Heat-Transfer LaboratoryIntroduction and ContextHeat transfer is the transport of thermal energy caused by a temperature difference. In mechanical engineering, experimental heat-transfer work connects governing equations with real surfaces, materials, sensors, fluid motion, radiation properties, and measurement uncertainty. Laboratory experiments are particularly important because ideal analytical models assume uniform properties, one-dimensional flow, perfect insulation, negligible contact resistance, and accurately known boundary conditions—conditions that are only approximately achieved in practice. This laboratory chapter develops experiments for steady and transient conduction, natural and forced convection, thermal radiation, extended surfaces (fins), and thermal contact resistance. The central task is to determine heat-transfer rates or coefficients from measurements and compare them with analytical or empirical predictions. Learning ObjectivesAfter completing this chapter, a student should be able to: - Explain the physical mechanisms of conduction, convection, and radiation.
- Apply Fourier’s law, Newton’s law of cooling, and the Stefan–Boltzmann law under appropriate assumptions.
- Distinguish steady from transient thermal behavior.
- Estimate thermal conductivity, convection coefficients, radiation properties, fin performance, and contact resistance from laboratory data.
- Select appropriate dimensionless correlations for natural and forced convection.
- Quantify uncertainty and identify dominant experimental errors.
- Compare measurements with models and justify discrepancies.
- Operate thermal laboratory equipment safely and professionally.
1. Experimental Basis of Heat Transfer1.1 Three modes of heat transferConduction is energy transfer through a stationary medium due to molecular and electronic interactions. For one-dimensional conduction through a plane wall, Q̇ = -kA dT/dx where Q̇ is heat-transfer rate (W), k is thermal conductivity (W m⁻¹ K⁻¹), A is area normal to heat flow (m²), and dT/dx is the temperature gradient (K m⁻¹). The negative sign indicates that heat flows toward decreasing temperature. For a plane wall of thickness L with constant k and uniform area, Q̇ = kA(T₁ − T₂)/L This relation assumes one-dimensional, steady conduction, constant thermal conductivity, and negligible heat generation. Convection is energy transfer between a surface and a moving fluid. Newton’s law of cooling is Q̇ = hAₛ(Tₛ − T∞) where h is the convection coefficient (W m⁻² K⁻¹), Aₛ is surface area, Tₛ is surface temperature, and T∞ is the bulk fluid temperature. The coefficient h is not a fluid property alone; it depends on geometry, fluid properties, velocity, flow regime, and temperature difference. Radiation is electromagnetic energy emitted by matter. For a surface exchanging radiation with large surroundings, Q̇rad = εσAₛ(Tₛ⁴ − Tsur⁴) where ε is emissivity, σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan–Boltzmann constant, and absolute temperatures must be in kelvin. This simplified expression assumes a diffuse-gray surface and large isothermal surroundings with view factor approximately equal to one. 1.2 Thermal resistanceHeat-transfer systems can often be represented by thermal resistances. For a plane wall and two convection films in series, Q̇ = (T∞,1 − T∞,2)/[1/(h₁A) + L/(kA) + 1/(h₂A)] The resistance analogy is useful for interpreting composite-wall and contact-resistance experiments. It must be used carefully when areas differ, properties vary strongly with temperature, or radiation and convection occur simultaneously. 1.3 Temperature measurementCommon instruments include thermocouples, resistance temperature detectors, thermistors, infrared sensors, and thermopiles. A thermocouple measures a voltage related to a temperature difference and requires suitable calibration and reference-junction compensation. Measurement errors may arise from sensor placement, conduction along sensor wires, poor surface contact, electrical noise, calibration drift, and thermal lag. A temperature should always be reported with its location, instrument type, unit, and estimated uncertainty. For example, “surface temperature at x = 0.10 m: 84.2 ± 0.5 °C” is more useful than an isolated numerical reading. 2. Laboratory Methodology and Data QualityA sound experiment begins with an energy balance. For a test section, Q̇in = Q̇useful + Q̇loss + Q̇storage At steady state, Q̇storage = 0. Electrical input to a heater is commonly estimated as Q̇elec = VI where V is voltage (V) and I is current (A). This equals useful heat only if electrical and environmental losses are accounted for. A practical steady-state criterion should be established before recording data. For example, readings may be considered stable when each monitored temperature changes by less than a specified value over a fixed interval. The criterion must be stated because “steady” is an operational judgment in a laboratory. Uncertainty and propagationIf a measured result y depends on independent variables x₁, x₂, …, the estimated standard uncertainty is commonly approximated by uᵧ = √[∑(∂y/∂xᵢ)²uᵢ²] where uᵢ is the standard uncertainty of xᵢ. For a calculated convection coefficient, h = Q̇/[Aₛ(Tₛ − T∞)] uncertainty may be dominated by the temperature difference when Tₛ and T∞ are close. Repeating measurements helps assess random variation but does not remove systematic errors such as a biased calibration. Good practice includes zero checks, calibration against known references, consistent sensor locations, repeated runs, a complete energy balance, and recording ambient pressure when it affects fluid properties. 3. Steady-State Conduction Experiment3.1 Objective and apparatusA typical apparatus consists of a heated metal bar or composite wall, a cooled end or heat sink, thermocouples at known positions, insulation, and a controlled electrical heater. The experiment may determine thermal conductivity or verify the temperature distribution. 3.2 Governing modelFor a uniform bar with constant cross-sectional area, no internal heat generation, negligible lateral loss, and one-dimensional steady conduction, the temperature distribution is linear: T(x) = T₁ − [(T₁ − T₂)/L]x The heat rate is Q̇ = kA(T₁ − T₂)/L Thus, if heat input is known, k = Q̇L/[A(T₁ − T₂)] If lateral heat loss is significant, the measured temperature profile becomes nonlinear and the simple equation overestimates or underestimates k depending on how Q̇ is represented. Guard heaters or insulation reduce this error. 3.3 Procedure- Measure specimen length, diameter or width, and cross-sectional area.
- Check thermocouple connections and heater supply.
- Apply a controlled heat input.
- Wait for the stated steady-state criterion.
- Record all temperatures, voltage, current, ambient temperature, and cooling conditions.
- Repeat at another heat input if safe and appropriate.
- Plot T against x and inspect linearity.
- Calculate k using the heat rate justified by the energy balance.
3.4 InterpretationA linear temperature profile supports the one-dimensional model. A curved profile may indicate heat loss, variable thermal conductivity, contact resistance, internal generation, or inaccurate sensor positions. The slope dT/dx is physically meaningful: a steeper slope indicates greater thermal resistance per unit heat rate. 4. Transient Conduction ExperimentTransient conduction concerns temperature changes with time. The governing heat equation for a homogeneous solid with constant properties and no heat generation is ∂T/∂t = α∇²T where α = k/(ρcₚ) is thermal diffusivity (m² s⁻¹), ρ is density (kg m⁻³), and cₚ is specific heat capacity (J kg⁻¹ K⁻¹). 4.1 Lumped-capacitance modelIf internal temperature gradients are small, the solid may be treated as having one uniform temperature. The energy balance is ρVcₚ dT/dt = −hAₛ(T − T∞) Integration gives (T − T∞)/(Tᵢ − T∞) = exp[−hAₛt/(ρVcₚ)] where Tᵢ is the initial temperature and t is time. The model is appropriate when the Biot number satisfies Bi = hL𝑐/k ≤ 0.1 Here L𝑐 = V/Aₛ is the characteristic length. The Biot number compares internal conduction resistance with external convection resistance. It is not the same as the Fourier number, Fo = αt/L𝑐² which measures the progress of transient diffusion. A linear plot of ln((T − T∞)/(Tᵢ − T∞)) against time has slope −hAₛ/(ρVcₚ). The slope method is often preferable to estimating h from one data point. 4.2 When lumped analysis failsIf Bi > 0.1, internal temperature gradients cannot be ignored. The body must be analyzed using spatial transient solutions, Heisler charts, numerical methods, or an experimentally fitted multi-node model. Applying lumped analysis outside its range can produce a systematically incorrect h. 5. Natural and Forced Convection5.1 Dimensionless groupsThe Nusselt number represents convection relative to conduction across a characteristic length: Nu = hL𝑐/kf The Reynolds number indicates the relative importance of inertia and viscosity in forced flow: Re = ρUL𝑐/μ = UL𝑐/ν The Prandtl number is Pr = μcₚ/kf = ν/α For natural convection, buoyancy is characterized by the Grashof number: Gr = gβ|Tₛ − T∞|L𝑐³/ν² and the Rayleigh number is Ra = GrPr Here ρ is density, U is characteristic velocity, μ is dynamic viscosity, ν is kinematic viscosity, kf is fluid thermal conductivity, β is volumetric expansion coefficient, g is gravitational acceleration, and α is thermal diffusivity. Properties should generally be evaluated at the film temperature, Tf = (Tₛ + T∞)/2, unless a correlation specifies otherwise. 5.2 CorrelationsExperimental correlations have the general form Nu = CReᵐPrⁿ for forced convection, or Nu = CRaⁿ for a specified natural-convection geometry and regime. Constants and exponents are not universal; they depend on geometry, boundary conditions, and flow range. A correlation must therefore be cited with its validity range. For internal flow, hydraulic diameter is Dₕ = 4A𝑐/P where A𝑐 is flow area and P is wetted perimeter. Using an inappropriate length scale can make an otherwise correct correlation appear inaccurate. 5.3 Experimental estimation of hAt steady state, an approximate convection coefficient is h = Q̇conv/[Aₛ(Tₛ − T∞)] The heat rate should be the heat actually transferred by convection, not automatically the electrical input. Radiation and conduction through supports may need to be subtracted. For a heated surface exposed to air, Q̇elec = Q̇conv + Q̇rad + Q̇support + Q̇other Comparisons with correlations should use consistent fluid properties and temperatures. 6. Thermal Radiation ExperimentA radiation experiment commonly uses a heated plate, thermocouples, a radiometer, or surfaces with different emissivities. The net exchange with large surroundings is Q̇rad = εσAₛ(Tₛ⁴ − Tsur⁴) Because the temperature is raised to the fourth power, kelvin must be used. Emissivity may be estimated by rearranging the equation if the net radiative heat rate is known: ε = Q̇rad/[σAₛ(Tₛ⁴ − Tsur⁴)] This estimate is sensitive to errors in heat rate, surface temperature, surroundings temperature, and area. In mixed-mode experiments, convection must be separated from radiation. A low-emissivity coating reduces radiation but may alter surface temperature and convection conditions. Real enclosures may require view factors and irradiation. For two diffuse-gray surfaces, the simple large-surroundings formula may be insufficient. Infrared cameras also require correct emissivity settings and attention to reflections; a camera reading is not automatically the true surface temperature. 7. Fins and Extended SurfacesFins increase heat transfer by increasing surface area. They are used in engine cooling, electronic heat sinks, air-cooled heat exchangers, and refrigeration equipment. For a straight fin of uniform cross-section with convection along its surface, constant k, one-dimensional conduction, and negligible radiation, define excess temperature θ = T − T∞. The fin equation is d²θ/dx² − m²θ = 0 where m = √(hP/kA𝑐) P is fin perimeter and A𝑐 is cross-sectional area. For an insulated-tip fin of length L, Q̇fin = √(hPkA𝑐) θᵦ tanh(mL) where θᵦ = Tᵦ − T∞ is base excess temperature. For an infinitely long fin, Q̇fin,∞ = √(hPkA𝑐) θᵦ A useful performance measure is fin efficiency: ηfin = Q̇fin/[hA𝑓θᵦ] where A𝑓 is fin surface area exposed to convection. Fin effectiveness is εfin = Q̇fin/[hAᵦθᵦ] where Aᵦ is the base area that would transfer heat without the fin. A fin is beneficial when effectiveness exceeds one, and practical designs usually require substantially greater effectiveness to justify manufacturing cost and pressure loss. Measured fin temperatures should decrease away from the base. A fit to the theoretical profile can provide h, but the result is affected by base contact resistance, tip losses, radiation, and uncertainty in k. 8. Thermal Contact ResistanceEven apparently smooth solid surfaces touch only at microscopic asperities. The actual contact area is much smaller than the apparent area, producing a temperature drop at the interface. Contact resistance is defined as R𝑐 = ΔT𝑐/Q̇ and contact conductance is h𝑐 = 1/(R𝑐A) where ΔT𝑐 is the interface temperature difference and A is the apparent contact area. Contact resistance depends on surface roughness, flatness, material hardness, contact pressure, oxidation, interstitial fluid, and thermal conductivity. In a laboratory stack of two specimens, measure temperatures sufficiently close to the interface and determine Q̇ from a justified conduction model. The interface resistance is then obtained after subtracting known bulk-wall temperature drops. This subtraction can amplify uncertainty, especially when ΔT𝑐 is small. Increasing clamping pressure usually increases contact conductance by enlarging real contact area, but the relationship is not necessarily linear. Thermal paste may reduce resistance by filling voids, although excessive thickness or poor application can increase resistance. 9. Worked Example: Lumped Transient CoolingA small copper sphere has diameter 20 mm. It initially is at 120 °C and cools in air at 25 °C. Take h = 20 W m⁻² K⁻¹, k = 390 W m⁻¹ K⁻¹, ρ = 8900 kg m⁻³, and cₚ = 385 J kg⁻¹ K⁻¹. Estimate its temperature after 60 s using lumped analysis. For a sphere, V = πD³/6 = π(0.020)³/6 = 4.19 × 10⁻⁶ m³ and Aₛ = πD² = π(0.020)² = 1.257 × 10⁻³ m² The characteristic length is L𝑐 = V/Aₛ = 0.00333 m Thus, Bi = hL𝑐/k = (20)(0.00333)/390 = 1.71 × 10⁻⁴ Since Bi < 0.1, lumped analysis is valid. The time constant factor is hAₛ/(ρVcₚ) = (20)(1.257 × 10⁻³)/[(8900)(4.19 × 10⁻⁶)(385)] = 0.00176 s⁻¹ Therefore, T − T∞ = (Tᵢ − T∞)e⁻⁰·⁰⁰¹⁷⁶×⁶⁰ T = 25 + 95e⁻⁰·¹⁰⁶ = 115.4 °C The small temperature decrease is physically plausible because copper has high volumetric heat capacity relative to the available air-side area and the convection coefficient is modest. 10. Illustrative Case: Diagnosing a Fin ExperimentBackground. A rectangular aluminum fin is mounted on a heated base and exposed to laboratory air. Thermocouples show that the measured temperature profile is consistently hotter than the analytical profile near the tip. Problem. The calculated fin heat rate, using the insulated-tip model, is 18 W, while the electrical energy balance suggests only 14 W is transferred through the fin assembly. Analysis. The discrepancy cannot be resolved by changing h arbitrarily. First, the student checks whether the electrical input includes heat conducted through supports and radiation from the base. Next, the tip is inspected: it is not insulated, so the insulated-tip boundary condition is inappropriate. The base-to-fin joint is also found to have a thin air gap, producing contact resistance. Finally, thermocouple attachment is examined because exposed beads may measure local air temperature rather than the solid temperature. Approach. The revised analysis includes a convective tip boundary condition, a measured or estimated contact resistance, and a separate radiation-loss estimate. A repeat experiment uses improved clamping and calibrated sensors. Data are compared with the corrected model over several heat inputs. Outcome. The corrected prediction falls within the combined experimental uncertainty of the energy-balance result. The remaining difference is attributed mainly to uncertain natural-convection conditions and radiation. Learning points. Model selection must follow the physical apparatus. An apparent disagreement may result from boundary conditions, parasitic heat paths, contact resistance, or instrumentation rather than from failure of heat-transfer theory. 11. Safety, Professional Practice, and LimitationsHeaters, hot surfaces, electrical supplies, rotating fans, compressed air, and fragile glass components create hazards. Students should use guards, insulated tools, appropriate gloves, eye protection, and emergency isolation procedures. Never touch a specimen to judge temperature, and allow sufficient cooling before dismantling equipment. Electrical power must be checked against equipment ratings. Fans and ducted systems require secure guards and unobstructed flow. Infrared sources may cause burns or eye injury. Hot oils and fluids require material-safety procedures and spill controls. Professional reporting requires raw data, calibration information, equations, assumptions, units, uncertainty estimates, sample calculations, graphs with labeled axes, and a discussion of discrepancies. Results should not be adjusted merely to agree with a correlation. If data are excluded, the exclusion criterion must be stated before or independently of the desired conclusion. Contemporary developmentsModern laboratories increasingly use computerized acquisition, infrared thermography, automated flow control, and numerical parameter estimation. These tools improve resolution but do not eliminate physical uncertainty. AI-assisted analysis can help identify trends, fit models, or detect anomalous readings; however, students must retain responsibility for sensor calibration, model assumptions, data provenance, and engineering judgment. A statistically excellent fit is not evidence that the underlying heat-transfer mechanism is correct. Chapter SummaryHeat-transfer laboratory experiments translate conduction, convection, radiation, and thermal-contact theory into measurable engineering quantities. Fourier’s law describes conduction, Newton’s law provides a practical convection model, and the Stefan–Boltzmann law describes net radiation under specified assumptions. Steady conduction experiments use temperature gradients and energy balances to estimate thermal conductivity. Transient experiments use thermal diffusivity, Biot number, Fourier number, and lumped-capacitance analysis to interpret temperature histories. Natural and forced convection require dimensionless groups and correlations whose validity ranges must be respected. Fins improve heat transfer by increasing area, but their effectiveness depends on conductivity, geometry, convection, contact quality, and radiation. Contact resistance arises from microscopic surface roughness and can substantially affect measured temperature drops. Across all experiments, reliable conclusions depend on calibration, energy balances, uncertainty analysis, physically appropriate boundary conditions, and transparent comparison between model and observation. Key TermsThermal conductivity: Material property measuring conduction ability, in W m⁻¹ K⁻¹. Convection coefficient: Proportionality coefficient between surface-to-fluid temperature difference and convective heat flux. Emissivity: Ratio of a real surface’s emitted radiation to blackbody emission at the same temperature. Steady state: Condition in which measured temperatures and energy storage are essentially time-independent. Transient conduction: Heat conduction during which temperatures vary with time. Thermal diffusivity: α = k/(ρcₚ), measuring the rate of temperature-field response. Biot number: Ratio of internal conduction resistance to external convection resistance. Fourier number: Dimensionless time for thermal diffusion. Nusselt number: Dimensionless convection coefficient. Reynolds number: Ratio of inertial to viscous effects in fluid flow. Prandtl number: Ratio of momentum diffusivity to thermal diffusivity. Grashof number: Dimensionless measure of buoyancy relative to viscous effects. Rayleigh number: Product of Grashof and Prandtl numbers. Fin: Extended surface used to increase heat-transfer area. Fin efficiency: Actual fin heat transfer divided by ideal transfer if the whole fin were at base temperature. Fin effectiveness: Fin heat transfer divided by heat transfer from the unextended base area. Thermal contact resistance: Temperature drop per unit heat rate at a contacting interface. Film temperature: Approximate mean of surface and bulk-fluid temperatures used for property evaluation. Systematic error: Consistent bias caused by calibration, method, or apparatus. Random uncertainty: Scatter that varies unpredictably between repeated measurements. Review QuestionsShort Answer Questions- State Fourier’s law for one-dimensional conduction and explain its negative sign.
- Why must absolute temperature be used in the Stefan–Boltzmann equation?
- What physical comparison is represented by the Biot number?
- Distinguish fin efficiency from fin effectiveness.
- Identify two causes of thermal contact resistance.
Long Answer Questions- Describe a steady conduction experiment for determining thermal conductivity, including apparatus, assumptions, measurements, calculations, and sources of error.
- Derive the lumped-capacitance cooling relation from an energy balance and state its validity criterion.
- Explain how natural-convection and forced-convection experiments differ in apparatus, governing dimensionless groups, and data reduction.
- Discuss the assumptions behind the straight-fin equation and explain how real fin behavior may depart from the model.
- Explain how radiation and convection can be separated in a heated-surface experiment using an energy balance.
Analytical / Application Questions- A specimen cools from 100 °C toward room air at 20 °C. Explain how a semilogarithmic temperature plot can be used to estimate h, and identify the measurements required.
- A measured fin profile is less steep than predicted, but the total heat rate is also lower than predicted. Analyze at least three possible physical or instrumentation causes and propose diagnostic tests.
- Two metal blocks show a temperature discontinuity at their interface under steady heat flow. Formulate a method to estimate contact resistance, including the necessary temperature measurements and uncertainty considerations.
Multiple Choice Questions with AnswersQuestion 1Which equation represents one-dimensional steady conduction through a plane wall of constant area and conductivity? A. Q̇ = hAₛ(Tₛ − T∞) B. Q̇ = kA(T₁ − T₂)/L C. Q̇ = εσAₛ(Tₛ⁴ − Tsur⁴) D. Q̇ = ρVcₚ dT/dt Correct Answer: B. Q̇ = kA(T₁ − T₂)/L Question 2For lumped-capacitance analysis, the usual validity criterion is: A. Bi ≥ 10 B. Bi = 1 C. Bi ≤ 0.1 D. Fo ≤ 0.1 Correct Answer: C. Bi ≤ 0.1 Question 3Which temperature scale must be used in the Stefan–Boltzmann equation? A. Celsius B. Fahrenheit C. Kelvin D. Rankine only Correct Answer: C. Kelvin Question 4The Reynolds number primarily compares: A. Radiation and convection B. Inertia and viscosity C. Conduction and radiation D. Buoyancy and thermal diffusion only Correct Answer: B. Inertia and viscosity Question 5For natural convection, the Rayleigh number is defined as: A. Re/Pr B. GrPr C. Nu/Re D. Pr/Gr Correct Answer: B. GrPr Question 6What is the main purpose of a fin? A. To reduce surface area B. To eliminate conduction C. To increase heat-transfer area D. To prevent all radiation Correct Answer: C. To increase heat-transfer area Question 7If a body has Biot number much smaller than 0.1, which approximation is generally appropriate? A. Uniform internal temperature B. Zero convection coefficient C. Infinite internal temperature gradient D. Radiation-only transfer Correct Answer: A. Uniform internal temperature Question 8Thermal contact resistance is calculated from: A. R𝑐 = Q̇/ΔT𝑐 B. R𝑐 = ΔT𝑐/Q̇ C. R𝑐 = kA/L D. R𝑐 = hA Correct Answer: B. R𝑐 = ΔT𝑐/Q̇ Question 9In a steady heated-surface experiment, electrical input may exceed convective heat transfer because of: A. Only measurement rounding B. Radiation and support losses C. The definition of temperature D. The use of kelvin Correct Answer: B. Radiation and support losses Question 10For a sphere of diameter 20 mm, the volume is approximately: A. 4.19 × 10⁻⁶ m³ B. 1.26 × 10⁻³ m³ C. 2.00 × 10⁻² m³ D. 8.38 × 10⁻⁶ m³ Correct Answer: A. 4.19 × 10⁻⁶ m³
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