Question 1

Single well-mixed tank (first-order) – transfer functions to inlet flow rate In a standard constant-area, linear-resistance single-tank system, the transfer function of both (i) outlet flow rate and (ii) liquid level to the inlet flow rate has the f…

Accepted Answer

Introduction / Context: The single-tank liquid system is the archetypal first-order process in process control. With linear flow resistance at the outlet and small perturbations, both the level and the outlet flow respond to inlet flow changes through identical first-order dynamics with the same time constant T. Given Data / Assumptions: Tank cross-sectional area is constant. Outlet obeys linear resistance: q_out ∝ head (level). Small-signal linearisation is valid around an operating point. Transfer functions are with respect to inlet flow as input. Concept / Approach: Mass balance gives A dh/dt = q_in − q_out. Linear resistance gives q_out = k h in deviation form. Taking Laplace transforms yields H(s)/Q_in(s) = 1/(T s + 1), where T = A/k. Because q_out = k h, the outlet flow also follows first-order dynamics with the same T relative to q_in, i.e., Q_out(s)/Q_in(s) = k * H(s)/Q_in(s) = (k)/(T s + 1). After normalisation, its dynamic form is also 1/(T s + 1) in shape. Step-by-Step Solution: Write balance: A dh/dt = q_in − k h.Take Laplace: (A s) H = Q_in − k H.Rearrange: H/Q_in = 1/(A s + k) = 1/(T s + 1).Since q_out = k h, Q_out/Q_in = k * (1/(A s + k)) → first-order with same pole. Verification / Alternative check: The step responses of level and outlet flow to inlet flow are exponential with time constant T; only the steady-state gains differ by k scaling. Why Other Options Are Wrong: Outlet only / Level only: Each alone is incomplete; both share first-order form. Neither: Contradicts the standard derivation. Common Pitfalls: Confusing dynamic shape with static gain; q_out has a different gain but identical pole location. Final Answer: Both (a) & (b)

Question 2

Optical pyrometry – disappearing filament type The working principle of the disappearing-filament optical pyrometer is based primarily on which radiation law?

Accepted Answer

Introduction / Context: Optical pyrometers measure high temperatures by comparing the brightness of a hot object to that of a calibrated filament. The disappearing-filament type is widely referenced in instrumentation courses and relies on the temperature dependence of monochromatic radiance. Given Data / Assumptions: Instrument: disappearing-filament optical pyrometer. Measurement: brightness matching at a selected wavelength. Hot body approximated as a gray body; calibration links brightness to temperature. Concept / Approach: Wien’s law (in its radiation form) captures how spectral radiance varies strongly with temperature at short wavelengths, enabling brightness temperature determination by matching the filament brightness to the target at a fixed wavelength. When the filament “disappears” against the background (same apparent brightness), the instrument reads the corresponding temperature. Seebeck and Peltier effects are thermoelectric phenomena, not radiation laws. Kirchhoff’s law relates emissivity and absorptivity but does not provide the operational brightness–temperature relation exploited for matching. Step-by-Step Solution: Use a red-filtered or monochromatic band to view the hot target and the lamp filament.Adjust filament current until the filament visually disappears (equal brightness).Convert the brightness at that wavelength to temperature via calibration based on Wien’s law. Verification / Alternative check: Instrument manuals describe “brightness temperature” and its reliance on the exponential temperature dependence from Wien’s radiation law at short wavelengths. Why Other Options Are Wrong: Seebeck / Peltier: Concern electromotive force or heat at junctions, not optical radiation. Kirchhoff's law: Important for emissivity–absorptivity equality at thermal equilibrium, but not the basis of the brightness comparison method. Common Pitfalls: Confusing Wien’s displacement law (peak wavelength shift) with the brightness law; the measurement uses the strong temperature dependence of spectral radiance at fixed wavelength. Final Answer: Wien's law

Question 3

Pressure measurement elements – Bourdon tube geometries Bourdon spring tubes are manufactured in several curved forms (C-shaped, helical, spiral). Which of the following configurations is never used as a practical Bourdon tube shape?

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Introduction / Context: Bourdon tubes are classic elastic elements that convert pressure into tip motion via cross-section ovalisation and curvature change. Their mechanical geometry strongly affects sensitivity, linearity, and manufacturability. Knowing which geometries are used (and which are not) is a standard instrumentation topic. Given Data / Assumptions: Typical shapes: C-shaped (semi-circular), helical, and spiral tubes are well-known. Material is an elastic metal with an initially curved form. We identify a configuration that is essentially never employed. Concept / Approach: All working Bourdon tubes require an open-ended arc so that pressure-induced ovalisation causes a measurable free-end displacement. A completely circular, closed-ring tube would constrain deformation symmetrically and prevent the free-tip motion required for pointer amplification, offering no practical displacement for measurement. Hence “circular (closed ring)” is not used, whereas semi-circular (C-shaped), helical, and spiral forms are standard. Step-by-Step Solution: Recall common catalog shapes: C-type, helical, spiral.Consider mechanics: closed ring gives little/no free-end displacement.Conclude circular (closed ring) is not a Bourdon tube configuration. Verification / Alternative check: Instrumentation handbooks illustrate C-type for low/medium pressures and helical/spiral for higher ranges, with no circular-ring designs. Why Other Options Are Wrong: Semi-circular: The archetypal Bourdon tube. Helical / Spiral: Used to increase sensitivity or pressure range. Common Pitfalls: Assuming any curved tube qualifies; without a free end, pointer motion cannot be generated effectively. Final Answer: Circular (closed ring)

Question 4

Thermodynamic Celsius scale – definition and interval Which statement correctly describes the thermodynamic Celsius scale used for practical temperature measurement?

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Introduction / Context: Although the modern SI definition uses the kelvin linked to the Boltzmann constant, the practical Celsius scale for teaching and many instruments still references the traditional fixed points: the ice point and the steam point under specified pressure. Understanding the wording matters for exam accuracy. Given Data / Assumptions: Ice point: equilibrium of ice–water at 1 atm. Steam point: equilibrium of water–steam at 1 atm (condensation/boiling point correspondence). Thermodynamic link to kelvin preserves 100 equal divisions between these points. Concept / Approach: The classical Celsius scale assigns 0 °C at the ice point and 100 °C at the steam point, establishing an interval of 100 degrees. The appropriate wording is “condensation (steam) temperature” for the upper fixed point. Therefore, statements (b) and (c) together correctly describe the traditional definition. Statement (a) uses “evaporation temperature,” which is less precise and not the standard phrasing for the upper fixed point in fixed-point calibration. Step-by-Step Solution: Confirm lower fixed point: melting point of ice → 0 °C.Confirm upper fixed point: condensation/boiling temperature → 100 °C at 1 atm.Recognise the 100-degree interval between the two points.Select “Both (b) and (c).” Verification / Alternative check: Classic ITS-90 descriptions use triple points and fixed points; historically, 0–100 °C between ice and steam points are the educational anchors. Why Other Options Are Wrong: (a): Ambiguous wording; the standard upper point is the steam (condensation) point. (b) or (c) alone: Each is correct but incomplete compared with the combined correct option. Common Pitfalls: Confusing evaporation with condensation terminology; the fixed point defines equilibrium of water with its saturated vapor at 1 atm. Final Answer: Both (b) and (c)

Question 5

Controller design – ideal proportional plus reset (PI) controller For an ideal proportional-plus-reset (integral) controller with reset time T, what is the transfer function Gc(s)?

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Introduction / Context: PI (proportional–integral) controllers are ubiquitous in process control, combining proportional action for immediate response with integral action to eliminate steady-state offset. Understanding the correct transfer function form is fundamental for tuning and stability analysis. Given Data / Assumptions: Proportional gain is Kc. Reset time (integral time constant) is T. Ideal PI form is intended (no derivative or filters). Concept / Approach: The time-domain ideal PI law is u(t) = Kc [ e(t) + (1/T) ∫ e(t) dt ]. Taking Laplace transforms yields Gc(s) = U(s)/E(s) = Kc [1 + 1/(T s)]. This distinguishes PI from PD (Kc(1 + T s)), pure integral (Kc/(T s)), and pure proportional (Kc). Step-by-Step Solution: Start from u(t) = Kc e(t) + Kc/T ∫ e(t) dt.Laplace: U/E = Kc + (Kc/T) * (1/s) = Kc [1 + 1/(T s)].Identify the correct option matching the derived expression. Verification / Alternative check: Block-diagram realisations of PI show a proportional branch in parallel with an integrator scaled by Kc/T, summing to the same transfer function. Why Other Options Are Wrong: Kc(1 + T s): This is PD, not PI. Kc/(T s): Pure integral action only. Kc: Pure proportional action only. Common Pitfalls: Mixing “reset time T” with “integral gain Ki”; note Ki = Kc/T so Gc(s) = Kc + Ki/s. Final Answer: Kc * (1 + 1/(T s))

Question 6

Dynamic behavior of a thermocouple installed in a thermowell A thermocouple housed inside a protective thermowell behaves, dynamically, as which kind of standard system?

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Introduction / Context: Response speed and fidelity of temperature sensors are influenced by sensor mass, protective hardware, and heat transfer paths. A bare thermocouple often approximates a first-order lag. However, adding a thermowell introduces additional thermal capacitances and resistances, changing the dynamic order and damping behavior. Given Data / Assumptions: Thermocouple is inserted into a metallic thermowell immersed in a flowing process fluid. Heat transfer involves convection (fluid → well), conduction (well → TC), and sensor thermal mass. No resonance or oscillatory mechanisms are present; thermal systems are typically heavily damped. Concept / Approach: The series of thermal resistances and capacitances (fluid boundary layer, well wall, fill material, sensor junction) gives at least two significant energy storage elements, yielding a second-order model. Because thermal systems dissipate energy strongly without oscillatory feedback, the effective damping ratio is large, resulting in an overdamped second-order response (sum of two exponentials) rather than oscillatory behavior. Step-by-Step Solution: Identify two or more thermal capacitances: well metal and sensor bead.Recognise corresponding resistances: convection and conduction paths.Model reduction: two dominant time constants → second-order overdamped approximation. Verification / Alternative check: Step tests on thermowell-protected probes show bi-exponential rise curves, well fit by an overdamped second-order model. Why Other Options Are Wrong: First order: Better matches bare-bead thermocouples, not well-protected probes. Multiple first order: While mathematically similar to second-order overdamped, the standard “true” description in exams is a second-order overdamped system. Second-order underdamped: Thermal systems seldom exhibit oscillatory underdamped responses without control loops. Common Pitfalls: Assuming all temperature sensors act as first-order; hardware significantly affects dynamics. Final Answer: Second order system (overdamped)

Question 7

Thermoelectric phenomena – heat absorption/evolution at junctions The relation connecting heat absorbed or evolved at a thermocouple junction with the electric current in the circuit is known as which thermoelectric effect?

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Introduction / Context: Thermoelectricity involves three closely related effects: Seebeck, Peltier, and Thomson. Distinguishing them is crucial for understanding thermocouple operation, active cooling devices, and temperature-gradient energy flows. Given Data / Assumptions: Peltier effect concerns heat exchange at a junction when an electric current passes. Seebeck effect concerns electromotive force generated by junctions at different temperatures. Thomson effect concerns heat absorption/evolution along a conductor with a temperature gradient and current. Concept / Approach: The question explicitly mentions heat absorption/evolution at the junctions related to current flow. This is the Peltier effect: Q̇ = Π I at the junction, where Π is the Peltier coefficient specific to the junction pair and I is current. Seebeck relates voltage to temperature difference (E = S ΔT), and Thomson relates distributed heating/cooling to current in a temperature gradient along a single conductor. Step-by-Step Solution: Identify the phenomenon location: at the junctions → points to Peltier.Identify the driver: electric current determines the heat rate → Peltier law.Select “Peltier.” Verification / Alternative check: Thermoelectric coolers (TEC) exploit the Peltier effect by driving DC current through junctions to pump heat. Why Other Options Are Wrong: Seebeck: Voltage due to temperature difference, not junction heat vs current. Thomson: Distributed along a conductor with gradient, not at the junction. None: Incorrect because the Peltier effect exactly matches the description. Common Pitfalls: Mixing up which effect produces EMF (Seebeck) versus which moves heat under current (Peltier). Final Answer: Peltier

Question 8

Gas storage modeling in process control: For an ideal gas contained in a rigid vessel, the (molar) capacitance of the vessel is defined as C = (∂n/∂P) at constant temperature. Using P = pressure, V = vessel volume, n = moles of gas, R = universal ga…

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Introduction / Context: Dynamic modeling of tanks and process vessels often uses the electrical analogy, where a vessel acts like a “capacitor” storing mass or moles. For gases, this storage capacity depends on how sensitive the contained moles are to changes in pressure at a fixed temperature. The question asks for the ideal-gas expression for this capacitance in a rigid vessel. Given Data / Assumptions: Ideal-gas behavior: P * V = n * R * T. Vessel is rigid: V is constant. Capacitance definition: C = ∂n/∂P at constant T and V. R is the universal gas constant; T is absolute temperature. Concept / Approach: The molar capacitance for an ideal gas is obtained by differentiating the ideal-gas equation with respect to pressure while holding temperature and volume constant. This directly provides how many moles “fit” per unit pressure rise, i.e., the storage property that governs pressure dynamics when inflow and outflow are unbalanced. Step-by-Step Solution: Start from P * V = n * R * T.Solve for n: n = (P * V) / (R * T).Differentiate with respect to P at constant V and T: ∂n/∂P = V / (R * T).Therefore, gas capacitance C = V / (R * T). Verification / Alternative check: If T increases, V/(R*T) decreases, meaning fewer moles per unit pressure rise (gas is “softer”), which matches physical intuition. Larger volumes proportionally increase storage, aligning with C ∝ V. Why Other Options Are Wrong: R * T / V: This is the inverse of the correct result; it would imply smaller vessels hold more per unit pressure, which is nonphysical. n * R / P or P * V / (n * R): These arise from rearrangements but do not equal ∂n/∂P at constant T and V. R / (P * V): Has incorrect units and dependence. Common Pitfalls: Confusing capacitance with compressibility, or forgetting the derivative must be taken at constant volume and temperature; using total derivatives that include T or V changes leads to wrong expressions. Final Answer: V / (R * T)

Question 9

Bode stability insight: According to the Bode stability criterion, a linear time-invariant feedback system is unstable if the open-loop amplitude ratio exceeds unity at the frequency where the phase lag equals:

Accepted Answer

Introduction / Context: The Bode stability criterion offers a frequency-domain way to infer closed-loop stability from open-loop frequency response data. It links gain and phase at critical frequencies to predict whether a system will oscillate or diverge after closing the loop. Given Data / Assumptions: Single-loop, negative-feedback, linear time-invariant system. Open-loop transfer L(jω) measured or modeled. Amplitude ratio refers to |L(jω)| and phase lag to ∠L(jω) (negative angle convention). Concept / Approach: The Bode criterion states that if, at the phase crossover (where the phase lag reaches 180°, i.e., ∠L = -180°), the magnitude |L| exceeds 1 (0 dB), the Nyquist plot encircles the -1 point and the closed loop becomes unstable. Equivalently, at the gain crossover (|L| = 1), sufficient negative phase margin (distance from -180°) is required to maintain stability. Step-by-Step Solution: Identify the phase-crossover frequency ω_pc where phase lag is 180°.Evaluate |L(jω_pc)|. If |L(jω_pc)| > 1, the loop lacks gain margin and is unstable.Therefore, the critical phase lag condition is 180°. Verification / Alternative check: Relating Bode and Nyquist: at -180° phase, a gain above unity places the Nyquist plot beyond the -1 point, implying an encirclement upon closure (instability). Why Other Options Are Wrong: 0°, 45°, 90°: These lags are not the Nyquist critical angle for sign reversal in feedback. 270°: Not the conventional critical lag for single-loop negative feedback using Bode margins. Common Pitfalls: Confusing gain crossover and phase crossover; the instability check at phase crossover uses the magnitude condition, and at gain crossover uses the phase margin condition. Final Answer: 180°

Question 10

Choosing a flowmeter: For fluids whose flow measurement should be largely insensitive to changes in density, viscosity, and velocity profile, the most suitable instrument is a:

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Introduction / Context: In process plants, flowmeters may suffer accuracy loss if fluid properties such as density and viscosity change. Selecting a technology that inherently decouples measurement from these properties saves recalibration effort and improves robustness. Given Data / Assumptions: Fluid is electrically conductive (requirement for magnetic flowmeters). Measurement range covers laminar to turbulent regimes. Objective: insensitive to density, viscosity, and velocity profile distortions. Concept / Approach: By Faraday’s law, a magnetic flowmeter induces a voltage proportional to volumetric flow rate (E ∝ B * L * v). Since volumetric flow does not require calibration against density or viscosity, and since the meter has no obstruction (full-bore design), profile effects are minimized with appropriate straight runs. Therefore, mag meters excel when property independence is desired. Step-by-Step Solution: Establish property goals: low sensitivity to density/viscosity/profile.Map technologies: DP meters and turbines depend on Reynolds number; pitot depends on density and profile.Select magnetic flowmeter as the robust choice for conductive liquids. Verification / Alternative check: Factory calibrations commonly remain valid over wide property variations for mag meters; many water/wastewater and slurry services rely on this advantage. Why Other Options Are Wrong: Pitot/Flow nozzle: Differential pressure scales with density and is viscosity/Re-dependent. Turbine: K-factor shifts with viscosity/Re and profile; moving parts add sensitivity. Rotameter: Float position depends on density/viscosity; profile variations matter. Common Pitfalls: Forgetting conductivity requirement; non-conductive fluids (e.g., hydrocarbons) are unsuitable for mag meters. Final Answer: Magnetic flowmeter

Question 11

Hydraulic controllers in industrial control: The working fluid most commonly used in hydraulic control systems (actuators and power elements) is:

Accepted Answer

Introduction / Context: Hydraulic control systems amplify signals into powerful, precise motions using incompressible fluids. The chosen working fluid must provide lubrication, corrosion protection, and stable viscosity across operating temperatures. Given Data / Assumptions: Industrial control valves, rams, and servoactuators are considered. Need for minimal compressibility and strong lubricity. Common ambient and equipment temperature ranges. Concept / Approach: Hydraulic oils are formulated mineral or synthetic oils with additives for anti-wear, oxidation resistance, rust prevention, and viscosity-index improvement. Compared with water (corrosion, poor lubrication), air/steam (compressible, unsuitable for hydraulic precision), oils deliver consistent, smooth power transmission and protect components. Step-by-Step Solution: List fluid requirements: low compressibility, lubrication, stability.Evaluate candidates against requirements.Select oil as the standard hydraulic working fluid. Verification / Alternative check: Hydraulic circuits, servo-hydraulic test systems, and heavy machinery universally specify hydraulic oil grades (e.g., ISO VG series). Why Other Options Are Wrong: Water: Corrosive, cavitation risk, poor lubricity (unless heavily conditioned). Steam/Air/CO2: Compressible; not suitable for precise hydraulic actuation. Common Pitfalls: Confusing pneumatic controllers (air) with hydraulic controllers; the former use compressible gas and deliver different performance characteristics. Final Answer: Oil

Question 12

Feedforward control concept: A feedforward controller primarily compensates for which class of changes acting on a process?

Accepted Answer

Introduction / Context: In control strategy design, feedforward is a proactive approach: disturbances are measured upstream and counteracted before they affect the controlled variable. This contrasts with feedback, which reacts to errors after they occur. Given Data / Assumptions: Process has measurable disturbance inputs (e.g., inlet composition, flow, temperature). Model (exact or approximate) relating disturbance to output exists. Controller can implement a compensating action based on disturbance measurement. Concept / Approach: Feedforward targets load disturbances. When the load changes, the controller injects a compensating move so the output barely deviates, ideally eliminating the need for corrective feedback. Set-point tracking is typically handled by the feedback loop; feedforward may be combined with feedback for robustness. Step-by-Step Solution: Identify disturbance channels (load).Use a process model to compute the required compensatory input.Apply compensation concurrently with the disturbance to preempt output error. Verification / Alternative check: In heat exchangers, measuring inlet temperature and adjusting steam flow preemptively is a classic feedforward application that reduces outlet temperature deviations. Why Other Options Are Wrong: Set-point changes: Primarily a feedback duty, though model-based controllers can do both. Both equally: The defining purpose of feedforward is disturbance rejection. Neither / Sensor bias: Not the principal intent of feedforward architecture. Common Pitfalls: Relying on feedforward without feedback; model mismatch and unmeasured disturbances still require a feedback loop for correction. Final Answer: Load disturbances

Question 13

Temperature measurement range selection: Which instrument is most suitable for measuring temperature in the range of -40°C to 425°C in typical industrial service?

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Introduction / Context: Selecting a temperature instrument depends on range, robustness, and environment. Many field applications need simple, durable, and reasonably accurate indication across sub-zero to several-hundred-degree ranges. Given Data / Assumptions: Target range: -40°C to 425°C. Preference for mechanical, line-mounted indicators with minimal maintenance. No requirement for non-contact high-temperature measurement. Concept / Approach: Bimetallic thermometers use the differential expansion of bonded metals to translate temperature into pointer motion. They are rugged, do not require power, and commonly cover about -50°C to 500°C. Mercury or alcohol thermometers have narrower safe ranges and fragility; radiation pyrometers are for much higher temperatures without contact; a bare thermistor typically cannot withstand 425°C and requires electronics. Step-by-Step Solution: Match the range requirement to instrument capabilities.Eliminate fragile or unsuitable technologies for 425°C service.Select bimetallic thermometer. Verification / Alternative check: Vendor datasheets list common bimetal ranges extending to 400–500°C, with various stem and case options for process duty. Why Other Options Are Wrong: Mercury/Alcohol: Range limitations and safety concerns; glass bulbs are fragile. Radiation pyrometer: Overkill for this range and requires emissivity considerations. Thermistor only: Electronics and material limits preclude 425°C. Common Pitfalls: Ignoring mechanical durability and readability in field conditions when choosing an instrument for wide-range service. Final Answer: Bimetallic thermometer

Question 14

Vapor-pressure thermometer construction: In a vapor-pressure thermometer, which component functions as the primary sensing element that directly experiences the process temperature?

Accepted Answer

Introduction / Context: Filled-system thermometers (liquid, gas, or vapor pressure types) use a sensing bulb connected via capillary to a pressure element. Understanding which part is the primary element clarifies how temperature changes get converted into an indication. Given Data / Assumptions: Instrument: vapor-pressure thermometer. Components: bulb, capillary, Bourdon tube (or bellows), linkage, pointer, dial. Concept / Approach: The bulb is the primary sensing element. It is immersed in the process and contains the volatile fluid whose saturation pressure varies strongly with temperature. This pressure is transmitted through the capillary to a Bourdon tube or bellows that converts pressure into motion for the pointer (a secondary element). Step-by-Step Solution: Identify which component first converts temperature to pressure.The bulb undergoes temperature change and sets vapor pressure accordingly.Therefore, the bulb is the primary element; Bourdon/pointer are secondary/display. Verification / Alternative check: Instrument handbooks classify sensing bulbs as primary elements irrespective of the downstream pressure-to-motion transducer type. Why Other Options Are Wrong: Pointer/Dial: Indication only, no sensing. Bourdon tube: Secondary conversion from pressure to motion. Capillary: Transmits pressure; not the sensor itself. Common Pitfalls: Assuming the Bourdon tube is primary because it “moves”; in filled systems, the bulb sets the pressure signal. Final Answer: Bulb

Question 15

Identifying the criterion: “The control system is unstable if the open-loop frequency response exhibits an amplitude ratio exceeding unity at the crossover frequency.” This statement corresponds to which stability criterion?

Accepted Answer

Introduction / Context: Control engineers use several classical criteria—Bode, Nyquist, Routh—to assess stability. Each articulates instability in distinct but related language. The quoted statement is a textbook description tied to gain/phase margins in Bode plots. Given Data / Assumptions: “Amplitude ratio exceeding unity” refers to |L(jω)| > 1 (0 dB). “Crossover frequency” implies phase crossover (∠L = -180°) or gain crossover depending on context; in Bode phrasing it links |L| and phase margins. Single-loop negative feedback system. Concept / Approach: The Bode criterion states that if at the phase crossover (∠L = -180°) the magnitude exceeds unity, the closed loop is unstable. Equivalently, at the gain crossover (|L| = 1), a negative phase margin (phase more negative than -180°) signals instability. This wording is characteristic of Bode analysis rather than the Nyquist encirclement rule or Routh’s algebraic test. Step-by-Step Solution: Interpret “amplitude ratio exceeding unity” as |L| > 1.Associate “crossover” with the phase crossover of -180°.Conclude the statement matches Bode’s stability condition. Verification / Alternative check: Nyquist gives an equivalent conclusion via encirclements of -1, but the phrasing with gain/phase margins is the hallmark of Bode plots. Why Other Options Are Wrong: Nyquist: Same physics but different statement form (encirclement count). Routh: Uses characteristic equation coefficients, not frequency response. None/Lyapunov: Not aligned with the exact wording. Common Pitfalls: Confusing the criteria because they are equivalent under certain assumptions; the language cues point to Bode. Final Answer: Bode stability

Question 16

Pressure measurement for high-pressure reactors: In an ammonia reactor operating at elevated pressure, which instrument is typically used for direct pressure indication?

Accepted Answer

Introduction / Context: Ammonia synthesis reactors run at high pressures (tens to hundreds of bar). Selecting an appropriate pressure instrument depends on the pressure regime and robustness needs in hazardous, high-temperature environments. Given Data / Assumptions: Process pressure far above atmospheric. Need for rugged, on-skid indication and compatibility with industrial conditions. Vacuum-level accuracy is not the requirement. Concept / Approach: The Bourdon gauge converts internal pressure into elastic deformation of a curved metallic tube, which is translated to pointer motion. It is the standard choice for medium to very high pressures in industrial service. Manometers are best for low to moderate pressures and lab settings; vacuum gauges (Pirani, McLeod) are specialized for low pressures, not high-pressure reactors. Step-by-Step Solution: Match pressure regime (high) to instrument capability (Bourdon).Eliminate instruments tailored to low or vacuum pressures.Select Bourdon gauge for robust reactor service. Verification / Alternative check: Plant P&IDs commonly show Bourdon-type gauges and transmitters (with diaphragm seals) on high-pressure ammonia loops. Why Other Options Are Wrong: U-tube/Inclined manometers: Bulky, fragile, suited to low differential pressures. Pirani/McLeod: Vacuum measurement devices, not applicable at high pressure. Common Pitfalls: Choosing vacuum instruments by mistake due to familiarity with lab equipment; reactor services demand rugged industrial devices. Final Answer: Bourdon gauge

Question 17

Inferring a process transfer function from a step test: A process is at steady state with output y = 1 for input u = 1. At t = 0, the input is stepped to 2. The measured output is y(t) = 1 + 2 t (a ramp with slope 2). What is the transfer function G…

Accepted Answer

Introduction / Context: Identifying simple process dynamics from step tests is fundamental in control engineering. An output ramp in response to a step input suggests integrating behavior. This question asks you to infer the transfer function from the given input–output relationship. Given Data / Assumptions: Initial steady state: y(0−) = 1 at u(0−) = 1. At t = 0, input steps from 1 to 2, so Δu = 1 (unit step). Observed response: y(t) = 1 + 2 t for t ≥ 0 (a ramp with slope 2). Assume linear time-invariant SISO system. Concept / Approach: For an integrator G(s) = K/s, a unit step input U(s) = 1/s produces output Y(s) = (K/s)*(1/s) = K/s^2. In time domain, y(t) = K * t (a ramp) plus any initial condition. The ramp slope equals K times the step amplitude. With Δu = 1 and observed slope 2, the process gain K is 2, hence G(s) = 2/s. Step-by-Step Solution: Compute input change: Δu = 2 − 1 = 1 (unit step).Match observed y(t) = 1 + 2 t ⇒ ramp slope dy/dt = 2.For integrator G(s) = K/s, slope = K * Δu ⇒ K = 2.Therefore, G(s) = 2/s. Verification / Alternative check: Differentiate y(t): dy/dt = 2 = K * Δu; consistent with integrator behavior. No exponential term appears, ruling out first-order lag forms like 2/(s+1). Why Other Options Are Wrong: 1/s: Would yield slope 1, not 2. 2/(s + 1): Step response is asymptotic, not a ramp. 2 s: Noncausal differentiation-like behavior; step input would produce an impulse, not a ramp. 1/(s^2): Step would give a parabolic output (t^2/2), not a ramp. Common Pitfalls: Using steady-state gain logic for an integrating process; integrators have infinite steady-state gain and require rate-based interpretation. Final Answer: 2/s

Question 18

Dynamic error for a critically damped second-order measuring instrument under a ramp input x(t) = A*t: what is the steady dynamic error expressed in terms of A and the instrument's time constant T (with two equal real roots)?

Accepted Answer

Introduction / Context: A “dynamic error” question checks whether you can relate standard test inputs (step, ramp, impulse) to the transient/steady tracking error of common instrument models. A critically damped second-order instrument (two equal first-order lags) responding to a ramp input x(t) = A*t exhibits a constant lag proportional to the slope A and the sum of its time constants. Given Data / Assumptions: Instrument model: critically damped second-order, equivalent to two identical first-order elements in series with time constant T each. Input: ramp x(t) = A * t (A is slope, t is time). Dynamic error ≡ input − output in steady tracking under a ramp. Concept / Approach: For a first-order element with time constant τ, the steady ramp tracking error is A * τ. Two first-order elements in series (critically damped, equal τ = T) yield a steady ramp error equal to A * (T + T) = 2 A T. Intuitively, each lag adds its own share of delay proportional to the ramp slope; the errors add linearly for cascaded lags. Step-by-Step Solution: Represent the critically damped second-order system as G(s) = 1 / (1 + T s)^2.For input X(s) = A / s^2 (ramp), steady ramp error of one lag is A*T.Two identical lags in series → total steady ramp error = A*T + A*T = 2 A T.Therefore, dynamic error = 2 A T. Verification / Alternative check: Using final value theorem on the error transfer function E(s) = X(s) [1 − G(s)]; with G(s) = 1/(1+T s)^2 and X(s) = A/s^2, the limit t→∞ of e(t) evaluates to 2 A T, confirming the time-constant additivity rule for ramp inputs across cascaded first-order lags. Why Other Options Are Wrong: 0.5 A T / A T / 1.5 A T: Underestimate the cumulative lag from two lags.A / T: Has wrong dimensions and ignores ramp-lag behavior. Common Pitfalls: Mixing up step error (which goes to zero for stable systems) with ramp error (nonzero for finite-order lags), or forgetting that for cascaded first-order lags the ramp errors add. Final Answer: 2 A T

Question 19

Pyrometry classification: which instrument is a contact-type pyrometer used by touching or inserting a sensing element into the hot medium?

Accepted Answer

Introduction / Context: Pyrometers are broadly divided into contact and non-contact types. Contact pyrometers sense temperature through physical contact (electrical resistance or thermoelectric effect), while non-contact pyrometers infer temperature from thermal radiation in the visible/infrared spectrum. Given Data / Assumptions: Choices include resistance, optical, radiation, and infrared instruments. “Contact” implies the sensor is inserted or touches the hot body. Concept / Approach: A resistance pyrometer (e.g., platinum resistance thermometer or high-temperature resistance element) measures temperature via change in electrical resistance upon contact. Optical/radiation/infrared pyrometers are line-of-sight devices that require no contact; they measure emitted radiation and apply emissivity corrections. Step-by-Step Solution: Identify which options require physical contact with the medium.Resistance pyrometer → contact; all others → non-contact radiation methods.Select “Resistance pyrometer.” Verification / Alternative check: Instrumentation manuals list resistance/thermocouple probes under contact methods and optical/radiative devices under non-contact pyrometry. Why Other Options Are Wrong: Optical/radiation/infrared/two-color pyrometers determine temperature from radiant energy without physical contact. Common Pitfalls: Assuming “optical” implies contact through a sighting tube; it still measures radiation remotely. Final Answer: Resistance pyrometer

Question 20

On–off (two-position) controllers are normally applied to which kind of industrial variable/process duty where cycling around a setpoint is acceptable?

Accepted Answer

Introduction / Context: On–off (two-position) control is the simplest feedback mode: the final element is either fully ON or fully OFF. It inherently produces cycling (limit cycles) around the setpoint. Understanding where this is acceptable helps in selecting the right controller for the job. Given Data / Assumptions: We need a process tolerant of cycling and actuator chatter. Deadband or differential is usually provided to limit frequent switching. No fine setpoint tracking or linearity is required. Concept / Approach: Temperature control of thermal masses (ovens, incubators, water baths, domestic HVAC) is the classic on–off application because thermal inertia smooths the cycling and exact continuous modulation is unnecessary. By contrast, flow, fast pressure, or servo position systems benefit from proportional/PI/PID control with continuous output modulation. Step-by-Step Solution: Identify processes with natural damping (thermal capacitance).Assess tolerance to limit cycles; temperature systems typically tolerate small oscillations.Choose temperature duty as the appropriate application. Verification / Alternative check: Textbook examples and commercial thermostats illustrate two-position control for temperature. Industrial flow and servo applications rarely use on–off except for interlocks. Why Other Options Are Wrong: Flow/pressure/servo/composition: Usually require continuous control (PI/PID) for stability and accuracy. Common Pitfalls: Using on–off where actuator wear or process quality cannot tolerate cycling; lacking differential (hysteresis) can cause rapid chattering. Final Answer: Low-inertia temperature control (e.g., ovens, baths)

Process Control and Instrumentation Questions