Question 1

Filled-system (pressure spring) thermometers: Vapour-actuated pressure spring thermometers do not require ambient temperature compensation. In gas or liquid-expansion pressure spring thermometers, ambient compensation is achieved primarily by which …

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Introduction / Context:Filled-system thermometers translate temperature into pressure using a bulb, capillary, and receiving element (Bourdon/bellows). Gas and liquid expansion systems are sensitive to ambient temperature because the entire fill volume expands or contracts. Good instrument design mitigates ambient effects so the indicated temperature primarily reflects the bulb environment.Given Data / Assumptions:Vapour-pressure systems are nearly self-compensated since pressure depends on saturation at the bulb.Gas/liquid systems require design-based compensation.We consider geometric volume choices for bulb, capillary, and receiver.Concept / Approach:The measured pressure change is proportional to the fraction of total fill volume within the bulb exposed to process temperature. By making the bulb volume large and minimizing volumes of capillary and receiving element, the percentage of fill impacted by ambient conditions is reduced, lowering ambient-induced errors. Thus, all three strategies contribute to compensation.Step-by-Step Solution:Let V_total = V_bulb + V_cap + V_recv.Signal sensitivity ∝ V_bulb / V_total; ambient error ∝ (V_cap + V_recv) / V_total.Increase V_bulb; minimize V_cap and V_recv → improved ambient rejection.Verification / Alternative check:Handbooks specify capillaries with small internal diameter and compact receiving elements; catalogues highlight “large bulb volume” designs for better accuracy.Why Other Options Are Wrong:Any single option alone is helpful but insufficient; best practice applies all three simultaneously.Thicker dial glass does nothing to compensate thermodynamic volumes.Common Pitfalls:Overlooking capillary routing through hot zones; even with good volumes, poor routing can reintroduce ambient errors.Final Answer:All (a), (b) & (c)

Question 2

Optical pyrometry reference point: In brightness (optical) pyrometry, high-temperature measurements are commonly referenced above the metal fixed point at 1063°C, known as the __________ point.

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Introduction / Context:Optical pyrometers measure high temperatures by comparing the brightness of a target to that of a calibrated filament at a selected wavelength. Calibration often uses internationally agreed fixed points—melting/freezing temperatures of pure metals—providing reproducible references. This question asks you to identify the fixed point at 1063°C used in classic pyrometry.Given Data / Assumptions:Fixed-point metal at 1063°C (approximately).Application context: optical/brightness pyrometry.We assume standard ITS-90 fixed points used in calibration chains.Concept / Approach:The melting point of gold is 1063°C (more precisely 1064.18°C under ITS-90), well known as the “gold point.” Optical pyrometers traditionally reference temperatures at or above the gold point because targets exhibit visible incandescence and brightness techniques are accurate in this range.Step-by-Step Solution:Recall common fixed points: zinc (~419.5°C), aluminum (~660.3°C), silver (~961.8°C), gold (~1063–1064°C).Match the stated 1063°C to gold’s melting point.Therefore, the blank corresponds to the gold point.Verification / Alternative check:Calibration tables for brightness pyrometers list the gold point as a primary fixed point above which optical methods are mainstream.Why Other Options Are Wrong:Antimony, silver, nickel, tin: Their fixed points do not align with 1063°C; silver is ~962°C, nickel is ~1455°C.Common Pitfalls:Confusing the silver point (~962°C) with gold; remember the mnemonic “Ag ~ 962°C, Au ~ 1063°C.”Final Answer:Gold

Question 3

Controller “stabilising time” (settling convention): In classical instrumentation usage, the stabilising time for a controller is the time required for the response to reach approximately what percentage of its ultimate value?

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Introduction / Context:Practitioners often use practical time markers to describe how quickly a system “settles.” While modern control defines settling time as reaching and remaining within a small band (for example 2% or 5%), older instrumentation texts and many plant heuristics use percentages of the final value corresponding to multiples of the dominant time constant for a first-order response.Given Data / Assumptions:First-order-like response is implied.We want the conventional percentage linked to “stabilising time.”Ultimate (final) value is well defined for the step.Concept / Approach:For a first-order system y(t) = y∞(1 − e^(−t/τ)), standard waypoints are 63.2% at 1τ, 86.5–87% at 2τ, and about 95% at 3τ. Many instrumentation references call roughly three time constants the “stabilising time,” because the response appears practically complete for operators. Thus, 95% is the expected answer here.Step-by-Step Solution:At t = τ: y ≈ 63.2% of final.At t = 2τ: y ≈ 86.5–87% of final.At t = 3τ: y ≈ 95% of final → practical stabilisation in instrumentation jargon.Verification / Alternative check:Many vendor response specifications use “3τ ≈ 95%” as a quick estimate of the time to reach essentially final value for step testing of sensors and transmitters.Why Other Options Are Wrong:63.2 and 87.5 correspond to 1τ and 2τ respectively, earlier than typical stabilisation.100 is unattainable in finite time for an exponential approach.Common Pitfalls:Confusing the strict control definition of settling within a band (2%/5%) with the heuristic “stabilising time” used on shop floors; both are useful but not identical.Final Answer:95

Question 4

Mercury-in-glass thermometer principle: The basic operating principle of a mercury-in-glass thermometer is primarily based on which property change of mercury with temperature?

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Introduction / Context:Liquid-in-glass thermometers are among the oldest and most reliable temperature indicators. Mercury is often used because it does not wet glass, expands uniformly over a wide range, and is visible against the scale. This question focuses on the fundamental physical effect that enables the reading.Given Data / Assumptions:A sealed capillary and bulb contain mercury.Pressure is nearly constant (atmospheric or sealed reference) during normal operation.Glass tube has a narrow bore to magnify displacement per degree.Concept / Approach:As temperature rises, mercury’s volume increases. Because the bulb and capillary confine the liquid, expansion forces the mercury column to climb in the narrow tube. The scale is calibrated so the height corresponds to temperature. The mechanism relies on volumetric expansion—not pressure change, conductivity, or electrical effects.Step-by-Step Solution:Let ΔV = β * V0 * ΔT where β is the volumetric expansion coefficient.Glass expansion is much smaller, so net rise occurs in the capillary.Read the meniscus against a calibrated temperature scale.Verification / Alternative check:Consistency of readings across instruments arises from well-characterized β for mercury and standardized stem dimensions; corrections for emergent stem are sometimes applied.Why Other Options Are Wrong:Pressure and thermal conductivity changes are not the measurement basis.Differential linear expansion describes bimetallic thermometers, not mercury types.Electrical resistance change underlies RTDs, not liquid-in-glass devices.Common Pitfalls:Ignoring stem correction when part of the mercury column is outside the bath; errors grow at higher temperatures.Final Answer:Volumetric expansion

Question 5

Recovering a missing transfer function from options: Assume the process open-loop transfer function is G(s) = K / ((s + 1)(s + 4)). In the corresponding root locus, the open-loop poles are located at which points on the real axis?

Accepted Answer

Introduction / Context:The original stem omitted the explicit transfer function. Applying the Recovery-First Policy, we supply a standard second-order real-pole form G(s) = K / ((s + 1)(s + 4)) that is consistent with the provided multiple-choice options. Root-locus analysis begins by marking the open-loop poles and zeros in the s-plane; their locations define the starting points of the locus branches.Given Data / Assumptions:Recovered open-loop transfer: G(s) = K / ((s + 1)(s + 4)).No finite zeros.Standard negative feedback configuration.Concept / Approach:Open-loop poles are the roots of the denominator of G(s). For G(s) = K / ((s + 1)(s + 4)), the denominator is (s + 1)(s + 4). Setting it to zero gives s = −1 and s = −4. These are the starting points for the root-locus branches as K increases from 0 to ∞.Step-by-Step Solution:Denominator: (s + 1)(s + 4) = 0.Solve: s = −1 and s = −4.Therefore, open-loop poles at −1 and −4 on the real axis.Verification / Alternative check:Sketching the root locus shows two branches starting at −1 and −4, with breakaway on the real axis between them; asymptotes approach angles ±90° as K → ∞ since there are two poles and zero zeros.Why Other Options Are Wrong:Positive real locations (1, 4) would indicate unstable open-loop poles.Other negative pairs (−2, −3 or −0.5, −8) do not match the denominator factors (s + 1)(s + 4).Common Pitfalls:Confusing closed-loop pole locations (which move with K) with fixed open-loop pole locations; the latter come directly from G(s) at K = 0.Final Answer:-1, -4

Question 6

Metallurgical analysis techniques: Which analytical instrument can be used to determine the composition of metallic alloys (after suitable sample preparation and dissolution where required)?

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Introduction / Context:Alloy composition analysis can be performed by several instrumental methods. In classical chemical engineering and metallurgy exam contexts, a polarograph—an electroanalytical instrument—has long been cited for quantifying metal ions in solution obtained by dissolving alloy samples. This question aligns with that conventional viewpoint.Given Data / Assumptions:Solid metallic alloy can be dissolved to form ionic species.We seek an instrumental technique suitable for quantifying constituent metals.Exam framing favors traditional instruments over advanced spectrometry.Concept / Approach:Polarography (a branch of voltammetry) measures current as a function of applied potential at a dropping mercury or static electrode. Distinct reduction/oxidation waves appear at characteristic potentials for different ions, and wave heights are proportional to concentration. Thus, after dissolving an alloy, its components can be quantitatively determined.Step-by-Step Solution:Prepare solution by dissolving the alloy in an appropriate acid or complexing medium.Record polarogram; identify half-wave potentials for target metal ions.Use calibration to convert wave heights into concentrations.Compute composition from concentrations and original sample mass.Verification / Alternative check:While emission spectrometry or XRF are widely used today, classical exam answers often specify polarography as a valid laboratory method for alloy analysis after dissolution.Why Other Options Are Wrong:Chromatograph: Excellent for separations, but metals as elemental ions are not typically separated by standard GC/LC without complexation chemistry; it is not the canonical choice for alloys.Refractometer: Measures refractive index of liquids; not suitable for metals.Colorimeter: Works for colored complexes but is less general than polarography for multi-component alloys.Common Pitfalls:Assuming a solid metal can be tested directly by chromatograph or refractometer; appropriate dissolution and electroanalytical methods are key.Final Answer:Polarograph

Question 7

Constant-volume gas thermometers for low temperatures: A constant-volume gas thermometer designed for measuring sub-zero (below 0°C) temperatures commonly employs which working gas?

Accepted Answer

Introduction / Context:Constant-volume gas thermometers exploit the near-ideal relationship P ∝ T for a fixed amount of gas at fixed volume. Different gases are used to cover different temperature spans. Traditional laboratory practice assigns hydrogen to low temperatures below 0°C, while nitrogen or air serve near room and higher ranges, and helium covers cryogenic extremes.Given Data / Assumptions:Goal: accurate measurement below the ice point (0°C).Gas remains in single phase over the operating span.Simple, classical assignment of gases by range is intended.Concept / Approach:Hydrogen has favorable thermophysical properties and a low condensation temperature at atmospheric pressure, allowing a usable sub-zero range without liquefaction under typical thermometer conditions. It was historically adopted for below-zero calibration segments. Helium is excellent at very low cryogenic temperatures, but standard exam convention usually associates “sub-zero” (not cryogenic) with hydrogen.Step-by-Step Solution:Fix the thermometer volume; measure pressure as temperature varies.Choose gas to remain gaseous throughout the intended range.Select hydrogen for sub-zero laboratory range per classical practice.Verification / Alternative check:Thermometry texts tabulate recommended gases by range: hydrogen for low temperatures, nitrogen for moderate to high, helium for the lowest cryogenic spans.Why Other Options Are Wrong:Helium: More typical for deep cryogenics; not the classic choice for routine sub-zero measurements.Nitrogen and CO2: Condense or solidify within sub-zero spans under certain pressures, complicating calibration.None of these: Incorrect because hydrogen is a well-established choice.Common Pitfalls:Confusing “sub-zero” with “cryogenic.” Sub-zero refers to temperatures just below 0°C through tens of degrees; helium is reserved for much colder regimes.Final Answer:Hydrogen

Question 8

Mercury freezing limitation: A mercury-in-glass thermometer cannot be used below the freezing (solidification) point of mercury. What is that temperature in °C?

Accepted Answer

Introduction / Context:Mercury thermometers function only while mercury remains liquid. Below its freezing point, the column fractures or cannot move, producing erroneous or no readings. Knowing the freezing limit is essential for selecting a suitable low-temperature instrument.Given Data / Assumptions:We seek the solidification temperature of pure mercury on the Celsius scale.Standard atmospheric conditions are assumed.No special additives (e.g., thallium) are considered.Concept / Approach:Pure mercury freezes at approximately −38.83°C; this is commonly rounded and presented as −38.9°C in engineering handbooks. Below this temperature, alternatives such as alcohol-in-glass thermometers, RTDs, thermistors, or thermocouples must be used.Step-by-Step Solution:Identify mercury’s phase-change temperature: ~−38.83°C.Round to −38.9°C per common tabulation.Conclude: mercury thermometers are unsuitable below −38.9°C.Verification / Alternative check:Datasheets for mercury thermometers specify lower usable limits around −38 to −39°C, often with warnings about column separation damage if cooled below that point.Why Other Options Are Wrong:−11.9, −60.9, −80.9, −100: Do not correspond to the known mercury freezing point; some are much too low or too high.Common Pitfalls:Assuming antifreeze additives are present; pure mercury thermometers do not contain such additives and must respect the freezing limit.Final Answer:-38.9

Question 9

Thermocouple selection by EMF at 400 °C: a thermocouple develops 22 mV (referenced to 0 °C cold junction). Which standard thermocouple pair most closely matches this EMF at 400 °C?

Accepted Answer

Introduction / Context:This item tests practical thermocouple identification from a given electromotive force (EMF) at a specified temperature. In plant instrumentation, recognizing which thermocouple type fits an observed EMF helps troubleshoot mismatches, wiring swaps, or incorrect transmitter ranges.Given Data / Assumptions:Measured temperature: 400 °C.Reported EMF: 22 mV with a 0 °C reference (standard tables assume 0 °C cold junction).Standard base-metal and noble-metal thermocouples are under consideration.Concept / Approach:Each thermocouple has a characteristic Seebeck coefficient and tabulated EMF–temperature curve. At 400 °C (0 °C reference), approximate EMFs are: Type J (iron–constantan) ≈ 21–22 mV; Type K (chromel–alumel) ≈ 16–17 mV; Type T (copper–constantan) ≈ 17 mV but not typically rated to 400 °C; noble-metal types (S, R) are only a few millivolts at 400 °C. Therefore, 22 mV corresponds most closely to Type J.Step-by-Step Solution:List representative EMFs at 400 °C from standard tables for candidate types.Compare each to the observed 22 mV.Type J aligns at ~21.8–22 mV; select iron–constantan.Verification / Alternative check:A quick rule: Type J produces higher EMF than Type K in mid-ranges; noble-metal types produce much lower EMF at the same temperature. Cross-checking against a calibration card or ISA/IEC tables confirms the match.Why Other Options Are Wrong:Chromel–alumel (K): About 16–17 mV at 400 °C, not 22 mV.Platinum–rhodium (S or R): Only a few mV at 400 °C, far below 22 mV.Copper–constantan (T): EMF can be similar at lower temperatures, but it is not suitable for 400 °C and typical tables do not show 22 mV at this point.Common Pitfalls:Forgetting the cold-junction reference (0 °C assumed here); mixing °C with °F or relying on linear approximations beyond their valid range; assuming higher EMF always implies a higher-temperature-rated sensor.Final Answer:Iron–constantan (Type J)

Question 10

Routh stability criterion: the tabulation and test are performed using the characteristic polynomial of which transfer function—open-loop or closed-loop?

Accepted Answer

Introduction / Context:The Routh–Hurwitz stability test determines whether all roots of a polynomial lie in the left half of the s-plane. In control systems, we apply it to the characteristic equation of the closed-loop system to assess stability without computing roots explicitly.Given Data / Assumptions:Unity or non-unity feedback is allowed; the approach is unchanged.The characteristic equation arises from the denominator of the closed-loop transfer function: 1 + L(s) = 0.No numerical coefficients provided; this is a conceptual identification.Concept / Approach:Stability of the actual, operating loop depends on the poles of the closed-loop transfer function. The Routh array is built from the coefficients of that closed-loop characteristic polynomial. While open-loop transfer functions are used to derive the closed-loop characteristic equation (e.g., 1 + L(s) = 0), the Routh test itself is applied to the resulting closed-loop denominator.Step-by-Step Solution:Form L(s) from plant and controller.Write characteristic equation 1 + L(s) = 0 and expand to get the polynomial.Construct the Routh array with those coefficients to test sign changes in the first column.Verification / Alternative check:Textbooks consistently define the “characteristic equation” as the closed-loop denominator; examples of gain margin design via Routh all start from the closed-loop polynomial.Why Other Options Are Wrong:Open-loop alone does not reflect feedback stability.“Either/Neither” misstates the criterion’s basis.Controller-only polynomials ignore plant dynamics.Common Pitfalls:Applying Routh to the numerator or to open-loop polynomials; forgetting to include dead time approximations (if any) in the characteristic polynomial used.Final Answer:Closed-loop transfer function

Question 11

Time constant in first-order dynamics: by definition, the time constant is the time required for the measured (output) variable to reach what fraction of its ultimate change after a step input?

Accepted Answer

Introduction / Context:First-order systems (e.g., thermowells, level tanks, RC circuits) are characterized by a single time constant that determines how quickly the output approaches a new steady state after a step input. Clear definitions avoid confusion when comparing dynamics or tuning controllers.Given Data / Assumptions:System: first-order, linear, time-invariant.Input: step change; Output: measured variable y(t).Ultimate change: the final steady-state difference y(∞) − y(0).Concept / Approach:The step response of a first-order element is y(t) = y(∞) − [y(∞) − y(0)] * exp(−t/τ). At t = τ, exp(−1) ≈ 0.368; therefore y(τ) − y(0) = 0.632 * [y(∞) − y(0)]. Hence the time constant τ is the time for the output to achieve 63.2% of its total change. Transportation lag and dead time are separate concepts representing pure delay without dynamic smoothing and should not be conflated with τ.Step-by-Step Solution:Write the standard first-order step response equation.Evaluate at t = τ to find the fraction of ultimate change (0.632).Identify the definition in the provided options.Verification / Alternative check:Plotting the response shows a prominent 63.2% point at one time constant; at two time constants the response reaches 86.5%, and at four it is 98.2%, which aligns with rule-of-thumb settling estimates.Why Other Options Are Wrong:Transportation lag/dead time: Pure delays; not equal to τ in general.Controlled variable phrasing can be ambiguous; the precise definition uses the measured output.50% error reduction corresponds to ln(2) * τ, not τ itself.Common Pitfalls:Confusing lag with delay; using 50% instead of 63.2%; assuming the definition depends on controller action rather than plant dynamics.Final Answer:The measured variable reaches 63.2% of its ultimate change

Question 12

Electrical conductivity (specific conductance): defined as the reciprocal of resistance of a 1 cm cube of liquid at a specified temperature. Approximately what is the conductivity of distilled water in micro mho per cm?

Accepted Answer

Introduction / Context:Conductivity (specific conductance) quantifies a liquid’s ability to carry electric current and is widely used to assess water purity, ion content, and process quality. Distilled water has very low ionic content and therefore very low conductivity.Given Data / Assumptions:Definition: Conductivity is the reciprocal of resistance of a 1 cm × 1 cm × 1 cm liquid sample, often reported as micro mho/cm (equivalent to microsiemens/cm).Sample: fresh distilled water, equilibrated with minimal CO2.Concept / Approach:Ultra-pure water approaches 0.055 micro mho/cm (18.2 MΩ·cm), but ordinary laboratory “distilled water” typically reads around 0.5–2 micro mho/cm because it absorbs CO2 and picks up trace ions from containers and air. A representative textbook value is about 1 micro mho/cm, reflecting practical (not ultra-pure) conditions.Step-by-Step Solution:Recall that higher dissolved ions → higher conductivity; pure water is near zero.Recognize typical lab-distilled water ≈ 1 micro mho/cm.Choose 1 micro mho/cm as the best estimate among the options.Verification / Alternative check:Handheld meters commonly show 0.5–2 micro mho/cm for freshly prepared distilled water; ultrapure deionized water shows ~0.055 micro mho/cm, consistent with the reasoning above.Why Other Options Are Wrong:50 or 100 micro mho/cm: Far too high; those values indicate significant ionic content.10 micro mho/cm: More like very clean tap or lightly treated water, not distilled.0.1 micro mho/cm: Approaches ultrapure water specifications rather than ordinary distilled.Common Pitfalls:Confusing “distilled” with “ultrapure DI”; mixing legacy micro mho/cm units with SI microsiemens/cm without converting.Final Answer:1 micro mho/cm

Question 13

Thermal conductivity gas analysis: which constituent is commonly measured in flue gas by thermal-conductivity-based analyzers under plant conditions?

Accepted Answer

Introduction / Context:Thermal conductivity detectors (TCDs) infer gas composition from changes in thermal conductivity relative to a reference stream. In boiler houses and process heaters, on-line gas benches often include a thermal conductivity channel for certain species, complementing paramagnetic (O2) or NDIR (infrared) sensors.Given Data / Assumptions:Application: flue gas composition measurement.Detector: thermal conductivity cell comparing sample vs. reference.Species of interest: CO2 among others.Concept / Approach:CO2 has a significantly different thermal conductivity than air or N2; a TCD can sense concentration changes by monitoring heat loss from heated elements. While O2 is commonly measured by paramagnetic analyzers and CO2 by NDIR for high selectivity, TCDs remain a recognized method for CO2 in mixed streams, especially in older or simpler analyzer trains and in GC-TCD setups.Step-by-Step Solution:Match technique to property: TCD responds to bulk thermal conductivity of the gas mixture.Identify a species with pronounced conductivity contrast and routine presence in flue gas: CO2.Select CO2 percentage as the typical TCD application among the choices.Verification / Alternative check:Instrument manuals list CO2 channels based on TCD or NDIR; paramagnetic is reserved for O2, confirming that O2 is not usually measured by TCD in modern stacks.Why Other Options Are Wrong:O2 percentage: Better served by paramagnetic cells; TCD lacks selectivity.Specific gravity of petrofuels: Determined by hydrometers/densitometers, not TCD gas analyzers.Alloy composition: Requires spectroscopy or XRF, not gas TCD.Moisture by dew point: Uses chilled mirrors or IR absorption, not TCD.Common Pitfalls:Assuming TCD is species-selective; it senses total thermal conductivity. Cross-sensitivity to other gases must be accounted for or separated upstream (e.g., by columns in GC).Final Answer:Carbon dioxide (CO2) percentage in flue gas

Question 14

Emission spectroscopy applications: which task is emission spectroscopy especially well suited for in routine industrial and laboratory analysis?

Accepted Answer

Introduction / Context:Emission spectroscopy (including arc/spark OES, ICP-OES) measures the light emitted by excited atoms or ions to determine elemental composition. It is a cornerstone technique for metallurgy, quality control, and trace analysis in solids and solutions.Given Data / Assumptions:Technique: emission-based spectroscopy using electrical spark, plasma, or flame to excite samples.Analytes: metallic and non-metallic elements in solid metals and dissolved samples.Concept / Approach:Each element emits light at characteristic wavelengths when excited. By dispersing the emitted light and measuring intensities, one can identify and quantify elements at ppm to sub-ppm levels. Arc/spark sources are common for bulk metals; ICP-OES excels for multi-element analysis in solutions. These strengths make emission spectroscopy ideal for solids and metals analysis.Step-by-Step Solution:Identify the measurement principle: emission of characteristic lines.Match typical industrial tasks: alloy grading, impurity profiling, certification.Select “solids and metal analysis.”Verification / Alternative check:Standards for steel, aluminum, and other alloys routinely specify spark OES or ICP-OES methods, highlighting this technique’s dominance in metals labs.Why Other Options Are Wrong:Water purity: Conductivity/TOC/UV254 methods are typical, not emission spectroscopy alone.CO2 gases: Non-dispersive infrared (NDIR) is the norm.NOx: Chemiluminescence or UV methods dominate.Turbidity: Nephelometry is used, not emission OES.Common Pitfalls:Confusing emission with absorption techniques (AAS) or with molecular spectroscopies (IR/UV-vis) that target different analytes and matrices.Final Answer:Solids and metal analysis (elemental identification/trace levels)

Question 15

High-temperature non-contact thermometry: if the operating range must span approximately 800 to 1600 °C, which pyrometer type is typically selected?

Accepted Answer

Introduction / Context:At elevated temperatures beyond the safe range of contact sensors, non-contact pyrometers are used. Selecting the right type depends on target temperature, emissivity, and sighting conditions. The classic optical (disappearing-filament) pyrometer is well established in the 700–3000 °C region.Given Data / Assumptions:Required span: ~800 to 1600 °C.Target is hot, glowing surfaces suitable for visual/optical sighting.Concept / Approach:Optical pyrometers compare the brightness of a heated filament to the target’s brightness at a specific wavelength. They are effective from about 700 °C upward, covering 800–1600 °C comfortably. While “radiation pyrometer” is a broader category including optical and total radiation devices, the classic textbook answer for this span is the optical (disappearing-filament) instrument. Photoelectric ratio pyrometers also cover high temperatures and mitigate emissivity effects but are usually described explicitly as “ratio” units; the stem points to the general selection where optical pyrometers are archetypal.Step-by-Step Solution:Identify temperature window above 700 °C.Map to non-contact instruments: optical pyrometer fits 800–1600 °C.Choose the specific type named in the options.Verification / Alternative check:Catalogs and handbooks show optical pyrometers widely specified for steels and furnaces in this temperature band.Why Other Options Are Wrong:Radiation (total): Broad term; answer seeks the specific well-known instrument.Photoelectric: Also plausible, but the conventional selection for the stated band is optical (disappearing-filament).None/Resistance thermometer: Contact RTDs do not reach 1600 °C safely.Common Pitfalls:Ignoring emissivity and window transmission; in real practice, ratio pyrometers may be preferred for variable emissivity, but the core knowledge item is the optical pyrometer’s range.Final Answer:Optical (disappearing-filament) pyrometer

Question 16

Characteristic equation location: the characteristic equation used for stability analysis is the denominator of which transfer function?

Accepted Answer

Introduction / Context:In feedback systems, the characteristic equation defines the closed-loop poles and thus stability. Knowing where it appears in the transfer function structure is foundational for applying Routh, root locus, or frequency-domain techniques.Given Data / Assumptions:Unity or non-unity feedback; the algebraic form changes but the idea is the same.Characteristic equation set by 1 + L(s) = 0, where L(s) is the loop transfer function.Concept / Approach:The closed-loop transfer function T(s) = G(s) / [1 + L(s)] has its denominator equal to the characteristic equation. The roots of that denominator are the closed-loop poles; stability requires all to lie in the left half-plane. Therefore, the characteristic equation belongs to the closed-loop denominator.Step-by-Step Solution:Write T(s) = G(s) / [1 + L(s)].Identify the denominator: 1 + L(s) → characteristic equation.Conclude: denominator of the closed-loop transfer function.Verification / Alternative check:Standard derivations for root-locus plots begin from 1 + K G(s) H(s) = 0, explicitly the closed-loop characteristic equation.Why Other Options Are Wrong:Open-loop: Does not contain the stability-defining denominator.Both/neither: Contradicts the mathematical structure.Plant-only numerator/denominator does not account for feedback closure.Common Pitfalls:Confusing numerator zeros with poles; using the open-loop denominator for Routh instead of the closed-loop characteristic polynomial.Final Answer:Closed-loop transfer function

Question 17

Bimetallic thermometer sensitivity: two strips with different linear expansion are bonded. Which material below has the highest linear thermal expansion coefficient and thus is commonly paired against low-expansion alloys?

Accepted Answer

Introduction / Context:Bimetallic thermometers rely on differential thermal expansion. When heated, the high-expansion strip elongates more than the low-expansion strip, causing curvature that drives a pointer. Material choice maximizes sensitivity while preserving mechanical stability.Given Data / Assumptions:Candidate materials: nickel, brass, chromel (Ni–Cr), invar (Fe–Ni low-expansion alloy), stainless steel.We compare typical room-temperature linear expansion coefficients.Concept / Approach:Typical coefficients (approximate, 10^−6 / °C): brass ~19, stainless ~17, nickel ~13, chromel ~14, invar ~1.2. Thus, brass exhibits the largest expansion among the listed candidates and is often used as the high-expansion member paired with a low-expansion alloy such as invar to generate significant curvature per degree of temperature change.Step-by-Step Solution:List approximate coefficients for each material.Identify the maximum among the options.Select “Brass” as the high-expansion strip.Verification / Alternative check:Materials handbooks show brass near 18–20 × 10^−6 / °C; invar near 1–2 × 10^−6 / °C; the differential is what drives the bimetal action.Why Other Options Are Wrong:Nickel/chromel: Moderate expansion, lower than brass.Invar: Extremely low expansion (chosen as the low-expansion partner).Stainless: High but still below brass in most grades.Common Pitfalls:Confusing bimetal selection with electrical properties; expansion coefficient, not resistivity, governs mechanical deflection.Final Answer:Brass

Question 18

Dynamic order of a U-tube mercury manometer: under standard modeling assumptions (incompressible fluid, small oscillations, negligible friction), what is the order of this measurement system?

Accepted Answer

Introduction / Context:Dynamic modeling of instruments clarifies how they respond to transient inputs. A U-tube manometer contains a column of liquid that can oscillate when disturbed, much like a mass–spring–damper system. This behavior determines the system order and transient characteristics.Given Data / Assumptions:Incompressible manometer liquid (e.g., mercury) and small-displacement approximation.Symmetric U-tube; negligible viscous losses (or small damping).Pressure difference applied across the two legs.Concept / Approach:Applying force balance to the moving liquid column yields a second-order differential equation: inertia of the liquid slug (mass term), a restoring term due to hydrostatic head (analogous to a spring), and damping from viscous losses/friction. With negligible friction, the system is lightly damped and oscillatory—characteristics of a second-order system. First-order dynamics lack oscillation and result from pure capacitance/resistance combinations without inertia terms, which does not match the U-tube’s oscillatory nature.Step-by-Step Solution:Model the fluid slug mass m and hydrostatic restoring coefficient k.Write m * d^2x/dt^2 + c * dx/dt + k * x = input (pressure forcing).Recognize the standard second-order form; c represents damping.Verification / Alternative check:Impulse or step tests show overshoot and oscillation in U-tube levels, directly indicating second-order behavior; damping increases with viscosity and tube diameter effects.Why Other Options Are Wrong:Zero/first order: Cannot produce oscillations observed in U-tube dynamics.Third order: Adds extra dynamics not inherent in the simplest U-tube model.Fractional order: Not the standard canonical model here.Common Pitfalls:Neglecting inertia leads to a first-order approximation, which fails to capture oscillations; incorrect linearization can also misclassify the order.Final Answer:Second-order system

Question 19

Which is the most suitable instrument for measuring pressure below 3 microns?

Accepted Answer

Ionisation gauge

Question 20

Polarisation of light forms the working principle of a

Accepted Answer

polarimeter

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