Question 1

Reading the psychrometric chart Which set of lines on a standard psychrometric chart are vertical and uniformly spaced, and what do they represent?

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Introduction / Context:Correctly interpreting the axes and families of lines on the psychrometric chart is essential for HVAC design and analysis.Given Data / Assumptions:Standard ASHRAE-style psychrometric chart for moist air.Horizontal axis is dry-bulb temperature T_db.Vertical axis is humidity ratio w (kg water/kg dry air).Concept / Approach:Lines of constant dry-bulb temperature are vertical because the x-axis is T_db. These are typically uniformly spaced to scale temperature directly. Other property lines (relative humidity, wet-bulb, enthalpy) are inclined or curved.Step-by-Step Solution:Identify axes: T_db on the x-axis, w on the y-axis.Therefore, a constant T_db corresponds to a vertical line.Uniform spacing reflects the linear temperature scale used.Verification / Alternative check:Compare with chart legends: isohumes (constant w) are horizontal; relative humidity curves are curved lines; wet-bulb lines are diagonally inclined.Why Other Options Are Wrong:Wet-bulb and relative humidity are not vertical; dew point is read at saturation for a given w and aligns horizontally with constant humidity ratio.Common Pitfalls:Confusing constant wet-bulb (slanted) with constant enthalpy lines; they are nearly but not exactly coincident in some ranges.Final Answer:Vertical, uniformly spaced lines represent dry-bulb temperature.

Question 2

Equivalent insulation thickness A furnace wall of red brick (thermal conductivity k₁ = 0.7 W/m·K) has thickness 0.50 m. For the same heat loss and the same temperature drop across the wall, what thickness of diatomite earth (thermal conductivity k₂ …

Accepted Answer

Introduction / Context:Thermal insulation design often requires replacing one material with another while keeping the same conductive heat loss for a given temperature difference. This is a straightforward conduction resistance equivalence problem.Given Data / Assumptions:Brick wall thickness t₁ = 0.50 m, conductivity k₁ = 0.7 W/m·K.Diatomite earth conductivity k₂ = 0.14 W/m·K; required thickness t₂ = ?One-dimensional steady conduction, same surface area A, and same temperature drop ΔT.Concept / Approach:For a plane wall, conductive heat flow Q = k * A * ΔT / t. For the same Q and ΔT with the same area, the ratio k/t must be equal for the two walls. Therefore, t is directly proportional to k when Q, A, and ΔT are held constant.Step-by-Step Solution:Set equivalence: k₁ / t₁ = k₂ / t₂.Solve for t₂: t₂ = t₁ * (k₂ / k₁).Compute ratio: k₂ / k₁ = 0.14 / 0.70 = 0.20.Therefore, t₂ = 0.50 * 0.20 = 0.10 m.Verification / Alternative check:Compare total resistances: R₁ = t₁/(k₁ A) = 0.50/(0.7A) ≈ 0.714/A; R₂ = 0.10/(0.14A) ≈ 0.714/A; equal as required.Why Other Options Are Wrong:0.50 m (same thickness) would lower Q by a factor of 0.14/0.70 = 0.2 (too much resistance). 0.20 m still over-insulates. 0.75 m and 1.00 m are unrelated distractors that significantly overestimate resistance.Common Pitfalls:Inverting the proportionality (using t ∝ 1/k) for a fixed Q; remember here we keep Q the same, thus k/t equality applies, leading to t ∝ k.Final Answer:0.10 m

Question 3

Relative performance comparison Compared with a typical domestic air-conditioner, the coefficient of performance (C.O.P.) of a domestic refrigerator is generally:

Accepted Answer

Introduction / Context:C.O.P. depends primarily on the temperature lift between evaporator and condenser. Domestic refrigerators and domestic air-conditioners operate across very different temperature ranges, leading to different typical C.O.P. values.Given Data / Assumptions:Typical refrigerator: evaporator near −5°C to −15°C, condenser near 35–45°C.Typical room air-conditioner: evaporator near 10–15°C, condenser near 35–45°C.Similar technology and comparable component efficiencies.Concept / Approach:For ideal cycles, C.O.P._R ≈ T_L / (T_H − T_L) in absolute units (Kelvin). A larger temperature lift (T_H − T_L) reduces C.O.P. Since refrigerators must maintain sub-zero or near-freezing temperatures, their lift is larger than that of comfort cooling systems, thus giving a lower C.O.P. in practice.Step-by-Step Solution:Estimate for refrigerator: T_L ≈ 268 K, T_H ≈ 308 K → C.O.P. ≈ 268 / 40 ≈ 6.7 (ideal).Estimate for air-conditioner: T_L ≈ 295 K, T_H ≈ 308 K → C.O.P. ≈ 295 / 13 ≈ 22.7 (ideal).Real systems operate at a fraction of these ideal values, but the relative ordering (refrigerator lower than A/C) holds.Verification / Alternative check:Seasonal efficiency ratings (e.g., EER/SEER vs. refrigerator labels) also reflect easier duty for A/C compared with low-temperature refrigeration.Why Other Options Are Wrong:“Higher” contradicts the temperature-lift argument; “about the same” ignores the different duty points; “zero” and “always double” are non-physical generalizations.Common Pitfalls:Comparing nameplate power rather than thermodynamic lift; C.O.P. must be compared at similar reference conditions in Kelvin.Final Answer:Lower

Question 4

Psychrometric processes — dehumidification path On a standard psychrometric chart, which line best represents an ideal dehumidification-at-constant-dry-bulb process (e.g., chemical dehumidification)?

Accepted Answer

Introduction / Context:Different dehumidification methods trace different paths on the psychrometric chart. Chemical (adsorption/absorption) dehumidification typically removes moisture at nearly constant dry-bulb temperature, unlike cooling-based dehumidification which lowers dry-bulb too.Given Data / Assumptions:Standard psychrometric chart with x-axis as dry-bulb temperature and y-axis as humidity ratio.Idealized chemical dehumidification maintains dry-bulb approximately constant.No external heat exchange that would raise or lower T_db significantly.Concept / Approach:At constant T_db, removing water vapour reduces humidity ratio while staying on the same vertical line. By contrast, cooling and dehumidifying with a cooling coil generally moves down and to the left toward the saturation curve, not a vertical path.Step-by-Step Solution:Identify constant dry-bulb: vertical line on the chart.Remove moisture: move downward (lower humidity ratio).Hence, ideal chemical dehumidification is a vertical downward line.Verification / Alternative check:Manufacturer data for desiccant wheels often show near-vertical movement on the chart for the process air path, confirming the idealization.Why Other Options Are Wrong:Horizontal line keeps humidity ratio constant (no dehumidification). Inclined and curved lines describe cooling/dehumidifying coils or constant relative humidity paths, not pure chemical dehumidification.Common Pitfalls:Assuming all dehumidification implies cooling; method matters. Chemical dehumidification can be nearly isothermal at the process air stream.Final Answer:Vertical line (constant dry-bulb temperature, decreasing humidity ratio)

Question 5

Modes of heat transfer — microscopic mechanism Heat transfer that occurs primarily through molecular collisions within a material (without bulk motion) is called:

Accepted Answer

Introduction / Context:Heat transfer occurs via three fundamental modes: conduction, convection, and radiation. Recognizing the mechanism at play is critical for modeling and design.Given Data / Assumptions:No bulk fluid motion is implied.Energy migrates because of temperature gradients within solids or stationary fluids.Microscopic interactions (molecules, electrons, phonons) dominate.Concept / Approach:Conduction is the transfer of thermal energy due to molecular activity and lattice vibrations in solids (and molecule collisions in stationary fluids). It is described macroscopically by Fourier’s law: q″ = −k * dT/dx, relating heat flux to the temperature gradient and thermal conductivity k.Step-by-Step Solution:Identify absence of bulk motion → rules out convection.No requirement for an intervening medium → radiation not the primary mode here.Hence, the heat transfer mechanism due to molecular collisions is conduction.Verification / Alternative check:In metals, free electrons carry a significant part of the conductive heat; in nonmetals, phonon transport dominates; in gases at rest, intermolecular collisions account for conduction.Why Other Options Are Wrong:Convection requires bulk fluid motion; radiation is electromagnetic emission and requires no medium; advection is transport by bulk flow, not molecular collisions; “thermal dispersion” is not a fundamental mode in classical heat transfer.Common Pitfalls:Confusing conduction within a boundary layer with convection of the overall boundary layer process; locally at the surface, heat still crosses the fluid film by conduction, but the macro mode is convection.Final Answer:Conduction

Question 6

Heat transfer from a hot cube to a constant-temperature bath: If the cube loses heat from the top, bottom, and the four side faces with surface heat transfer coefficients h1, h2, and h3 respectively, what is the area-averaged overall heat transfer c…

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Introduction / Context:This question tests understanding of convective heat transfer from composite surfaces. A cube immersed in a constant-temperature bath can have different local heat transfer coefficients on different faces. The correct “average” coefficient must be area-weighted, not an arithmetic or geometric mean chosen blindly.Given Data / Assumptions:Cubic body; each face has equal area A.Top face coefficient = h1; bottom face coefficient = h2; each of four side faces coefficient = h3.Surroundings are isothermal; film coefficients are uniform over each face.Concept / Approach:The overall or average heat transfer coefficient h_avg for multiple parallel heat transfer paths over distinct areas is the area-weighted mean. Total heat rate is the sum of facewise convection rates. Divide total by total area times the same driving temperature difference to define h_avg.Step-by-Step Solution:Total area of cube faces = 6A.Heat from top = h1 * A * ΔT; bottom = h2 * A * ΔT; four sides = 4 * (h3 * A * ΔT).Total heat rate Q_total = (h1 + h2 + 4h3) * A * ΔT.Define h_avg by Q_total = h_avg * (6A) * ΔT, giving h_avg = (h1 + h2 + 4h3)/6.Verification / Alternative check:If all faces had the same coefficient h, formula gives h_avg = (h + h + 4h)/6 = h, confirming consistency.Why Other Options Are Wrong:h1 + h2 + h3 ignores multiplicity of side faces and total area.(h1h2h3)^(1/3) is an unjustified geometric mean; convection adds linearly over areas.2(h1 + h2 + h3)/3 is an arbitrary average and miscounts side faces.Common Pitfalls:Forgetting that there are four side faces; using simple averages instead of area-weighted sums.Final Answer:(h1 + h2 + 4*h3) / 6

Question 7

Psychrometrics check: During sensible cooling of air (no moisture removed), the relative humidity of the air stream will __________.

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Introduction / Context:Sensible cooling lowers the dry-bulb temperature without intentionally removing moisture. Understanding how relative humidity (RH) responds is fundamental in HVAC design and comfort analysis.Given Data / Assumptions:Cooling coil surface is above the air stream dew-point, so no condensation occurs.Humidity ratio (specific humidity) is constant during the process.Pressure approximately constant (near atmospheric).Concept / Approach:Relative humidity is the ratio of the partial pressure of water vapour to the saturation vapour pressure at the same temperature. During sensible cooling with constant humidity ratio, the saturation vapour pressure decreases as temperature drops; hence the ratio (RH) increases.Step-by-Step Solution:Hold humidity ratio w constant (no moisture removal).Lower dry-bulb temperature, which lowers saturation pressure p_ws(T).With vapour partial pressure essentially unchanged, RH = p_w / p_ws increases.Conclusion: RH rises until dew-point is reached; below that, condensation would start.Verification / Alternative check:On a psychrometric chart, the process is a horizontal move left (constant w); lines of constant RH slope upward to the left, so RH increases.Why Other Options Are Wrong:Decreasing or constant RH contradicts chart behavior; becoming 100% immediately or zero is nonphysical unless the coil is at or below dew-point.Common Pitfalls:Confusing relative humidity with humidity ratio; only when the coil temperature dips below dew-point does moisture removal begin.Final Answer:Yes, it increases

Question 8

Radiation fundamentals (Wien’s displacement): The wavelength corresponding to the peak (maximum) emissive power of a blackbody varies inversely with absolute temperature (λ_max * T = constant). The blank in “wavelength corresponding to __________” s…

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Introduction / Context:Wien’s displacement law links blackbody temperature to the spectral position of peak emission. It is often tested by asking what physical quantity defines the peak.Given Data / Assumptions:Ideal blackbody emission.Absolute temperature T measured in kelvin.λ_max is the wavelength at which emissive power is highest.Concept / Approach:Wien’s displacement states λ_max * T = constant. Thus as T increases, λ_max decreases. The “peak” refers to the wavelength of maximum monochromatic emissive power, i.e., maximum energy in the spectral distribution.Step-by-Step Solution:Identify the required phrase: “wavelength corresponding to maximum emissive power.”Recall the inverse relation: λ_max is inversely proportional to T.Therefore the correct fill is “maximum energy (peak emissive power).”Other phrases do not represent the spectral peak that Wien’s law addresses.Verification / Alternative check:Plot of blackbody spectral curves at various temperatures shows peak shift toward shorter wavelengths as T increases, confirming the inverse relationship.Why Other Options Are Wrong:“Minimum energy/minimum emissive power/average” do not define the spectral peak; “net radiative exchange” is a system-level quantity, not a spectral peak descriptor.Common Pitfalls:Confusing Wien’s law (peak position) with Stefan–Boltzmann law (total emission proportional to T^4) or Planck’s law (full spectral distribution).Final Answer:maximum energy (peak emissive power)

Question 9

Refrigeration cycles comparison: The coefficient of performance (COP) of a vapour-compression refrigeration system is generally __________ than that of an air (gas) refrigeration system operating between similar temperature limits.

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Introduction / Context:Vapour-compression systems (using phase change) are the workhorse of refrigeration. Air refrigeration (reversed Brayton) relies on gas expansion without phase change. COP comparisons are a staple concept in refrigeration engineering.Given Data / Assumptions:Similar evaporator and condenser (or sink/source) temperatures.Idealized cycles for conceptual comparison.Neglecting pressure losses and non-idealities for clarity.Concept / Approach:COP = refrigeration effect / work input. Two-phase systems exploit large latent heat at near-constant temperature, yielding high refrigeration effect per unit work. Gas cycles produce smaller temperature drops and refrigeration effects for the same work, so COP is typically lower than vapour-compression.Step-by-Step Solution:Note that latent heat in evaporator provides large heat absorption at nearly constant temperature.Compression of vapour near saturation requires less specific work for comparable temperature lift.Therefore, for similar temperature limits, COP_vapour-compression > COP_air.Hence, the correct choice is “high.”Verification / Alternative check:Textbook examples show typical COPs: domestic vapour-compression refrigerators often achieve COP 2–4+, while simple air cycles are lower under similar spans.Why Other Options Are Wrong:“Low/about the same/zero/indeterminate” contradict fundamental thermodynamic behaviors and practical performance data.Common Pitfalls:Comparing COPs without holding temperature limits roughly similar; ignoring non-ideal compressor efficiencies that still do not reverse the general trend.Final Answer:high

Question 10

Sensible heating coil performance: For air heated sensibly across a coil, the coil efficiency (or effectiveness) in terms of the by-pass factor (B.P.F.) is:

Accepted Answer

Introduction / Context:By-pass factor (B.P.F.) quantifies the fraction of air that effectively “misses” contact with the coil surface temperature, while coil efficiency (or apparatus dew-point effectiveness for sensible heating) tells how close the outlet approaches the coil surface temperature.Given Data / Assumptions:Sensible heating (no moisture change).Well-mixed outlet air.Coil surface temperature is uniform.Concept / Approach:For sensible processes, coil efficiency ε = (T_out - T_in) / (T_surface - T_in). The by-pass factor is defined as B.P.F. = (T_surface - T_out) / (T_surface - T_in). From these definitions, ε + B.P.F. = 1.Step-by-Step Solution:Start with B.P.F. = (T_s - T_out) / (T_s - T_in).Define ε = (T_out - T_in) / (T_s - T_in).Add the two: ε + B.P.F. = [(T_out - T_in) + (T_s - T_out)] / (T_s - T_in) = 1.Hence ε = 1 - B.P.F.Verification / Alternative check:Limiting cases: perfect coil contact gives B.P.F. → 0, so ε → 1; fully bypassed air gives B.P.F. → 1 and ε → 0, which is consistent.Why Other Options Are Wrong:Other algebraic forms contradict the standard definitions; for example, B.P.F. - 1 is negative for normal coils, which is nonphysical for an efficiency.Common Pitfalls:Mixing signs or inverting temperatures; forgetting that efficiency must lie between 0 and 1.Final Answer:1 - B.P.F.

Question 11

Psychrometric limit: At 100% relative humidity, the wet-bulb temperature is __________ the dry-bulb temperature.

Accepted Answer

Introduction / Context:Wet-bulb temperature (WBT) reflects evaporative cooling potential. At saturation (RH = 100%), air cannot accept more moisture, so evaporation from a wetted wick ceases and WBT converges to dry-bulb temperature (DBT).Given Data / Assumptions:Air–water vapour mixture at atmospheric pressure.Thermometer bulb properly wetted and ventilated.State point lies on the saturation curve (RH = 100%).Concept / Approach:At saturation, the partial pressure of water vapour equals the saturation pressure at that temperature. No net evaporation occurs from the wick, eliminating evaporative cooling. Therefore, WBT = DBT at the saturation state.Step-by-Step Solution:RH = 100% implies state on the saturation line.Evaporation rate from the wick → 0; latent cooling → 0.Thermometer stabilizes at the ambient dry-bulb temperature.Thus WBT equals DBT.Verification / Alternative check:On a psychrometric chart, lines of constant WBT meet the saturation curve exactly at the same point as DBT, confirming equality at RH = 100%.Why Other Options Are Wrong:“Lower” or “higher” occur only for unsaturated air; “unrelated” and “always 0°C” are physically incorrect.Common Pitfalls:Assuming some evaporation persists at 100% RH; in ideal measurement, none does.Final Answer:the same as

Question 12

Air-conditioning process metric: The sensible heat factor (SHF) for an air-conditioning process is defined as (use S.H. = sensible heat and L.H. = latent heat over the process):

Accepted Answer

Introduction / Context:The sensible heat factor (SHF) quantifies the proportion of sensible heat change in the total heat exchange of an HVAC process. It helps classify processes (pure sensible, cooling and dehumidification, humidification with heating, etc.).Given Data / Assumptions:S.H. is net sensible heat added or removed.L.H. is net latent heat added or removed (due to moisture change).Total heat = S.H. + L.H.Concept / Approach:By definition, SHF = S.H. / (S.H. + L.H.). A purely sensible process has L.H. = 0 and SHF = 1; pure latent process has S.H. = 0 and SHF = 0.Step-by-Step Solution:Write total heat over process = S.H. + L.H.Divide S.H. by total to obtain SHF.Therefore SHF = S.H. / (S.H. + L.H.).Interpret values: 0 ≤ SHF ≤ 1, indicating the share of sensible change.Verification / Alternative check:On the psychrometric chart, lines of constant SHF have characteristic slopes tied to the apparatus dew-point method.Why Other Options Are Wrong:Other ratios invert or subtract the correct relationship and do not yield normalized values between 0 and 1 for general processes.Common Pitfalls:Confusing SHF with room sensible heat factor (RSHF) or grand sensible heat factor (GSHF); those include additional loads but keep the same ratio idea.Final Answer:S.H. / (S.H. + L.H.)

Question 13

Reading the psychrometric chart: The horizontal, non-uniformly spaced lines on a standard psychrometric chart correspond to which property?

Accepted Answer

Introduction / Context:Being fluent with psychrometric charts is essential in HVAC. Different families of lines represent different properties; recognizing their orientation and spacing avoids reading errors.Given Data / Assumptions:Standard ASHRAE-style chart orientation.Pressure approximately 1 atm.Chart axes: humidity ratio (vertical) vs. dry-bulb temperature (horizontal).Concept / Approach:On typical charts: dry-bulb are vertical straight lines; humidity ratio lines are horizontal and generally uniformly spaced; wet-bulb are diagonally sloped lines; relative humidity are curved lines; dew-point temperature corresponds to horizontal projection to the saturation curve and is shown as horizontal, non-uniformly spaced markings along saturation.Step-by-Step Solution:Identify horizontal families: both humidity ratio and dew-point relate to horizontal moves.Humidity ratio lines are typically uniformly spaced grid lines.Dew-point scales along the saturation curve appear as horizontal references that are non-uniformly spaced due to non-linear saturation properties.Therefore, the description “horizontal and non-uniformly spaced” best matches dew-point temperature scales.Verification / Alternative check:Cross-check with a standard chart legend; RH lines are curved, DBT vertical, WBT sloped, w horizontal, and T_dp indicated along saturation with non-uniform spacing.Why Other Options Are Wrong:Dry-bulb are vertical; wet-bulb are slanted; specific humidity lines are horizontal but typically near-uniform in spacing; RH are curved.Common Pitfalls:Confusing humidity ratio horizontal lines with dew-point markings; remember dew-point is read at the saturation boundary.Final Answer:Dew-point temperature

Question 14

Cooling and dehumidification requirement: To cool and dehumidify a moist air stream across a coil, the coil surface temperature must be set __________ the incoming air’s dew-point temperature.

Accepted Answer

Introduction / Context:Dehumidification requires condensing moisture from air. The condition for condensation is governed by the dew-point temperature of the incoming stream and the coil surface (or apparatus dew-point).Given Data / Assumptions:Uniform coil surface temperature.Well-mixed outlet air leaving the coil.Steady operation near atmospheric pressure.Concept / Approach:Condensation begins when a surface is at or below the dew-point temperature of the adjacent moist air. For continuous cooling and dehumidification, the coil must be colder than the incoming air’s dew-point so that water vapour condenses on the fins/tubes.Step-by-Step Solution:Find the incoming dew-point T_dp from the psychrometric state.Set coil surface temperature T_s Result: both sensible cooling and latent removal occur.If T_s > T_dp, only sensible cooling occurs and humidity ratio remains unchanged.Verification / Alternative check:On the chart, cooling and dehumidification follows a line trending down and left toward the saturation curve, which only happens when the coil is below T_dp.Why Other Options Are Wrong:Temperatures between DBT and WBT may not be low enough to reach dew-point; adiabatic saturation temperature is unrelated to coil surface requirement.Common Pitfalls:Assuming any cooling causes dehumidification; unless T_s ≤ T_dp, no condensation occurs.Final Answer:lower than the incoming dew-point temperature

Question 15

Refrigerant safety comparison: Which of the following refrigerants is both highly toxic and flammable under typical service conditions?

Accepted Answer

Introduction / Context:Refrigerant selection weighs thermodynamic performance against safety and environmental criteria. Toxicity and flammability classifications (e.g., ISO/ASHRAE) guide application choices.Given Data / Assumptions:Standard industrial interpretations of toxicity and flammability.Normal HVAC&R service pressures and temperatures.Generalized safety classes (e.g., ammonia A3/B2L type considerations).Concept / Approach:Ammonia (R-717) is efficient but toxic and mildly flammable; it requires stringent safety measures (ventilation, detectors, restricted occupancy). CO2 (R-744) is non-toxic and non-flammable but operates at high pressure. SO2 is toxic but not commonly classified as flammable in HVAC usage. R-12 and R-134a are non-flammable and of low acute toxicity (environmental concerns aside).Step-by-Step Solution:Assess toxicity: Ammonia and SO2 are toxic; CO2 and common fluorocarbons have low acute toxicity.Assess flammability: Ammonia is flammable in certain concentration ranges; SO2 is not typically treated as flammable in refrigeration contexts.Therefore, the only option that is both toxic and flammable is ammonia.Hence, choose Ammonia (R-717).Verification / Alternative check:Safety standards place ammonia systems in machinery rooms with ventilation and leak detection; CO2 systems focus more on pressure management and asphyxiation risks, not flammability.Why Other Options Are Wrong:CO2 is non-flammable; SO2 is toxic but not flammable; R-12 and R-134a are non-flammable and lower acute toxicity (though they have environmental impacts).Common Pitfalls:Confusing toxicity with asphyxiation risk; CO2 can displace oxygen but is not flammable.Final Answer:Ammonia (R-717)

Question 16

HVAC duct sizing — choose the most widely used design approach In practical air-distribution design for buildings, several sizing methods exist (velocity reduction, equal friction, static regain, and dual methods). Which method is most commonly adop…

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Introduction / Context:Duct design in heating, ventilation, and air-conditioning (HVAC) must balance airflow delivery, noise, energy use, and cost. Several textbook methods exist, but one is favored in routine building design for its practicality and speed.Given Data / Assumptions:Goal: select the method most commonly used for initial duct sizing.Air distribution is for comfort applications at typical velocities and pressures.Fan total pressure and allowable friction loss per unit length guide choices.Concept / Approach:The equal friction method sets a uniform friction rate (pressure drop per unit length) for all main ducts. This leads to diameters that naturally taper with flow, limiting excessive velocities and simplifying balancing. It offers a good trade-off between accuracy and design time.Step-by-Step Solution:Choose a friction rate (e.g., 0.8–1.2 Pa/m) based on noise/energy criteria.Use charts or software to size each duct segment for the assigned flow at that friction.Iterate branches so that the network remains near the chosen friction rate.Check velocities and make minor adjustments to meet noise limits.Verification / Alternative check:Compare with static regain. While static regain can yield lower fan power for large systems, it needs more iteration and is less common for small to medium buildings. Velocity reduction is a rule-of-thumb start but not as systematic for pressure control.Why Other Options Are Wrong:Static regain method: energy-efficient for large high-velocity systems, but more complex and not the most common.Velocity reduction method: simple starting point, lacks consistent network pressure control.Dual or double method: specialized for dual-duct systems, not general sizing.Constant velocity method: oversimplifies and can mismanage pressure losses.Common Pitfalls:Choosing an unrealistically low friction rate (oversized ducts, high capital cost) or too high a rate (noise and energy penalty). Always confirm terminal velocities and sound criteria.Final Answer:equal friction method

Question 17

Refrigerant selection — identify the undesirable property For a refrigerant used in vapor-compression systems, which of the following is an undesirable thermophysical or safety property for practical HVAC/R applications?

Accepted Answer

Introduction / Context:Refrigerant choice affects efficiency, component size, environmental impact, and safety. Key properties include normal boiling point, flammability, toxicity, chemical stability, and compatibility with materials.Given Data / Assumptions:Comparing single properties in isolation for typical HVAC/R equipment.Focus on practical desirability in compressors, condensers, and evaporators.Concept / Approach:A desirable refrigerant should boil at a low temperature at near-atmospheric pressure to achieve evaporating temperatures below ambient while keeping suction pressures moderate. Safety attributes (non-toxic, non-flammable, non-explosive) and chemical stability are also desirable. Therefore, a high boiling point is undesirable because it forces high evaporator pressures and limits low-temperature operation.Step-by-Step Solution:Recall: low normal boiling point → effective cooling at moderate pressures.High boiling point → poor low-temperature capability or excessive pressure ratios.Hence, among the listed properties, “high boiling point” is undesirable.Verification / Alternative check:Examine pressure-temperature charts: refrigerants with low boiling points achieve sub-zero evaporator temperatures without extreme suction pressures; high-boiling fluids would require impractical conditions.Why Other Options Are Wrong:Non-toxic, non-flammable, non-explosive, and chemically stable are all desirable safety and reliability attributes.Common Pitfalls:Confusing normal boiling point with critical temperature; both matter, but the normal boiling point directly relates to evaporator operating ranges.Final Answer:high boiling point

Question 18

Radiation fundamental — name the law: Q = σ A T^4 Identify the classical radiation relation that states total emissive power of a black surface varies with the fourth power of absolute temperature.

Accepted Answer

Introduction / Context:Thermal radiation differs from conduction and convection because it requires no medium. A core result links total radiant emission from an ideal black surface to temperature.Given Data / Assumptions:Q = σ * A * T^4 where σ is the Stefan–Boltzmann constant, A is area, and T is absolute temperature.Relation applies to an ideal blackbody; real surfaces use Q = ε * σ * A * T^4 with emissivity ε.Concept / Approach:The statement describes the Stefan–Boltzmann law, a cornerstone of radiation heat transfer. It shows a strong temperature dependence: small increases in T cause large increases in radiative emission.Step-by-Step Solution:Recognize the T^4 dependence → radiation, not conduction or convection.Associate symbol σ with Stefan–Boltzmann constant.Therefore, select “Stefan–Boltzmann equation.”Verification / Alternative check:Fourier’s law uses q = −k * dT/dx (linear gradient); Newton’s law uses q = h * A * ΔT (linear in ΔT). Neither has a T^4 term.Why Other Options Are Wrong:Fourier: conduction law, no T^4.Newton–Rikhmann: convection correlation, linear in temperature difference.Joseph–Stefan: not a standard heat-transfer law.Planck’s law: spectral distribution, not total emissive power.Common Pitfalls:Confusing total emissive power (Stefan–Boltzmann) with wavelength-dependent emission (Planck). For real surfaces, include emissivity.Final Answer:Stefan–Boltzmann equation

Question 19

Heat rejection factor (HRF) for R-12/R-22 at 40°C condenser and 5°C evaporator In typical comfort-cooling applications with R-12 or R-22 and operating conditions Tc ≈ 40°C and Te ≈ 5°C, the heat rejection factor (ratio of condenser heat rejected to …

Accepted Answer

Introduction / Context:The heat rejection factor (HRF) relates the condenser heat load to the net refrigeration effect (evaporator load). It helps size condensers and cooling towers from known refrigeration capacities.Given Data / Assumptions:Refrigerants: R-12 or R-22 in legacy/typical textbook examples.Operating conditions: condenser about 40°C, evaporator about 5°C.Steady vapor-compression cycle with moderate superheat/subcooling.Concept / Approach:By definition HRF = Q_H / Q_L = 1 + W/Q_L. For many standard conditions with R-12 or R-22, the compressor work fraction W/Q_L is around 0.2–0.3, giving HRF ≈ 1.2–1.3. Thus, a commonly cited design value near these temperatures is about 1.25.Step-by-Step Solution:Start from HRF = (Q_L + W) / Q_L = 1 + (W/Q_L).At the stated temperatures, typical W/Q_L ≈ 0.25.Therefore HRF ≈ 1 + 0.25 = 1.25.Verification / Alternative check:Manufacturer data for similar conditions often show condenser loads ≈ 1.2–1.35 times the evaporator load, supporting 1.25 as a round design figure.Why Other Options Are Wrong:2.15 and 5.12 are far too high for these conditions, implying unrealistically large compressor work.1.00 would mean zero compressor work, impossible.1.60 is above typical range at these temperatures.Common Pitfalls:Confusing HRF with compressor heat ratio or using kW/ton multipliers incorrectly. Always compute HRF from energy balance.Final Answer:1.25

Question 20

Psychrometrics concept check “The heating of air without any change in its specific humidity is called sensible heating.” State whether the statement is correct.

Accepted Answer

Introduction / Context:Air-conditioning processes are classified by how temperature and moisture content change. Distinguishing sensible heating from humidification/dehumidification is foundational for coil selection and load calculations.Given Data / Assumptions:Sensible heating: temperature increases while moisture content (specific humidity or humidity ratio) remains constant.No moisture is added or removed by the process equipment.Concept / Approach:On a psychrometric chart, sensible heating tracks horizontally to the right at essentially constant humidity ratio. Relative humidity decreases because warmer air can hold more moisture, but the absolute moisture content does not change.Step-by-Step Solution:Define specific humidity w = mass of water vapour / mass of dry air.Sensible heating → increase in dry-bulb temperature, no change in w.Thus the statement is correct.Verification / Alternative check:Compare with sensible cooling (horizontal left) and cooling & dehumidification (left and down across saturation curve). Only when crossing saturation or contacting condensate does w decrease.Why Other Options Are Wrong:“False” and conditional variants impose restrictions that are not part of the definition; sensible heating does not require saturation or special flow conditions.Common Pitfalls:Confusing relative humidity change (it does change) with specific humidity (it does not). Remember: “sensible” modifies temperature only.Final Answer:True

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