Question 1

Among the listed thermocouples, identify the pair capable of measuring the highest temperatures (extreme high-temperature applications under protective atmospheres).

Accepted Answer

Introduction / Context: Thermocouple upper limits depend on metallurgy and service environment. Noble-metal thermocouples operate reliably in oxidizing hot zones; refractory-metal combinations extend to even higher temperatures but require inert or vacuum atmospheres to avoid oxidation. Given Data / Assumptions: Comparison is for maximum attainable temperatures, not routine stability in air. Protective atmosphere or vacuum is available for refractory pairs. Standard industrial pairings are implied. Concept / Approach: Chromel–alumel (Type K) and iron–constantan (Type J) serve low-to-mid ranges. Platinum–rhodium (Types S/R/B) reach ~1600–1700°C in oxidizing conditions. Tungsten-based refractory-metal pairs (e.g., W–Re or W–Mo) can measure up to ~2300–2600°C under vacuum or inert gas, which surpasses the noble-metal limits. Thus, for absolute maximum temperature capability, tungsten–molybdenum class leads. Step-by-Step Solution: Rank the families by metallurgy and oxidation resistance at extreme heat.Note that refractory metals out-range noble metals under inert/vacuum conditions.Choose tungsten–molybdenum as the highest-temperature option among the list. Verification / Alternative check: High-temperature furnace instrumentation commonly uses W-based thermocouples for temperatures above noble-metal capabilities when oxygen is excluded. Why Other Options Are Wrong: Platinum–rhodium: excellent to ~1600°C, but below refractory limits.Chromel–alumel, iron–constantan: far lower maximum temperatures. Common Pitfalls: Assuming “best in air” equals “highest temperature overall.” Refractory pairs demand special atmospheres. Final Answer: Tungsten–molybdenum (refractory metal)

Question 2

Analytical instrumentation application: Gas chromatography (GC) is primarily used to measure which quantity in a sample stream under appropriate calibration and detection?

Accepted Answer

Introduction / Context: Gas chromatography separates volatile components based on differential partitioning between stationary and mobile phases. After separation, detectors quantify each component, enabling compositional analysis essential to process control, quality assurance, and emissions monitoring. Given Data / Assumptions: GC system with appropriate column and detector (e.g., FID, TCD). Calibration curves relate detector response to amount. Sample properly conditioned for volatility and stability. Concept / Approach: GC produces chromatographic peaks whose areas (or heights, with caution) are proportional to moles or mass of each component injected. With calibration, peak areas convert to concentration or mole fraction in the original stream. Temperature and pressure are controlled system variables, not the measurand. Flow rate is a carrier-gas or process parameter that influences retention but is not what the GC reports as the analytical result. Step-by-Step Solution: Identify GC’s function: separation + detection.Link detector response to calibrated concentration for each component.Select “Concentration” as the correct measured quantity. Verification / Alternative check: Typical GC outputs list peak areas and calculated wt%/mol%/ppm for each analyte following calibration standards. Why Other Options Are Wrong: Temperature/pressure: operating conditions, not analytical outputs.Flow rate: an instrument parameter; GC does not inherently measure process flow rate. Common Pitfalls: Confusing detector signal with absolute mass without calibration; quantitation requires standards and response factors. Final Answer: Concentration

Question 3

Materials of construction for thermistors: identify what thermistors are typically made of, given their large negative temperature coefficient and semiconductor behavior.

Accepted Answer

Introduction / Context: Thermistors are temperature-sensitive resistors widely used for precise temperature measurement, compensation, and control. Their hallmark is a strong, often negative, temperature coefficient of resistance (NTC), enabling high sensitivity over limited ranges compared with metal RTDs. Given Data / Assumptions: Focus on common NTC thermistors used in instrumentation. Behavior: semiconductor with exponential R–T characteristic. Construction involves sintered materials and passivation. Concept / Approach: Typical thermistors are formed from sintered mixtures of transition-metal oxides such as manganese, nickel, cobalt, copper, or iron oxides. These ceramic semiconductors exhibit strong temperature-dependent conductivity due to carrier concentration changes. In contrast, ultra-pure metals (e.g., platinum) are used in RTDs, not thermistors. Iron–copper or nickel–chromium are alloy systems for specific resistors or thermocouples (not thermistors). Step-by-Step Solution: Recognize thermistor behavior as ceramic semiconductor.Recall common compositions: Mn–Ni–Co–Cu oxide blends.Select “Metal oxides” as the correct material class. Verification / Alternative check: Manufacturer datasheets list compositions and sintering processes for NTC thermistors; their Beta constants characterize the R–T curve. Why Other Options Are Wrong: Ultra-pure metals: used for platinum RTDs (PT100), not thermistors.Iron–copper, nickel–chromium: metallic alloys for other sensors or heaters; not the basis of thermistors. Common Pitfalls: Confusing thermistors with metal RTDs or with thermocouples; each uses different materials and principles. Final Answer: Metal oxides

Question 4

Closed-loop dynamics: a unity-feedback system has an integrating process Gp(s) = 1/(2 s). The controller considered is purely integral with Gc(s) = 1/(T1 s). For a step change in set point, what qualitative response will the controlled variable exhi…

Accepted Answer

Introduction / Context: Controller–process pairings profoundly affect closed-loop behavior. An integrating process already contributes one pole at the origin. An integral controller adds another pole at the origin. The combined loop dynamics can lead to borderline stability and sustained oscillations if not shaped by proportional or derivative action. Given Data / Assumptions: Process: Gp(s) = 1/(2 s), unity feedback. Controller: Gc(s) = 1/(T1 s), integral only. Set-point (servo) step input. Concept / Approach: The open-loop transfer function L(s) = Gc(s)*Gp(s) = 1/(2 T1 s^2). The closed-loop characteristic equation is 1 + L(s) = 0 → 1 + 1/(2 T1 s^2) = 0. Multiplying by s^2 gives s^2 + 1/(2 T1) = 0. The roots are purely imaginary: s = ±j * sqrt(1/(2 T1)). Purely imaginary poles imply sustained oscillations with no decay or growth—an undamped response—under ideal linear assumptions. Step-by-Step Solution: Form the loop transfer L(s) = 1/(2 T1 s^2).Write the characteristic: 1 + L(s) = 0.Solve s^2 + 1/(2 T1) = 0 → purely imaginary roots.Conclude the step response is undamped (sustained oscillation). Verification / Alternative check: Root-locus from the origin for integral-only control on an integrator shows loci on the imaginary axis; adding proportional or derivative shifts poles to the left for damping. Why Other Options Are Wrong: Overdamped/underdamped: require negative real parts; not the case here.Unstable: would require positive real part poles; here they are purely imaginary (marginally stable). Common Pitfalls: Calling marginal (sustained) oscillation “unstable.” In linear theory, it is neutrally stable but practically unacceptable. Final Answer: Undamped response

Question 5

Frequency-response fundamentals: a Bode diagram consists of which plots for a linear system's transfer function when expressed versus frequency?

Accepted Answer

Introduction / Context: Bode diagrams are the cornerstone of single-input single-output frequency-response analysis. They allow engineers to read gain margins, phase margins, and asymptotic behavior directly from log-scale plots, aiding robust control design and filter synthesis. Given Data / Assumptions: AR denotes amplitude ratio (magnitude) of the transfer function. Φ denotes phase angle in degrees (or radians). Frequency axis is logarithmic to span decades efficiently. Concept / Approach: A complete Bode plot has two subplots sharing the same log-scale frequency axis: the magnitude (in decibels, 20 log10|AR|) versus log(f), and the phase Φ versus log(f). Using a logarithmic frequency axis linearizes slope interpretations and makes corner frequencies and asymptotes easy to visualize across wide ranges. Step-by-Step Solution: Identify required axes: log frequency for both subplots.Plot magnitude as log of amplitude ratio (typically in dB) vs log(f).Plot phase angle vs log(f) on a second subplot. Verification / Alternative check: Classical control texts show straight-line asymptotes for poles/zeros on the magnitude plot and characteristic phase transitions around break frequencies. Why Other Options Are Wrong: log(AR) vs f or AR vs log(f): mixes linear and log incorrectly; standard is log–log for magnitude and log frequency for phase.None of these: incorrect because the standard definition is provided. Common Pitfalls: Forgetting to convert magnitude to decibels; while raw |AR| can be plotted, the “Bode magnitude” is conventionally in dB on a log frequency axis. Final Answer: log(AR) vs log(f) and Φ vs log(f)

Question 6

For a first-order process characterized by a resistance R and a capacitance C (electrical or analogous thermal/fluid storage), what is the time constant τ expressed in terms of R and C?

Accepted Answer

Introduction / Context: First-order dynamics appear in RC circuits, well-mixed tanks with linear resistances, and thermal systems with lumped capacitance. The time constant τ determines how quickly the system output approaches its new steady state after a step input. Given Data / Assumptions: Linear first-order model with a single energy/mass storage element. R denotes resistance to flow/heat/electric current. C denotes capacitance (storage capacity). Concept / Approach: In standard form, the first-order differential equation is τ dy/dt + y = K u, where τ is the time constant. For an RC circuit, τ = R * C in seconds. This generalizes via analogies: hydraulic (R = flow resistance, C = tank capacity), thermal (R = thermal resistance, C = heat capacity), etc. The product RC sets the exponential rise/decay speed: y(t) reaches about 63.2% of its final value in one τ. Step-by-Step Solution: Write the governing equation for an RC charge/discharge: v(t) = V(1 - e^{-t/(RC)}).Identify τ as the denominator in the exponential → τ = RC.Conclude that the time constant equals the product of resistance and capacitance. Verification / Alternative check: Dimensional analysis: [R]*[C] gives seconds in both electrical and analogous domains, confirming consistency. Why Other Options Are Wrong: R + C or R - C: mixing unlike units; not dimensionally valid.1/RC: appears in the exponent numerator; τ is the reciprocal of that coefficient. Common Pitfalls: Confusing τ with 1/(RC); the exponential form must be carefully read to avoid inversion errors. Final Answer: RC

Question 7

Chemical process nonlinearity: choose the system that typically exhibits nonlinear dynamic behavior due to temperature–rate coupling and energy balance effects.

Accepted Answer

Introduction / Context: Nonlinearity in chemical process models often arises from reaction-rate temperature dependence (Arrhenius) and multiplicative terms in mass/energy balances. Recognizing inherently nonlinear units is essential for selecting appropriate control strategies and for safe operation near multiple steady states or thermal runaway regions. Given Data / Assumptions: CSTR stands for continuous stirred-tank reactor. Isothermal means reactor temperature is held constant; non-isothermal couples species and energy balances. Reaction rates typically follow k = k0 * exp(-E/RT), introducing strong nonlinearity. Concept / Approach: In a non-isothermal CSTR, the energy balance links temperature to heat generation (reaction) and heat removal (cooling). Because reaction rate depends exponentially on temperature, small temperature changes strongly affect conversion, which in turn changes heat release—a feedback that creates nonlinear dynamics, possible multiple steady states, and oscillations. In contrast, an isothermal CSTR with a first-order reaction can often be linearized to an LTI form. An ideal mixer without reaction is typically linear in concentration with constant density assumptions. Step-by-Step Solution: Identify Arrhenius dependence as the core nonlinearity source.Note coupling of mass and energy balances in non-isothermal operation.Select “Non-isothermal CSTR” as the nonlinear example. Verification / Alternative check: Phase-plane and bifurcation analyses for exothermic CSTRs show S-shaped steady-state curves and ignition/extinction points—canonical nonlinear phenomena. Why Other Options Are Wrong: Isothermal CSTR: with fixed temperature and simple kinetics, behavior is often linear (or easily linearized).Mixer: linear mass balance if density and volume are constant and no reaction occurs.None of these: incorrect because a clear nonlinear system is provided. Common Pitfalls: Assuming all reactors are equally nonlinear; temperature coupling is the key differentiator here. Final Answer: Non-isothermal CSTR

Question 8

Radiation pyrometry: select the condition under which accurate temperature measurement performance of a radiation pyrometer is not adversely affected among the listed scenarios.

Accepted Answer

Introduction / Context: Radiation pyrometers infer temperature from thermal radiation. Accuracy depends on emissivity knowledge, optical path integrity, and background reflections. Recognizing which conditions degrade accuracy and which conditions are benign helps practitioners set up reliable non-contact measurements. Given Data / Assumptions: Pyrometer aligned to a target with some background in the field of view. Sighting path may include gases, dust, or windows. Object emissivity may vary with surface state and wavelength. Concept / Approach: Reflected background radiation is a source of error when the surroundings differ significantly in temperature from the target. If the object and surroundings are at nearly the same temperature, the reflected component has nearly the same radiance as the object's own emission, so the measurement error due to reflection becomes minimal. By contrast, attenuation or enhancement in the sighting path (e.g., due to absorbing or emitting gases/particles) alters the received radiance; varying emissivity changes the proportionality between temperature and radiance; and transparency means the instrument sees through to the background, all of which can cause significant errors. Step-by-Step Solution: List potential error sources: path attenuation/emission, emissivity uncertainty, reflections, transparency.Identify the special case where reflections are benign: surroundings at same temperature as object.Select the condition that does not adversely affect accuracy. Verification / Alternative check: Radiation thermometry guidance notes that reflected radiation bias is minimized when background and target temperatures match, reducing net error. Why Other Options Are Wrong: Enhancement/attenuation in path: changes the signal before it reaches the detector → error.Varying emissivity: invalidates fixed-emissivity calibration → error.Transparent object: instrument may “see” background instead of the object → error. Common Pitfalls: Ignoring the field-of-view and reflective surfaces; shielding or emissivity coatings can mitigate many issues. Final Answer: Object and surroundings are at almost the same temperature

Question 9

Thermocouple alloys — identify the exception: Which of the following commonly used thermocouple materials does not contain nickel in its composition?

Accepted Answer

Introduction / Context: Industrial temperature measurement often uses thermocouples made from specific metal alloys. Knowing the principal constituents of these alloys helps with selection, compatibility, and troubleshooting. This question tests recognition of nickel-containing thermocouple materials and asks you to spot the exception—if any—among standard names used in plants and labs. Given Data / Assumptions: Alumel, Chromel, and Constantan are standard thermocouple or lead wire alloys. The task is to determine which listed material lacks nickel in its composition. We consider conventional compositions used in K-, J-, and T-type systems. Concept / Approach: Recall the compositions: Chromel is primarily nickel with chromium (typical K-type positive leg). Alumel is a nickel-based alloy containing aluminum (plus manganese/silicon) and serves as the negative leg for K-type. Constantan is a copper–nickel alloy (about 55% Cu and 45% Ni) widely used with copper (T-type) or iron (J-type) in some configurations. Since all three contain nickel, none of them is nickel-free. Step-by-Step Solution: Identify Chromel → Ni + Cr → contains nickel.Identify Alumel → Ni-rich with Al (and minor Mn/Si) → contains nickel.Identify Constantan → Cu + Ni → contains nickel.Therefore, among the listed choices, there is no nickel-free item → choose “None of these.” Verification / Alternative check: Thermocouple reference tables list K-type as Chromel–Alumel (both Ni-based), and Constantan’s shorthand (Cu–Ni) confirms nickel content. This cross-check eliminates ambiguity. Why Other Options Are Wrong: Alumel — nickel-based, so not nickel-free.Chromel — nickel–chromium alloy, contains nickel.Constantan — copper–nickel alloy; contains nickel.Copper (extra distractor) — not a thermocouple alloy name given in the stem list and not correct for the asked set. Common Pitfalls: Misremembering Constantan as copper-only. It is consistently a copper–nickel alloy, hence does include nickel. Final Answer: None of these

Question 10

Control systems — unit step response at t = 0⁺: For the transfer function G(s) = (s + 1) / (2s + 1), what is the initial value (at t = 0⁺) of the unit step response?

Accepted Answer

Introduction / Context: Evaluating the very first value of a time response is a common check in control engineering. The initial value theorem provides a fast way to compute y(0⁺) directly from the Laplace-domain representation without fully inverting transforms. This problem reinforces comfort with that theorem for a stable, proper transfer function driven by a unit step. Given Data / Assumptions: Transfer function: G(s) = (s + 1) / (2s + 1). Input: unit step → U(s) = 1 / s. System is linear time-invariant; initial rest is assumed. Concept / Approach: For a unit step input, Y(s) = G(s) * 1/s. The initial value theorem states y(0⁺) = lim_{s→∞} s * Y(s) provided poles are in the left half-plane and the limit exists. This avoids full partial fractions and time-domain inversion for the initial value alone. Step-by-Step Solution: Form the output transform: Y(s) = [(s + 1)/(2s + 1)] * (1/s).Apply the theorem: y(0⁺) = lim_{s→∞} s * Y(s) = lim_{s→∞} (s + 1) / (2s + 1).Evaluate the limit by dividing numerator and denominator by s: lim_{s→∞} (1 + 1/s) / (2 + 1/s) = 1 / 2.Therefore, the initial value is 1/2. Verification / Alternative check: One can also compute the full step response via partial fractions to see the exponential term vanish at t = 0⁺, leaving the same initial value of 0.5. Both approaches agree. Why Other Options Are Wrong: 0 — would require numerator to be of lower order in s at infinity; not the case here.1 — corresponds to lim_{s→∞} (s + 1)/(s + 1), which is not our ratio.2 — larger than the high-frequency gain; inconsistent with the computed limit.Undefined — the limit exists and equals 0.5. Common Pitfalls: Forgetting to multiply Y(s) by s in the initial value theorem, or mistakenly applying the final value theorem for y(0⁺). Ensure stability assumptions hold when using these theorems. Final Answer: 1/2

Question 11

Remote indication — choose the unsuitable thermometer: Which of the following is not suitable for distant (remote) reading over lengths up to about 60 metres in industrial practice?

Accepted Answer

Introduction / Context: Many temperature instruments are installed at process equipment while indications must be read from a control room tens of metres away. Certain technologies are inherently suitable for remote indication via capillaries or electrical leads, while others require local visual reading. This question checks awareness of which is not appropriate for distant reading up to ~60 m. Given Data / Assumptions: Distance to indicator is on the order of tens of metres. No special optical viewing system is assumed. Standard industrial mounting and cabling/capillary use cases apply. Concept / Approach: Mercury-in-glass thermometers require direct visual inspection of the mercury column at the bulb’s location, unless paired with special optical devices, which are not implied here. In contrast, filled-system (vapour pressure) thermometers can drive remote dials over long capillaries; RTDs transmit electrical signals over cable lengths of this scale; constant-volume gas systems can also be configured with remote manometry. Hence the traditional mercury-in-glass type is unsuitable for routine 60 m remote readout. Step-by-Step Solution: Identify instruments that inherently support remote indication (capillary or wires).Note that mercury-in-glass requires local observation of the glass stem scale.Conclude it is not suitable for the specified remote distance. Verification / Alternative check: Plant standards frequently prefer RTDs/thermocouples or filled systems when the operator panel is far from the measuring point, underscoring the limitation of simple glass thermometers. Why Other Options Are Wrong: Vapour-pressure — designed for remote dials via capillary.Constant-volume gas — transmits pressure to a remote indicator.RTD — uses electrical leads; readily read at long distances.Bimetallic remote dial — also uses a capillary/remote head arrangement. Common Pitfalls: Assuming that any thermometer can be “made remote” without additional mechanisms. The glass stem type is inherently local-read unless augmented with special optics. Final Answer: Mercury-in-glass thermometer

Question 12

Open-channel metering — small channels: For measuring the liquid flow rate in a small open channel, which primary measuring device is the most appropriate?

Accepted Answer

Introduction / Context: Measuring flow in open channels requires devices that relate upstream head to discharge under free-surface conditions. For small channels in civil/chemical process settings—lab flumes, plant trenches—simplicity, accuracy, and ease of installation are key considerations. This item compares common instruments and selects the best match for a small open-channel scenario. Given Data / Assumptions: Flow is free-surface (open channel), not pressurised pipe flow. Channel is small; permanent head loss and cost should be modest. Steady, subcritical flow is assumed near the device. Concept / Approach: Weirs (sharp-crested, V-notch, rectangular) are standard for small-channel measurement. The discharge Q relates to upstream head h via empirical formulae, allowing practical calibration. Pitot tubes, venturimeters, or orifice plates are designed for closed, pressurised conduits; vane meters likewise suit pipelines, not open channels. Therefore, a weir is the preferred choice here. Step-by-Step Solution: Recognise open-channel condition → need head–discharge relation over a crest.Identify weir types (e.g., 90° V-notch) suited to small discharges.Reject pipeline-only instruments (pitot, venturi, vane, orifice) for open channels. Verification / Alternative check: Standard hydraulics texts list V-notch weirs as highly sensitive for low flows, exactly the use case for small channels. Why Other Options Are Wrong: Pitot/venturi/orifice/vane — rely on closed-conduit pressure/velocity relations; not intended for open channels without modification. Common Pitfalls: Confusing “small channel” with “small pipe” and defaulting to orifice/venturi meters, which are inappropriate for free-surface flow. Final Answer: Weir

Question 13

Pressure measurement fundamentals: A barometer is an instrument that measures which kind of pressure?

Accepted Answer

Introduction / Context: Correctly distinguishing absolute, gauge, and differential pressure is essential when selecting instruments and interpreting readings. The classic barometer is one of the earliest pressure-measuring devices and establishes an absolute reference using a vacuum space above a column of fluid (mercury in traditional designs). Given Data / Assumptions: Barometer reading is referenced to a near vacuum (Torricellian vacuum) rather than atmospheric pressure. No dynamic (velocity) pressure component is implied. Steady ambient conditions are assumed. Concept / Approach: Absolute pressure is measured relative to perfect vacuum. A barometer balances atmospheric pressure against a column of liquid in a sealed tube with vacuum above. The height directly reflects the absolute atmospheric pressure. In contrast, gauge pressure instruments (e.g., Bourdon gauges) compare process pressure to the surrounding atmosphere. Dynamic pressure pertains to kinetic energy per unit volume of a moving fluid and is not what a barometer senses. Step-by-Step Solution: Identify reference: barometer uses a vacuum reference → absolute.Recognise gauge pressure would require atmospheric reference → not a barometer.Exclude dynamic and “both” as inconsistent with barometer construction. Verification / Alternative check: Standard definitions: P_absolute = P_gauge + P_atmospheric. A barometer yields P_atmospheric directly as an absolute value. Why Other Options Are Wrong: Gauge — measures relative to atmosphere (e.g., Bourdon gauge), not a barometer.Both — barometer does not toggle references.Dynamic — relates to velocity head; barometer is static.Differential only — not applicable; barometer’s reference is vacuum. Common Pitfalls: Assuming all industrial dials are “gauge” and misclassifying barometric readings; remember the vacuum reference. Final Answer: absolute

Question 14

Non-contact high-temperature measurement: Compared with thermocouples, which statement best characterises radiation pyrometers used in industry?

Accepted Answer

Introduction / Context: Temperature sensors fall into contact (thermocouples, RTDs) and non-contact (radiation/optical pyrometers) categories. Selection depends on temperature level, response needs, accessibility, and process environment. This item asks for the most distinguishing advantage of radiation pyrometers relative to thermocouples. Given Data / Assumptions: Typical industrial use over a wide temperature span. Direct line-of-sight is available for the pyrometer. Comparison focuses on principal capability rather than niche models. Concept / Approach: Radiation pyrometers infer temperature from thermal radiation, enabling measurement at very high temperatures beyond the safe or practical limits of contact sensors. They are also suitable for moving or rotating targets since no contact is required. While response speed can vary by design, modern pyrometers are often very fast; their headline advantage remains access to higher temperature ranges and non-contact operation. Step-by-Step Solution: Identify thermocouple limits: drift/oxidation and sheath limits constrain the top measurable temperature.Recognise pyrometer advantage: non-contact technique tolerates incandescent/high-temperature targets.Select the statement capturing this key benefit: “can measure higher temperature.” Verification / Alternative check: Specifications for optical/IR pyrometers routinely extend well above 1000°C, where many thermocouples face life or accuracy limitations without elaborate protection. Why Other Options Are Wrong: Slower response — not universally true; many pyrometers are very fast.Cannot measure moving objects — incorrect; non-contact is ideal for moving targets.More affected by corrosive atmosphere — false; non-contact and remote optics reduce exposure.Require physical contact — the opposite of how pyrometers work. Common Pitfalls: Overgeneralising response time; model specifics matter. The most robust, broadly true differentiator is the higher measurable temperature range with non-contact operation. Final Answer: can measure higher temperature

Question 15

Process control concepts — spot the incorrect statement: Which of the following statements is wrong in the context of common modelling and control choices for process units?

Accepted Answer

Introduction / Context: Control engineers routinely classify systems as lumped-parameter (finite states described by ordinary differential equations) or distributed-parameter (fields described by partial differential equations). They also match controllers to process characteristics. This question checks conceptual clarity by asking you to pick the incorrect statement among mostly correct assertions. Given Data / Assumptions: CSTR refers to a well-mixed reactor approximation. Level control practice typically uses proportional (or PI) controllers. Distributed-parameter modelling involves spatial variation. On–off control switches output and is non-linear. Concept / Approach: A CSTR, by definition, assumes perfect mixing, making its dominant dynamics lumped and represented by ordinary differential equations. Conversely, distributed-parameter systems (e.g., heat conduction in a rod) require partial differential equations. Therefore, claiming a CSTR is distributed is wrong in the usual modelling context. The other statements reflect standard practice and definitions. Step-by-Step Solution: Check (a): Proportional/PI level control is indeed common → correct.Check (b): CSTR is lumped, not distributed → this is the wrong statement.Check (c): Distributed models give PDEs → correct.Check (d): On–off is non-linear → correct.Check (e): “Always linearises any process completely” is an overstatement, but it is not part of the original four; the selected wrong item remains (b). Verification / Alternative check: Standard reactor dynamics texts derive CSTR mass and energy balances as ODEs under the well-mixed assumption, consistent with a lumped-parameter model. Why Other Options Are Wrong (if chosen): (a) Would be wrong to reject; proportional/PI level control is industry standard.(c) Rejecting this would contradict the mathematics of distributed models.(d) Rejecting this would ignore the inherent switching non-linearity. Common Pitfalls: Confusing “large volume” with “distributed.” A big tank is still lumped if well-mixed; spatial gradients are neglected by design. Final Answer: A continuously stirred tank reactor (CSTR) can be considered as a distributed-parameter system.

Question 16

Filled-system expansion thermometers — lowest temperature capability: Which of the following filled-system/expansion thermometers can measure to the lowest temperatures reliably?

Accepted Answer

Introduction / Context: Expansion thermometers rely on the thermal expansion of a contained medium. The choice of medium sets the usable temperature range. For very low temperatures, the working fluid must remain liquid and responsive without freezing or becoming too viscous. This question examines which common type reaches the lowest temperatures in practice. Given Data / Assumptions: Compare standard media: mercury, alcohols, liquid metals. Focus is on the lower limit of measurement. Glass or steel construction affects durability but not the fundamental freezing point. Concept / Approach: Mercury freezes at about −39°C, so mercury-based thermometers cannot measure far below this. Alcohols (e.g., ethanol, toluene-based fluids) remain liquid well below −50°C (certain mixtures near −100°C), permitting significantly lower temperature measurement. Sodium or potassium metal thermometers target high-temperature ranges, not low. Thus, alcohol-in-glass thermometers are preferred for very low temperatures among the listed options. Step-by-Step Solution: Assess low-temperature limits: alcohol < mercury << fused metals (high-temp only).Recognise container material (glass vs steel) does not change the medium’s freezing point.Select alcohol-in-glass as the type capable of lowest temperatures. Verification / Alternative check: Calibration charts for laboratory thermometers show alcohol-filled devices reading well below −50°C, confirming their suitability for low-temperature work. Why Other Options Are Wrong: Mercury-in-glass/steel — limited by mercury’s freezing point around −39°C.Fused metal (Na/K) — used for high temperatures; not a low-temperature instrument.Gas-filled (N2) — useful but typically not as low-reaching as specialty alcohol mixtures among the common expansion types listed. Common Pitfalls: Assuming “steel” inherently expands the range; the fill medium dominates the temperature limits. Final Answer: Alcohol-in-glass thermometer

Question 17

Gas constant-volume thermometry — hydrogen range: What is the typical operating temperature range (°C) for a hydrogen gas constant-volume thermometer?

Accepted Answer

Introduction / Context: Gas thermometers form the basis of thermodynamic temperature scales because gas properties can be related accurately to temperature. In constant-volume gas thermometry, pressure changes with temperature are measured. Different gases are chosen for different ranges to maintain sensitivity and avoid condensation or chemical reactivity issues. This item asks for the typical range for a hydrogen constant-volume thermometer (CVGT). Given Data / Assumptions: Hydrogen is used as the working gas at suitable pressures. We refer to practical ranges commonly cited in instrumentation texts. Temperature is expressed in degrees Celsius. Concept / Approach: Hydrogen remains stable and gives good sensitivity at elevated temperatures and is often assigned to the higher portion of gas-thermometer service, whereas helium/nitrogen are used differently depending on the low-temperature needs. A commonly taught practical range for a hydrogen CVGT is about 100 to 500°C, covering the mid-to-high span where hydrogen behaves reliably and the apparatus materials can be managed safely. Step-by-Step Solution: Recall that water vapour and mercury thermometers cover low-to-moderate ranges; gas thermometers extend range and accuracy.Associate hydrogen CVGT with elevated-temperature service in many references.Select 100 to 500°C as the representative range among the options. Verification / Alternative check: Tabulated instrument ranges in classic metrology sources indicate hydrogen’s use in higher-temperature gas thermometry, roughly spanning hundreds of degrees Celsius. Why Other Options Are Wrong: −10 to 20 and 0 to 100 — too low; simpler liquid-in-glass instruments cover these ranges.400 to 1000 — pushes beyond typical CVGT apparatus limits for routine use.−200 to −50 — belongs to cryogenic ranges better served by helium or special setups. Common Pitfalls: Confusing constant-volume gas thermometers with resistance/thermocouple ranges; gas thermometers are reference-grade but not universally used due to practical constraints. Final Answer: 100 to 500

Question 18

First-order systems under sinusoidal excitation: When a sinusoidal input passes through a linear first-order system, how does the output amplitude and frequency change?

Accepted Answer

Introduction / Context: The frequency response of linear time-invariant (LTI) systems is foundational for control and signal processing. A first-order process behaves like a low-pass filter. Understanding what such a system does to a sinusoidal input—without solving differential equations every time—helps predict attenuation and phase lag across frequencies. Given Data / Assumptions: System is linear and time-invariant, first order with transfer function form K / (1 + sτ) or similar. Input is sinusoidal at some angular frequency ω. Steady-state sinusoidal response is considered. Concept / Approach: For LTI systems, a sinusoidal input at frequency ω produces a sinusoidal output at the same frequency, scaled in magnitude by |G(jω)| and shifted in phase by ∠G(jω). For a first-order low-pass, |G(jω)| = K / sqrt(1 + (ωτ)^2), which is ≤ K, showing attenuation that increases with ω. The frequency does not change; only amplitude is reduced and phase lags. Step-by-Step Solution: Represent the input as sin(ωt) and evaluate the steady-state output using G(jω).Compute magnitude ratio: |G(jω)| = K / sqrt(1 + (ωτ)^2) → less than or equal to K.Conclude amplitude decreases with increasing ω; frequency remains ω (unchanged). Verification / Alternative check: Bode plots of a first-order system show a 0 dB plateau at low ω and a −20 dB/decade roll-off past the corner at 1/τ, confirming attenuation at higher frequencies. Why Other Options Are Wrong: Frequency increases/decreases — LTI systems do not change input frequency.Gets amplified — a passive first-order low-pass does not amplify; it attenuates above the corner.Frequency doubles while amplitude unchanged — violates linear frequency invariance. Common Pitfalls: Confusing transient behaviour with steady-state sinusoidal response; only amplitude and phase change at the same frequency in steady state. Final Answer: Gets attenuated (magnitude decreases)

Question 19

Control systems – classify inputs by boundedness From the following standard test inputs (given by their Laplace transforms), identify which input is unbounded as time increases: 1, 1/S, 1/S^2, and 1/(S^2 + 1).

Accepted Answer

Introduction / Context: In control engineering, standard test inputs—impulse, step, ramp, and sinusoid—are used to probe system behavior. A key classification is whether an input is bounded in amplitude over time. This question asks you to spot which of the commonly used inputs grows without limit as time progresses. Given Data / Assumptions: Impulse input: Laplace transform 1. Step input: Laplace transform 1/S. Ramp input: Laplace transform 1/S^2. Unit sine input: Laplace transform 1/(S^2 + 1). Boundedness refers to the magnitude of the time-domain signal for t ≥ 0. Concept / Approach: An input is called bounded if |u(t)| remains finite for all t ≥ 0. The ramp increases linearly with time (u(t) = t for t ≥ 0), so it is unbounded. The step is a constant (bounded). The sinusoid oscillates between fixed limits (bounded). The ideal impulse, though having a Dirac spike at t = 0, is a generalised function with finite area; in the usual boundedness discussion for test signals, “unbounded over time” identifies signals whose magnitude grows without limit as t increases—namely the ramp. Step-by-Step Solution: Map Laplace forms to time-domain signals: 1 → impulse, 1/S → step, 1/S^2 → ramp, 1/(S^2+1) → sine.Check time behavior: step = constant (bounded); sine = oscillatory (bounded); ramp = grows linearly (unbounded).Select 1/S^2 as the unbounded input. Verification / Alternative check: Plot u(t) = t, u(t) = 1, and u(t) = sin(t). Only u(t) = t diverges as t → ∞, confirming unboundedness. Why Other Options Are Wrong: 1/S: Step is finite for all t. 1/(S^2+1): Sinusoid has bounded amplitude. 1: Ideal impulse is a distribution with finite area; the question targets signals that grow without limit over time, which the impulse does not. Common Pitfalls: Confusing mathematical idealisations (impulse spike) with long-time unbounded growth; unbounded “over time” points to ramp behavior. Final Answer: 1/S^2

Question 20

Thermometry basics – which temperature scale assigns 0° to the ice point and 80° to the steam point at standard pressure?

Accepted Answer

Introduction / Context: Multiple historical temperature scales exist in thermal engineering. While SI practice relies on kelvin and Celsius, older or regional literature sometimes references the Réaumur (Reaumur) and Fahrenheit scales. Recognising their fixed points prevents misinterpretation of legacy data, calibration plates, or examination items. Given Data / Assumptions: “Ice point” means the equilibrium temperature of pure ice and water at 1 atm. “Steam point” means the equilibrium temperature of saturated water and steam at 1 atm. We seek the scale that assigns 0° at the ice point and 80° at the steam point. Concept / Approach: By definition, the Reaumur scale sets 0°R (Re) at the ice point and 80°R (Re) at the steam point, creating an 80-division interval. The Celsius scale assigns 0 °C and 100 °C to the same points. Fahrenheit assigns 32 °F and 212 °F. Rankine is an absolute scale with degree size equal to Fahrenheit degrees and zero at absolute zero, not at the ice point. Step-by-Step Solution: List fixed points: Reaumur 0–80, Celsius 0–100, Fahrenheit 32–212, Rankine 491.67–671.67 for the ice/steam points (approx.).Match the “0 and 80” requirement with Reaumur.Select “Reaumur.” Verification / Alternative check: Conversion formulas: C = (5/4) Re; hence Re = (4/5) C, consistent with 100 °C → 80 °Re. Why Other Options Are Wrong: Celsius: Steam point is 100 °C, not 80. Fahrenheit: Fixed points 32 and 212. Rankine: Absolute scale; does not place 0 at the ice point. Common Pitfalls: Spelling confusions (Celcius/Farenhite/Reumur) and mixing up absolute vs relative scales. Final Answer: Reaumur

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