Q: Thermistors and material class: thermistors have a very high (negative or positive) temperature coefficient of resistivity. They belong to which class of solids?

Introduction / Context: Thermistors are temperature-sensitive resistors widely used for precision measurement, compensation, and control. Their defining feature is a large temperature coefficient of resistance, typically negative (NTC) but sometimes positive (PTC). Correctly classifying their material type clarifies why their resistance changes so strongly with temperature. Given Data / Assumptions: Thermistors show large dR/dT magnitudes. Common compositions include metal oxides sintered into beads or chips. Applications range from temperature sensing to inrush current limiting. Concept / Approach: Thermistors are made from semiconductor materials. In NTC thermistors, increasing temperature excites more charge carriers, reducing resistance sharply. In PTC thermistors (such as doped barium titanate), a phase transition or band structure effect increases resistance with temperature. These behaviors are characteristic of semiconducting solids, not good conductors or pure insulators. Step-by-Step Solution: Relate strong temperature dependence of resistivity to semiconductor physics (carrier concentration).Identify typical thermistor construction: metal oxide semiconductors like Mn, Ni, Co oxides.Conclude class: semiconductors. Verification / Alternative check: Device datasheets plot log(R) versus 1/T (Arrhenius-like), a hallmark of semiconductor conduction mechanisms. Why Other Options Are Wrong: Dielectrics/Insulators: Very high resistivity with relatively small temperature dependence in normal ranges.Conductors: Metals have positive but modest temperature coefficients, far smaller than thermistors’.Superconductors: Exhibit zero resistance below critical temperature, unrelated to thermistor behavior. Common Pitfalls: Assuming “resistor” implies metallic conductor; in thermistors, the dominant physics is semiconducting carrier behavior. Final Answer: Semiconductors

Q: Resistance temperature detectors (RTDs): which material listed is never used as a sensing element in industrial resistance thermometers?

Introduction / Context: Resistance temperature detectors use the predictable change of electrical resistance with temperature in certain metals and alloys. Material selection dictates accuracy, stability, and usable temperature range. This question checks familiarity with common RTD elements and highlights one metal that is unsuitable for this purpose. Given Data / Assumptions: Industrial RTDs emphasize linearity, stability, and repeatability. Common RTD materials: platinum (standard), nickel and copper (legacy/industrial), and certain nickel–iron alloys. We seek the material that is not used as an RTD element. Concept / Approach: Platinum (Pt100, Pt1000) is the international standard due to stability and near-linear response. Nickel (Ni120) and copper RTDs exist for specific ranges and cost considerations. Alloys such as 70% nickel + 30% iron (Balco-type) are used in instrumentation circuits. Lead, however, is soft, has poor mechanical properties, low melting point, and an unsuitable temperature coefficient for precision RTD service, making it inappropriate as a sensing element. Step-by-Step Solution: List typical RTD materials: platinum, nickel, copper, and certain Ni–Fe alloys.Evaluate lead’s properties: low melting point, creep, instability.Conclude: lead is never used as an RTD sensing element. Verification / Alternative check: Standards (IEC 60751) specify platinum characteristics; vendor catalogs also offer nickel/copper variants, but none use lead for sensing elements. Why Other Options Are Wrong: Nickel: Historically used, good sensitivity; limited high-temp stability vs Pt.Copper: Good linearity near ambient; used in some HVAC/instrumentation.30% iron + 70% nickel: Employed in measurement circuits and specialized sensors.Platinum: The standard industrial RTD material. Common Pitfalls: Confusing lead as “wiring material” with “sensing element”; lead may appear in solder, not as an RTD element. Final Answer: Lead

Q: Distributed versus lumped parameter systems: which process behaves as a distributed parameter system with spatial gradients in state variables?

Introduction / Context: Systems are modeled as lumped or distributed depending on whether state variables (temperature, concentration, pressure) can be assumed uniform within control volumes. Recognizing distributed systems helps select the correct mathematical tools (partial differential equations) and control strategies. Given Data / Assumptions: Tubular reactor: plug flow behavior with axial (and possibly radial) gradients. CSTR: ideally well-mixed; state variables uniform within the tank. On–off controller: a control law, not a physical distributed system. Concept / Approach: Distributed parameter systems require PDEs because variables depend on both time and space (e.g., T(z,t), C_A(z,t)). A tubular reactor develops concentration and temperature profiles along its length; thus it is inherently distributed. By contrast, a CSTR assumes complete mixing so ordinary differential equations suffice, making it lumped. Step-by-Step Solution: Identify the presence of spatial gradients: tubular reactor has axial profiles.Classify according to modeling framework: PDEs → distributed.Select tubular reactor as the correct example. Verification / Alternative check: Introductory reaction engineering texts derive the PFR material and energy balances as PDEs (or steady ODE in space), contrasting them with lumped CSTR balances in time only. Why Other Options Are Wrong: CSTR: Uniform state by assumption → lumped parameter.On–off controller: Not a spatially distributed physical medium.None of these: Incorrect because a tubular reactor is a classic distributed example. Common Pitfalls: Assuming any large vessel is distributed; the key is the mixing model, not size. Final Answer: Tubular reactor

Q: Fixed-point reference in thermometry: the internationally recognized silver point (melting point of pure silver at standard pressure) corresponds approximately to which temperature in °C?

Introduction / Context: In temperature metrology, metal fixed points (freezing/melting points of pure metals) provide precise calibration references. The silver point is one of the high-temperature fixed points used to calibrate radiation and contact thermometers. Given Data / Assumptions: Pressure is at or near 1 atm (standard atmospheric pressure). Pure silver in thermal equilibrium at its melting/freezing plateau. Approximate tabulated values are acceptable for MCQ selection. Concept / Approach: The silver point is commonly cited near 961 °C. For selection among rounded MCQ choices, 960.5 °C is the closest reasonable figure to the accepted value (about 961 °C), reflecting the temperature plateau observed during melting/freezing of pure silver. Step-by-Step Solution: Identify the fixed point of interest: silver (Ag) melting/freezing point.Recall approximate accepted value: ≈ 961 °C.Choose the closest option: 960.5 °C. Verification / Alternative check: Calibration tables for ITS-90 list the silver point very close to 961 °C; rounding differences explain the offered 960.5 °C choice. Why Other Options Are Wrong: 760.5 °C and 860.5 °C: too low for silver; no standard metal fixed point at those temperatures for silver.1060.5 °C: closer to the melting point of gold (≈ 1064 °C), not silver. Common Pitfalls: Confusing the silver point (≈ 961 °C) with the gold point (≈ 1064 °C) or the antimony point (≈ 631 °C). Final Answer: 960.5

Q: Platinum resistance thermometer (PRT) upper fixed point: up to which fixed point can a standard PRT be used—the antimony point (solid–liquid equilibrium of Sb) at approximately what temperature in °C?

Introduction / Context: Platinum resistance thermometers (PRTs, often Pt100/Pt25) are primary contact sensors for precision thermometry. Their practical upper limit for high-accuracy use is commonly tied to metal fixed points; the antimony point is a well-established reference. Given Data / Assumptions: Standard PRT in a protective sheath operating at atmospheric pressure. Fixed point considered is the antimony (Sb) melting/freezing point. Goal is to identify the approximate temperature of this fixed point. Concept / Approach: The antimony point occurs at approximately 630.74 °C. PRTs are frequently calibrated from the triple point of water up to the zinc or antimony points depending on grade; beyond these, errors from self-heating, oxidation, and material limits increase, and non-contact methods are often preferred. Step-by-Step Solution: Recall fixed point values: antimony ≈ 630.74 °C; silver ≈ 961 °C; gold ≈ 1064 °C.Map feasible PRT range to antimony point for high-accuracy operation.Select 630.74 °C as the antimony point value. Verification / Alternative check: ITS-90 documentation lists the antimony fixed point near 631 °C; exact decimal may vary slightly with reference tables. Why Other Options Are Wrong: 961.93 °C: near silver point, higher than standard PRT high-accuracy range.1064.43 °C: gold point; typically requires radiation pyrometry.1261.93 °C: not a standard metal fixed point for PRT use. Common Pitfalls: Mixing up the hierarchy of fixed points; memorize antimony (≈ 631 °C), silver (≈ 961 °C), and gold (≈ 1064 °C). Final Answer: 630.74

Q: Standard feedback control block diagram (conceptual): identify the incorrect statement about variable/transfer-function labeling in a simple single-loop system (controller, process, measuring element, feedback).

Introduction / Context: In textbook single-loop diagrams, signals and transfer functions have conventional labels: the controller C(s) (or Gc), the process G2(s) (or Gp), the measuring element H(s), the set point R (or X1), the load/disturbance X, the controlled output Y (or C), and the manipulated variable m (or u). Understanding these conventions helps avoid misinterpretation when writing closed-loop transfer functions. Given Data / Assumptions: Simple unity-feedback style diagram with controller, process, and measuring element. Common notation: C(s) for controller TF, G2(s) for process TF, H(s) for sensor TF. Output variable often denoted C or Y; manipulated variable often denoted m or u. Concept / Approach: The manipulated variable (MV) is the actuator output that drives the process (e.g., valve position, flow). In common notation, C usually designates the controlled variable (CV), not the MV. Therefore, a statement claiming “Manipulated variable is represented by C” is incorrect in the standard labeling scheme. Step-by-Step Solution: Map symbols: H → measuring element, G2 → process.Recognize C(s) often denotes controller transfer function; C (without s) or Y denotes controlled output.Identify that MV is typically m or u, not C; hence the statement is wrong. Verification / Alternative check: Any standard process control text shows CV labeled y or c, MV labeled u or m; controller is Gc(s). The mismatch highlights the incorrect statement. Why Other Options Are Wrong (or Not Chosen): Options (b) and (c) reference generic forms of overall transfer functions; although not fully written, they are not inherently incorrect by label usage.Option (a) aligns with common assignment: H for sensor, G2 for process. Common Pitfalls: Confusing the symbol C for “controller” vs. “controlled variable.” Context (function C(s) vs. signal C) matters. Final Answer: Manipulated variable is represented by C.

Q: Applications of bimetal strips: identify where bimetal strips are not used among the listed devices and controls.

Introduction / Context: Bimetal strips exploit the different thermal expansion coefficients of two bonded metals; when heated, they bend. This simple principle is used for temperature indication and on–off switching in numerous devices. Thermocouples, in contrast, rely on a thermoelectric effect and do not use bimetal bending. Given Data / Assumptions: Bimetallic behavior: differential expansion causes deflection. Thermocouple behavior: Seebeck effect generates a millivolt signal at a junction of dissimilar metals. Devices listed include both thermal-mechanical and electro-thermoelectric types. Concept / Approach: Bimetallic thermometers and thermostats directly use the bending of a bimetal strip. Relays and cut-outs often include bimetal elements to actuate electrical contacts with temperature. A thermocouple, however, contains no moving bimetal element; it is a junction producing EMF proportional to temperature difference. Step-by-Step Solution: Classify each device by principle.Recognize thermocouples measure temperature via EMF, not mechanical bending.Conclude that bimetal strips are not used in thermocouples. Verification / Alternative check: Instrument catalogs clearly distinguish bimetal thermometers/thermostats (mechanical) from thermocouples (electrical EMF sensors). Why Other Options Are Wrong: Bimetallic thermometers: core use case of bimetal strips.Thermostats and relays: common bimetal applications for switching with temperature. Common Pitfalls: Assuming any “two metals” device is bimetallic; thermocouples use two metals but for Seebeck voltage, not bending. Final Answer: Thermocouples

Q: Vacuum instrumentation: the McLeod gauge is specifically used to measure which of the following quantities in a system?

Introduction / Context: The McLeod gauge is a classic laboratory instrument for measuring low absolute pressures in the vacuum range. It compresses a known quantity of gas using a mercury column and infers the original pressure from Boyle’s law. Given Data / Assumptions: Applicable range: typically from about 1 Torr down to 10^-3 Torr (order of magnitude; depends on design). Clean, non-condensing gases only due to gas compression. Manual, non-continuous indication. Concept / Approach: By sealing and compressing a trapped gas sample, the McLeod gauge creates a measurable pressure and volume change. Using P1V1 = P2V2, the unknown low pressure (vacuum) is calculated. This method directly measures vacuum (very low absolute pressure), not velocity or flow. Step-by-Step Solution: Trap a known gas volume.Raise mercury to compress the gas to a measurable pressure.Apply Boyle’s law to compute the original vacuum pressure. Verification / Alternative check: Instrument manuals specify it as a primary standard-type gauge for vacuum calibration in the millitorr region. Why Other Options Are Wrong: Point velocity and flow rate: require anemometry or DP devices, not a McLeod gauge.Pressure (generic): while vacuum is a form of pressure, the instrument is specifically for low absolute pressure (vacuum) measurement. Common Pitfalls: Using McLeod on condensable vapors; readings become erroneous. Final Answer: Vacuum

Question 1

Constant-volume gas thermometry below 0 °C: which fill gas is preferred for sub-zero temperature measurement to avoid condensation and ensure accuracy?

Accepted Answer

Introduction / Context: Gas thermometers rely on the relationship between gas pressure/volume and temperature. For sub-zero measurements, the choice of fill gas is critical to avoid liquefaction and to maintain a nearly ideal-gas behavior over the range of interest. This question probes the practical selection of a fill gas for constant-volume thermometers operating below 0 °C. Given Data / Assumptions: Temperature range: below 0 °C (e.g., −50 to 0 °C). Instrument: constant-volume gas thermometer. Goal: stable, non-condensing gas with good sensitivity. Concept / Approach: Hydrogen has a very low normal boiling point and remains gaseous at low temperatures encountered in typical sub-zero industrial measurements. It behaves close to ideal and provides adequate sensitivity to temperature changes. While helium also remains gaseous to extremely low temperatures, traditional constant-volume thermometer practice often specifies hydrogen for the sub-zero region, whereas nitrogen is common at moderate positive temperatures but can present larger non-ideality or risk of liquefaction at deeper cryogenic conditions. Step-by-Step Solution: Identify risk: gas must not liquefy within the operating range.Compare candidates: hydrogen and helium both remain gaseous; nitrogen is less favored for deep sub-zero.Select the conventional answer used in constant-volume gas thermometry for sub-zero: hydrogen. Verification / Alternative check: Handbook tables pair gas choices with ranges: nitrogen (around ambient and moderate highs), hydrogen (sub-zero), and helium (very low cryogenic work). The sub-zero selection commonly points to hydrogen as the standard fill gas. Why Other Options Are Wrong: Nitrogen: Widely used but less ideal for deeper sub-zero thermometry in this classical context.Helium: Technically suitable but more associated with extreme cryogenics and specialized laboratory setups.Argon/None: Argon is heavier and less favored; “none” ignores practical instrument design. Common Pitfalls: Assuming any inert gas suffices; overlooking boiling points and gas non-idealities at the targeted temperature and pressure. Final Answer: Hydrogen

Question 2

Electrical analogy for a single-tank liquid system: in the standard hydraulic–electrical analogy, which variable corresponds to electrical voltage?

Accepted Answer

Introduction / Context: Analogies help engineers visualize dynamic behavior across domains. The hydraulic–electrical analogy maps variables in fluid systems to electrical circuit counterparts. Understanding these mappings is valuable for intuition about transient responses and controller design. Given Data / Assumptions: System: single tank with inlet and outlet. Electrical analogy: voltage ↔ pressure head; current ↔ flow; capacitance ↔ storage capacity. Level is proportional to hydrostatic pressure head. Concept / Approach: In the analogy, the potential difference driving current in a circuit corresponds to pressure difference driving flow in a hydraulic system. For a tank, the hydrostatic head is directly related to liquid level. Therefore, tank level (head) is analogous to electrical voltage; flow rate corresponds to current; and the tank’s cross-sectional area acts like capacitance storing “charge” in the form of volume per unit head. Step-by-Step Solution: Identify electrical voltage as potential energy per unit charge.Map to hydraulic potential: pressure head due to level.Conclude that liquid level (head) corresponds to electrical voltage in the analogy. Verification / Alternative check: Mathematically, continuity gives A * dh/dt = q_in − q_out, which mirrors capacitor equation C * dv/dt = i_in − i_out, where A ↔ C and h (level) ↔ v (voltage). Why Other Options Are Wrong: Heat content: Belongs to thermal analogies, not directly to hydraulic–electrical mapping.Liquid volume: Closer to electrical charge (C * V), not voltage itself.Flow rate: Corresponds to current, not voltage.Area: Corresponds to capacitance, not voltage. Common Pitfalls: Confusing storage (capacitance) with potential (voltage) and mixing thermal–electrical with hydraulic–electrical analogies. Final Answer: Liquid level (pressure head)

Question 3

Damping ratio and system classification: when the damping coefficient (ξ) equals 1, how is the second-order system classified and what does it imply about transient response?

Accepted Answer

Introduction / Context: The damping ratio ξ is a key parameter for second-order dynamics. It dictates how the system responds to a step input—whether it overshoots, oscillates, or returns smoothly to steady state. Correct classification is fundamental to tuning controllers and predicting transient behavior. Given Data / Assumptions: Standard second-order model with natural frequency ω_n and damping ratio ξ. Input: step change; Output: typical process variable. Question case: ξ = 1. Concept / Approach: For a canonical second-order system: 0 < ξ < 1 → underdamped (oscillatory); ξ = 1 → critically damped (fastest non-oscillatory return, no overshoot); ξ > 1 → overdamped (no overshoot, slower). Thus, ξ = 1 denotes the critically damped condition that achieves the quickest possible settling without overshoot. Step-by-Step Solution: Identify ξ = 1 boundary between underdamped and overdamped.Recall transient traits: no oscillation, no overshoot, fastest aperiodic response.Select classification: critically damped. Verification / Alternative check: Compare step responses: ξ = 0.7 shows overshoot; ξ = 1 shows monotonic rise to setpoint; ξ = 2 is slower with larger rise time and no overshoot. Plots confirm the qualitative behavior. Why Other Options Are Wrong: Overdamped: Requires ξ > 1 and yields slower response than critical.Underdamped: Requires 0 < ξ < 1 and shows oscillation/overshoot.Highly fluctuating/sustained: Implies oscillatory or unstable dynamics, not ξ = 1.Aperiodic unstable: Not applicable; ξ relates to stable, real or complex-conjugate pole patterns. Common Pitfalls: Confusing ξ = 1 with ξ = 0 (undamped) or thinking critical damping is “slow”; in reality it is the fastest non-overshooting case. Final Answer: Critically damped

Question 4

Continuous flue gas composition monitoring for O2 and CO2: which instrument provides on-line, automatic, continuous analysis in furnace/boiler stacks?

Accepted Answer

Introduction / Context: Combustion optimization and environmental compliance require continuous, on-line flue gas analysis. The goal is to track oxygen and carbon dioxide in real time for air–fuel ratio control and efficiency calculations. Understanding which instruments are truly continuous as opposed to manual or batch is essential for proper selection. Given Data / Assumptions: Target species: O2 and CO2 in flue gas. Requirement: continuous, automated operation suitable for industrial stacks. Sampling system with conditioning is assumed. Concept / Approach: A process chromatograph (on-line GC) cycles continuously and produces regular, automated readings for multiple components, including O2 and CO2, by separating species before detection. While modern CEMS often combine a zirconia O2 probe with an NDIR CO2 analyzer to achieve continuous monitoring, among the provided single-instrument choices, a chromatograph is the one explicitly designed for automated, continuous multi-component analysis (beyond a single property). Orsat is manual and intermittent. A single thermal conductivity cell lacks selectivity and typically cannot deliver simultaneous specific O2 and CO2 readings without prior separation. A zirconia probe alone measures O2 only, not CO2. Step-by-Step Solution: Eliminate manual methods (Orsat) as non-continuous.Note that zirconia provides only O2; TCD lacks selectivity.Select the instrument capable of automated multi-component separation and measurement: chromatograph. Verification / Alternative check: Typical plant CEMS racks may mix technologies (paramagnetic/zirconia for O2 and NDIR for CO/CO2). However, if a single instrument type must cover both continuously from the list, an on-line GC qualifies due to its automated cyclic analysis capability. Why Other Options Are Wrong: Orsat apparatus: Manual, laboratory/field portable—not continuous.Thermal conductivity cell: Non-selective without separation; prone to cross-sensitivity.Zirconia probe: Measures O2 only.Paramagnetic + NDIR combo: Effective in practice but not a single named instrument choice here. Common Pitfalls: Assuming “continuous” for any sensor that can be left in-line; true continuous multi-component analysis requires selectivity or separation. Final Answer: Chromatograph

Question 5

Thermistors and material class: thermistors have a very high (negative or positive) temperature coefficient of resistivity. They belong to which class of solids?

Accepted Answer

Introduction / Context: Thermistors are temperature-sensitive resistors widely used for precision measurement, compensation, and control. Their defining feature is a large temperature coefficient of resistance, typically negative (NTC) but sometimes positive (PTC). Correctly classifying their material type clarifies why their resistance changes so strongly with temperature. Given Data / Assumptions: Thermistors show large dR/dT magnitudes. Common compositions include metal oxides sintered into beads or chips. Applications range from temperature sensing to inrush current limiting. Concept / Approach: Thermistors are made from semiconductor materials. In NTC thermistors, increasing temperature excites more charge carriers, reducing resistance sharply. In PTC thermistors (such as doped barium titanate), a phase transition or band structure effect increases resistance with temperature. These behaviors are characteristic of semiconducting solids, not good conductors or pure insulators. Step-by-Step Solution: Relate strong temperature dependence of resistivity to semiconductor physics (carrier concentration).Identify typical thermistor construction: metal oxide semiconductors like Mn, Ni, Co oxides.Conclude class: semiconductors. Verification / Alternative check: Device datasheets plot log(R) versus 1/T (Arrhenius-like), a hallmark of semiconductor conduction mechanisms. Why Other Options Are Wrong: Dielectrics/Insulators: Very high resistivity with relatively small temperature dependence in normal ranges.Conductors: Metals have positive but modest temperature coefficients, far smaller than thermistors’.Superconductors: Exhibit zero resistance below critical temperature, unrelated to thermistor behavior. Common Pitfalls: Assuming “resistor” implies metallic conductor; in thermistors, the dominant physics is semiconducting carrier behavior. Final Answer: Semiconductors

Question 6

Resistance temperature detectors (RTDs): which material listed is never used as a sensing element in industrial resistance thermometers?

Accepted Answer

Introduction / Context: Resistance temperature detectors use the predictable change of electrical resistance with temperature in certain metals and alloys. Material selection dictates accuracy, stability, and usable temperature range. This question checks familiarity with common RTD elements and highlights one metal that is unsuitable for this purpose. Given Data / Assumptions: Industrial RTDs emphasize linearity, stability, and repeatability. Common RTD materials: platinum (standard), nickel and copper (legacy/industrial), and certain nickel–iron alloys. We seek the material that is not used as an RTD element. Concept / Approach: Platinum (Pt100, Pt1000) is the international standard due to stability and near-linear response. Nickel (Ni120) and copper RTDs exist for specific ranges and cost considerations. Alloys such as 70% nickel + 30% iron (Balco-type) are used in instrumentation circuits. Lead, however, is soft, has poor mechanical properties, low melting point, and an unsuitable temperature coefficient for precision RTD service, making it inappropriate as a sensing element. Step-by-Step Solution: List typical RTD materials: platinum, nickel, copper, and certain Ni–Fe alloys.Evaluate lead’s properties: low melting point, creep, instability.Conclude: lead is never used as an RTD sensing element. Verification / Alternative check: Standards (IEC 60751) specify platinum characteristics; vendor catalogs also offer nickel/copper variants, but none use lead for sensing elements. Why Other Options Are Wrong: Nickel: Historically used, good sensitivity; limited high-temp stability vs Pt.Copper: Good linearity near ambient; used in some HVAC/instrumentation.30% iron + 70% nickel: Employed in measurement circuits and specialized sensors.Platinum: The standard industrial RTD material. Common Pitfalls: Confusing lead as “wiring material” with “sensing element”; lead may appear in solder, not as an RTD element. Final Answer: Lead

Question 7

Distributed versus lumped parameter systems: which process behaves as a distributed parameter system with spatial gradients in state variables?

Accepted Answer

Introduction / Context: Systems are modeled as lumped or distributed depending on whether state variables (temperature, concentration, pressure) can be assumed uniform within control volumes. Recognizing distributed systems helps select the correct mathematical tools (partial differential equations) and control strategies. Given Data / Assumptions: Tubular reactor: plug flow behavior with axial (and possibly radial) gradients. CSTR: ideally well-mixed; state variables uniform within the tank. On–off controller: a control law, not a physical distributed system. Concept / Approach: Distributed parameter systems require PDEs because variables depend on both time and space (e.g., T(z,t), C_A(z,t)). A tubular reactor develops concentration and temperature profiles along its length; thus it is inherently distributed. By contrast, a CSTR assumes complete mixing so ordinary differential equations suffice, making it lumped. Step-by-Step Solution: Identify the presence of spatial gradients: tubular reactor has axial profiles.Classify according to modeling framework: PDEs → distributed.Select tubular reactor as the correct example. Verification / Alternative check: Introductory reaction engineering texts derive the PFR material and energy balances as PDEs (or steady ODE in space), contrasting them with lumped CSTR balances in time only. Why Other Options Are Wrong: CSTR: Uniform state by assumption → lumped parameter.On–off controller: Not a spatially distributed physical medium.None of these: Incorrect because a tubular reactor is a classic distributed example. Common Pitfalls: Assuming any large vessel is distributed; the key is the mixing model, not size. Final Answer: Tubular reactor

Question 8

Glass-electrode pH measurement principle: when two aqueous solutions with different hydrogen ion concentrations are separated by a thin specialized glass membrane, a millivolt-level electromotive force (mV) is generated across the membrane. This phe…

Accepted Answer

Introduction / Context: The glass electrode is the most widely used electrochemical sensor for measuring acidity or alkalinity of solutions. When it separates two solutions of different hydrogen ion activity (typically the test solution and an internal reference solution), a small electromotive force (on the order of millivolts) develops due to selective H+ exchange at the hydrated glass surface. This EMF is interpreted by a high-impedance meter to report pH directly. Given Data / Assumptions: Two solutions with different hydrogen ion concentrations are separated by a thin glass wall. An EMF of the order of millivolts is produced. Electrochemical measurement uses a reference electrode plus a glass indicator electrode. Concept / Approach: The Nernst equation predicts that the potential across a selective membrane varies linearly with the logarithm of ion activity. For H+ at 25°C, the theoretical slope is about 59.16 mV per pH unit. A pH meter measures the potential difference between the glass electrode and a reference electrode (e.g., Ag/AgCl), converts that voltage to pH, and displays the result after temperature compensation and calibration with buffer standards. Step-by-Step Solution: Recognize that a hydrogen-ion-selective glass membrane generates a potential dependent on pH.Note that the EMF magnitude is in millivolts per pH unit, consistent with Nernstian behavior.Identify the instrument that reads this EMF and reports pH: the pH meter. Verification / Alternative check: Practical pH meters show approximately 59 mV change per pH at 25°C; multi-point buffer calibration aligns the meter to real electrode behavior. Why Other Options Are Wrong: Polarimeter: measures optical rotation, not electrochemical potential.Chromatograph: separates mixture components; unrelated to H+ membrane EMF.Polarograph: electroanalytical voltammetry using dropping electrodes; not pH glass membrane EMF. Common Pitfalls: Confusing pH glass electrode (ion selective) with general redox electrodes; pH response is specific to H+ activity and requires temperature compensation. Final Answer: pH meter

Question 9

Fixed-point reference in thermometry: the internationally recognized silver point (melting point of pure silver at standard pressure) corresponds approximately to which temperature in °C?

Accepted Answer

Introduction / Context: In temperature metrology, metal fixed points (freezing/melting points of pure metals) provide precise calibration references. The silver point is one of the high-temperature fixed points used to calibrate radiation and contact thermometers. Given Data / Assumptions: Pressure is at or near 1 atm (standard atmospheric pressure). Pure silver in thermal equilibrium at its melting/freezing plateau. Approximate tabulated values are acceptable for MCQ selection. Concept / Approach: The silver point is commonly cited near 961 °C. For selection among rounded MCQ choices, 960.5 °C is the closest reasonable figure to the accepted value (about 961 °C), reflecting the temperature plateau observed during melting/freezing of pure silver. Step-by-Step Solution: Identify the fixed point of interest: silver (Ag) melting/freezing point.Recall approximate accepted value: ≈ 961 °C.Choose the closest option: 960.5 °C. Verification / Alternative check: Calibration tables for ITS-90 list the silver point very close to 961 °C; rounding differences explain the offered 960.5 °C choice. Why Other Options Are Wrong: 760.5 °C and 860.5 °C: too low for silver; no standard metal fixed point at those temperatures for silver.1060.5 °C: closer to the melting point of gold (≈ 1064 °C), not silver. Common Pitfalls: Confusing the silver point (≈ 961 °C) with the gold point (≈ 1064 °C) or the antimony point (≈ 631 °C). Final Answer: 960.5

Question 10

Platinum resistance thermometer (PRT) upper fixed point: up to which fixed point can a standard PRT be used—the antimony point (solid–liquid equilibrium of Sb) at approximately what temperature in °C?

Accepted Answer

Introduction / Context: Platinum resistance thermometers (PRTs, often Pt100/Pt25) are primary contact sensors for precision thermometry. Their practical upper limit for high-accuracy use is commonly tied to metal fixed points; the antimony point is a well-established reference. Given Data / Assumptions: Standard PRT in a protective sheath operating at atmospheric pressure. Fixed point considered is the antimony (Sb) melting/freezing point. Goal is to identify the approximate temperature of this fixed point. Concept / Approach: The antimony point occurs at approximately 630.74 °C. PRTs are frequently calibrated from the triple point of water up to the zinc or antimony points depending on grade; beyond these, errors from self-heating, oxidation, and material limits increase, and non-contact methods are often preferred. Step-by-Step Solution: Recall fixed point values: antimony ≈ 630.74 °C; silver ≈ 961 °C; gold ≈ 1064 °C.Map feasible PRT range to antimony point for high-accuracy operation.Select 630.74 °C as the antimony point value. Verification / Alternative check: ITS-90 documentation lists the antimony fixed point near 631 °C; exact decimal may vary slightly with reference tables. Why Other Options Are Wrong: 961.93 °C: near silver point, higher than standard PRT high-accuracy range.1064.43 °C: gold point; typically requires radiation pyrometry.1261.93 °C: not a standard metal fixed point for PRT use. Common Pitfalls: Mixing up the hierarchy of fixed points; memorize antimony (≈ 631 °C), silver (≈ 961 °C), and gold (≈ 1064 °C). Final Answer: 630.74

Question 11

Standard feedback control block diagram (conceptual): identify the incorrect statement about variable/transfer-function labeling in a simple single-loop system (controller, process, measuring element, feedback).

Accepted Answer

Introduction / Context: In textbook single-loop diagrams, signals and transfer functions have conventional labels: the controller C(s) (or Gc), the process G2(s) (or Gp), the measuring element H(s), the set point R (or X1), the load/disturbance X, the controlled output Y (or C), and the manipulated variable m (or u). Understanding these conventions helps avoid misinterpretation when writing closed-loop transfer functions. Given Data / Assumptions: Simple unity-feedback style diagram with controller, process, and measuring element. Common notation: C(s) for controller TF, G2(s) for process TF, H(s) for sensor TF. Output variable often denoted C or Y; manipulated variable often denoted m or u. Concept / Approach: The manipulated variable (MV) is the actuator output that drives the process (e.g., valve position, flow). In common notation, C usually designates the controlled variable (CV), not the MV. Therefore, a statement claiming “Manipulated variable is represented by C” is incorrect in the standard labeling scheme. Step-by-Step Solution: Map symbols: H → measuring element, G2 → process.Recognize C(s) often denotes controller transfer function; C (without s) or Y denotes controlled output.Identify that MV is typically m or u, not C; hence the statement is wrong. Verification / Alternative check: Any standard process control text shows CV labeled y or c, MV labeled u or m; controller is Gc(s). The mismatch highlights the incorrect statement. Why Other Options Are Wrong (or Not Chosen): Options (b) and (c) reference generic forms of overall transfer functions; although not fully written, they are not inherently incorrect by label usage.Option (a) aligns with common assignment: H for sensor, G2 for process. Common Pitfalls: Confusing the symbol C for “controller” vs. “controlled variable.” Context (function C(s) vs. signal C) matters. Final Answer: Manipulated variable is represented by C.

Question 12

Applications of bimetal strips: identify where bimetal strips are not used among the listed devices and controls.

Accepted Answer

Introduction / Context: Bimetal strips exploit the different thermal expansion coefficients of two bonded metals; when heated, they bend. This simple principle is used for temperature indication and on–off switching in numerous devices. Thermocouples, in contrast, rely on a thermoelectric effect and do not use bimetal bending. Given Data / Assumptions: Bimetallic behavior: differential expansion causes deflection. Thermocouple behavior: Seebeck effect generates a millivolt signal at a junction of dissimilar metals. Devices listed include both thermal-mechanical and electro-thermoelectric types. Concept / Approach: Bimetallic thermometers and thermostats directly use the bending of a bimetal strip. Relays and cut-outs often include bimetal elements to actuate electrical contacts with temperature. A thermocouple, however, contains no moving bimetal element; it is a junction producing EMF proportional to temperature difference. Step-by-Step Solution: Classify each device by principle.Recognize thermocouples measure temperature via EMF, not mechanical bending.Conclude that bimetal strips are not used in thermocouples. Verification / Alternative check: Instrument catalogs clearly distinguish bimetal thermometers/thermostats (mechanical) from thermocouples (electrical EMF sensors). Why Other Options Are Wrong: Bimetallic thermometers: core use case of bimetal strips.Thermostats and relays: common bimetal applications for switching with temperature. Common Pitfalls: Assuming any “two metals” device is bimetallic; thermocouples use two metals but for Seebeck voltage, not bending. Final Answer: Thermocouples

Question 13

Exothermic CSTR temperature control with cooling jacket: select the recommended fail-safe valve action and controller action pairing for stable, safe regulation using cooling water.

Accepted Answer

Introduction / Context: For an exothermic continuous stirred-tank reactor (CSTR), cooling reliability is critical to prevent thermal runaway. The control loop usually manipulates cooling water flow via a control valve on the jacket or coil. Choosing the correct valve fail action and controller action ensures both safety and proper loop direction. Given Data / Assumptions: Process: exothermic CSTR with water-jacket cooling. Controlled variable: reactor temperature (high risk if it increases). Manipulated variable: cooling water flow through a valve. Air failure should favor increased cooling (fail-safe). Concept / Approach: An air-to-close valve (fail-open) will open upon loss of air, increasing cooling water flow and enhancing safety. Loop direction: when temperature rises above set point, the controller must open the cooling valve. For an air-to-close valve, increasing controller output pressure would close the valve; therefore the controller must be indirect (reverse) acting so that a rise in measured temperature decreases controller output, opening the valve. Step-by-Step Solution: Select fail-open hardware: air-to-close valve on cooling line.Match loop direction: rising temperature → more cooling → valve should open.Choose reverse (indirect) acting controller so higher temperature yields lower output pressure and thus opens an air-to-close valve. Verification / Alternative check: Typical P&IDs for exothermic services show fail-open cooling valves with reverse acting temperature controllers to achieve the correct sign. Why Other Options Are Wrong: Air-to-open with direct acting: air failure would shut cooling (unsafe).Air-to-open with reverse acting: wrong fail-safe and may complicate loop direction.Air-to-close with direct acting: sign mismatch—hotter reactor would drive valve to close. Common Pitfalls: Overlooking fail-safe hardware intent; always design for safe state on instrument failure. Final Answer: Air-to-close valve with the controller indirect (reverse) acting

Question 14

Vacuum instrumentation: the McLeod gauge is specifically used to measure which of the following quantities in a system?

Accepted Answer

Introduction / Context: The McLeod gauge is a classic laboratory instrument for measuring low absolute pressures in the vacuum range. It compresses a known quantity of gas using a mercury column and infers the original pressure from Boyle’s law. Given Data / Assumptions: Applicable range: typically from about 1 Torr down to 10^-3 Torr (order of magnitude; depends on design). Clean, non-condensing gases only due to gas compression. Manual, non-continuous indication. Concept / Approach: By sealing and compressing a trapped gas sample, the McLeod gauge creates a measurable pressure and volume change. Using P1V1 = P2V2, the unknown low pressure (vacuum) is calculated. This method directly measures vacuum (very low absolute pressure), not velocity or flow. Step-by-Step Solution: Trap a known gas volume.Raise mercury to compress the gas to a measurable pressure.Apply Boyle’s law to compute the original vacuum pressure. Verification / Alternative check: Instrument manuals specify it as a primary standard-type gauge for vacuum calibration in the millitorr region. Why Other Options Are Wrong: Point velocity and flow rate: require anemometry or DP devices, not a McLeod gauge.Pressure (generic): while vacuum is a form of pressure, the instrument is specifically for low absolute pressure (vacuum) measurement. Common Pitfalls: Using McLeod on condensable vapors; readings become erroneous. Final Answer: Vacuum

Question 15

Nonlinear system behavior: with respect to steady states, a nonlinear dynamic process can exhibit how many distinct steady-state values under the same set of inputs and parameters?

Accepted Answer

Introduction / Context: Unlike linear systems, nonlinear processes can possess multiple equilibria (steady states) for the same external conditions. This has major implications for start-up, control design, and safety, especially in reactors and separation systems with exothermic/endothermic behavior and strong internal feedbacks. Given Data / Assumptions: Deterministic, time-invariant nonlinear model. Inputs and parameters are held fixed. Steady state defined by f(x*) = 0 (no time variation). Concept / Approach: Nonlinear algebraic equations can have multiple real solutions. Each solution corresponds to a steady state, which may be stable or unstable. Processes like CSTRs with heat release and heat removal often display S-shaped temperature–conversion curves, yielding three steady states (two stable, one unstable) for certain conditions. Therefore, the correct general statement is that a nonlinear system can have more than one steady state. Step-by-Step Solution: Form steady-state equations by setting derivatives to zero.Solve the resulting nonlinear algebraic equations; multiple roots may exist.Classify stability to understand which equilibria are attainable in practice. Verification / Alternative check: Phase-plane analysis or continuation methods demonstrate multiplicity of equilibria in many nonlinear models. Why Other Options Are Wrong: One/two/three: place an artificial cap; nonlinear systems can have any number, including none. Common Pitfalls: Assuming uniqueness from linear intuition; multiplicity requires careful operating procedures to avoid unwanted equilibria. Final Answer: More than one

Question 16

Low-temperature thermocouple selection: which of the following thermocouples is well suited for measuring about −50 °C with good sensitivity and stability?

Accepted Answer

Introduction / Context: Thermocouples are classified into noble-metal (high-temperature) and base-metal (general purpose, including low-temperature) types. Selecting a sensor for −50 °C requires attention to usable range, sensitivity (Seebeck coefficient), and stability at low temperatures. Given Data / Assumptions: Target temperature near −50 °C. Ambient industrial/lab environment without strong electromagnetic noise. Standard thermocouple types and common usage ranges. Concept / Approach: Type T (copper–constantan) thermocouples are specifically recommended for low temperatures (about −200 to +350 °C) with good stability and relatively high Seebeck coefficient in the sub-zero region, making them excellent for −50 °C. Type J (iron–constantan) typically specifies −40 °C as a lower bound in many datasheets, making −50 °C marginal. Type K (chromel–alumel) can reach −50 °C but has lower sensitivity and higher error at low temperatures compared with Type T. Noble-metal thermocouples (e.g., platinum–platinum+rhodium, Type S/R) are intended for high temperatures and are not preferred for −50 °C measurements. Step-by-Step Solution: Identify low-temperature-optimized thermocouple: Type T (Cu–Con).Compare ranges: T (−200 to 350) vs. J (−40 to 750) vs. K (−200 to 1250).Select copper–constantan for best performance near −50 °C. Verification / Alternative check: Calibration tables show Type T with strong, linear Seebeck response near −50 °C and good repeatability. Why Other Options Are Wrong: Platinum–platinum+rhodium: high-temp; poor choice for sub-zero work.Chromel–alumel: can measure −50 °C but with lower sensitivity; not the best choice.Iron–constantan: typical lower limit −40 °C; −50 °C is outside many standard specs. Common Pitfalls: Confusing “capable of” with “optimal”; Type T is preferred for accuracy and stability at −50 °C. Final Answer: Copper–constantan

Question 17

Photoelectric (optical) pyrometer application range: select the typical non-contact temperature range in °C over which photoelectric/brightness pyrometers are suitable in industrial practice.

Accepted Answer

Introduction / Context: Photoelectric (brightness) pyrometers determine temperature from the spectral brightness of hot objects, typically comparing radiation at a specified wavelength to a calibrated source and using optics plus photodetectors. They are used where contact sensors fail due to temperature, moving targets, or contamination. Given Data / Assumptions: Opaque targets with reasonably stable emissivity in the band of interest. Line-of-sight access and minimal intervening absorption. Industrial conditions with hot metals, furnaces, or glass streams. Concept / Approach: Brightness/optical pyrometers operate effectively at higher temperatures where thermal radiation is intense and visible/near-infrared detection is practical. A commonly cited usable span is about 800 to 2500 °C, covering many metallurgical and kiln processes. Below ~800 °C, emitted brightness becomes low for visible-band devices, and total-radiation or long-wavelength IR methods are preferred. Step-by-Step Solution: Identify that brightness pyrometry requires strong incandescent emission.Recall typical industrial span: approximately 800–2500 °C.Choose 800–2500 °C as the appropriate range. Verification / Alternative check: Manufacturer datasheets for optical/brightness pyrometers list lower limits near 700–800 °C depending on optics and detector sensitivity. Why Other Options Are Wrong: 400–1000 °C or 400–1600 °C: lower end too cool for standard brightness pyrometry without special detectors/filters.800–1600 °C: upper end too low; many applications exceed 1600 °C. Common Pitfalls: Ignoring emissivity; even within range, a poor emissivity estimate can bias temperature significantly. Final Answer: 800-2500

Question 18

Region of convergence for the Laplace transform of exp(a t) with a > 0 In continuous-time process control, the unilateral Laplace transform of f(t) = exp(a t) exists only for certain values of the complex variable s. For which condition on s is the …

Accepted Answer

Introduction / Context: Laplace transforms are a cornerstone of control systems and process dynamics. They convert time-domain signals into the complex s-domain, enabling straightforward algebraic manipulation of differential equations. However, the transform only exists where the defining integral converges. This question tests your understanding of the region of convergence for the classic exponential input f(t) = exp(a t) with a > 0. Given Data / Assumptions: Unilateral Laplace transform definition: F(s) = ∫0∞ f(t) * exp(−s t) dt. Function of interest: f(t) = exp(a t), with a > 0. We consider real s for convergence insight (extension to complex s uses Re{s}). Concept / Approach: Substituting f(t) = exp(a t) gives F(s) = ∫0∞ exp((a − s) t) dt. An improper integral of the form ∫ exp(α t) dt from 0 to ∞ converges only when α < 0. Therefore, we require a − s < 0, i.e., s > a. In complex analysis terms, the region of convergence is Re{s} > a. Only within this half-plane does the integral produce a finite result. Step-by-Step Solution: Start from F(s) = ∫0∞ exp(a t) exp(−s t) dt = ∫0∞ exp((a − s) t) dt.Recall convergence rule: ∫0∞ exp(β t) dt converges iff β < 0.Set β = a − s ⇒ require a − s < 0 ⇒ s > a.Evaluate integral under this condition: F(s) = 1 / (s − a) for s > a. Verification / Alternative check: Differentiate F(s) w.r.t. s or evaluate a numerical approximation for s slightly greater/less than a; you will see divergence as s approaches a from above, confirming the boundary Re{s} = a. Why Other Options Are Wrong: “the function is exponential” — true but irrelevant to convergence; many exponentials diverge.“initial values only for s > a” — the issue is integral convergence, not initial conditions.“piece-wise continuous only if s > a” — continuity pertains to f(t), not s.“s = a exactly” — at s = a the integral behaves like ∫ 1 dt, which diverges. Common Pitfalls: Confusing existence of a closed-form result with convergence; F(s) = 1/(s − a) is valid only where the integral converges. Also, forgetting that for general complex s the condition becomes Re{s} > a, not simply s > a on the real line. Final Answer: the Laplace transform integral of exp(a t) has finite value only when s > a

Question 19

Elastic pressure gauges: which sensing element form is never used as the primary elastic member?

Accepted Answer

Introduction / Context: Elastic pressure gauges convert fluid pressure into mechanical deflection using elastic elements. Common types include Bourdon tubes, diaphragms, and bellows. Recognizing which geometries are actually employed helps avoid design misconceptions and ensures correct selection for pressure range and media. Given Data / Assumptions: An “elastic element” must flex under pressure-induced stress and translate it into measurable motion. Industrial families: Bourdon tube (C-type, helical, spiral), diaphragm, bellows. We are selecting the form that is not used as the elastic pressure-sensing element. Concept / Approach: Bourdon tubes are curved hollow sections that uncoil with internal pressure. Diaphragms are thin circular membranes that deflect with differential pressure. Bellows are convoluted cylinders expanding axially under pressure. A simple “strip” (solid bar) does not provide the hollow, pressure-containing geometry required to translate internal pressure into useful elastic deformation for measurement; it lacks an internal cavity and suitable sensitivity. Step-by-Step Solution: List standard pressure elements: Bourdon, diaphragm, bellows.Evaluate “strip”: no internal pressurised volume; poor pressure-to-deflection transduction.Conclude “strip” is never used as the primary elastic pressure element. Verification / Alternative check: Instrument catalogs and standards (e.g., ASME/EN) classify pressure elements exclusively as Bourdon, diaphragm, or bellows (and variants), confirming that “strip” is not a category. Why Other Options Are Wrong: Bellows — widely used for low pressure/differential pressure.Diaphragm — standard for DP and low-pressure gages.Bourdon tube — most common for medium/high pressures.Helical Bourdon — a Bourdon variant for extended range/sensitivity. Common Pitfalls: Confusing temperature-sensing bimetal “strips” with pressure-sensing elastic elements; they are different instrument classes. Final Answer: strip

Question 20

Definition check: what is a “stable system” in control engineering (BIBO stability)?

Accepted Answer

Introduction / Context: Stability is the bedrock of controller design. The most common and practically meaningful definition for linear time-invariant (LTI) systems is bounded-input bounded-output (BIBO) stability. If a plant or closed loop is not BIBO-stable, even small input disturbances can lead to runaway outputs and unsafe operation. Given Data / Assumptions: We refer to continuous-time LTI systems under standard feedback. “Bounded” means the magnitude remains finite for all time. No special input class (sinusoid only, etc.) is assumed; any bounded input qualifies. Concept / Approach: A system is BIBO-stable if every bounded input produces a bounded output. This is equivalent (for LTI systems) to having all closed-loop poles strictly in the left half of the complex plane. Other statements (e.g., “servo conditions” or selective input classes) are neither necessary nor sufficient as a general definition. Step-by-Step Solution: Recall BIBO criterion: bounded input ⇒ bounded output.Recognise equivalence for LTI: closed-loop poles must have negative real parts.Choose the statement that exactly matches the BIBO definition. Verification / Alternative check: Nyquist and Routh–Hurwitz methods test whether all closed-loop poles lie in the left half-plane, ensuring BIBO stability and hence satisfying the definition. Why Other Options Are Wrong: Unbounded response to bounded input — by definition unstable.Servo conditions — concern tracking a reference, not the fundamental stability definition.“None of these” — incorrect because option (a) is precisely the definition.Sinusoidal-only statements — stability must hold for any bounded input class. Common Pitfalls: Confusing asymptotic stability with BIBO stability in nonlinear contexts; for LTI systems, BIBO and internal stability line up under standard conditions. Final Answer: for which the output response is bounded for all bounded input

Process Control and Instrumentation Questions