated liquid to downstream units.

Solution for Case 2 [2022-23 Q6(c) & 2018-19 Q2(c) — Flash Drum with 5 Control Valves]:

To achieve the lowest temperature and pressure in the flash drum and prevent liquid from flooding downstream equipment during an air failure:

1. Valve 1 (Steam Inlet Line to Heating Coil): Fail-Close (FC / Air-to-Open A/O).
   Justification: Stops heat input into the flash drum upon air failure, ensuring temperature and vapor generation drop to the safest minimum level.
2. Valve 2 (Steam Condensate Exit Line from Coil): Fail-Open (FO / Air-to-Close A/C).
   Justification: Drains remaining hot steam/condensate out of the heating coil rapidly so residual heat is removed and coil water-hammer is avoided.
3. Valve 3 (Overhead Vapor Exit Line controlled by PC): Fail-Open (FO / Air-to-Close A/C).
   Justification: Opens fully to depressurize the flash drum by venting vapor, achieving the lowest possible vessel pressure.
4. Valve 4 (Bottom Liquid Product Exit Line controlled by LC): Fail-Close (FC / Air-to-Open A/O) (or Fail-Open if preventing vapor blow-through vs. preventing liquid to downstream).
   Justification: Condition (ii) states that "liquid flow to a downstream equipment can cause a hazardous situation" (either via the bottom liquid line or via liquid overfill into the overhead vapor line). If downstream liquid line flow is hazardous, Valve 4 must be Fail-Close (FC / A/O) (and since Valve 5 shuts off the feed, the drum cannot overfill into the vapor line!). Conversely, if condition (ii) refers to preventing liquid overflow into the vapor line (3), Valve 4 could be FO, though shutting off Feed Valve 5 (FC) already prevents overfilling. (State both clearly to get full credit!).
5. Valve 5 (Liquid Feed Inlet Line to Flash Drum): Fail-Close (FC / Air-to-Open A/O).
   Justification: Stops incoming liquid feed to the flash drum during an emergency, preventing vessel overfilling (which would send liquid into the downstream vapor equipment) and minimizing inventory.

Solution for Case 3 [2021-22 Q4(c) — Exothermic Reactor with Preheater Furnace (3 Valves)]:

1. Valve 1 — Fuel Valve to Furnace (controlled by FC-103): Fail-Close (FC / Air-to-Open A/O).
   Justification: On instrument air failure, fuel flow to the furnace burners must immediately shut off to prevent furnace overheating, tube rupture, or fire/explosion, and to stop preheating the feed to the exothermic reactor.
2. Valve 2 — Process Stream (Reactant Feed) Valve (entering Furnace): Fail-Close (FC / Air-to-Open A/O) (or Fail-Open briefly if furnace residual heat requires tube cooling, but for an exothermic reactor downstream, stopping reactant feed is critical).
   Justification: Since the downstream reactor is exothermic, shutting off reactant feed (Fail-Close) stops the chemical reaction and prevents reactor thermal runaway.
3. Valve 3 — Cooling Water (C.W.) Valve to Exothermic Reactor Jacket: Fail-Open (FO / Air-to-Close A/C).
   Justification: Since the reaction in the reactor is exothermic, maximum cooling water flow must be maintained upon air failure to remove heat of reaction and prevent a dangerous thermal runaway explosion.

Solution for Case 4 [2019-20 Q8(a) — Three Industrial Valve Cases]:

1. (i) Flare stack line of an ammonia tank (controlling tank pressure): Air-to-Close (A/C) / Fail-Open (FO).
   Justification: If instrument air fails, the valve on the flare stack line must fail open to relieve excess ammonia vapor to the flare system and prevent the storage tank from rupturing due to overpressure.
2. (ii) Cooling water line of a jacketed exothermic reactor: Air-to-Close (A/C) / Fail-Open (FO).
   Justification: On air failure, the cooling water valve must open fully to provide maximum cooling capacity and prevent thermal runaway of the exothermic reaction.
3. (iii) Feed line of a crude oil distillation column: Air-to-Open (A/O) / Fail-Close (FC).
   Justification: On air failure, crude oil feed to the column must be shut off to prevent column flooding, overpressurization, and accumulation of flammable hydrocarbons.

Step-by-Step Detailed Solution:

1. Solution for Case A [2023-24 Q2(b)]: $G(s) = \frac{5(0.5s + 1)e^{-0.5s}}{(20s + 1)(4s + 1)}$

Step 1: Substitute $s = j\omega$ into $G(s)$:

Step 2: Find the Magnitude and Phase of Each Individual Factor:

• Constant gain $G_a = 5$: $|5| = 5$, $\angle 5 = 0^\circ$
• Numerator zero $G_b = (0.5 j\omega + 1)$: $|0.5 j\omega + 1| = \sqrt{1 + (0.5\omega)^2} = \sqrt{1 + 0.25\omega^2}$, $\angle(0.5 j\omega + 1) = +\tan^{-1}(0.5\omega)$
• Time delay $G_c = e^{-0.5 j\omega} = \cos(0.5\omega) - j\sin(0.5\omega)$: $|e^{-0.5 j\omega}| = 1$, $\angle e^{-0.5 j\omega} = -0.5\omega\text{ [rad]} = -0.5\omega\left(\frac{180^\circ}{\pi}\right)$
• Denominator pole $G_1 = (20 j\omega + 1)$: $|20 j\omega + 1| = \sqrt{1 + (20\omega)^2} = \sqrt{1 + 400\omega^2}$, $\angle(20 j\omega + 1) = +\tan^{-1}(20\omega)$
• Denominator pole $G_2 = (4 j\omega + 1)$: $|4 j\omega + 1| = \sqrt{1 + (4\omega)^2} = \sqrt{1 + 16\omega^2}$, $\angle(4 j\omega + 1) = +\tan^{-1}(4\omega)$

Step 3: Combine Using the Product/Quotient Rule (Eqs. 13-24a & 13-24b):

2. Solution for Case B [2020-21 Q3(a)]: $G(s) = \frac{(2s + 1)e^{-2s}}{(5s + 1)(10s + 1)}$

Substituting $s = j\omega$:

Applying the magnitude and phase rules:

Qualitative Bode Plot Sketch Description:
• Log-Log Plot of $AR$ vs. $\omega$: At low frequencies ($\omega \to 0$), $AR \to 1$ (horizontal asymptote at $1$). Break frequencies occur at $\omega_{b1} = 1/10 = 0.1\text{ rad/time}$ (slope becomes $-1$), $\omega_{b2} = 1/5 = 0.2\text{ rad/time}$ (slope becomes $-2$), and $\omega_{b3} = 1/2 = 0.5\text{ rad/time}$ (zero brings high-frequency asymptotic slope back to $-1$).
• Semi-Log Plot of $\phi$ (deg) vs. $\omega$: At $\omega \to 0$, $\phi \to 0^\circ$. As $\omega$ increases, the two lags and the time delay $-2\omega(180^\circ/\pi)$ cause $\phi$ to decrease continuously past $-180^\circ, -270^\circ, -\infty$.

3. Solution for Case C [2022-23 Q6(b), 2019-20 Q5(c), & 2018-19 Q2(b)]: FOPTD Process $G(s) = \frac{K e^{-\theta s}}{\tau s + 1}$

Substitute $s = j\omega$ and use Euler's identity $e^{-j\omega\theta} = \cos(\omega\theta) - j\sin(\omega\theta)$:

By the product rule:

• Amplitude Ratio ($AR$): $$AR = |G(j\omega)| = \left|\frac{K}{1 + j\omega\tau}\right| \cdot \left|e^{-j\omega\theta}\right| = \frac{K}{\sqrt{1 + \omega^2\tau^2}} \cdot \sqrt{\cos^2(\omega\theta) + \sin^2(\omega\theta)} = \frac{K}{\sqrt{\omega^2\tau^2 + 1}}$$
• Phase Angle ($\phi$): $$\phi = \angle G(j\omega) = \angle\left(\frac{K}{1 + j\omega\tau}\right) + \angle\left(e^{-j\omega\theta}\right) = -\tan^{-1}(\omega\tau) - \omega\theta\text{ [rad]} = -\tan^{-1}(\omega\tau) - \omega\theta\left(\frac{180^\circ}{\pi}\right)\text{ [deg]}$$

Qualitative Bode Plot of a FOPTD System (see Fig. 14.7 on Slide 15 of Chapter 14 PDF):
• Top Graph ($\log AR$ vs. $\log \omega$): Horizontal low-frequency asymptote at $AR = K$ for $\omega \ll 1/\tau$; corner/break frequency at $\omega_b = 1/\tau$ (where $AR = 0.707 K$); high-frequency asymptote with slope $-1$ ($AR \approx \frac{K}{\omega\tau}$) for $\omega \gg 1/\tau$.
• Bottom Graph ($\phi$ in deg vs. $\log \omega$): Starts at $0^\circ$ for small $\omega$, passes through $-\left(45^\circ + \frac{\theta}{\tau}\frac{180^\circ}{\pi}\right)$ at $\omega = 1/\tau$, and drops sharply without bound below $-180^\circ, -270^\circ, -360^\circ\dots$ at high frequencies due to the $-\omega\theta$ delay term.

Step-by-Step Detailed Solution:

1. Steady-State Gain ($K$) [5 Marks]:

Since $\lim_{\omega \to 0} AR(\omega) = \lim_{s \to 0} G(s) = K$, we read the value of the low-frequency horizontal asymptote of the $AR$ curve at $\omega = 0.01\text{ rad/min}$ in Fig. Q2(c).
On the logarithmic vertical axis between $1$ and $10$, the flat low-frequency curve starts halfway up on the log scale at:

2. Gain Margin ($GM$) [5 Marks]:

• First, locate $\phi = -180^\circ$ on the lower phase plot of Fig. Q2(c). Moving horizontally to the phase curve and reading down to the frequency axis gives the critical (phase crossover) frequency: $$\omega_c \approx 1.5\text{ rad/min}$$
• Moving vertically up to the $AR$ curve at $\omega_c = 1.5\text{ rad/min}$, we read the critical amplitude ratio: $$AR_c = AR(\omega_c) \approx 0.10$$
• Therefore, the Gain Margin is: $$GM = \frac{1}{AR_c} = \frac{1}{0.10} = 10.0$$

3. Phase Margin ($PM$) [5 Marks]:

• First, locate $AR = 1$ on the upper amplitude ratio plot of Fig. Q2(c). Moving horizontally to the $AR$ curve and reading down to the frequency axis gives the gain crossover frequency: $$\omega_g \approx 0.35\text{ rad/min} \quad (\text{between } 0.3 \text{ and } 0.4\text{ rad/min})$$
• Moving vertically down to the phase curve at $\omega_g = 0.35\text{ rad/min}$, we read the phase angle: $$\phi_g = \phi(\omega_g) \approx -105^\circ \quad (\text{just below } -90^\circ)$$
• Therefore, the Phase Margin is: $$PM = 180^\circ + \phi_g = 180^\circ + (-105^\circ) = 75^\circ$$

4. Maximum Additional Time Delay ($\Delta\theta_{\max}$) Before Instability [5 Marks]:

We show both the exact gain-crossover formula ($\omega_g$) and the textbook Slide 18 formula ($\omega_c$):

• Using Gain Crossover Frequency $\omega_g = 0.35\text{ rad/min}$ (Exact Frequency Where $AR = 1$): $$\Delta\theta_{\max} = \left(\frac{PM}{\omega_g}\right)\left(\frac{\pi}{180^\circ}\right) = \left(\frac{75^\circ}{0.35\text{ rad/min}}\right)\left(\frac{\pi}{180^\circ}\right) = \frac{1.309\text{ rad}}{0.35\text{ rad/min}} = 3.74\text{ min}$$
• Using Critical Frequency $\omega_c = 1.5\text{ rad/min}$ (Eq. 14-14 on Slide 18 of Chapter 14 PDF): $$\Delta\theta_{\max} = \left(\frac{PM}{\omega_c}\right)\left(\frac{\pi}{180^\circ}\right) = \left(\frac{75^\circ}{1.5\text{ rad/min}}\right)\left(\frac{\pi}{180^\circ}\right) = \frac{1.309\text{ rad}}{1.5\text{ rad/min}} = 0.873\text{ min}$$

Step-by-Step Detailed Solution:

Step 1: Derive Normalized Amplitude Ratio ($AR_N$) and Phase Angle ($\phi$)

Substitute $s = j\omega$ (noting that $s^2 = (j\omega)^2 = -\omega^2$) into the second-order transfer function:

Multiply numerator and denominator by the complex conjugate $\left(1 - \omega^2\tau^2\right) - j\left(2\zeta\omega\tau\right)$:

• Amplitude Ratio ($AR$) and Normalized Amplitude Ratio ($AR_N = AR / K$): $$AR = \sqrt{R^2 + I^2} = \frac{K}{\sqrt{\left(1 - \omega^2\tau^2\right)^2 + \left(2\zeta\omega\tau\right)^2}}$$ $$\implies AR_N = \frac{AR}{K} = \frac{1}{\sqrt{\left(1 - \omega^2\tau^2\right)^2 + \left(2\zeta\omega\tau\right)^2}}$$
• Phase Difference ($\phi$): $$\phi = \tan^{-1}\left(\frac{I}{R}\right) = \tan^{-1}\left(\frac{-2\zeta\omega\tau}{1 - \omega^2\tau^2}\right)$$

Step 2: Key Features for Sketching the Bode Plot for $\zeta = 2, \; 1,$ and $0.5$ (Figure 13.3, Slide 15)

• Low-Frequency Asymptote ($\omega\tau \ll 1$): For all three values of $\zeta$, $AR_N \to 1$ and $\phi \to 0^\circ$.
• High-Frequency Asymptote ($\omega\tau \gg 1$): For all three values of $\zeta$, $AR_N \approx \frac{1}{(\omega\tau)^2}$ (a straight line of slope $= -2$ on the log-log plot) and $\phi \to -180^\circ$.
• Behavior Near the Corner/Natural Frequency ($\omega\tau = 1$):
  At $\omega\tau = 1$, for all values of $\zeta$: $$AR_N(\omega\tau = 1) = \frac{1}{2\zeta}, \qquad \phi(\omega\tau = 1) = -90^\circ$$
  • For $\zeta = 2.0$ (Overdamped): $AR_N(\omega\tau=1) = \frac{1}{4} = 0.25$ (lowest curve, most attenuated near $\omega\tau = 1$; phase transition from $0^\circ$ to $-180^\circ$ is the most gradual/spread out).
  • For $\zeta = 1.0$ (Critically Damped): $AR_N(\omega\tau=1) = \frac{1}{2} = 0.50$ (middle curve, monotonic decrease).
  • For $\zeta = 0.5$ (Underdamped, $\zeta < 0.707$): $AR_N(\omega\tau=1) = \frac{1}{1} = 1.0$, and exhibits a slight resonant peak ($AR_{N,\max} = \frac{1}{2\zeta\sqrt{1-\zeta^2}} = 1.155$ at $\omega\tau = \sqrt{1-2\zeta^2} = 0.707$) before dropping with slope $-2$; its phase curve has the steepest drop through $-90^\circ$ at $\omega\tau = 1$.

Step-by-Step Detailed Solution:

Step 1: Determine the Gain Margin ($GM$) from the Nyquist Plot

On a Nyquist plot (Figure 14.10, Chapter 14 PDF), the critical frequency $\omega_c$ corresponds to the point where the polar curve $G_{OL}(j\omega)$ intersects the negative real axis ($\text{Im} = 0, \; \phi_{OL} = -180^\circ$).
Inspecting Figure 6(b) on the horizontal line $\text{Im} = 0$ to the left of the origin $(0,0)$:

• The curve crosses the negative real axis exactly at $\text{Re} = -0.50$.
• Therefore, the open-loop amplitude ratio at the critical frequency is: $$AR_c = |\text{Re}_{\text{intercept}}| = 0.50$$
• The Gain Margin ($GM$) is: $$GM = \frac{1}{AR_c} = \frac{1}{0.50} = 2.0$$

Step 2: Determine the Phase Margin ($PM$) from the Nyquist Plot

The gain crossover frequency $\omega_g$ corresponds to the point where the Nyquist curve intersects the unit circle (circle centered at the origin with radius $R = \sqrt{\text{Re}^2 + \text{Im}^2} = 1.0$).
Inspecting Figure 6(b) in the third quadrant ($\text{Re} < 0, \; \text{Im} < 0$):

• The circle of radius $1.0$ intersects the Nyquist curve at approximately: $$\text{Re} \approx -0.75, \qquad \text{Im} \approx -0.66 \qquad \left(\text{Check: } \sqrt{(-0.75)^2 + (-0.66)^2} = \sqrt{0.5625 + 0.4356} = 1.0\right)$$
• The angle made by this point with the negative real axis is the Phase Margin ($PM$): $$PM = \tan^{-1}\left(\frac{|\text{Im}|}{|\text{Re}|}\right) = \tan^{-1}\left(\frac{0.66}{0.75}\right) = 41.3^\circ \quad (\approx 41^\circ, \text{ or } 0.722\text{ rad})$$ (Note: Depending on visual reading of the unit circle intersection between $(-0.71, -0.71)$ and $(-0.77, -0.64)$, any value between $40^\circ$ and $45^\circ$ is completely accurate).

Step 3: Maximum Additional Time Delay ($\Delta\theta_{\max}$) Before Instability

The maximum additional time delay $\Delta\theta_{\max}$ that the system can tolerate before becoming unstable is related to the Phase Margin $PM$ by Eq. (14-14):

(Note: Since frequency values $\omega$ are not explicitly printed along the Nyquist locus in Fig. 6(b), if $\omega_c = 1.0\text{ rad/min}$ is assumed as a reference, then $\Delta\theta_{\max} = 0.722\text{ min}$).

Step-by-Step Detailed Solution:

Parts A & B [Definitions of $\omega_c, \omega_g, GM, PM$ & Usefulness in Controller Design]:
Please refer to Rank 3 Theory Section 2 above for the exact definitions and equations. For "How are they useful in designing controllers?", write:
1. They provide a measure of relative stability (degree of robustness) rather than just a yes/no stability answer, and are exact for systems with time delays.
2. They guide controller tuning: a well-tuned PID controller is designed to achieve a Gain Margin between $1.7$ and $4.0$ and a Phase Margin between $30^\circ$ and $45^\circ$ (representing an optimal tradeoff between fast performance and robustness against model uncertainties/time delay changes).

Part C [2019-20 Q5(b)]: Complete Derivation of First-Order Sinusoidal Response (Slide 1 & 5, Chapter 13 PDF)

Step 1: Set Up the Laplace-Domain Output $Y(s)$
Given a first-order transfer function $G(s) = \frac{Y(s)}{X(s)} = \frac{K}{\tau s + 1}$ forced by a sinusoidal input $x(t) = A\sin(\omega t)$ with Laplace transform $X(s) = \frac{A\omega}{s^2 + \omega^2}$:

Step 2: Perform Partial Fraction Expansion

Multiply both sides by $(\tau s + 1)(s^2 + \omega^2)$:

Equating coefficients of like powers of $s$:

• At $s = -1/\tau$: $\alpha_1\left(\frac{1}{\tau^2} + \omega^2\right) = K A \omega \implies \alpha_1 = \frac{K A \omega \tau^2}{\omega^2 \tau^2 + 1}$
• From $s^2$ coefficient: $\alpha_1 + \alpha_2 \tau = 0 \implies \alpha_2 = -\frac{\alpha_1}{\tau} = -\frac{K A \omega \tau}{\omega^2 \tau^2 + 1}$
• From $s^1$ coefficient: $\alpha_2 + \alpha_3 \tau = 0 \implies \alpha_3 = -\frac{\alpha_2}{\tau} = \frac{K A \omega}{\omega^2 \tau^2 + 1}$

Substitute $\alpha_1, \alpha_2, \alpha_3$ back into $Y(s)$ (as shown on Slide 1 of Chapter 13 PDF):

Step 3: Take the Inverse Laplace Transform $\mathcal{L}^{-1}[Y(s)]$

Using $\mathcal{L}^{-1}\left[\frac{1}{s + 1/\tau}\right] = e^{-t/\tau}$, $\mathcal{L}^{-1}\left[\frac{s}{s^2+\omega^2}\right] = \cos(\omega t)$, and $\mathcal{L}^{-1}\left[\frac{\omega}{s^2+\omega^2}\right] = \sin(\omega t)$ gives Eq. (5-25):

Step 4: Simplify Using Trigonometric Identity for Long-Time Response ($t \to \infty$)

Using the trigonometric identity $a\cos(\omega t) + b\sin(\omega t) = \sqrt{a^2 + b^2}\sin(\omega t + \phi)$ with $a = -\omega\tau$ and $b = 1$ (so $\sqrt{a^2 + b^2} = \sqrt{\omega^2\tau^2 + 1}$ and $\phi = \tan^{-1}(- \omega\tau / 1) = -\tan^{-1}(\omega\tau)$):

Substituting this into $y(t)$ yields:

Step 5: What Can Be Inferred From It? (3 Key Inferences — Slide 5 of Chapter 13 PDF)

1. After the exponential transient $e^{-t/\tau}$ dies out ($t \to \infty$), the output signal $y_\ell(t)$ is a sinusoid with the exact same frequency $\omega$ as the input signal $x(t) = A\sin(\omega t)$.
2. The amplitude of the output signal $\hat{A} = \frac{KA}{\sqrt{\omega^2\tau^2 + 1}}$ is a function of the input frequency $\omega$: at low frequencies ($\omega \to 0$), $\hat{A} \to KA$, whereas at high frequencies ($\omega \to \infty$), $\hat{A} \to 0$ (the first-order process attenuates high-frequency fluctuations/noise).
3. The output signal lags behind the input signal by a phase shift (phase lag) $\phi = -\tan^{-1}(\omega\tau)$, which varies from $0^\circ$ (at $\omega \to 0$) to $-90^\circ$ (at $\omega \to \infty$).

Complete Exam-Ready Solutions (Chapter 8 PDF):

1. Proportional (P), Integral (I), and Derivative (D) Modes: Equations, Merits, Demerits & Remedies

2. What is Integral Windup (Reset Windup) and How to Eliminate It? (Slides 17, 19–20, 31)

• Definition: When a large set-point change or sustained disturbance occurs, the error $e(t)$ remains nonzero for a long period, causing the integral term $\frac{K_c}{\tau_I}\int_0^t e(t^*)dt^*$ to build up to a very large value so that the controller output $p(t)$ saturates the control valve (at $0\%$ or $100\%$ open). The continued accumulation (buildup) of the integral term while the final control element is already saturated is called reset windup (or integral windup). It causes a large, prolonged overshoot because the error must change sign for a long time to "unwind" the integral term.
• How to Eliminate / Avoid It (Anti-Reset Windup):
  1. Conditional Integration (Clamping): Halt (freeze) the integration calculation whenever the controller output or control valve reaches its saturation limit ($p_{\max}$ or $p_{\min}$).
  2. Use Digital Velocity Form of PID: The velocity algorithm ($\Delta p_k = p_k - p_{k-1}$) inherently prevents reset windup because the summation of errors is not explicitly calculated.
  3. External Reset Feedback / Back-Calculation: Feed the actual measured actuator/valve position back to reset the integral term.

3. Elimination of Derivative Kick and Proportional Kick / Beta-Gamma ($\beta$-$\gamma$) Controller (Slides 25–26)

• Cause of Kick: A sudden step change in setpoint $y_{sp}(t)$ causes an instantaneous jump in error $e(t) = y_{sp}(t) - y_m(t)$, making the proportional term $K_c e(t)$ jump (proportional kick) and making the derivative $\frac{de}{dt}$ momentarily infinite (derivative kick), which slams the control valve.
• Eliminating Derivative Kick Only (Eq. 8-17): Since $\frac{de}{dt} = \frac{dy_{sp}}{dt} - \frac{dy_m}{dt}$, replace $\frac{de}{dt}$ with $-\frac{dy_m}{dt}$ (derivative on measurement only): $$p(t) = \bar{p} + K_c\left[e(t) + \frac{1}{\tau_I}\int_0^t e(t^*)\,dt^* - \tau_D \frac{dy_m(t)}{dt}\right]$$
• Eliminating Both Proportional Kick and Derivative Kick — The Beta-Gamma ($\beta$-$\gamma$) Controller (Eqs. 8-18 to 8-21):
  Weight the setpoint $y_{sp}(t)$ in the proportional and derivative terms using nonnegative constants $\beta$ and $\gamma$ ($0 \le \beta \le 1$, $0 \le \gamma \le 1$): $$p(t) = \bar{p} + K_c\left(e_P(t) + \frac{1}{\tau_I}\int_0^t e(t^*)\,dt^* + \tau_D \frac{de_D(t)}{dt}\right)$$ where: $$e_P(t) \triangleq \beta y_{sp}(t) - y_m(t), \qquad e(t) \triangleq y_{sp}(t) - y_m(t), \qquad e_D(t) \triangleq \gamma y_{sp}(t) - y_m(t)$$ Setting $\beta = 0$ and $\gamma = 0$ completely eliminates both proportional kick and derivative kick upon a setpoint step change!

4. On-Off Controller: Equation, Dead Band Chart, Example, Merits & Demerits (Slides 2–10)

• Working Principle & Equations: An On-Off controller (also called a two-position or bang-bang controller) is the simplest feedback controller where the output $p(t)$ has only two possible values: $p_{\max}$ ("ON" / $100\%$ open) and $p_{\min}$ ("OFF" / $0\%$ closed).
  • Ideal Case: $$p(t) = \begin{cases} p_{\max} & \text{if } e > 0 \\ p_{\min} & \text{if } e < 0 \end{cases}$$
  • Practical Case with Dead Band ($\pm\delta$) (to prevent rapid chattering due to noise): $$p(t) = \begin{cases} p_{\max} & \text{for } e > +\delta \\ p_{\min} & \text{for } e < -\delta \end{cases}$$ (Within $-\delta \le e \le +\delta$, the controller stays in its current state, forming a hysteresis loop).
• Example: Domestic refrigerator thermostat, room space heater, or electric water heater (Slide 4–5).
• Advantages (Merits): (1) Very simple in construction and operation, (2) Inexpensive (cheap), (3) Requires no complex tuning.
• Disadvantages (Demerits): (1) Causes continuous cycling (hunting) of the controlled variable around the setpoint, (2) Causes excessive mechanical wear and tear on the control valve/actuator, (3) Poor control quality for industrial chemical processes.

5. Derivation of Digital Velocity Form of PID Controller & Its Advantages Over Position Form (Slides 28–31)

• Step 1 — Continuous Parallel PID Form (Eq. 8-13): $$p(t) = \bar{p} + K_c\left[e(t) + \frac{1}{\tau_I}\int_0^t e(t^*)\,dt^* + \tau_D \frac{de(t)}{dt}\right]$$
• Step 2 — Finite Difference Approximations (with sampling period $\Delta t$ at $k$-th instant): $$\int_0^t e(t^*)\,dt^* \approx \sum_{j=1}^k e_j \Delta t, \qquad \frac{de}{dt} \approx \frac{e_k - e_{k-1}}{\Delta t}$$
• Step 3 — Digital Position Form at Step $k$ and Step $(k-1)$ (Eqs. 8-26 & 8-27): $$p_k = \bar{p} + K_c\left[e_k + \frac{\Delta t}{\tau_I}\sum_{j=1}^k e_j + \frac{\tau_D}{\Delta t}(e_k - e_{k-1})\right] \qquad \text{--- (Eq. 8-26)}$$ $$p_{k-1} = \bar{p} + K_c\left[e_{k-1} + \frac{\Delta t}{\tau_I}\sum_{j=1}^{k-1} e_j + \frac{\tau_D}{\Delta t}(e_{k-1} - e_{k-2})\right] \qquad \text{--- (Eq. 8-27)}$$
• Step 4 — Subtract Eq. (8-27) from Eq. (8-26) to Obtain the Velocity Form (Eq. 8-28):
  Noting that $\sum_{j=1}^k e_j - \sum_{j=1}^{k-1} e_j = e_k$ and $\bar{p} - \bar{p} = 0$: $$\Delta p_k = p_k - p_{k-1} = K_c\left[(e_k - e_{k-1}) + \frac{\Delta t}{\tau_I} e_k + \frac{\tau_D}{\Delta t}\left(e_k - 2e_{k-1} + e_{k-2}\right)\right]$$
• Three Advantages of Velocity Form Over Position Form (Slide 31):
  1. Inherently contains anti-reset windup because the summation of errors is not explicitly calculated.
  2. The incremental output $\Delta p_k$ can be used directly by incremental final control elements, such as a control valve driven by a pulsed stepping motor.
  3. Bumpless transfer from Manual to Automatic mode: Does not require initialization of the bias $\bar{p}$ when switching from manual to automatic mode.

Complete Exam-Ready Solutions (Chapter 12 PDF, Slides 39–50):

Method 1: Ziegler-Nichols (Z-N) Continuous Cycling Method (Slides 39–43)

• Step-by-Step Procedure:
  1. Step 1: Bring the process to steady state. Turn off integral and derivative action by setting $\tau_D = 0$ and $\tau_I = \infty$ (largest possible value) so the controller acts as a Proportional-only (P) controller.
  2. Step 2: Set $K_c$ to a small value (e.g., $0.5$) and place the controller in automatic mode.
  3. Step 3: Introduce a small, momentary set-point change and gradually increase $K_c$ in small increments until the controlled variable exhibits continuous cycling (sustained oscillation with constant amplitude). Record this critical gain as the Ultimate Gain ($K_{cu}$) and the period of sustained oscillation as the Ultimate Period ($P_u$) (Fig. 12.12).
  4. Step 4: Calculate the P, PI, or PID controller parameters using the Ziegler-Nichols (or Tyreus-Luyben) tuning formulas (Table 12.6):

     Controller Type	Ziegler-Nichols ($K_c, \tau_I, \tau_D$)	Tyreus-Luyben ($K_c, \tau_I, \tau_D$)
     P	$K_c = 0.5 K_{cu}$	—
     PI	$K_c = 0.45 K_{cu}, \quad \tau_I = \frac{P_u}{1.2}$	$K_c = 0.31 K_{cu}, \quad \tau_I = 2.2 P_u$
     PID	$K_c = 0.6 K_{cu}, \quad \tau_I = \frac{P_u}{2}, \quad \tau_D = \frac{P_u}{8}$	$K_c = 0.45 K_{cu}, \quad \tau_I = 2.2 P_u, \quad \tau_D = \frac{P_u}{6.3}$

  5. Step 5: Evaluate the settings with a small set-point change and fine-tune if necessary.
• Merits (Advantages):
  • Does not require an explicit mathematical/transfer function model of the process.
  • Less sensitive to analog controller calibration errors because $K_c$ is adjusted directly during the closed-loop test.
• Demerits (Disadvantages — Slides 42–43):
  1. Time-consuming trial-and-error procedure, especially for slow processes, which can reduce production or cause off-spec product.
  2. Hazardous / Objectionable: Pushes the process right to its stability limit, which can cause runaway or equipment damage if $K_c$ slightly exceeds $K_{cu}$.
  3. Not applicable to integrating or open-loop unstable processes (which are unstable at both low and high $K_c$).
  4. First- and second-order processes without time delays have no ultimate gain ($K_{cu}$ does not exist).

Method 2: Relay Auto-Tuning Method (Åström and Hägglund, 1984 — Slides 44–46)

• Working Principle & Procedure:
  1. In the relay auto-tuning method, the PID feedback controller is temporarily switched out and replaced by an On-Off controller (relay) with a dead zone of amplitude $\pm d$ (total relay swing $= 2d$, as shown in Fig. 12.14 on Slide 45).
  2. When the control loop is closed, the relay switches between $+d$ and $-d$, automatically causing the process output $y(t)$ to exhibit a small, controlled sustained oscillation of amplitude $a$ and period $P_u$ (the ultimate period).
  3. A small dead band in the relay prevents false switching due to measurement noise.
  4. From the relay amplitude $d$ and measured output oscillation amplitude $a$, the Ultimate Gain ($K_{cu}$) is calculated by principal harmonic approximation:
     $$K_{cu} = \frac{4 d}{\pi a}$$
  5. Once $K_{cu}$ and $P_u$ are found from this single test, the PID settings ($K_c, \tau_I, \tau_D$) are calculated using the standard Ziegler-Nichols or Tyreus-Luyben relations (Table 12.6), and the switch is returned to the PID controller.
• Merits (Advantages Over Continuous Cycling — Slide 46):
  1. Only a single experimental test is required instead of a lengthy trial-and-error procedure.
  2. Safe & Controlled Amplitude: The oscillation amplitude $a$ of the process output can be strictly restricted by adjusting the relay amplitude $d$.
  3. The process is not forced to an uncontrolled stability limit.
  4. Easily automated in commercial digital PID controllers ("one-button auto-tuning").
• Demerits (Disadvantages):
  1. Formula $K_{cu} = \frac{4d}{\pi a}$ is an approximation (assumes higher harmonics are filtered out by the process).
  2. Still introduces a temporary oscillation during the test and can be affected by large external load disturbances occurring during the test.

Complete Exam-Ready Solutions:

1. Justification & Sketch of 3-Element Boiler Drum Level Control [2023-24 Q4(a), 2020-21 Q1(b), 2018-19 Q4(b)]

• What are the "3 Elements" (3 Measured Variables)?
  1. Element 1: Boiler Drum Liquid Level (measured by LT — Controlled Variable).
  2. Element 2: Steam Outlet Flow Rate (measured by FT-steam — Major Disturbance Variable).
  3. Element 3: Boiler Feedwater Inlet Flow Rate (measured by FT-feedwater — Manipulated Variable in a slave flow loop).
• Why 1-Element (Simple Feedback) and 2-Element Control Are Insufficient (Justification for 3-Element Control):
  1. Small Liquid Capacity & Rapid Steam Demand Changes: Boiler drums have a small liquid holdup relative to high steam generation rates, making the level extremely sensitive to sudden downstream steam demand changes.
  2. Inverse Response ("Shrink and Swell" Phenomenon): When steam demand suddenly increases, drum pressure drops slightly, causing steam bubbles trapped in the boiling water to expand rapidly ("swell"). This causes the apparent drum level to initially rise even though mass inventory is decreasing! A simple 1-element feedback controller (LT $\to$ LC $\to$ valve) gets fooled by this swell and reduces feedwater flow, risking boiling the drum dry!
  3. Feedwater Supply Pressure Fluctuations: Even if a 2-element (Feedforward + Feedback, Fig. 15.4) controller sums the Steam FT signal (FFC) and Drum Level LC signal, fluctuations in feedwater header pressure will cause the actual feedwater flow rate to deviate from the demanded valve signal.
• How 3-Element Control Solves All Three Problems (Working Principle & Sketch):
  • The Feedforward Controller (FFC) measures the Steam Flow Rate (FT-1) and adds its signal at a summing junction $(+)$ with the output of the Feedback Level Controller (LC, from LT) (which provides feedback trim for blowdown/model errors).
  • The summed output serves as the Remote Setpoint (RSP) to a secondary Cascade Slave Flow Controller (FC) that measures the actual Feedwater Flow Rate (FT-2) and manipulates the feedwater control valve!
  (To sketch: Draw Fig. 15.4 on Slide 7 of Chapter 15 PDF, and add an inner Flow Transmitter FT + Flow Controller FC loop on the Feedwater line between the summing junction and the feedwater control valve!).

2. Steady-State Feedforward Controller Design for Distillation Column [2020-21 Q1(d) & 2018-19 Q1(c)] (Chapter 15 PDF, Slides 9–11)

• Step 1 — Write Overall and Component Steady-State Mass Balances:
  For the binary distillation column with Feed $(F, z)$, Distillate $(D, y)$, and Bottoms $(B, x)$: $$F = D + B \implies B = F - D \qquad \text{--- (Eq. 15-4)}$$ $$F z = D y + B x \qquad \text{--- (Eq. 15-5)}$$
• Step 2 — Eliminate Unmeasured Bottoms Flow Rate $B$:
  Substitute $B = F - D$ from Eq. (15-4) into Eq. (15-5): $$F z = D y + (F - D)x = D(y - x) + F x$$ Solving for the manipulated distillate flow rate $D$: $$D = \frac{F(z - x)}{y - x} \qquad \text{--- (Eq. 15-6)}$$
• Step 3 — Replace Unmeasured Compositions $x$ and $y$ with Their Setpoints ($x_{sp}, y_{sp}$):
  Since online measurements of $x$ and $y$ are not available, replace $x$ and $y$ with their desired setpoints $x_{sp}$ and $y_{sp}$ to obtain the Steady-State Feedforward Control Law (Eq. 15-7): $$D = \frac{F\left(z - x_{sp}\right)}{y_{sp} - x_{sp}}$$ (Note: If Bottoms flow $B$ or Steam is also manipulated to control bottoms composition $x$ as in 2018-19 Q1(c), solving for $B = F - D$ gives $B = \frac{F(y_{sp} - z)}{y_{sp} - x_{sp}}$).
• Step 4 — Sketch of the Feedforward Control Loop:
  On the distillation column diagram (Fig. 15.8), draw a Flow Transmitter (FT) and Composition Transmitter (AT) on the Feed stream ($F, z$). Send both measured signals ($F_m, z_m$) into a Feedforward Controller (FFC) (which computes $D_{sp} = \frac{F(z - x_{sp})}{y_{sp} - x_{sp}}$ using setpoints $x_{sp}, y_{sp}$), and connect the FFC output to the control valve on the Distillate stream ($D$)!

3. Short Notes on Ratio Control, Split-Range Control, and Auctioneering (Selective) Control

• (a) Ratio Control (Chapter 15 PDF, Slides 18–26):
  • Justification & Objective: A special type of feedforward control used to maintain the ratio of two flow rates ($R = \frac{u}{d}$) at a specified desired value $R_d$, where $d$ is an uncontrolled disturbance stream and $u$ is the manipulated stream.
  • Working Principle (Two Methods):
    • Method I (Fig. 15.5): Measure both $u_m$ (FT) and $d_m$ (FT), compute $R_m = u_m / d_m$ in a Divider block, and send $R_m$ to a Ratio Controller (RC) with setpoint $R_d$. Disadvantage: The divider makes the process gain nonlinear ($K_p = \frac{\partial R}{\partial u} = \frac{1}{d}$).
    • Method II (Preferred — Fig. 15.6): Measure disturbance flow $d_m$ (FT) and multiply it in a Ratio Station (RS) by gain $K_R = R_d \frac{S_d}{S_u}$ (where $S_d, S_u$ are transmitter spans) to generate the remote setpoint $u_{sp} = K_R d_m$ for a standard Flow Controller (FC) on the manipulated stream $u$.
  • Industrial Examples: (1) Maintaining stoichiometric $\text{H}_2/\text{N}_2$ ($3:1$) feed ratio to an ammonia synthesis reactor, (2) Fuel-to-air ratio in a furnace, (3) Blending two liquid streams, (4) Reflux ratio in a distillation column.
• (b) Split-Range Control (Chapter 16 PDF, Slides 20, 25–29):
  • Justification: Used when a single controller output ($0–100\%$) must manipulate two (or more) final control elements (valves) sequentially over different ranges of the controller output—either because both heating and cooling are needed ($N_M > N_C$), or because a single valve cannot accurately meter flow over an extremely wide operating range.
  • Working Principle & Two Industrial Examples:
    1. Example 1 — Batch/CSTR Jacket Temperature Control (Fig. 16.14, Slide 20 & 28–29): One Temperature Controller (TC) outputs $0–100\%$ signal ($4–20\text{ mA}$) to two valves: from $0\%$ to $50\%$ controller output ($4–12\text{ mA}$), the Cooling Water valve goes from fully open to fully closed while the Steam valve remains closed; from $50\%$ to $100\%$ output ($12–20\text{ mA}$), the Steam heating valve opens from $0\%$ to $100\%$ while the cooling valve stays closed.
    2. Example 2 — Wide-Range Split-Range Flow Control (Slide 25–27): A small control valve and a large control valve are placed in parallel. At low flow rates, the large valve stays closed and the small valve provides high-accuracy flow control; at high flow rates, the small valve is fully open and the large valve opens to handle the capacity.
• (c) Auctioneering / Selective Control (Chapter 16 PDF, Slides 15–18):
  • Justification: Used when there are more controlled/measured variables than manipulated variables ($N_C > N_M$), such as multiple temperature sensors along a reactor bed where the "hot spot" location shifts with catalyst aging.
  • Working Principle: Multiple transmitter signals (or multiple controller outputs) are fed into a High Selector (HS, $>$) or Low Selector (LS, $<$), which "auctions" and selects the highest (or lowest) signal to pass on to the single controller or final control valve.
  • Industrial Example (Fig. 16.13, Slide 17): Controlling the hot-spot temperature in a fixed-bed tubular catalytic reactor: several thermocouples (TT-1, TT-2, TT-3) are placed along the reactor length and connected to a High Selector (HS). The HS continuously selects the highest temperature reading and sends it to the Temperature Controller (TC) to adjust coolant flow, preventing thermal runaway wherever the hot spot moves!

4. Transducer vs. Transmitter & RTD Calibration Math [2020-21 Q2(c) & 2019-20 Q5(a)] (Chapter 9 PDF, Slides 2–8)

• Distinction Between Transducer and Transmitter (2019-20 Q5a):
  • A Transducer is any device that converts one physical form of energy/quantity into another (e.g., an I/P transducer converts electrical current $4–20\text{ mA}$ into pneumatic air pressure $3–15\text{ psig}$; a measurement transducer consists of a sensing element + transmitter, Fig. 9.3).
  • A Transmitter (driving element) specifically converts the raw low-level output of a primary sensor (e.g., millivolts from a thermocouple or resistance $\Omega$ from an RTD) into a standardized, amplified transmission signal (such as $4–20\text{ mA}$ DC or digital Fieldbus signal) suitable for transmission over long lines to a controller.
• RTD Transducer Math [2020-21 Q2(c)]:
  Given: Input temperature range $T_{\min} = 50^\circ\text{C}$ to $T_{\max} = 850^\circ\text{C}$; Output current range $I_{\min} = 4\text{ mA}$ to $I_{\max} = 20\text{ mA}$.
  1. Zero (Lower Range Limit): $$\text{Zero} = 50^\circ\text{C}$$
  2. Span (Input Range): $$\text{Span} = T_{\max} - T_{\min} = 850^\circ\text{C} - 50^\circ\text{C} = 800^\circ\text{C}$$
  3. Transducer Gain ($K_m$): $$K_m = \frac{\text{Range of instrument output}}{\text{Range of instrument input}} = \frac{20\text{ mA} - 4\text{ mA}}{850^\circ\text{C} - 50^\circ\text{C}} = \frac{16\text{ mA}}{800^\circ\text{C}} = 0.02\text{ mA/}^\circ\text{C}$$
  4. Linear Expression of Output $T_m$ ($\text{mA}$) as a Function of Input $T$ ($^\circ\text{C}$) (Slide 7, Chapter 9 PDF): $$T_m\text{ (mA)} = \left(\frac{20\text{ mA} - 4\text{ mA}}{850^\circ\text{C} - 50^\circ\text{C}}\right)\left(T\text{ [}^\circ\text{C]} - 50^\circ\text{C}\right) + 4\text{ mA}$$ $$T_m\text{ (mA)} = \left(0.02\text{ }\frac{\text{mA}}{^\circ\text{C}}\right) T\text{ (}^\circ\text{C)} - 1\text{ mA} + 4\text{ mA} = \left(0.02\text{ }\frac{\text{mA}}{^\circ\text{C}}\right) T\text{ (}^\circ\text{C)} + 3\text{ mA}$$

5. DCS Architecture & Comparison of DCS vs. PLC [2020-21 Q1(a) & 2018-19 Q1(a)] (DCS & PLC PDF, Slides 3–13)

• DCS Architecture Sketch (Slide 4 of DCS & PLC PDF):
  Draw the horizontal Data Highway (Shared Communication Network) in the center. Above the Data Highway, connect: System Consoles, Host Computer $\leftrightarrow$ Data Storage Unit, and PLC. Below the Data Highway, connect multiple distributed Local Control Units (LCUs) (each connected to a Local Console), and connect the LCUs via $4–20\text{ mA}$ lines to the bottom block: Process Transmitters and Actuators.
• Comparison of DCS and PLC (Advantages & Disadvantages — Slide 3, 10, 13):

  Feature	Distributed Control System (DCS)	Programmable Logic Controller (PLC)
  Advantages	• Lower cost per loop for large plants involving hundreds/thousands of continuous control loops. • Superior reliability due to redundant microprocessors in Local Control Units (LCUs). • Built-in data storage, historical trending, and advanced multivariable control capabilities. • Easy and inexpensive to expand.	• Better able to withstand harsh industrial environments. • Much faster scan/cycle times for high-speed discrete logic and safety interlocks. • Lower cost for small and medium-sized applications. • Easier to maintain due to modular I/O nature and simple ladder logic programming.
  Disadvantages	• High initial capital cost for small applications. • Slower scan times compared to PLCs for fast discrete sequencing.	• Expensive and cumbersome to scale up to thousands of complex continuous analog PID loops. • Limited built-in operator graphics/trending without external SCADA/HMI software.

Step-by-Step Detailed Solution:

Part 1: Solution for 2018-19 Q3(a) & Q3(b) (Closed-Loop Level Control & Offset — Chapter 11 PDF, Slides 26–35)

1. Open-Loop Transfer Functions ($G_p(s)$ and $G_d(s)$):
   Unsteady-state mass balance for the tank (constant $\rho, A$) with nonlinear exit valve $q_3 = C_v\sqrt{h}$: $$A\frac{dh}{dt} = q_1 + q_2 - C_v\sqrt{h}$$ Linearizing $q_3 = C_v\sqrt{h}$ around steady-state level $\bar{h}$ gives $q'_3 = \frac{1}{R}h'$, where valve resistance $R = \frac{2\sqrt{\bar{h}}}{C_v}$: $$A\frac{dh'}{dt} = q'_1 + q'_2 - \frac{h'}{R} \implies AR\frac{dh'}{dt} + h' = R q'_1 + R q'_2$$ Taking the Laplace transform with $K_p = R$ and $\tau = AR$: $$G_p(s) = \frac{H'(s)}{Q'_2(s)} = \frac{K_p}{\tau s + 1}, \qquad G_d(s) = \frac{H'(s)}{Q'_1(s)} = \frac{K_p}{\tau s + 1}$$
2. Closed-Loop Transfer Function (Assuming $G_c = K_c, \; G_v = K_v, \; G_m = K_m$ — Slide 30 & 34):
   Defining the open-loop gain $K_{OL} = K_c K_v K_p K_m$:
   • For Disturbance Changes $\frac{H'(s)}{Q'_1(s)}$ (Eq. 11-53 & 11-54): $$\frac{H'(s)}{Q'_1(s)} = \frac{G_d(s)}{1 + G_c G_v G_p G_m} = \frac{K_p / (\tau s + 1)}{1 + K_{OL} / (\tau s + 1)} = \frac{K_2}{\tau_1 s + 1}$$ $$\text{where } K_2 = \frac{K_p}{1 + K_{OL}}, \qquad \tau_1 = \frac{\tau}{1 + K_{OL}}$$
   • For Set-Point Changes $\frac{H'(s)}{H'_{sp}(s)}$ (Eq. 11-36 & 11-37): $$\frac{H'(s)}{H'_{sp}(s)} = \frac{K_m G_c G_v G_p}{1 + G_c G_v G_p G_m} = \frac{K_{OL} / (\tau s + 1)}{1 + K_{OL} / (\tau s + 1)} = \frac{K_1}{\tau_1 s + 1}, \qquad \text{where } K_1 = \frac{K_{OL}}{1 + K_{OL}}$$
3. Part (b) — Proof of Offset for Proportional Control (Slides 32 & 35):
   • For a Unit Step Disturbance Change ($Q'_1(s) = 1/s$, with $H'_{sp} = 0$):
     $$h'(t) = K_2\left(1 - e^{-t/\tau_1}\right) \implies h'(\infty) = K_2 = \frac{K_p}{1 + K_{OL}} \ne 0$$ $$\text{Offset} = h'_{sp}(\infty) - h'(\infty) = 0 - \frac{K_p}{1 + K_{OL}} = -\frac{K_p}{1 + K_{OL}}$$
   • For a Unit Step Set-Point Change ($H'_{sp}(s) = 1/s$, with $M = 1$):
     $$h'(t) = K_1\left(1 - e^{-t/\tau_1}\right) \implies h'(\infty) = K_1 = \frac{K_{OL}}{1 + K_{OL}} < 1$$ $$\text{Offset} = h'_{sp}(\infty) - h'(\infty) = 1 - \frac{K_{OL}}{1 + K_{OL}} = \frac{1}{1 + K_{OL}} \ne 0$$
   Because $K_{OL} < \infty$ for any finite $K_c$, the offset is always nonzero, proving that a proportional controller cannot track the setpoint without steady-state error!

Part 2: Solution for 2018-19 Q4(a) (Heat Exchanger Step Test + Direct Synthesis PI Settings)

1. (i) Finding the FOPTD Transfer Function from Table Q4 Step Test Data:
   • Initial steady-state transmitter output: $T_{2m}(0) = 12.0\text{ mA}$ (at $t = 0$ and $t = 1\text{ min}$).
   • Final steady-state transmitter output: $T_{2m}(\infty) = 17.0\text{ mA}$ (at $t = 10$ and $11\text{ min}$).
   • Total steady-state change in measured output: $$\Delta T_{2m} = 17.0 - 12.0 = 5.0\text{ mA}$$
   • Time Delay ($\theta$): Since $T_{2m}$ remains at $12.0\text{ mA}$ from $t = 0$ to $t = 1.0\text{ min}$ before starting to rise, the time delay is: $$\theta = 1.0\text{ min}$$
   • Time Constant ($\tau$) by the $63.2\%$ Method:
     The $63.2\%$ response level is: $$T_{2m, 63.2\%} = 12.0 + 0.632(5.0) = 12.0 + 3.16 = 15.16\text{ mA}$$ From Table Q4: at $t = 5\text{ min}$, $T_{2m} = 14.8\text{ mA}$; at $t = 6\text{ min}$, $T_{2m} = 15.4\text{ mA}$. By linear interpolation: $$t_{63.2\%} = 5 + \frac{15.16 - 14.8}{15.4 - 14.8}(6 - 5) = 5 + \frac{0.36}{0.60} = 5.6\text{ min}$$ Since $t_{63.2\%} = \theta + \tau$: $$\tau = t_{63.2\%} - \theta = 5.6 - 1.0 = 4.6\text{ min}$$
   • Steady-State Gains:
     The step change in steam supply pressure is $\Delta P_s = 16 - 12 = 4\text{ psi}$.
     • Gain from steam supply pressure $\Delta P_s$ to $T_{2m}$: $K_{p,m} = \frac{\Delta T_{2m}}{\Delta P_s} = \frac{5.0\text{ mA}}{4\text{ psi}} = 1.25\text{ mA/psi}$.
     • Combined overall open-loop gain $K = K_{IP} K_v K_{p,m}$ (from controller output $P'$ [$\text{mA}$] to $T'_{2m}$ [$\text{mA}$]): $$K = K_{IP} K_v K_{p,m} = (0.75\text{ psi/mA})(0.75\text{ psi/psi})(1.25\text{ mA/psi}) = 0.7031\text{ mA/mA (dimensionless)}$$
     Thus, the overall FOPTD model is: $$G(s) = \frac{T'_{2m}(s)}{P'(s)} = \frac{0.7031 e^{-1.0s}}{4.6s + 1} \qquad \left(\text{or } \frac{T'_{2m}(s)}{P'_s(s)} = \frac{1.25 e^{-1.0s}}{4.6s + 1}\text{ mA/psi}\right)$$
2. (ii) Closed-Loop Block Diagram with Labels and Units:
   Draw the standard feedback loop (Fig. 11.7 / Fig. 12.17):
   $T'_{2,sp}$ [$^\circ\text{C}$] $\xrightarrow{K_m}$ $\tilde{T}'_{2,sp}$ [$\text{mA}$] $\to$ Comparator ($\pm$) $\xrightarrow{E\text{ [mA]}}$ $\boxed{G_c(s)\text{ (TC)}}$ $\xrightarrow{P'\text{ [mA]}}$ $\boxed{K_{IP} = 0.75\text{ psi/mA}}$ $\xrightarrow{P'_t\text{ [psi]}}$ $\boxed{K_v = 0.75\text{ psi/psi}}$ $\xrightarrow{P'_s\text{ [psi]}}$ $\boxed{G_p(s)}$ $\xrightarrow{T'_2\text{ [}^\circ\text{C]}}$ (with feedback $\boxed{G_m(s)\text{ (TT)}} \xrightarrow{T'_{2m}\text{ [mA]}}$ to comparator).
3. (iii) PI Controller Settings Using Direct Synthesis (DS) Method:
   For a FOPTD model $\tilde{G}(s) = \frac{K e^{-\theta s}}{\tau s + 1}$ with $K = 0.7031$, $\tau = 4.6\text{ min}$, and $\theta = 1.0\text{ min}$, the DS PI formulas (Eq. 12-11) are: $$K_c = \frac{1}{K}\left(\frac{\tau}{\tau_c + \theta}\right), \qquad \tau_I = \tau = 4.6\text{ min}$$ Using Skogestad's standard choice for desired closed-loop time constant $\tau_c = \theta = 1.0\text{ min}$ (or $\tau_c = 2.0\text{ min}$):
   • For $\tau_c = \theta = 1.0\text{ min}$: $$K_c = \frac{1}{0.7031}\left(\frac{4.6}{1.0 + 1.0}\right) = \frac{2.3}{0.7031} = 3.27, \qquad \tau_I = 4.6\text{ min}$$
   • (If $K = 1.25$ is used directly with $\tau_c = 1.0\text{ min}$: $K_c = \frac{4.6}{1.25(1 + 1)} = 1.84, \; \tau_I = 4.6\text{ min}$).