Overview
This unit studies the three modes of heat transfer: conduction, convection and radiation. It explains how thermal energy moves through solids, fluids and across empty space, and how material properties and geometry affect transfer rates. The unit covers Fourier’s law for conduction, thermal resistance and conductance, steady and transient conduction in one dimension, Newton’s law of cooling for convection, boundary layers, free and forced convection, and Stefan–Boltzmann law for thermal radiation including emissivity and absorptivity. Practical topics include combining modes, insulating materials, heat exchangers, and simple calculations used in engineering design. Understanding these modes is essential for thermal management in engines, buildings, electronics and industrial processes. Mastery helps in selecting materials, estimating temperature changes, designing insulation, and predicting heat loss or gain. The unit also introduces problem-solving techniques: making assumptions, using appropriate formulas, setting up energy balances, and checking units and limits. Overall, this unit builds the foundation needed to analyse thermal systems, design for safety and efficiency, and understand everyday phenomena like warming by sunlight, cooling by breeze, or why metals feel colder than wood at the same temperature.
Learning Objectives
- Describe the basic mechanisms of heat transfer by conduction, convection and radiation.
- Apply Fourier’s law to calculate heat flow in one-dimensional steady conduction.
- Compute thermal resistance and use it to model composite walls and layered systems.
- Solve simple transient conduction problems using lumped capacitance and basic time constants.
- Use Newton’s law of cooling to estimate convective heat transfer and recognise free versus forced convection.
- Apply correlations for convective heat transfer coefficients for simple flows and geometries.
- Use the Stefan–Boltzmann law to compute radiative heat exchange between surfaces and account for emissivity.
- Combine conduction, convection and radiation to analyse practical heat transfer situations and perform basic design calculations.
Topics in this chapter
19 topics · tap a topic title to jump straight to it.
Introduction to heat transfer and basic concepts
Heat transfer is the movement of thermal energy between regions that are at different temperatures. In engineering, we treat heat transfer as a process that tries to reduce temperature differences until thermal equilibrium is reached. The three classical mechanisms are conduction, convection and radiation. Each mechanism has distinct causes and dominant situations. Conduction is internal to a material and caused by microscopic motion and interactions of particles. Convection appears when fluids move and carry heat with them, combining conduction in boundary layers with bulk advection. Radiation is electromagnetic emission from surfaces and can transmit energy across a vacuum.
Key quantities used in heat transfer are temperature, heat flux and heat transfer rate. Temperature measures thermal state; heat flux is heat flow per unit area with units W m−2; and heat rate is power in watts (W). The temperature gradient, the spatial rate of change of temperature, is the driving factor for conduction. Material properties such as thermal conductivity k, density ρ and specific heat c determine how heat moves and is stored. Thermal diffusivity α = k/(ρ c) determines the speed of temperature diffusion in transient problems.
Engineering models often simplify reality: one-dimensional conduction, steady-state approximations, constant material properties, or linearised radiation. Each assumption narrows applicability but makes problems solvable by hand. Another powerful idea is thermal resistance, which treats heat transfer analogously to electrical current flow. Resistances in series or parallel represent conduction layers and convective films and allow quick calculation of heat rates. The overall heat transfer coefficient U is useful for assemblies and building elements because it connects heat rate directly to temperature difference via Q̇ = U A ΔT.
Boundary conditions are essential: they specify temperatures, heat fluxes, or convective/radiative exchanges at surfaces. Correct identification of boundary conditions determines the mathematical solution. Units and careful bookkeeping prevent common mistakes: check that areas, lengths and temperatures are in SI units and that temperatures are converted to Kelvin when using radiation laws. In practice, engineers combine analytic formulas, empirical correlations and numerical methods to model real systems. The rest of this unit develops these tools, shows how to choose appropriate models, and gives exercises relevant to thermal design and everyday engineering problems.
- Touching a metal spoon and a wooden spoon at the same temperature — metal feels colder because it conducts heat faster.
- Sunlight warming a classroom — radiation transfers energy through glass and air.
- Feeling cooler in a breeze — moving air increases convective heat transfer from your skin.
- Heat transfer occurs from high to low temperature; driven by ΔT.
- Thermal conductivity: k (W m−1 K−1).
- Heat flux q = heat rate/area (W m−2).
Conduction: microscopic origin and Fourier’s law
Conduction transfers heat within materials by microscopic mechanisms: in metals free electrons carry significant energy, scattering and transporting thermal energy; in insulators lattice vibrations (phonons) dominate. These microscopic processes collectively produce macroscopic heat flow that depends on local temperature gradients. Fourier’s law summarises this macroscopic behaviour: heat flux at a point is proportional to the negative of the temperature gradient, meaning heat flows from hotter to colder regions.
In one-dimensional form Fourier’s law is written q_x = -k dT/dx. Here q_x is heat flux (W m−2), k is thermal conductivity (W m−1 K−1) and dT/dx is spatial gradient of temperature. The negative sign shows the flux direction is opposite to the gradient sign. For a bar of uniform cross-section A, the heat rate Q̇ = -k A dT/dx. If conductivity k is constant and steady conditions hold, temperature varies linearly and Q̇ simplifies to Q̇ = k A (T1 - T2)/L for a slab of thickness L with end temperatures T1 and T2.
Thermal conductivity is a material property that can vary strongly with temperature, especially in metals and semiconductors. In many engineering problems k is assumed constant over the temperature range for simplicity, but when accuracy is needed include temperature dependence or use mean values evaluated at film temperature. Materials can be isotropic (same k in all directions) or anisotropic (different k along different axes); layered composites require careful treatment of directional conductivities and series/parallel paths.
Conduction is governed by the same mathematics as other diffusion processes. Combining Fourier’s law with energy conservation yields the heat equation. In steady state, the second derivative of temperature equals zero in regions without internal heat generation, leading to linear profiles in simple geometries. With internal heat generation (for example electrical heating distributed in a resistor) the steady profile becomes parabolic in one dimension. For cylindrical and spherical geometries the changing area with radius leads to logarithmic or reciprocal solutions after integrating Fourier’s law; these are important for pipes and spherical tanks.
Understanding Fourier’s law also guides practical decisions: metals transfer heat quickly so they are good for heat sinks; insulators have low k and reduce heat loss. The electrical analogy (treating thermal conductivity as inverse resistance) simplifies many multi-layer conduction problems and connects naturally to combined conduction–convection analyses where surface films add resistance terms. Always state assumptions when applying Fourier’s law so the solution remains valid for the intended engineering case.
- Calculate heat flow through a plane wall: Q̇ = k A (T1 − T2)/L for a uniform slab.
- Explain why a copper rod heats faster along its length than an insulating rod when one end is heated.
- Fourier’s law (1D): q_x = -k dT/dx
- Heat rate through slab: Q̇ = k A (T1 - T2)/L
Thermal resistance and heat flow through composite walls
Thermal resistance treats heat transfer similar to electrical current flow: temperature difference drives heat rate just as voltage difference drives current. For a single plane layer of thickness L, cross-sectional area A and conductivity k, the conductive thermal resistance is R_cond = L/(k A). Its unit is K W−1. When layers are stacked in series, resistances add: R_total = Σ_i L_i/(k_i A). The heat rate then becomes Q̇ = (T_hot - T_cold)/R_total, a form convenient for practical design.
Convective heat transfer at surfaces can be included as boundary resistances. For a surface with convective coefficient h, R_conv = 1/(h A). Thus, a wall exchanging heat with fluids on both sides uses R_total = 1/(h1 A) + L/(k A) + 1/(h2 A). This algebraic form is widely used in building heat loss calculations and in simple equipment design where one-dimensional heat flow assumption holds.
Cylindrical and spherical shells require different expressions because area varies with radius. For radial conduction through a cylinder of length L between radii r1 and r2, R_cyl = ln(r2/r1)/(2πkL). For a spherical shell between r1 and r2, R_sph = (1/(4πk)) (1/r1 - 1/r2). These forms come from integrating Fourier’s law across a radial coordinate and are essential when insulating pipes or spherical tanks.
Parallel heat paths such as studs in a wall create parallel resistances. Compute each path resistance and combine with 1/R_eq = Σ 1/R_i. This is important in real wall assemblies where thermal bridging can significantly increase overall heat transfer. Contact resistance at interfaces and thin air gaps may also introduce additional resistance terms and should be included when accuracy is required.
The resistance method also yields the overall heat transfer coefficient U by U = 1/(R_total A). U is useful when comparing different assemblies and for heat exchanger calculations. Keep in mind the method assumes steady-state, one-dimensional flow and uniform properties. When multi-dimensional effects or transient changes matter, more detailed analysis or numerical methods are necessary. Nevertheless, thermal resistance is a powerful engineering tool that simplifies preliminary design and helps identify which layers or changes most effectively reduce heat loss.
- Composite wall with two layers: R_total = L1/(k1A) + L2/(k2A); then Q̇ = ΔT/R_total.
- Heat loss through a pipe with insulation: use cylindrical resistance R = ln(r2/r1)/(2πkL) plus convective resistances at surfaces.
- Plane layer conductive resistance: R = L/(kA)
- Convective resistance: R = 1/(hA)
- Cylindrical conduction: R = ln(r2/r1)/(2πkL)
- Series resistances: R_total = Σ R_i
Steady one-dimensional conduction in plane walls
Steady one-dimensional conduction in plane walls assumes temperature varies only across the thickness and does not change with time. This idealisation applies when wall area is large compared to thickness and heat flows perpendicular to the faces, so edge effects can be neglected. For constant thermal conductivity and no internal heat generation, the temperature distribution is linear between face temperatures; the heat rate is Q̇ = k A (T1 - T2)/L.
When surfaces are in contact with fluids the surface temperatures are not known a priori. We include convective heat transfer coefficients h1 and h2 to model exchange with fluids at temperatures T∞1 and T∞2. The combined expression becomes Q̇ = (T∞1 - T∞2)/[1/(h1 A) + L/(k A) + 1/(h2 A)]. This places the conductive resistance of the wall between two convective resistances and is directly comparable to an electrical circuit with resistors in series. It helps compute heat loss from building envelopes and the required insulation thickness to meet energy targets.
If thermal conductivity varies with temperature, integrate 1/k(T) across the thickness: Q̇ = A (T1 - T2)/∫(dx/k(x)). For walls with uniform internal heat generation q''' (W m−3), the steady-state differential equation becomes k d2T/dx2 + q''' = 0; solving gives a parabolic temperature profile with a maximum inside the wall. Boundary conditions (specified temperature or flux at faces) determine constants of integration and the resulting heat flow.
Practical checks include ensuring the assumption of one-dimensional heat flow is valid (edges or corners may cause multidimensional effects) and that steady-state is appropriate (no strong time dependence). For thin layered walls with different k values use the resistance method, while for problems with significant two- or three-dimensional effects use numerical methods or more advanced analytical solutions. For teaching and examination problems this steady one-dimensional model provides a robust foundation from which to understand more complex behaviours.
Design exercises commonly ask for heat loss through a wall, interior surface temperatures given outside conditions, or insulation thickness required to restrict heat flow to a specified value. Always show the resistance network, state assumptions, and check units to avoid common mistakes. This steady one-dimensional approach is widely used in preliminary engineering estimates and building heat loss calculations.
- A plane wall 0.2 m thick, k = 0.5 W m−1 K−1, area 10 m2, with temperatures 40 °C and 10 °C: Q̇ = kAΔT/L.
- Wall with convection on both sides: include R_conv and R_cond to find Q̇.
- Heat flow plane wall: Q̇ = k A (T1 - T2)/L
- Combined: Q̇ = (T∞1 - T∞2)/[1/(h1A) + L/(kA) + 1/(h2A)]
Steady conduction in cylindrical and spherical coordinates
Radial conduction in cylindrical and spherical geometries differs from plane conduction because the area available for heat flow changes with radius. For a cylinder of length L, inner radius r1 and outer radius r2, steady radial conduction integration of Fourier’s law yields Q̇ = 2πkL (T1 - T2)/ln(r2/r1). The corresponding thermal resistance per unit length is R' = ln(r2/r1)/(2πk). These equations assume radial symmetry and constant k, and are widely used for pipes, electrical cables and insulating jackets.
For a spherical shell between radii r1 and r2, integrate over spherical area 4πr2 to obtain Q̇ = 4πk (T1 - T2)/(1/r1 - 1/r2). The thermal resistance is R = (1/(4πk)) (1/r1 - 1/r2). Compared to cylinders, spheres often have smaller resistance for a given volume because the area increases rapidly with radius; this makes small spheres harder to cool per unit area than large ones.
When dealing with thin-walled cylinders where thickness t = r2 - r1 is much smaller than r1, the plane approximation Q̇ ≈ k A ΔT / t may be used for simplicity, but the exact logarithmic form is recommended when accuracy matters. For multi-layer cylindrical insulation add resistances in series: R_total per length = Σ ln(r_{i+1}/r_i)/(2πk_i). Include convective resistances at inner and outer surfaces to find total heat loss per unit length: R_total = 1/(h_i 2π r_i L) + Σ ln(r_{i+1}/r_i)/(2πk_i L) + 1/(h_o 2π r_o L).
Applications include heat loss from steam pipes, design of thermal jackets and cryogenic tanks, and analysis of spherical pressure vessel heat transfer. Always use radii in metres, length in metres and thermal conductivity in proper SI units. Check limiting behaviours: as r2 → r1 the logarithm tends to t/r which recovers plane form; as k → ∞ resistance tends to zero and the body is nearly isothermal. Mastery of cylindrical and spherical conduction formulas is essential for correct analysis of many common engineering components.
- Heat loss per metre from a hot pipe with insulation: use R = ln(r2/r1)/(2πkL) plus convective resistances.
- Temperature difference across a spherical cavity wall computed from Q̇ = 4πk (T1 - T2)/(1/r1 - 1/r2).
- Cylindrical conduction: Q̇ = 2πkL (T1 - T2)/ln(r2/r1)
- Cylindrical resistance: R = ln(r2/r1)/(2πkL)
- Spherical conduction: Q̇ = 4πk (T1 - T2)/(1/r1 - 1/r2)
- Spherical resistance: R = (1/(4πk)) (1/r1 - 1/r2)
Lumped system analysis and transient conduction basics
Transient conduction looks at how temperatures change with time. The lumped capacitance model greatly simplifies many transient problems by assuming the temperature inside a body is spatially uniform at any instant. This is valid when internal conduction is fast relative to surface heat transfer, which the Biot number quantifies: Bi = h L_c / k. If Bi ≤ 0.1, internal gradients are small and lumped analysis is acceptable.
Derive the lumped solution from energy balance: rate of decrease of internal energy equals convective heat loss, m c dT/dt = -h A (T - T∞). Separate variables and integrate to get an exponential decay: (T - T∞)/(T_i - T∞) = exp(-t/τ), where τ = m c/(h A) is the characteristic time constant. After one time constant the temperature difference has fallen to 37% of its initial value; after about 3τ it is near 5% and considered close to steady. This result lets engineers estimate cooling or heating times for small parts, food items, or electronic components.
If Bi > 0.1, internal conduction cannot be ignored and temperature varies within the body. Then the heat equation must be solved. For simple shapes (infinite slab, infinite cylinder, sphere) analytical series solutions exist; the nondimensional Fourier number Fo = α t / L_c^2 measures diffusion progress and appears in these solutions. Early-time behaviour (small Fo) differs from later-time behaviour and Heisler charts or transient tables are practical tools to obtain solutions without summing infinite series.
Semi-infinite solids represent another important class: when thermal penetration is small compared to object size, the semi-infinite approximation yields solutions based on the error function. These are used for problems like sudden heating of a wall surface or short pulses on a solid. For complex geometries or coupling with convection and radiation, numerical methods (finite difference, finite element) are used. Always check Bi before choosing lumped method and compute τ to understand time scales of heating or cooling in your system.
- A small metal sphere cooling in air: check Bi; if Bi ≤ 0.1 use lumped method to find cooling time.
- Estimate time for a hot plate to cool to within 1% of ambient using τ = m c/(h A).
- Biot number: Bi = h L_c / k
- Fourier number: Fo = α t / L_c^2
- Lumped-capacitance solution: (T - T∞)/(T_i - T∞) = exp(-t/τ)
- Time constant: τ = m c/(h A)
General heat diffusion (the heat equation)
The heat equation, also called the diffusion equation, governs transient conduction. For a homogeneous material with thermal diffusivity α = k/(ρ c), the one-dimensional form is ∂T/∂t = α ∂2T/∂x2. In three dimensions the Laplacian ∇2 appears: ∂T/∂t = α ∇2T. If internal volumetric heat generation q''' exists, add a source term: ∂T/∂t = α ∇2T + q'''/(ρ c). Solving this partial differential equation requires initial temperature distribution and boundary conditions (prescribed temperature, heat flux, convective boundary, or symmetry).
Analytical solutions are available for several canonical problems. For an infinite slab with uniform initial temperature and convective boundaries, solutions use eigenfunction expansions and series terms that decay exponentially with time. For semi-infinite bodies subjected to a sudden surface temperature change or a prescribed surface heat flux, the error function solution is standard and gives temperature as a function of depth and time. These analytical solutions are useful for verification, insight and constructing approximate formulas.
Dimensionless numbers make solutions general. The Fourier number Fo = α t / L_c^2 measures relative diffusion progress; small Fo indicates early times, large Fo indicates near steady-state. The Biot number Bi helps determine whether lumped analysis applies. Solutions are often expressed in nondimensional form so they scale to different sizes and materials by substitution. Heisler charts present nondimensional temperatures for common geometries as functions of Fo and Bi and are practical for hand calculations.
When geometry is complex, properties vary, or boundary conditions are non-linear (radiation), numerical methods are used. Finite difference and finite element methods discretise space and time to approximate derivatives. Numerical approaches require stability and convergence checks and often produce large algebraic systems solved by computers. In engineering practice combine analytic solutions where applicable with numerical methods for full designs, and always interpret results physically to ensure they are plausible and meet design constraints.
- Semi-infinite solid suddenly exposed to a different surface temperature — use error function solution to find temperature vs depth and time.
- Long bar with internal heat generation leads to steady parabolic temperature profile in one dimension.
- Heat equation (1D): ∂T/∂t = α ∂2T/∂x2
- Thermal diffusivity: α = k/(ρ c)
Convection: physical mechanism and Newton’s law of cooling
Convection is heat transfer between a solid surface and a moving fluid. It involves two processes: microscopic conduction in the very thin layer adjacent to the surface (the thermal boundary layer) and bulk transport of heat by moving fluid (advection). Near the surface, velocity and temperature change rapidly; the thickness of the velocity boundary layer and the thermal boundary layer determines local heat transfer rates.
Newton’s law of cooling models convective heat flux from a surface with q_conv = h (T_s - T∞), where h is the convective heat transfer coefficient, T_s is surface temperature and T∞ is free-stream temperature. The coefficient h depends on fluid properties, flow velocity, surface geometry and orientation. It is not a pure material property and must be obtained from experiments or empirical correlations. Because h often varies with position on a surface, engineers typically use an average h for design calculations.
Convection is divided into forced and free (natural) regimes. Forced convection occurs when an external device such as a fan or pump moves fluid over the surface; forced flows generally yield higher h values because fresh fluid is continuously supplied. Free convection arises when buoyancy forces due to density differences (caused by temperature differences) drive the flow; it is important for passive cooling and natural ventilation. Mixed convection combines both effects when both buoyancy and forced motion are significant.
To relate h to flow properties we use dimensionless numbers. Nusselt number Nu = h L / k non-dimensionalises heat transfer and is correlated with Reynolds and Prandtl numbers for forced flows (Nu = f(Re, Pr)) and with Grashof and Prandtl numbers for natural convection (Nu = f(Gr, Pr)). Characteristic length L depends on geometry; for a flat plate it is typically the plate length, for a cylinder the diameter, and for ducts the hydraulic diameter. Practical design proceeds by identifying flow regime, selecting appropriate correlation, computing Nu and hence h, and then using q = h A ΔT for heat rate calculation.
- Estimate heat loss from a heated plate in air with known velocity: use Nu correlations to find h and apply q = hAΔT.
- Cup of hot tea cooling naturally: free convection around the cup reduces temperature according to convective heat loss.
- Newton’s law of cooling: q_conv = h (T_s - T∞)
- Nusselt number: Nu = h L / k
Nusselt, Reynolds, Prandtl and Grashof numbers
Dimensionless numbers are central to convective heat transfer because they collapse many physical parameters into combinations that describe flow and heat transfer regimes. The Reynolds number Re = ρ V L / μ compares inertial and viscous forces. A low Re indicates viscous-dominated, laminar flow; a high Re implies inertia-dominated, turbulent flow. For flow in pipes the common thresholds are roughly Re < 2300 for laminar and Re > 4000 for turbulent flow, with a transition region in between.
The Prandtl number Pr = μ c_p / k measures the relative thickness of velocity and thermal boundary layers by comparing momentum diffusivity (ν = μ/ρ) to thermal diffusivity (α = k/(ρ c)). Low-Pr fluids like liquid metals have very thin thermal boundary layers compared to their velocity layers; high-Pr fluids like oils have thicker thermal layers. Many empirical correlations for heat transfer include Pr as an exponent because it affects how temperature couples with flow.
The Grashof number Gr = g β (ΔT) L^3 / ν^2 measures the importance of buoyancy-driven flow relative to viscous forces and is the natural convection analogue of Re. In natural convection correlations Gr is often multiplied by Pr to form Gr Pr which appears in Nusselt correlations of the form Nu = C (Gr Pr)^n.
The Nusselt number Nu = h L / k non-dimensionalises the convective coefficient h. It expresses how strong convective transport is relative to conduction over the same characteristic length. Empirical correlations are often written as Nu = C Re^m Pr^n for forced convection and Nu = C (Gr Pr)^n for natural convection. Examples: Dittus–Boelter for turbulent internal flow Nu = 0.023 Re^{0.8} Pr^{0.4} (valid in certain ranges), and laminar flat-plate average Nu_L = 0.664 Re_L^{1/2} Pr^{1/3} for local heat transfer with constant wall temperature in laminar boundary layers. Always ensure the correlation chosen matches the flow geometry, property ranges and flow regime; use film temperature for property evaluation and check units carefully.
Using these dimensionless numbers allows engineers to select appropriate correlations, scale experiments, and generalise results across fluids and sizes. They also help estimate whether forced or natural convection dominates, and whether boundary layers are thin or thick relative to object dimensions—information that is vital for accurate heat transfer predictions.
- Compute Re and determine if flow in a pipe is laminar or turbulent: Re < 2300 laminar, Re > 4000 turbulent (for circular pipes).
- Use Dittus–Boelter to estimate Nu and then h for turbulent flow in a heated pipe.
- Reynolds: Re = ρ V L / μ
- Prandtl: Pr = μ c_p / k
- Grashof: Gr = g β (ΔT) L^3 / ν^2
- Nusselt: Nu = h L / k
Forced convection over flat plates and external flows
Forced convection over flat plates is a classical topic illustrating boundary layer development and the use of local and average Nusselt numbers. When a uniform free-stream velocity V flows parallel to a flat plate, a boundary layer forms and grows downstream. Local behaviour depends on the local Reynolds number Re_x = ρ V x / μ based on distance x from the leading edge. For small Re_x the boundary layer is laminar; beyond a critical Re_x it transitions to turbulence where heat transfer characteristics change.
Boundary-layer theory provides local Nusselt number relations. For laminar flow over a plate with constant surface temperature, the local Nusselt number Nu_x = C Re_x^{1/2} Pr^{1/3} (with C from theory). Integrating Nu_x from x = 0 to L gives the average Nu_L over plate length. For laminar region the classic result for average Nusselt number is Nu_L = 0.664 Re_L^{1/2} Pr^{1/3} under suitable conditions. For turbulent flow, empirical correlations derived from experiments are used; one common average result for turbulent boundary layers is Nu_L ≈ 0.037 Re_L^{4/5} Pr^{1/3} for moderate Pr and fully turbulent regions.
External flows around bluff bodies like cylinders and spheres are more complex because flow separation and wake formation affect heat transfer. Empirical correlations for average Nusselt number around cylinders as functions of Re and Pr are available in handbooks. Surface roughness, free-stream turbulence and surface curvature change heat transfer and must be considered when choosing a correlation. For example, a smooth cylinder in cross-flow has lower heat transfer than a roughened one at the same Re, because surface roughness can promote earlier transition to turbulence and raise local h.
Designers select correlations that match geometry and flow conditions, compute Nu to find h via h = Nu k / L, and then compute heat transfer Q̇ = h A (T_s - T∞). Film temperature is used to evaluate fluid properties. For short plates or near the leading edge, entrance effects matter and local Nu_x should be integrated to obtain average values. Forced convection generally yields higher h than natural convection for the same ΔT because the moving fluid continually removes thermal boundary layers and supplies fresh fluid to the surface.
- Flat plate in air: use Nu_L = 0.664 Re_L^{1/2} Pr^{1/3} for laminar region to compute average h.
- Cylinder in cross-flow: use appropriate empirical correlation for Nu vs Re to estimate h and Q̇.
- Local Re: Re_x = ρ V x / μ
- \[Laminar flat plate average Nu: Nu_L = 0.664 Re_L^{1/2} Pr^{1/3} (for laminar flow\]\[constant surface temperature)\]
- \[Turbulent flat plate average Nu: Nu_L = 0.037 Re_L^{4/5} Pr^{1/3} (for turbulent flow\]\[approximate)\]
Natural (free) convection from surfaces
Natural convection is driven by buoyancy forces created by temperature-induced density variations in a fluid. When a surface is hotter than the surrounding fluid, the adjacent fluid becomes lighter and rises, while cooler fluid moves in to replace it. This results in a flow pattern and a thermal boundary layer. For a vertical heated plate, this produces rising plumes adjacent to the surface; for horizontal plates the geometry of rising flow differs and can produce weaker or stronger convective currents depending on orientation.
The Grashof number Gr = g β (ΔT) L^3 / ν^2 measures the strength of buoyancy relative to viscous forces. In natural convection correlations Gr is combined with Pr to form Gr Pr which appears in Nusselt correlations: Nu = C (Gr Pr)^n for appropriate ranges. For a vertical plate in laminar natural convection a commonly used average correlation is Nu_L = 0.59 (Gr_L Pr)^{1/4} for certain GrPr ranges; for turbulent conditions the exponent and constant change. The characteristic length L is chosen based on geometry: plate height for vertical plates, diameter for horizontal cylinders, or characteristic dimension for complex shapes.
Surface orientation affects the flow strongly. A hot horizontal plate facing upward develops rising plumes and can have different Nu compared to a plate facing downward where hot fluid tends to stay attached and may produce weaker convection or even flow separation. Complex geometries and inclined surfaces require use of specialized correlations or experiments. Also, property variations with temperature can influence results, so evaluating fluid properties at film temperature T_f = (T_s + T∞)/2 is common practice.
Natural convection heat transfer coefficients are generally smaller than forced convection coefficients for the same ΔT, making forced convection preferable when high heat removal is needed. However, natural convection is essential for passive cooling systems, building ventilation without fans, and safety systems that must operate without power. Engineers often start with order-of-magnitude h values (e.g., 5–25 W m−2 K−1 for air in many natural convection cases) and then refine using appropriate correlations for the geometry and GrPr range.
- Estimate h for a vertical heated plate of height L in air by computing Gr_L Pr and using Nu_L correlation.
- Explain why a horizontal hot plate facing upward has weaker convection than a vertical heated plate.
- Grashof: Gr = g β (ΔT) L^3 / ν^2
- Natural convection correlation form: Nu = C (Gr Pr)^n
Convective heat transfer in internal flows and heat exchangers
Internal flows in pipes and ducts are central to heat exchanger design and many industrial processes. The heat transfer behaviour depends on flow regime, entrance effects and wall boundary conditions. In fully developed laminar flow inside a circular tube the Nusselt number is constant: Nu = 3.66 for constant wall temperature or Nu = 4.36 for constant heat flux. These values result from analytical solutions of the energy equation with no axial conduction and are valid when the thermal entry length is small compared to tube length.
For turbulent internal flows, empirical correlations are used. The Dittus–Boelter correlation Nu = 0.023 Re^{0.8} Pr^{0.4} is widely used for turbulent flow in smooth tubes for heating or cooling, within specified Re and Pr ranges. For more accurate work, other correlations and corrections for entry effects, viscosity variation and surface roughness may be needed. Entrance regions require integration of local Nu_x to obtain average values; for short tubes this can dominate overall heat transfer.
Heat exchanger analysis combines convective resistances of the two fluids, conduction through tube walls, and any fouling resistance. The overall heat transfer coefficient U is defined by Q̇ = U A LMTD for steady counterflow or parallel-flow exchangers. For design the log-mean temperature difference LMTD accounts for changing temperature differences between streams. When outlet temperatures are unknown, the effectiveness–NTU method uses NTU = UA/C_min and the capacity ratio to find exchanger effectiveness and outlet temperatures without requiring LMTD directly.
Pressure drop is an important trade-off: increasing turbulence or velocity raises h but also increases pumping power. Designers balance heat transfer performance and pressure drop by selecting tube diameters, lengths, and flow arrangements. Fouling reduces heat transfer over time; include fouling resistances in calculations and provide cleaning access. Using film properties at mean temperature and ensuring units consistency helps avoid calculation errors. Overall, combining convective correlations, conduction through walls and exchanger methods allows practical design of heat exchangers for industrial use.
- Use Dittus–Boelter to estimate convective heat transfer coefficient inside a turbulent pipe and compute heat transfer per metre.
- Compute required area of a counter-flow heat exchanger using LMTD for given inlet and outlet temperatures and heat duty.
- \[Dittus–Boelter: Nu = 0.023 Re^{0.8} Pr^{0.4}\]
- Fully developed laminar Nu for constant wall temperature in a circular tube: Nu = 3.66
- Overall resistance for tube: 1/(U A) = 1/(h_i A_i) + R_wall + 1/(h_o A_o)
Radiation: electromagnetic nature and blackbody concept
Radiation transfers energy by electromagnetic waves emitted by bodies due to their temperature. It does not require a medium and therefore can occur through vacuum. The ideal emitter is the blackbody, which absorbs all incident radiation and emits the maximum possible energy at each wavelength. The total emission from a blackbody per unit area follows the Stefan–Boltzmann law: j* = σ T^4, where σ is the Stefan–Boltzmann constant and T is absolute temperature.
Actual surfaces emit less than a blackbody; emissivity ε (0 ≤ ε ≤ 1) describes this reduction. The emitted power per unit area from a real surface is j = ε σ T^4. Emissivity depends on material, surface finish, temperature and wavelength. For many engineering problems the grey-surface assumption (ε independent of wavelength) and diffuse emission (uniform in all directions) are used to simplify calculations.
When two surfaces exchange radiation directly the net heat flow depends on their temperatures, emissivities and geometry. For two large parallel plates of equal area, the net radiative heat flux is q = σ (T1^4 - T2^4)/(1/ε1 + 1/ε2 - 1). For more complex geometries view factors (configuration factors) must be used; they represent the fraction of radiation leaving one surface that reaches another. View factors depend only on geometry and satisfy the reciprocity relation A_i F_{i→j} = A_j F_{j→i}.
Radiation becomes especially important at high temperatures because of its T^4 dependence. In vacuum applications, like spacecraft, radiation is the dominant mode of heat transfer. In terrestrial engineering radiation competes with convection and conduction; designers often linearise T^4 around an operating temperature to combine radiative and convective heat transfer as an effective heat transfer coefficient h_rad ≈ 4 ε σ T_m^3 for small ΔT. Good thermal control may use low-emissivity coatings or reflective foils to reduce radiative exchange, and selective surfaces to enhance absorption or emission in specific wavelength ranges.
- Compute radiative heat loss from a hot plate with emissivity ε facing the environment using q = ε σ (T^4 - T_env^4).
- Compare radiative and convective heat losses for a hot object at 500 K in air; radiation often dominates at high T.
- Stefan–Boltzmann law: j* = σ T^4
- Emissive power for real surface: j = ε σ T^4
- Net radiative exchange for two large parallel plates: q = σ (T1^4 - T2^4) / (1/ε1 + 1/ε2 - 1)
View factors and radiation exchange between surfaces
View factors, also called configuration factors, quantify the fraction of radiation leaving one surface that strikes another directly. They depend only on geometry and relative orientation. Important properties include reciprocity: A_i F_{i→j} = A_j F_{j→i}, and the summation rule Σ_j F_{i→j} = 1 for an enclosure where surface i sees all other surfaces and itself. View factors for simple geometries are tabulated in heat transfer references and can be computed by integration for more complex shapes.
In radiative exchange problems we use radiosity J (total leaving radiation including emitted and reflected portions) and irradiation G (incident radiation). For grey, diffuse surfaces, J_i = ε_i σ T_i^4 + (1 - ε_i) G_i and G_i = Σ_j F_{i→j} J_j. These linear relations allow setting up a system of linear equations for J_i which can be solved to obtain net heat flows. This method is analogous to solving nodal circuits, with radiative resistances replacing electrical resistances.
A practical simplification arises when two surfaces exchange predominantly with each other, such as two large parallel plates, or when one surface is much larger and can be approximated as a sink. In those cases closed-form expressions for net heat transfer exist. For a two-surface enclosure the net heat flux between surfaces i and j may be expressed as q = (σ (T_i^4 - T_j^4))/(1/(ε_i A_i) + 1/(ε_j A_j) + 1/(A_i F_{i→j})). Carefully choose the correct area associated with the view factor in that expression to avoid mistakes.
In complex multi-surface enclosures construct a matrix of view factors and radiosities and solve numerically. In engineering practice combine radiative networks with conductive and convective resistances to obtain total heat transfer. Applications include furnace design, thermal control of enclosures, spacecraft radiation analysis and building energy models where radiative exchange between surfaces can substantially affect temperatures and comfort.
- Compute view factor between two equal parallel rectangular plates facing each other at small separation — use tabulated formula or approximate as 1 for large plate area compared to gap.
- Two concentric cylinders: use A_i F_{i→j} = A_j F_{j→i} to find unknown view factor when one is known.
- \[Reciprocity: A_i F_{i→j} = A_j F_{j→i}\]
- \[Summation rule: Σ_j F_{i→j} = 1\]
- \[Rad radiative resistance form for two surfaces: q = σ (T1^4 - T2^4) / (1/(ε1 A1) + 1/(ε2 A2) + 1/(A1 F_{1→2}))\]
Combined heat transfer problems and examples
Real systems often involve conduction, convection and radiation acting together. The engineering approach is to identify each heat transfer path, model it with the appropriate equation, and combine the contributions. For steady problems with conduction through layers followed by convective and radiative losses at the outer surface, thermal resistance networks simplify the analysis: add series conductive resistances for layers and include surface convective resistances; radiation from the outer surface may be handled as a separate branch or linearised into an equivalent convective coefficient for combination with convection.
Consider a hot pipe with insulation losing heat to the surrounding air. Step one is to compute conductive resistance of the insulation using the cylindrical formula R_cond = ln(r2/r1)/(2πkL). Next include the convective resistance at the outer surface R_conv = 1/(h A_o). Solve Q̇ = (T_i - T∞)/(R_cond + R_conv) for heat rate Q̇ and for outer surface temperature T_s. If radiation from the outer surface is significant, compute radiative loss q_rad = ε σ A_o (T_s^4 - T∞^4). One practical tactic is to linearise radiation around a mean temperature T_m to get h_rad ≈ 4 ε σ T_m^3 and combine with h_conv to an effective h_total = h_conv + h_rad. Then handle conduction and combined surface transfer as a series resistance network.
Another example is a building wall with windows: conduction through wall layers, convection on both sides, and radiative exchange between interior surfaces and windows all influence comfort and energy use. In transient combined problems conduction in walls couples with time-varying convective boundary conditions; numerical methods usually solve these coupled behaviours. Dimensionless analysis compares magnitudes of modes: compare h (convective) with linearised radiative coefficient 4 ε σ T_m^3 to see which mode dominates for a given temperature range.
Engineering practice uses conservative assumptions in preliminary design: assume higher h to ensure adequate cooling, include safety factors, and check sensitivity to material parameters. When combining modes, document assumptions, compute intermediate variables like surface temperatures, and verify final answers with limiting-case checks such as k → ∞ or ε → 0. These combined analyses underpin insulation design, heat exchanger performance estimates and thermal management of equipment and buildings.
- Hot pipe with insulation: compute conduction resistance, find T_surface, then compute convective and radiative losses and sum to get total heat loss per unit length.
- Window heat loss: conduction through glass plus convection at surfaces and radiative exchange with room and outdoors.
- Total heat loss = conduction-limited Q̇ through layers then convective and radiative losses at outer surface; combine as needed.
- Linearised radiative coefficient (approx): h_rad ≈ 4 ε σ T_m^3
Insulating materials and design considerations
Insulation reduces heat flow by adding thermal resistance between hot and cold regions. Material selection depends on thermal conductivity k, density, mechanical strength, moisture resistance, flammability and cost. Common building and industrial insulators include mineral wool, fibreglass, expanded polystyrene and polyurethane foam. Their thermal conductivities are typically low (for example 0.03–0.05 W m−1 K−1 for many foams), making them effective at reducing conduction. For high-temperature applications ceramic fibres and refractory bricks are used because they retain insulating properties at elevated temperatures.
Designing insulation involves choosing thicknesses and materials to meet performance goals such as limiting heat loss, controlling surface temperature for safety, preventing freezing, or achieving energy efficiency targets. For planar walls use R = L/(k A) to compute required thickness for a desired resistance. For pipes use the cylindrical resistance formula R = ln(r2/r1)/(2πkL). Adding insulation yields diminishing returns: the first layers often give the greatest reduction in heat loss, and beyond a certain thickness further reduction per added unit thickness becomes small relative to cost and weight. This is important when calculating payback periods and economic thickness for retrofit projects.
Moisture and vapour control are critical because water infiltration increases effective thermal conductivity and can lead to condensation and degradation. Use vapour barriers and appropriate seals in building applications to prevent moisture ingress. Thermal bridging through studs, fasteners or supports creates low-resistance paths that bypass insulation; include thermal breaks or continuous insulation layers to mitigate bridging. For fire safety select materials that are fire-resistant or include protective cladding; some insulating foams are flammable and require special treatment.
Installation quality affects performance: gaps, compression, and misalignment lower effective R-values. For process piping and equipment consider mechanical strength, durability and access for maintenance. Environmental concerns such as global warming potential of blowing agents or recyclability may influence material choice. Standards and building codes provide minimum insulation requirements; use them and consider life-cycle energy savings and payback when specifying insulation thickness. Testing and verification ensure installed performance meets design expectations.
- Calculate required insulation thickness on pipe to limit heat loss to a set value using cylindrical resistance formula.
- Compare heat loss through two insulating materials of same thickness by computing R = L/(kA).
- Resistance of insulation layer: R = L/(kA) (plane), R = ln(r2/r1)/(2πkL) (cylinder)
- Required thickness found by solving R_total = ΔT/Q̇_target
Practical measurement and experimental methods
Measuring heat transfer properties and coefficients requires careful experimental technique and control of boundary conditions. Thermal conductivity k of solids is often measured using steady-state methods like the guarded hot plate, where a constant heat flux is driven through a specimen and temperature difference across a known thickness is measured. Guarding minimises lateral losses. Transient methods such as the laser flash technique measure thermal diffusivity α by applying a short heat pulse to the front face and recording the rear-face temperature rise; with measured density and specific heat, conductivity is k = α ρ c.
Convective heat transfer coefficients h are typically obtained by heating a surface with a known power input, measuring surface and fluid temperatures, and solving h = Q̇/(A (T_s - T∞)). Experimental setups include heated cylinders in wind tunnels, flat-plate facilities, and flow loops for internal flows. Careful measurement of flow velocity, temperature and power input is essential. Flow visualisation techniques such as smoke or particle-image velocimetry help understand boundary layer behaviour and validate correlations.
Measuring emissivity involves comparative or calorimetric methods, often using a known blackbody or reference surface. Infrared thermography gives surface temperature maps but requires knowledge of emissivity and may need calibration. For radiation exchange experiments ensure enclosure geometry and view factors are controlled and analyse reflected as well as emitted components. Use of error analysis and repeated trials improves confidence in reported coefficients.
Uncertainty quantification is essential: estimate instrument errors, propagate uncertainties through calculations and report ranges. Parasitic heat losses to supports or through wiring can bias results; design experiments to minimise or measure such losses. When reporting measured h or k specify fluid or material conditions, test geometry and temperature range so results can be correctly used. Practical measurement skills build intuition about magnitudes of heat transfer rates and validate theoretical models used in engineering design.
- Guarded hot plate experiment to measure thermal conductivity of a slab: apply steady heat flux and measure temperature difference across known thickness.
- Determine convective coefficient for a heated cylinder in cross-flow by measuring power input and surface temperature and solving for h.
- Compute h from experiment: h = Q̇/(A (T_s - T∞))
- Thermal conductivity from steady test: k = Q̇ L /(A ΔT)
Safety, standards and common engineering applications
Heat transfer principles are applied in many engineering fields: HVAC, process industries, electronics cooling, power generation, automotive systems and aerospace thermal control. Standards and codes guide safe and efficient design: building codes specify minimum U-values for walls and roofs, industrial standards set insulation requirements for piping and equipment, and safety standards limit surface temperatures to prevent burns. Following standards ensures regulatory compliance, occupant safety and predictable performance.
In electronics cooling, heat sinks, thermal interface materials and forced-air cooling are combined to keep components within allowable temperatures; manufacturers often follow standards for thermal testing and classification. For power plants and chemical processing, heat exchangers are designed according to standard methods that consider both thermal performance and mechanical safety. In aerospace, thermal control relies heavily on radiation and certified thermal coatings; standards govern qualification testing, thermal vacuum testing and material outgassing to ensure long-term reliability.
Fire safety and material regulations influence insulation choices. Many industrial and building applications require non-combustible materials or fire-retardant treatments; certified fire tests define maximum allowable surface temperatures and smoke/toxicity limits. Thermal insulation used on hot equipment must be rated for operational temperatures and for exposure to moisture and chemicals; codes often require specific claddings or shields to meet workplace safety rules and to protect personnel.
Standards also prescribe measurement and test methods so results are reproducible and comparable. Examples include standard test procedures for thermal conductivity, emissivity, and convective coefficient determination. Certification bodies and product standards provide labels and data sheets that engineers use to select materials and components. Compliance with energy-efficiency regulations (such as minimum insulation levels or appliance efficiency ratings) often drives design choices and economic assessments.
Practical engineering includes life-cycle and environmental considerations. Designers evaluate operational energy savings against upfront costs, consider durability, maintenance needs and recyclability of insulating materials, and account for possible degradation of thermal properties over time (fouling, moisture, compression). Monitoring and maintenance plans (periodic thermal imaging inspections, scheduled cleaning of heat exchangers) help ensure systems remain efficient and safe. Ultimately, combining sound heat transfer analysis with adherence to standards and safety practices yields reliable, efficient and sustainable engineering solutions.
- Choosing heat sink and fan for a power electronics module based on allowed maximum junction temperature and estimated power dissipation.
- Specifying insulation for hot process lines to limit surface temperature to safe values and minimise heat loss.
- U-value for building element: U = 1/(Σ R_i)
- Heat duty in heat exchanger: Q̇ = U A LMTD
Problem-solving strategies and common pitfalls
Successful problem solving in heat transfer follows a clear sequence: read the problem carefully and draw a labelled sketch; list knowns and unknowns; state assumptions (steady vs transient, one-dimensionality, constant properties); choose governing equations or correlations; perform algebra and compute; then check results for units and physical plausibility. Documenting assumptions helps examiners and future reviewers understand the solution path.
Dimensionless numbers help choose methods: compute Bi to decide lumped vs distributed transient models, Re and Pr to choose convective correlations, and Fo to understand transient progress. Use film temperature T_f = (T_s + T∞)/2 to evaluate fluid properties when applying correlations. For radiation problems always work in kelvin for T^4 calculations; mixing Celsius and Kelvin causes errors. When combining modes do not forget to convert radiative heat to equivalent linear form only when ΔT is small and linearisation is justified.
Common mistakes include using incorrect characteristic lengths for Re or Nu, forgetting perimeter vs area distinctions in internal flow problems, neglecting contact resistances or thermal bridges in composite assemblies, and applying correlations outside their stated validity ranges. In heat exchanger problems misuse of LMTD occurs when streams change phase or flow arrangement differs; use effectiveness–NTU methods in such cases. For transient analyses avoid applying steady formulas and always check limiting cases: as k → ∞ the body should become isothermal, as h → ∞ the surface should reach fluid temperature, and as ε → 0 radiation vanishes.
Practical tips: estimate orders of magnitude first to catch impossible results; check units at every step; use consistent SI units; and perform a sensitivity check to see which parameters most affect the outcome. Where possible validate with simple experiments or compare to handbook examples. Repeated practice with problems of varying complexity builds intuition and reduces common errors, preparing students for board-style questions and engineering applications.
- A student forgot Kelvin conversion in radiation problem and got a tiny heat flux; correcting to Kelvin yields correct large value.
- Choosing L_c incorrectly in Biot number led to rejecting lumped method; recomputing with correct L_c showed lumped method valid.
- Film temperature: T_f = (T_s + T∞)/2
- Linearised radiative coefficient: h_rad ≈ 4 ε σ T_m^3 (for small ΔT around T_m)
Key Concepts
- Heat transfer
- Movement of thermal energy from one region to another due to temperature difference.
- Conduction
- Heat transfer within a material caused by microscopic collisions and energy carriers with Fourier’s law describing the flux.
- Convection
- Heat transfer between a surface and a moving fluid that combines conduction and bulk fluid motion.
- Radiation
- Heat transfer by electromagnetic waves that can occur through vacuum and depends on surface temperature to the fourth power.
- Fourier’s law
- A law stating heat flux is proportional to the negative temperature gradient, q = -k dT/dx.
- Thermal conductivity (k)
- A material property measuring ability to conduct heat, in W m−1 K−1.
- Thermal resistance
- A quantity R = L/(kA) representing opposition to heat flow through a layer.
- Biot number (Bi)
- Dimensionless ratio Bi = h L_c / k that indicates internal versus surface resistance to heat transfer.
- Fourier number (Fo)
- Dimensionless time Fo = α t / L_c^2 measuring relative importance of diffusion over time.
- Nusselt number (Nu)
- Dimensionless heat transfer coefficient Nu = h L / k linking convective heat transfer to conduction.
- Reynolds number (Re)
- Dimensionless ratio Re = ρ V L / μ indicating the relative importance of inertial to viscous forces in a flow.
- Prandtl number (Pr)
- Dimensionless ratio Pr = μ c_p / k representing momentum diffusivity relative to thermal diffusivity.
- Stefan–Boltzmann law
- Law stating blackbody radiative emission per unit area is j* = σ T^4 with σ the Stefan–Boltzmann constant.
- Emissivity (ε)
- Ratio of radiation emitted by a real surface to that emitted by a blackbody at same temperature.
- View factor (F)
- Geometric fraction of radiation leaving one surface that directly reaches another surface.
- Lumped capacitance model
- An approximation assuming uniform internal temperature when Bi ≤ 0.1 leading to exponential transient behaviour.
- Log-mean temperature difference (LMTD)
- A temperature difference measure used to calculate heat transfer in heat exchangers with varying temperatures.
- Thermal diffusivity (α)
- Material property α = k/(ρ c) indicating rate at which temperature disturbances diffuse through a material.
Practice Questions
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A metal rod 1 m long and 0.01 m2 cross-sectional area has thermal conductivity 60 W m−1 K−1. One end is at 150 °C and the other at 50 °C. Calculate the steady heat flow through the rod. / एक धातु की छड़ जिसकी लंबाई 1 m और अनुप्रस्थ कट 0.01 m2 है, और ताप चालकता 60 W m−1 K−1 है। एक सिरा 150 °C और दूसरा 50 °C है। स्थिर स्थिति में छड़ से होने वाला ऊष्मा प्रवाह ज्ञात कीजिए।
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Heat rate Q̇ = k A (T1 - T2)/L = 60 × 0.01 × (150 - 50)/1 = 60 × 0.01 × 100 = 60 W. / ऊष्मा दर Q̇ = k A (T1 - T2)/L = 60 × 0.01 × (150 - 50)/1 = 60 W।
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Explain the meaning of the Biot number and state when lumped capacitance method is applicable. / बायोट संख्या का अर्थ समझाइए और बताइए कि लम्प्ड कैपेसिटेन्स पद्धति कब लागू होती है।
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Biot number Bi = h L_c / k compares surface convective resistance to internal conductive resistance. If Bi ≤ 0.1 the internal conduction is fast and temperature within the body can be assumed uniform, so the lumped capacitance method applies. / बायोट संख्या Bi = h L_c / k सतही संवहन प्रतिरोध और आंतरिक चालन प्रतिरोध की तुलना करती है। यदि Bi ≤ 0.1 तो आंतरिक चालन तेज़ है और वस्तु के अंदर तापमान समान माना जा सकता है, अतः लम्प्ड कैपेसिटेन्स पद्धति लागू होती है।
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A cylindrical pipe has inner radius 0.05 m and outer radius 0.075 m, length 2 m and k = 0.2 W m−1 K−1. Find the conductive resistance per metre length. / एक बेलनाकार पाइप का आंतरिक त्रिज्या 0.05 m और बाहरी त्रिज्या 0.075 m, लंबाई 2 m और k = 0.2 W m−1 K−1 है। एक मीटर लंबाई का चालनात्मक प्रतिरोध ज्ञात कीजिए।
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Cylindrical resistance per length R' = ln(r2/r1)/(2πk) = ln(0.075/0.05)/(2π × 0.2). Compute ln(1.5)=0.4055 so R' = 0.4055/(1.2566 × 0.2)=0.4055/0.2513≈1.613 K W−1 per metre. / बेलनाकार प्रतिरोध प्रति लंबाई R' = ln(r2/r1)/(2πk) = ln(0.075/0.05)/(2π × 0.2). ln(1.5)=0.4055 से R' = 0.4055/(1.2566 × 0.2)=≈1.613 K W−1 प्रति मीटर।
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State Fourier’s law and use it to explain why metals feel colder than wood at the same temperature. / फौरियर का नियम बताइए और इसमें समझाइए कि एक ही तापमान पर धातु लकड़ी से ठंडी क्यों महसूस होती है।
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Fourier’s law: q = -k dT/dx, heat flux is proportional to temperature gradient and conductivity. Metals have much higher thermal conductivity k than wood, so when touched they draw heat from the skin faster, producing a larger heat flux and a stronger sensation of cold. / फौरियर का नियम: q = -k dT/dx; ऊष्मा प्रवाह तापीय ढाल और चालकता के आनुपातिक ہوتا ہے। धातुओं की चालकता लकड़ी से बहुत अधिक होती है, इसलिए वे त्वचा से ऊष्मा तेज़ी से खींचती हैं, जिससे ठंडा अधिक महसूस होता है।
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Calculate the net radiative heat flux between two large parallel plates at 800 K and 300 K with emissivities 0.8 and 0.7 respectively. / दो बड़े समांतर प्लेट जिनका तापमान क्रमशः 800 K और 300 K है तथा विमिसिविटी 0.8 और 0.7 है, उनके बीच शुद्ध विकिरण ऊष्मा प्रवाह घनत्व ज्ञात कीजिए।
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Net q = σ (T1^4 - T2^4) / (1/ε1 + 1/ε2 - 1). Compute T1^4 = (800)^4, T2^4 = (300)^4. Using σ = 5.670×10^-8 W m−2 K−4, numerator = σ(800^4 - 300^4). Denominator = 1/0.8 + 1/0.7 -1 = 1.25 +1.4286 -1 =1.6786. Numerically 800^4=4.096×10^11, 300^4=8.1×10^9, difference ≈4.015×10^11. Multiply by σ: 5.67e-8 × 4.015e11 ≈ 2.278×10^4 W m−2. Divide by 1.6786 gives q ≈ 1.358×10^4 W m−2. / शुद्ध q = σ (T1^4 - T2^4) / (1/ε1 + 1/ε2 - 1). यहाँ σ = 5.67×10^-8। T1^4 - T2^4 ≈ 4.015×10^11। σ×差 ≈ 2.278×10^4 W m−2। भाजक = 1/0.8 + 1/0.7 -1 = 1.6786। अन्त में q ≈ 1.358×10^4 W m−2।
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A small sphere of mass 0.5 kg and specific heat 900 J kg−1 K−1 at initial temperature 200 °C is suddenly placed in air at 20 °C. Its surface area is 0.01 m2 and h = 10 W m−2 K−1. Using lumped method, estimate the time constant τ and temperature after 5 minutes. / 0.5 kg द्रव्यमान और विशिष्ट ऊष्मा 900 J kg−1 K−1 वाला एक छोटा गोला जिसकी प्रारंभिक ताप 200 °C है, हवा 20 °C में डाला जाता है। उसका सतह क्षेत्र 0.01 m2 और h = 10 W m−2 K−1 है। लम्प्ड पद्धति का उपयोग करके समय स्थिरांक τ और 5 मिनट के बाद तापमान अनुमानित कीजिए।
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τ = m c/(h A) = 0.5 × 900 /(10 × 0.01) = 450 /(0.1) = 4500 s ≈ 75 min. After t = 300 s (5 min), temperature ratio (T - T∞)/(T_i - T∞) = exp(-t/τ) = exp(-300/4500)=exp(-0.0667)≈0.9355. Initial excess ΔT_i = 200 - 20 = 180 K, so ΔT = 0.9355 × 180 ≈ 168.4 K. Thus T ≈ T∞ + ΔT = 20 + 168.4 ≈ 188.4 °C. / τ = m c/(h A) = 0.5×900/(10×0.01)=4500 s ≈ 75 मिनट। t=300 s पर (T - T∞)/(T_i - T∞)=exp(-300/4500)≈0.9355। प्रारंभिक ΔT=180 K → ΔT(t)≈168.4 K। अतः T≈20+168.4≈188.4 °C।
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Why is the Nusselt number useful and how is it related to the convective coefficient h? / नुस्सेल्ट संख्या उपयोगी क्यों है और यह संवहन गुणांक h से कैसे सम्बन्धित है?
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Nusselt number Nu = h L / k non-dimensionalises convective heat transfer, comparing convective transport to conductive transport across a characteristic length. It allows use of universal correlations Nu = f(Re, Pr, Gr) to find h for different fluids and geometries by rearranging h = Nu k / L. / Nu = h L / k संवहन ऊष्मा स्थानान्तरण को गैर-आयामी बनाती है और इसे स्थानीय प्रवाह नम्बरोँ के साथ सम्बन्धित करने में सहायक होती है; h को पाने के लिए h = Nu k / L प्रयोग किया जाता है।
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Describe how you would model heat loss from a hot pipe including conduction through insulation, convection and radiation to surroundings. / एक गरम पाइप से होने वाले ऊष्मा ह्रास का मॉडल कैसे बनायेंगे जिसमें इन्सुलेशन के माध्यम से चालन, परिवहन और परिवेश को विकिरण शामिल हों?
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Model steps: (1) draw radial cross-section and identify layers; (2) compute conductive resistance of insulation R_cond = ln(r2/r1)/(2πkL); (3) include convective resistance at outer surface R_conv = 1/(h A_o); (4) compute outer surface temperature by solving Q̇ = (T_i - T∞)/(R_cond + R_conv) iteratively if radiation significant; (5) include radiative loss q_rad = ε σ A_o (T_s^4 - T∞^4) and add to convective loss if treating separately, or include radiation by adding equivalent radiative resistance using linearised h_rad ≈ 4 ε σ T_m^3 to combine with h. Sum convective and radiative heat losses for total. / मॉडल: (1) त्रिज्या के क्रॉस-सेक्शन बनायें; (2) इन्सुलेशन का चालन प्रतिरोध R_cond = ln(r2/r1)/(2πkL) निकालें; (3) बाहरी सतह पर संवहन प्रतिरोध R_conv = 1/(h A_o) जोड़ें; (4) यदि विकिरण महत्वपूर्ण हो तो सतह ताप T_s के लिए परिकलन करें; (5) विकिरण q_rad = ε σ A_o (T_s^4 - T∞^4) निकालकर संवहन से जोड़ें या विकिरण को रैखिकीकृत कर h_rad ≈ 4 ε σ T_m^3 के रूप में शामिल करें। अंततः कुल ऊष्मा ह्रास प्राप्त करें।
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A plate at 400 K and emissivity 0.9 faces the surroundings at 300 K. Estimate radiative heat loss per unit area. / 400 K ताप वाला और विमिसिविटी 0.9 वाला एक प्लेट 300 K परिवेश का सामना कर रहा है। प्रति इकाई क्षेत्र विकिरणीय ऊष्मा ह्रास अनुमानित कीजिए।
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q = ε σ (T^4 - T∞^4) = 0.9 × 5.67×10^-8 × (400^4 - 300^4). Compute 400^4=2.56×10^10, 300^4=8.1×10^9 difference=1.75×10^10. Multiply: 5.67e-8 ×1.75e10≈992.25 W m−2; times 0.9 gives ≈893 W m−2. / q = ε σ (T^4 - T∞^4) = 0.9×5.67×10^-8×(400^4 - 300^4) ≈ 893 W m−2।
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Explain why in vacuum radiation is the only mode of heat transfer and give one practical example where this matters. / निर्वात में विकिरण ही केवल ऊष्मा स्थानान्तरण का तरीका क्यों है और एक व्यावहारिक उदाहरण दीजिए जहाँ यह महत्वपूर्ण है।
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In vacuum there are no material particles to conduct or convect heat, so only electromagnetic radiation can transfer energy. Practical example: thermal control of spacecraft relies on radiation for both heating and cooling since space is vacuum and there is no air for convection. / निर्वात में पदार्थिक कण नहीं होते अतः चालन या परिवहन संभव नहीं; केवल इलेक्ट्रोमैग्नेटिक विकिरण ऊष्मा पहुँचाता है। व्यावहारिक उदाहरण: अंतरिक्षयान का तापीय नियंत्रण विकिरण पर निर्भर करता है।