Question 1

Steady-state offset among classical controllers: For a unit-step load/change on a stable process, which controller structure tends to have the largest steady-state offset (residual error)?

Accepted Answer

Introduction / Context: Offset (steady-state error) is a key performance metric for regulatory control. Controller structures differ in how they remove persistent error after a step disturbance or set-point change. Understanding which structures inherently eliminate offset guides practical selection in plants. Given Data / Assumptions: Stable plant, unity feedback conceptual model. Step-type change and steady-state considered. Reasonable tuning, no actuator saturation or integrator windup. Concept / Approach: A pure proportional controller leaves nonzero steady-state error for typical process types because the controller output must balance the process load, requiring a finite error to generate a finite output. Adding integral action (PI or PID) introduces a pole at the origin in the open loop, driving the steady-state error to zero for step inputs. PD alone does not change the steady-state gain (derivative term vanishes at steady state), so it behaves like P with respect to offset. Hence, among the listed, P shows the largest offset; PI and PID remove it (ideally), PD behaves similar to P for offset. Step-by-Step Solution: P: steady-state error > 0 for step inputs on most plants.PI/PID: integral action accumulates error → drives offset → 0.PD: derivative term → 0 at steady state → offset persists. Verification / Alternative check: Final value theorem applied to standard unity-feedback loops confirms that integral action guarantees zero steady-state error for step inputs when the closed loop is stable. Why Other Options Are Wrong: PI and PID — integral term eliminates offset for a step.PD — does not remove offset; comparable to P in steady state.I-only — removes offset but is not in the original list of common choices here. Common Pitfalls: Confusing transient improvement from derivative action with steady-state accuracy. Only integral action addresses offset. Final Answer: Proportional (P) controller

Question 2

Level measurement for cryogenic liquid in a pressurised tank: Liquid argon level in a pressurised storage tank (about 3 kg/cm²) should be measured using which primary method?

Accepted Answer

Introduction / Context: Measuring level in cryogenic, pressurised service (e.g., liquid argon) requires accuracy, safety, and reliability. The instrument must tolerate low temperatures, avoid icing hazards, and work on a closed, pressurised tank. This drives the choice toward indirect, sealed methods rather than direct sighting devices. Given Data / Assumptions: Service: liquid argon, cryogenic, tank pressure ≈ 3 kg/cm². Closed vessel without open venting to atmosphere at the measurement point. Industrial-grade instrumentation options. Concept / Approach: Differential pressure (DP) transmitters infer level from hydrostatic head: ΔP = ρ * g * h (after compensating for any vapour-space pressure via a high-pressure tap). With proper wet legs or remote seals rated for cryogenic use, DP level is standard and robust. Gauge glasses are unsafe and impractical (frosting, breakage hazard) on cryogenic pressured service. External floats can suffer from icing, leakage paths, and poor accuracy under pressure. Thus, DP is preferred in practice for cryogenic pressurised tanks. Step-by-Step Solution: Identify the need for a closed, sealed measurement approach.Apply hydrostatic principle: ΔP tracks liquid head independent of vessel pressure when referenced properly.Select DP level with cryogenic-rated remote seals/capillaries. Verification / Alternative check: Vendor application notes for cryogenic storage (LIN, LOX, LAr) consistently recommend DP with remote seals or weigh systems; sight glasses are avoided for safety. Why Other Options Are Wrong: Gauge glass — icing and safety risks; not typical on pressurised cryogenic service.External float — sealing/icing issues degrade performance.None of these — incorrect because a standard method exists (DP).Ultrasonic open-air stick — not suited inside closed pressurised vessels. Common Pitfalls: Ignoring vapour-space pressure compensation; DP must reference top/bottom taps correctly to cancel vessel pressure and read only the head of liquid. Final Answer: Differential pressure (DP) level measurement

Question 3

Optical radiation pyrometer lower limit: What is the typical starting (lower) temperature for reliable measurement by an optical radiation pyrometer in industry?

Accepted Answer

Introduction / Context: Optical (radiation) pyrometers measure temperature from emitted thermal radiation. Their usable range is limited at the low end by weak radiation intensity and background/emissivity uncertainties. A common rule-of-thumb lower bound for classic optical pyrometers is around 700–800°C, depending on design and target emissivity. Given Data / Assumptions: Conventional optical pyrometer (disappearing-filament or equivalent) and industrial targets. Normal emissivity and line-of-sight conditions. “Starting temperature” refers to practical, not absolute theoretical, limit. Concept / Approach: Below roughly 800°C, emitted visible radiation is weak and errors become large for traditional optical approaches. Infrared pyrometers expanded usable ranges downward in modern practice, but the classic value taught for an optical radiation pyrometer’s lower bound remains about 800°C, which aligns with standard exam expectations. Step-by-Step Solution: Relate radiative intensity to T using Stefan–Boltzmann/Wien laws conceptually.Note sensitivity limits of visual/optical comparison methods.Select 800°C as the practical lower threshold. Verification / Alternative check: Typical vendor datasheets for disappearing-filament instruments cite lower limits near 700–800°C depending on optics and target properties. Why Other Options Are Wrong: 400°C — too low for classic optical methods.1200°C and 1500°C — within range, but not the lower limit; they are well above the start point.250°C — far too low for optical comparison pyrometers. Common Pitfalls: Confusing modern IR pyrometers (which can measure lower) with traditional optical radiation pyrometers used in foundational instrumentation courses. Final Answer: 800°C

Question 4

Why magnetic flowmeters avoid hydrocarbons: Magnetic (electromagnetic) flowmeters are generally not used for measuring hydrocarbon flow/velocity because hydrocarbons typically have which limiting property?

Accepted Answer

Introduction / Context: Electromagnetic flowmeters employ Faraday’s law: a conductive fluid moving through a magnetic field induces a voltage proportional to velocity. They are popular for slurries and conductive liquids, but they require a minimum electrical conductivity to function. Hydrocarbons, being non-aqueous and largely non-ionic, fall well below that threshold. Given Data / Assumptions: Standard magmeter with typical conductivity requirement (often > 1–5 μS/cm). Measured fluid class: petroleum hydrocarbons (gasoline, diesel, kerosene, crude). No special electrodes or injected conductivity enhancers are used. Concept / Approach: For magmeters, induced EMF E = k * B * L * v * σ_effective, where σ (conductivity) must be high enough to produce a measurable signal. Hydrocarbons have extremely low σ (near insulating), so the induced voltage is negligible relative to noise, rendering measurements unreliable. Their viscosity, flash point, or thermal conductivity are unrelated to the sensing principle and do not preclude other meter technologies (turbine, PD, Coriolis, ultrasonic clamp-on, etc.). Step-by-Step Solution: Identify sensing principle (Faraday’s law) and its dependence on conductivity.Classify hydrocarbons as poor electrical conductors (nearly insulators).Conclude magmeters are unsuitable primarily due to low electrical conductivity. Verification / Alternative check: Manufacturer selection charts explicitly exclude low-conductivity fluids like oils and fuels from standard electromagnetic flowmeter applications. Why Other Options Are Wrong: Low thermal conductivity — unrelated to EMF sensing.High viscosity index — affects meter sizing, not the fundamental EMF requirement.Low flash point — a safety property, not a measurement constraint for magmeters specifically.High vapour pressure — may cause flashing/cavitation but not the magmeter’s primary limitation. Common Pitfalls: Confusing conductivity prerequisites with other physical properties; only conductivity is decisive for magmeter feasibility. Final Answer: Low electrical conductivity

Question 5

Turbine flowmeter application spectrum: Turbine meters are especially suitable for which of the following application domains?

Accepted Answer

Introduction / Context: Turbine flowmeters use a rotor whose angular speed is proportional to volumetric flow, yielding excellent linearity over wide, well-conditioned ranges. Their strengths include fast response, good repeatability, and suitability for clean, low-viscosity fluids and gases—including cryogenic services—making them a good fit for aerospace contexts where weight and response matter. Given Data / Assumptions: Clean service with properly filtered fluids or gases. Appropriate upstream/downstream straight runs and flow conditioning. Temperature/pressure ratings compatible with the chosen meter build. Concept / Approach: Turbine meters are widely used for cryogenic fluids (e.g., liquid nitrogen, liquid oxygen, liquid hydrogen, RP-1 kerosene under controlled conditions) and for aerospace/airborne fuel flow and test stands. They are not restricted to “very limited range”; rather, they have a good turndown (often 10:1 or more). They are not designed for highly viscous slurries where drag and particle impact compromise accuracy and life. Step-by-Step Solution: Match meter principle to application needs: low viscosity, cleanliness, fast dynamics.List typical domains: cryogenic propellants, aviation fuel flow, propulsion test benches.Select “both (b) and (c)” to capture these domains. Verification / Alternative check: Vendor catalogs and aerospace standards document turbine meters across cryogenic and airborne systems, corroborating their suitability. Why Other Options Are Wrong: Very limited flow ranges — incorrect; practical turndown is substantial.Only highly viscous slurries — opposite of suitable conditions for turbines. Common Pitfalls: Deploying turbine meters without adequate filtration or ignoring viscosity effects, which can bias calibration. Final Answer: Both (b) and (c)

Question 6

PI control (proportional + integral) in process control When integral (I) action is added to proportional (P) control in a feedback loop, which key performance outcome is enabled under steady disturbances and setpoint changes?

Accepted Answer

Introduction / Context: In classical feedback control for process industries, proportional (P) control provides immediate corrective action, but it typically leaves a steady-state error (known as offset) in the presence of constant disturbances or load changes. Adding integral (I) action creates PI control, the workhorse of refinery, chemical, and utility loops. This question asks what benefit I-control brings when used alongside P-control. Given Data / Assumptions: Loop is linearized around an operating point. Disturbance is constant (step-like) or setpoint is a step. Controller is ideal PI with finite tuning and a stable plant. Concept / Approach: Integral action accumulates error over time and continues to adjust the controller output until the error becomes zero. In steady state under a constant load, the only way integral of error can stop growing is if the error itself is driven to zero. Therefore, PI control eliminates offset for constant disturbances in stable, properly tuned loops. P-only cannot do this unless process static gain is infinite (which it is not in real systems). Step-by-Step Solution: Recognize that steady-state offset = nonzero error after transients decay with P-only.Integral action integrates error → controller output ramps until error → 0.Hence, PI achieves zero steady-state error for step-type loads in stable systems. Verification / Alternative check: Final value theorem applied to closed-loop transfer with an integrator in the controller shows steady error term goes to zero for step inputs/disturbances. Why Other Options Are Wrong: Reduction of offset: Under proper tuning, the target is elimination, not mere reduction. Reduction of stability time: Integral action often slows response if not tuned; it is not the defining benefit. None of these: Incorrect because PI's hallmark is offset elimination. Common Pitfalls: Assuming integral always speeds up response; it primarily removes steady error but can lengthen settling if overused. Final Answer: Elimination of offset

Question 7

First-order plus time delay (FOPDT) model Select the correct transfer function form for a first-order process with pure time delay (dead time) θ, process gain K, and time constant τ.

Accepted Answer

Introduction / Context: The first-order plus dead time (FOPDT) model is the most widely used low-order approximation in process control, capturing dominant lag and transport delay for many thermal, flow, and composition loops. Recognizing its canonical transfer function form is essential for tuning (e.g., Ziegler–Nichols, Cohen–Coon) and for controller design heuristics. Given Data / Assumptions: Process gain: K. Time constant: τ. Pure time delay (dead time): θ. Linearity and small-signal behavior around an operating point. Concept / Approach: A dead time element multiplies any transfer function by e^{−θ s} (unity magnitude, frequency-dependent phase). A first-order lag is 1/(τ s + 1). Therefore, the FOPDT model is K * e^{−θ s} / (τ s + 1). Higher-order denominators or extra 1/s factors indicate different dynamics (second order or integrating behavior) and are not correct for a simple FOPDT. Step-by-Step Solution: Identify first-order lag → denominator (τ s + 1).Incorporate pure delay → multiplicative factor e^{−θ s} in the numerator.Multiply by steady-state gain K → K * e^{−θ s} / (τ s + 1). Verification / Alternative check: Frequency response of e^{−θ s} has |G| = 1 and phase = −ω θ, while 1/(τ s + 1) gives −tan^{-1}(ω τ) phase and 1/√(1 + (ω τ)^2) magnitude roll-off, matching FOPDT characteristics. Why Other Options Are Wrong: K/(τ s + 1): No dead time included. K * e^{−θ s} /(τ s + 1)^2: Second-order lag, not first order. K /(s(τ s + 1)): Integrating plus first-order, not a simple FOPDT. Common Pitfalls: Modeling delay as extra lag terms; true transport delay is exponential in s. Final Answer: K * e^{-θ s} / (τ s + 1)

Question 8

Open-channel discharge measurement Liquid discharge from a tank or reservoir (free surface) cannot be measured using which of the following primary devices?

Accepted Answer

Introduction / Context: Measuring volumetric flow (discharge) from tanks, sumps, and reservoirs often involves free-surface hydraulics rather than pressurized pipe flow. Selection of a suitable primary element is crucial for accuracy and practicality. This question focuses on which device is not used for such open-channel or free-discharge situations. Given Data / Assumptions: Flow leaves a tank/reservoir with a free surface (non-pressurized). Head-discharge relationships are applied in open-channel context. We consider standard textbook devices. Concept / Approach: Weirs and notches are classic open-channel metering devices placed in flumes or at channel/tank exits, using head over crest to compute discharge. Mouthpieces (short tubes attached to orifices on a tank wall) are also associated with tank discharge measurement via empirical coefficients. In contrast, orificemeters are differential-pressure flow meters installed in closed, pressurized pipes with well-developed flow; they are not used on free-surface tanks/reservoirs for direct discharge metering. Step-by-Step Solution: Identify open-channel devices: weirs, notches → suitable.Recognize mouthpieces as tank-wall discharge elements with known correlations → suitable.Orificemeter requires a pressurized pipeline and DP taps → unsuitable here. Verification / Alternative check: Hydraulics texts classify weirs/notches under open-channel measurement and orificemeters under incompressible pipe-flow DP meters. Why Other Options Are Wrong: Weirs / Notches: Designed precisely for free-surface discharge measurement. Mouthpieces: Applied on tank outlets with empirical discharge coefficients. Common Pitfalls: Confusing a tank orifice (with or without mouthpiece) with an orificemeter assembly used in pipes with pressure taps and flange taps. Final Answer: Orificemeters

Question 9

Frequency response signature – constant magnitude, linear phase lag A dynamic element exhibits unit (constant) magnitude at all frequencies and a negative phase that increases linearly in absolute value with frequency. Which transfer function matche…

Accepted Answer

Introduction / Context: Frequency-domain fingerprints help identify elementary dynamic elements without time-domain tests. A pure time delay (dead time) has a distinctive Bode plot: flat magnitude and linearly increasing phase lag with frequency. This question asks you to map that signature to the correct transfer function. Given Data / Assumptions: Magnitude ratio is constant (unity) over all frequencies. Phase is negative and proportional to ω (linear in frequency). No amplitude roll-off or resonant features observed. Concept / Approach: A pure time delay of amount τ has transfer function G(s) = e^{−τ s}. In the frequency domain, G(jω) has magnitude |G| = |e^{−j ω τ}| = 1 for all ω and phase ∠G = −ω τ (radians), a straight line through the origin with slope −τ. By contrast, first-order lag 1/(τ s + 1) shows decreasing magnitude and asymptotic −90° phase; second-order terms add curvature and possible resonance; an integrator K/s has magnitude decreasing 20 dB/decade and −90° phase, not constant magnitude. Step-by-Step Solution: Match constant magnitude → only delay term e^{−τ s} meets this exactly.Check phase: ∠ = −ω τ (linear) → confirms pure delay.Exclude other elements due to magnitude roll-off or fixed phase limits. Verification / Alternative check: Plot Bode: delay line shows 0 dB across, with a straight-line phase lag increasing with ω; others do not. Why Other Options Are Wrong: 1/(τ s + 1): Low-pass magnitude and saturating phase. 1/(τ^2 s^2 + 1): Second-order behavior, not constant magnitude. K/s: Magnitude falls with frequency; constant −90° phase, not linear unbounded lag. Common Pitfalls: Approximating delay by extra poles; only e^{−τ s} preserves unit magnitude. Final Answer: e^{-τ s}

Question 10

Barometry and pressure types A mercury barometer is designed to measure which type of pressure directly?

Accepted Answer

Introduction / Context: Understanding pressure types—atmospheric, absolute, gauge, and vacuum—is fundamental to instrumentation. Devices differ in their reference: some measure relative to ambient, others to absolute zero. A mercury barometer is a classical instrument whose operation is often misunderstood in this regard. Given Data / Assumptions: The barometer consists of a mercury column with an evacuated space above the column. Ambient air pressure acts on the reservoir at the base. Temperature and local gravity affect the reading but not the reference concept. Concept / Approach: A barometer balances atmospheric pressure against a column of mercury. Because the space above the mercury column is (ideally) vacuum, the column height directly reflects the atmospheric pressure. While atmospheric pressure itself is an absolute pressure, the conventional answer distinguishes the measurand: the device specifically measures the current atmospheric pressure value, not gauge pressure (which is relative to atmosphere on both sides) nor “vacuum” level of an enclosure. Thus, the correct response is atmospheric pressure. Step-by-Step Solution: Recognize the evacuated headspace → top reference is near absolute zero pressure.Ambient acts on reservoir → column balances atmospheric force.Therefore the instrument reads atmospheric pressure. Verification / Alternative check: Standard meteorology expresses barometric readings in mmHg or hPa as atmospheric pressure. Why Other Options Are Wrong: Gauge: Gauge instruments compare to atmosphere; barometer compares atmosphere to vacuum. Vacuum: Barometer does not measure vacuum level inside process vessels. Absolute: Although atmospheric pressure is absolute, the accepted identification for this instrument is “atmospheric.” Common Pitfalls: Confusing the reference concept with the quantity measured and mixing terminology in exam settings. Final Answer: Atmospheric

Question 11

Principle of polarography – which sample state is analyzed? Polarography (dropping-mercury-electrode voltammetry) is primarily used to analyze which type of samples?

Accepted Answer

Introduction / Context: Polarography is an electroanalytical technique employing a dropping mercury electrode (or its variants) to record current–voltage curves and quantify electroactive species. Knowing the physical state and preparation of samples suited to polarography helps avoid conceptual errors about where the method applies. Given Data / Assumptions: Technique: polarography/voltammetry at liquid interfaces. Working electrode immersed in an electrolyte solution. Analytes must be soluble and electroactive within the potential window. Concept / Approach: Polarography fundamentally operates in the liquid phase—aqueous or non-aqueous solutions—where ions or molecules undergo reduction/oxidation at the electrode surface. While gases and solids can be studied indirectly (after dissolution or via other techniques), the direct application is to liquid solutions. The “isotonic solutions” option is a specific subset of liquids and not the general answer expected in a fundamentals question. Step-by-Step Solution: Identify that electrodes and supporting electrolyte imply a solution phase.Recognize electroactive species must be dissolved for mass transport and kinetics to be meaningful.Select “Liquids” as the general, correct category. Verification / Alternative check: Analytical chemistry texts position polarography under solution electrochemistry with calibration curves for dissolved ions and organics. Why Other Options Are Wrong: Isotonic solutions: Too narrow; polarography applies to many liquid solutions, not only isotonic ones. Gases / Solids: Not directly analyzed without dissolution or different instrumentation. Common Pitfalls: Assuming the mercury drop implies mercury vapor/gas analysis; the electrode simply provides a renewable liquid metal surface in a solution. Final Answer: Liquids

Question 12

Measuring sub-zero (°C) temperatures in industry Which instrument type is most commonly used for accurate and reliable industrial temperature measurement below 0 °C?

Accepted Answer

Introduction / Context: Industrial processes, cold storage, cryogenic lines, and environmental chambers often require accurate temperature measurement below 0 °C. Several sensor families exist, but their practicality, accuracy, and robustness differ significantly in this range. This question asks which is most commonly used in industry for sub-zero Celsius measurements. Given Data / Assumptions: Routine industrial environments (not laboratory primary standards). Temperatures from roughly −200 °C to 0 °C are of interest. Need accuracy, repeatability, and maintainability. Concept / Approach: Platinum resistance thermometers (RTDs, e.g., Pt100) offer excellent accuracy, linearity, and stability from about −200 °C to +600 °C. They are commonly deployed in industry for sub-zero applications. Thermocouples are rugged and cover very high temperatures, but their accuracy and stability (especially at low temperature with small EMFs) are typically inferior to RTDs unless carefully compensated and calibrated. Gas thermometers are laboratory reference devices, not industrial workhorses. Bi-metallic thermometers are mechanical, inexpensive, and simple, but comparatively coarse and less accurate for precise sub-zero monitoring. Step-by-Step Solution: Compare sensor classes vs requirements: accuracy + stability → RTDs lead.Note thermocouples: broad range but lower accuracy at low temperatures without special care.Exclude gas and bi-metal thermometers for common industrial use below 0 °C. Verification / Alternative check: Industry standards and vendors specify Pt100 RTDs as preferred for refrigeration, HVAC, pharmaceutical storage, and environmental test chambers. Why Other Options Are Wrong: Thermocouples: Adequate for ruggedness, less precise at low temperature. Gas thermometers: Reference/lab instruments. Bi-metallic: Limited precision and calibration stability. Common Pitfalls: Assuming thermocouples are universally best due to range; for accuracy below 0 °C, RTDs are favored. Final Answer: Resistance thermometers (RTDs)

Question 13

Mechanical vs manometric pressure elements Which of the following is not a mechanical elastic pressure sensing element?

Accepted Answer

Introduction / Context: Pressure measurement employs two broad families of primary elements: (1) mechanical elastic elements that deflect (Bourdon tubes, diaphragms, bellows), and (2) hydrostatic columns/manometers (U-tubes, inclined manometers). Distinguishing these is foundational for selecting transmitters and gauges. Given Data / Assumptions: Focus on the sensing element itself, not the dial or transducer. Mechanical elastic elements store strain energy and translate it into displacement. Manometers rely on fluid columns and hydrostatic balance. Concept / Approach: Bourdon tubes, diaphragms, and bellows are elastic metal elements that deform under pressure and are linked to pointers or transducers. A U-tube manometer is not an elastic element; it measures pressure from the height difference of a fluid column relative to a reference. Therefore, among the choices, the U-tube is not a mechanical elastic pressure element. Step-by-Step Solution: List elastic elements: Bourdon, diaphragm, bellows → mechanical deflection.Recognize U-tube uses hydrostatic head, not elastic deformation.Conclude U-tube does not belong to the mechanical elastic family. Verification / Alternative check: Instrumentation catalogs group Bourdon/diaphragm/bellows gauges separately from manometers. Why Other Options Are Wrong: Bourdon / Diaphragm / Bellows: All are classic elastic elements. Common Pitfalls: Calling any non-electronic device “mechanical”; the criterion here is elastic deflection versus hydrostatic balance. Final Answer: U-tube

Question 14

Vacuum measurement – Pirani gauge application range A Pirani gauge is typically used to measure which pressure regime?

Accepted Answer

Introduction / Context: Pirani gauges are common thermal-conductivity-based vacuum gauges used in vacuum systems for packaging, metallurgy, and research. Knowing where they apply on the vacuum scale guides proper selection and prevents misuse outside their calibrated range. Given Data / Assumptions: Gauges infer pressure from gas thermal conductivity changes. The sensing filament is heated electrically; cooling depends on gas density. Calibration is gas-dependent and typically referenced to air or nitrogen. Concept / Approach: A Pirani gauge covers the high vacuum through rough/medium vacuum region (commonly ~10^−3 to a few tens of mbar, depending on design), where thermal conductivity varies significantly with pressure. It is not used for measuring very high pressure, nor is it a level instrument. For ultra-high vacuum (<10^−7 mbar), ionization gauges are used instead. Step-by-Step Solution: Relate sensing principle (thermal conductivity) to gas density.Note the working range corresponds to low absolute pressures (high vacuum).Eliminate options unrelated to vacuum measurement. Verification / Alternative check: Vacuum technology references place Pirani gauges in the rough to high vacuum range; beyond that, cold cathode/hot cathode gauges take over. Why Other Options Are Wrong: Very high pressure: Outside operating principle and construction. Liquid level options: Different measurement category entirely. Common Pitfalls: Assuming “high vacuum” means UHV; Pirani covers up to high/medium vacuum, not UHV. Final Answer: High vacuum

Question 15

Psychrometry and humidity measurement in process engineering: What parameter does a psychrometer directly determine when used with gas streams in drying or air-conditioning operations?

Accepted Answer

Introduction / Context: Psychrometry deals with the thermodynamic properties of gas–vapour mixtures, most commonly air–water vapour. In chemical and mechanical engineering, a psychrometer is a simple, robust instrument used to assess how much vapour is carried by a gas stream. This question checks whether you can link the instrument to the property it is designed to measure in day-to-day plant practice. Given Data / Assumptions: A psychrometer uses two thermometers: a dry-bulb and a wet-bulb. The gas is well mixed and flows adequately over both bulbs during measurement. Standard psychrometric relations (charts or formulas) connect the readings to humidity measures. Concept / Approach: The wet-bulb thermometer cools by evaporation; the rate of evaporation (and therefore the temperature depression from dry-bulb to wet-bulb) depends on the vapour content of the gas. By combining both readings, one can determine relative humidity, humidity ratio (kg water per kg dry gas), and other related properties such as enthalpy and dew point. The essential outcome is the humidity of gases; other properties (dew point, enthalpy) are secondary results derived from the same data. Step-by-Step Solution: Identify instrument: a psychrometer consists of dry-bulb and wet-bulb thermometers.Relate wet-bulb depression to vapour content via mass and heat transfer.Use charts/equations to compute humidity measures from the two temperatures.Conclude that the primary purpose is determining humidity of the gas. Verification / Alternative check: A sling or aspiration psychrometer yields consistent readings in the field; values map to standard ASHRAE or chemical-engineering psychrometric charts for humidity ratio and relative humidity, confirming the measurement’s purpose. Why Other Options Are Wrong: Moisture content of solids: Requires oven drying or moisture analysers, not a psychrometer. Water of crystallisation: Determined by thermal analysis or chemical tests. Hygroscopic nature of solids: Inferred from sorption isotherms, not directly by psychrometry. Dew point of steel surfaces only: Psychrometers are general gas-phase instruments, not surface-specific tools. Common Pitfalls: Confusing humidity (gas property) with moisture in solids; also assuming a single thermometer can give humidity without a wet-bulb reference. Final Answer: Humidity of gases

Question 16

Characteristic equation in feedback control: For a closed-loop transfer function written as G1 / (1 + G2), the characteristic equation is 1 + G2 = 0. What does this imply about dependence, stability assessment, and applicability to set-point vs load…

Accepted Answer

Introduction / Context: In classical control theory, the characteristic equation captures the closed-loop poles, which entirely determine stability and transient behaviour. This question probes your understanding of how the characteristic equation relates to the open-loop dynamics and whether it changes with the type of excitation (set-point versus load disturbance). Given Data / Assumptions: Closed-loop form is written so that the denominator is 1 + G2. Linear time-invariant system with negative feedback. Set-point and load disturbances enter through standard summing junctions. Concept / Approach: The characteristic equation is obtained by setting the closed-loop denominator to zero. It depends only on the open-loop transfer function combination that appears in the denominator (typically the product of controller and plant). Because closed-loop poles are independent of the input path (set-point or load), the same characteristic equation governs stability for any excitation. Hence, it both depends only on open-loop dynamics and determines stability, and it is common to all input cases. Step-by-Step Solution: Write closed-loop denominator: 1 + G2.Set 1 + G2 = 0 to obtain the characteristic equation.Note that poles (roots) depend on open-loop elements only.Conclude: same characteristic equation applies to set-point and load analyses. Verification / Alternative check: Block-diagram manipulations or Mason’s gain formula show that closed-loop poles do not depend on input placement, confirming invariance across set-point and disturbance cases. Why Other Options Are Wrong: Choosing only one of (a), (b), or (c) ignores the full role of the characteristic equation. “None of these” contradicts standard control results. Common Pitfalls: Confusing the characteristic equation (poles) with the full closed-loop transfer function (which includes zeros and may differ for set-point vs load paths). Final Answer: All (a), (b) & (c)

Question 17

Frequency response of a first-order system under sinusoidal forcing: For a first-order system subjected to a sinusoidal input, what is the ratio of output amplitude to input amplitude (the sinusoidal steady-state gain)?

Accepted Answer

Introduction / Context: First-order systems are ubiquitous (thermal lags, level dynamics, simple RC networks). In frequency response, their magnitude declines with frequency. This question checks your qualitative understanding of sinusoidal steady-state behaviour without resorting to detailed calculations. Given Data / Assumptions: Standard first-order transfer: G(jω) = K / (1 + jωτ). Sinusoidal forcing with angular frequency ω. We ask about amplitude ratio |G(jω)|. Concept / Approach: The magnitude of a first-order transfer is |G(jω)| = K / sqrt(1 + (ωτ)^2). For unity static gain (K = 1), this is always less than or equal to 1, with equality only at ω → 0. At nonzero frequencies, the denominator exceeds 1, so the amplitude ratio is strictly less than 1. Even for K ≠ 1, the frequency-dependent factor is 1 / sqrt(1 + (ωτ)^2) < 1, so the dynamic portion reduces amplitude relative to the static gain. Step-by-Step Solution: Write magnitude: |G(jω)| = K / sqrt(1 + (ωτ)^2).Note that sqrt(1 + (ωτ)^2) ≥ 1, equality only when ω = 0.Therefore, for finite ω, amplitude ratio is less than K; for K = 1 it is < 1.Conclude that the ratio is less than 1 for practical sinusoidal testing. Verification / Alternative check: Bode magnitude plots for first-order lags show 0 dB at low frequency rolling off to −20 dB/decade past the corner frequency ωc = 1/τ, confirming amplitude reduction. Why Other Options Are Wrong: 1: Occurs only at zero frequency (DC), not for a sinusoid with finite ω. > 1: First-order lag cannot amplify sinusoidal amplitude. None of these / Exactly 0: Contradicted by standard frequency response. Common Pitfalls: Confusing static gain with dynamic gain; forgetting that phase lag and magnitude roll-off start as soon as ω > 0. Final Answer: < 1

Question 18

Alphatron low-pressure measurement range: The Alphatron instrument (an ionisation-type vacuum gauge) measures absolute pressure most effectively in which microns of mercury range?

Accepted Answer

Introduction / Context: Different vacuum gauges cover different orders of magnitude of pressure. The Alphatron (a form of ionisation/partial-pressure gauge) is used for extremely low absolute pressures in high-vacuum service. Knowing the valid measuring window prevents gross errors and instrument misuse. Given Data / Assumptions: Pressure is stated in microns of mercury (1 micron = 0.001 mm Hg). The Alphatron is intended for high-vacuum ranges. We select the option that best reflects its practical operating window. Concept / Approach: Ionisation-type gauges infer pressure from electron/ion currents in rarefied gases and are effective at very low pressures (high vacuum). The Alphatron’s usable range is typically within a few microns of Hg and below; therefore, a limit such as “less than 3 microns” captures its normal application window more accurately than broader or higher ranges. Step-by-Step Solution: Classify gauge: Alphatron → high-vacuum/ionisation type.Match to vacuum range expressed in microns of Hg.Select the option indicating a few microns and below: < 3 microns. Verification / Alternative check: Vacuum technology references list Alphatron-class instruments among gauges used in the 10^0 to 10^-1 micron region, consistent with “less than 3 microns.” Why Other Options Are Wrong: > 3 or 10 to 100: These are too high for ionisation gauges; Pirani or capacitance manometers are better there. < 0.1: Possible for some gauges but overly restrictive relative to typical Alphatron use. < 1: Also feasible, but “< 3” better matches the commonly cited exam window. Common Pitfalls: Confusing microns with millimetres of Hg; mixing up gauge families (Pirani vs ionisation) across ranges. Final Answer: < 3

Question 19

Assessing stability from transfer-function denominators: Which of the following single-input single-output systems is (asymptotically) stable based on its denominator polynomial?

Accepted Answer

Introduction / Context: Continuous-time linear systems are asymptotically stable if and only if all poles lie strictly in the left-half complex plane (negative real parts). Recognising stability by inspection of the characteristic polynomial is a core skill for controls and process dynamics. Given Data / Assumptions: Stability means responses decay to zero for bounded inputs after transients. We assess the sign of real parts of the roots of each denominator. Time delays (e.g., exp(−20s)) do not move poles by themselves but can worsen margins; denominator roots still dictate asymptotic stability. Concept / Approach: For a quadratic s^2 + 2ζω_ns + ω_n^2, stability requires ζ > 0 and ω_n > 0. Alternatively, compute roots directly and check real parts. Imaginary-axis poles indicate marginal stability (undamped oscillation) rather than asymptotic stability. Right-half-plane poles imply instability. Step-by-Step Solution: (a) s^2 + 2 = 0 ⇒ s = ± j√2 → marginal, not asymptotically stable.(b) s^2 − 2s + 3 = 0 ⇒ s = 1 ± j√2 → positive real part → unstable.(c) s^2 + 2s + 2 = 0 ⇒ s = −1 ± j → negative real part → stable.(d) s^2 + 2s − 1 = 0 ⇒ s = −1 ± √2 → one root at 0.414… > 0 → unstable (delay does not fix this).(e) s^2 − 1 = 0 ⇒ s = ±1 → right-half-plane pole → unstable. Verification / Alternative check: Routh–Hurwitz for (c): coefficients [1, 2, 2] are positive with no sign changes, confirming stability. Why Other Options Are Wrong: (a): Sustained oscillations violate asymptotic stability. (b), (d), (e): Each has at least one pole with positive real part. Common Pitfalls: Calling imaginary-axis poles “stable”; they are only marginally stable and sensitive to disturbances. Final Answer: 1 / (s^2 + 2s + 2)

Question 20

Measuring ~2000°C on incandescent mantles: Which instrument is most appropriate to measure the very high temperature (around 2000°C) of an incandescent gas mantle or filament?

Accepted Answer

Introduction / Context: At temperatures approaching 2000°C, contact sensors face survivability and measurement-error challenges. Non-contact optical methods that rely on visible incandescence are preferred for filaments and mantles. This question distinguishes between optical, radiation, and contact approaches at extreme temperatures. Given Data / Assumptions: Target is a bright, incandescent surface with high radiance. Line-of-sight measurement is available. Emissivity is reasonably stable or can be accounted for by the technique. Concept / Approach: An optical pyrometer (disappearing-filament type) compares the brightness of a calibrated filament to the target’s brightness at a specific wavelength. When the filament “disappears” against the target, their brightness temperatures match. This technique works well from roughly 700°C up to 3000°C and is ideal for incandescent mantles. Total-radiation pyrometers are also high-temperature capable but are generally used for larger, opaque targets; classic exam convention associates filaments/mantles with optical pyrometers. Step-by-Step Solution: Identify temperature regime: ~2000°C suggests non-contact optical methods.Select instrument tailored to brightness comparison at a wavelength: optical pyrometer.Avoid contact thermocouples due to survivability and radiative losses. Verification / Alternative check: Manufacturer datasheets for optical pyrometers list upper ranges near or above 2500°C, confirming suitability for incandescent sources. Why Other Options Are Wrong: Special thermocouples: Possible with protection, but intrusive and less accurate for tiny bright sources. Total-radiation pyrometer: Valid for very hot bodies, but the standard pairing for filaments is optical (brightness) pyrometry. Mercury thermometer: Impossible at such temperatures. Common Pitfalls: Assuming any radiation pyrometer is equivalent; brightness (optical) pyrometers are textbook choices for incandescent mantles. Final Answer: Optical pyrometer

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