Plant Control Systems and Automation — October/November 2023

Question 1(a)

(i) Define control system with reference to plant automation. (ii) List four components of open loop control systems.

(5 marks)

Answer

A control system is an arrangement of connected elements that directs or regulates a plant process to achieve a specified output.

1. Reference/input command.
2. Controller or control unit.
3. Actuator/final control element.
4. Plant or process producing the output.

In an open-loop arrangement, the measured process output is not fed back to correct the command.

Question 1(b)

A heating control system is used to regulate the temperature of a room. The room has ventilation vents. (i) Draw a block diagram representation of the control system. (ii) Describe its operation.

(9 marks)

Answer

1. Set the desired room temperature T_ref; measure the actual room temperature T with a sensor.
2. Compare the feedback temperature with the set-point: e = T_ref − T.
3. When the room is too cold, the thermostat/controller increases heater output or switches the heater on. When the upper switching threshold is reached, it reduces output or switches the heater off.
4. Ventilation exchanges room air with outside air and changes heat loss; it enters as a disturbance to the room thermal process.
5. Negative feedback detects the resulting temperature change and adjusts heating. A practical on/off thermostat uses hysteresis to avoid rapid switching.

Question 1(c)

Classify the following as open-loop or closed-loop control system; stating the reason in each case: (i) room air conditioner; (ii) electric hand drier; (iii) automatic washing machine.

(6 marks)

Answer

Classification must name the controlled variable. Automation or the mere presence of a sensor does not by itself prove closed-loop regulation of the intended output.

Question 2(a)

(i) State three merits of open-loop control systems. (ii) Differentiate between manual and automatic control techniques with respect to closed loop control system.

(5 marks)

Answer

1. Simple construction and straightforward operation.
2. Lower cost because output sensing and feedback hardware can be omitted.
3. No instability caused by an added feedback loop, although the plant itself still has to be stable.

In manual closed-loop control, a person observes the measured output, compares it with the target and adjusts the control input. In automatic closed-loop control, the sensor, comparator and controller perform that feedback correction without continuous human adjustment. Both are closed-loop when the output information is used to correct the input.

Question 2(b)

With the aid of a block diagram of a canonical control system, derive the expression for its transfer function.

(7 marks)

Answer

For a positive-feedback comparator the corresponding denominator is 1−GH. The feedback sign must be read from the diagram.

Question 2(c)

Figure 1 shows a block diagram of a control system. Determine its: (i) canonical form representation; (ii) overall transfer function.

(8 marks)

Answer

Question 3(a)

(i) Explain the need for control system modelling. (ii) Outline the steps involved in control systems modelling.

(6 marks)

Answer

A model represents the relationships between system inputs, states and outputs. It enables prediction of response and stability, comparison of controller choices, and testing of designs before physical implementation.

1. Define the control objective, system boundary, input, output and disturbances.
2. Choose physical assumptions and a suitable level of detail; identify parameters.
3. Apply relevant physical laws, such as force balance, energy balance or Kirchhoff’s laws, to derive equations.
4. Linearise about a stated operating point when appropriate, then obtain a transfer-function or state-space representation.
5. Estimate/measure parameters and compare model predictions with observed behaviour; revise the model if it is inadequate.

Question 3(b)

Figure 2 shows an electric network model of a control system. Determine the expression for the transfer function, Eₒ(s)/Eᵢ(s).

(6 marks)

Answer

Question 3(c)

Figure 3 shows a mechanical system model of an accelerometer. (i) Write the differential equation for the system. (ii) Determine the transfer function, X(s)/A(s). (iii) If Eₒ(s) = K₁X(s), where K₁ is proportionality constant of LVDT, determine the expression, Eₒ(s)/A(s).

(8 marks)

Answer

Define y(t) as housing displacement, z(t) as the absolute mass displacement, a(t) = d²y/dt², and x(t) = z(t)−y(t) as mass displacement relative to the housing, measured in the same positive direction. Work about static equilibrium so gravity is already balanced.

The sign is a coordinate convention, not a change in the accelerometer dynamics. For frequencies well below √(K/M), the magnitude approaches M/K and the output is approximately proportional to acceleration.

Question 4(a)

State four merits of d.c. servomotors.

(4 marks)

Answer

1. High available starting torque for a suitable motor and drive.
2. Motor torque is approximately proportional to armature current when field flux is held constant.
3. Convenient speed control over a useful range by controlling the armature supply.
4. Rapid controlled acceleration, deceleration and reversal in low-inertia servo designs.

Question 4(b)

Figure 4 shows an equivalent circuit of armature controlled d.c. servomotor. (i) Write the differential equation for the: I. armature circuit input, eₐ; II. torque, Tₘ. (ii) Determine the laplace transform for the equations in b(i).

(8 marks)

Answer

Question 4(c)

(i) Draw a labelled diagram of a double-acting cylinder connected to a 4/3 spool control valve. (ii) Describe the operation of the set-up in (c)(i).

(8 marks)

Answer

1. In the extend position, P connects to A and B connects to T. Pressure acts on the cap end and the rod extends; displaced rod-end fluid returns to tank.
2. In the retract position, P connects to B and A connects to T. Pressure acts on the rod end and the rod retracts.
3. For the closed-centre valve illustrated, the neutral position blocks P, T, A and B, tending to hold the cylinder position, subject to leakage.
4. The pressure source needs an appropriate relief/unloading arrangement. Other 4/3 centre configurations are possible; their neutral behaviour must be stated separately.

Question 5(a)

Differentiate between each of the following with respect to process control: (i) step input and transient state; (ii) set-point and controlled variable.

(4 marks)

Answer

Question 5(b)

Figure 5 shows a milk heating process. It is desired to maintain the milk temperature at 60° C. Identify the following: (i) set-point; (ii) manipulated variable; (iii) controlled variable; (iv) final control element; (v) feedback element.

(5 marks)

Answer

The printed sensor appears in the water jacket. As drawn, it directly senses bath temperature, which is only a proxy for milk temperature. Directly maintaining milk at 60 °C would require milk-temperature feedback or a validated relationship and suitable bath setting; the two temperatures are not automatically identical.

Question 5(c)

Figure 6 shows an analogue controller. Determine the: (i) proportional gain, kp; (ii) output voltage, Vₒ.

(4 marks)

Answer

Question 5(d)

(i) Draw a labelled circuit diagram of operational amplifier based proportional Plus Integral plus Derivative (PID) controller. (ii) Write the expression for each of the following with respect to the PID controller in d(i): I. proportional gain, Kp; II. integral gain, Ki; III. derivative gain, Kd.

(7 marks)

Answer

The final summer gives vₒ = Kp e + Ki∫e dt + Kd de/dt. These are the gains of the illustrated parallel form. In the alternative ideal form Kp[e + (1/Ti)∫e dt + Td de/dt], Ki = Kp/Ti and Kd = KpTd. Practical circuits limit differentiator bandwidth and integrator saturation.

Question 6(a)

State four merits of industrial robots.

(4 marks)

Answer

1. Repeatable execution of programmed operations with consistent quality.
2. High productivity on suitable repetitive tasks.
3. Removal of people from hazardous or unpleasant operations.
4. Reprogrammability for different products or task sequences, within the robot’s physical capabilities.

Question 6(b)

Describe each of the following with respect to robot performance: (i) work envelope; (ii) degree of freedom; (iii) speed of movement.

(6 marks)

Answer

1. Work envelope: the region of space the robot’s end effector can reach, subject to joint limits, geometry and tool configuration.
2. Degree of freedom: an independent joint or coordinate of motion needed to describe the robot configuration, such as a rotary or linear axis.
3. Speed of movement: the rate of joint or end-effector motion, stated in units such as degrees/s, rad/s or mm/s and subject to payload, acceleration and path limits.

Question 6(c)

The arm of an industrial robot obtains a total range of rotation of 160°. The arm fully extends to 1500 mm and fully retracts to 750 mm from the pivot joint. The robot has 10-bit memory. Determine the: (i) control resolution of rotation; (ii) control resolution of translation.

(6 marks)

Answer

Interpret “10-bit memory” as a 10-bit position command/encoding for each axis. A memory-capacity statement by itself would not define position resolution. For N discrete endpoint-inclusive positions, there are N−1 intervals.

Resolution describes the command increment. It is not a guarantee of mechanical accuracy or repeatability.

Question 6(d)

Explain each of the following with respect to robots: (i) tactile sensing; (ii) vision.

(4 marks)

Answer

1. Tactile sensing detects contact-related information, such as touch, pressure, force distribution or slip at a gripper. The controller can use it to regulate grasping or confirm contact.
2. Vision uses cameras and image processing to identify, locate or inspect objects and guide robot action. Calibration relates image information to the robot/workspace coordinates.

Question 7(a)

State: (i) two features of SCADA software; (ii) three types of communication system in a SCADA system.

(5 marks)

Answer

1. Real-time graphical monitoring and supervisory display of process variables.
2. Alarm/event handling and historical data logging/trending.
3. Wired serial or industrial communication links, such as RS-485-based systems.
4. Wired Ethernet/IP networks, including suitable fibre links.
5. Wireless communication links, such as licensed radio or cellular telemetry.

The communication categories describe the medium/network; the application protocol and reliability requirements must also be selected.

Question 7(b)

Figure 7 shows a Programmable Logic Controller (PLC) ladder diagram program. (i) Describe the control operation; (ii) Write the equivalent instruction listing program.

(10 marks)

Answer

Read the contacts exactly as printed. In the first rung X100, X101 and Y5 are all drawn as normally open logic contacts. Y110 and the 60-second on-delay timer T₁ are parallel outputs. In the second rung the T₁ contact is in parallel with Y5, followed by X102 and output coil Y5.

Equivalent generic instruction listing for the printed circuit (the exact timer mnemonic and time-base encoding depend on the PLC):

LD   X100
AND  X101
AND  Y5
OUT  Y110
TON  T1, T#60s
LD   T1
OR   Y5
AND  X102
OUT  Y5
END

Conditional intended sequence: if the first-rung Y5 contact was intended to be normally closed, replace only that AND Y5 by AND NOT Y5 (often ANI Y5). With X102 = 1 as the healthy/reset-permissive state, motion and the enabled toggle then start Y110 and T₁. At 60 s, T₁ sets and seals Y5, the danger lamp comes on, and the inverted Y5 contact turns the warning lamp/timer off on the next scan. Opening the X102 permissive resets the latch. This is an explicitly corrected interpretation, not the circuit as printed.

Question 7(c)

(i) Draw a PLC ladder symbol for NC coil. (ii) Describe the following states of coil of a PLC output relay: I. NC state; II. NO state.

(5 marks)

Answer

The paper uses “NC coil”. Strictly, a physical relay coil is energised or de-energised; normally closed and normally open describe its contacts. A negated software coil is commonly drawn as a coil symbol with a slash and sets Y = NOT(rung result). It must be distinguished from an NC contact instruction.

The normal state refers to the unoperated relay. A software contact tests a bit; its symbol alone does not specify how a field switch is physically wired.

Question 8(a)

Define each of the following 8051 micro controller registers: (i) stack pointer; (ii) data pointer; (iii) program counter.

(3 marks)

Answer

1. Stack pointer (SP): the 8-bit register identifying the current internal-RAM stack top; PUSH/CALL increment it before storing.
2. Data pointer (DPTR): the 16-bit address register formed from DPH and DPL, used for indexed/external-memory addressing.
3. Program counter (PC): the 16-bit register holding the address used for instruction sequencing in program memory.

Question 8(b)

Describe each of the following elements of 8051 micro controller internal architecture: (i) Central Processing Unit (CPU); (ii) Read Only Memory (ROM).

(6 marks)

Answer

1. CPU: fetches and decodes instructions, performs arithmetic/logic operations and controls data movement and peripheral activity. Its working elements include the ALU, registers and instruction-control circuitry.
2. ROM: non-volatile program storage read during execution. The original 8051 includes 4 KB on-chip program ROM; other family members have different code-memory sizes or technologies. Program memory is logically distinct from data memory.

Question 8(c)

State the special function of each of the following I/O pins of port 3 in 8051 micro controller: (i) RXD; (ii) INT̅₀; (iii) T₀; (iv) W̅R̅; (v) R̅D̅.

(5 marks)

Answer

Question 8(d)

Explain what the following 8051 Instructions accomplishes, stating the machine Hexadecimal code of each: (i) MOV A, 50 H; (ii) ADD A, @R0; (iii) MOV A, #47.

(6 marks)

Answer

Opcode lookup: E5 is MOV A,direct; 26 is ADD A,@R0; 74 is MOV A,#immediate. The operand byte follows where required. If an assembler’s radix is explicitly changed, the unsuffixed #47 depends on that setting. If the intended literal were #47H, the bytes would instead be 74 47; H is not printed in this third instruction.