SSRAO

Explanation:
Fourier's law: Q = -kA(dT/dx). The negative sign indicates heat flows in the direction of decreasing temperature (from hot to cold). The rate of heat transfer is proportional to the area A, thermal conductivity k, and the magnitude of the temperature gradient (dT/dx). The negative sign ensures Q is positive when heat flows in the positive x-direction with dT/dx < 0.

Explanation:
Thermal conductivity k is defined from Fourier's law: k = Q/(A × |dT/dx|), i.e., heat flux per unit temperature gradient. Units: W/(m·K). It is a material property indicating how well a material conducts heat. Metals have high k (copper: 385 W/m·K), insulators have low k (glass wool: 0.04 W/m·K).

Explanation:
Newton's law of cooling: Q = hA(Ts - T∞), where h is the convective heat transfer coefficient (W/m²·K), A is the surface area, Ts is the surface temperature, and T∞ is the fluid temperature. The heat flux q = Q/A = h(Ts - T∞) is linearly proportional to the temperature difference. This is an empirical law, not derived from first principles.

Explanation:
Stefan-Boltzmann law: Eb = σT⁴, where Eb is the blackbody emissive power (W/m²), σ = 5.67 × 10⁻⁸ W/(m²·K⁴) is the Stefan-Boltzmann constant, and T is the absolute temperature in Kelvin. This T⁴ dependence makes radiation dominant at high temperatures. For a real surface: E = εσT⁴, where ε is emissivity (0 ≤ ε ≤ 1).

Explanation:
Radiation is electromagnetic wave propagation and does not require a material medium — it can occur in vacuum (e.g., solar radiation reaching Earth through space). Conduction requires molecular contact (solid or fluid medium). Convection requires a fluid medium for bulk motion. Radiation is the only mode that can transfer heat across a vacuum.

Explanation:
Thermal resistance for conduction: R_cond = L/(kA), analogous to electrical resistance R = ρL/A. Heat flow Q = ΔT/R = kA(T1-T2)/L (Fourier's law). For a composite wall, resistances in series add: R_total = R1 + R2 + ... = L1/(k1A) + L2/(k2A) + ... This analogy simplifies complex heat transfer problems.

Explanation:
A blackbody is an ideal radiator that: (1) absorbs all incident radiation (absorptivity α = 1), (2) emits maximum possible radiation at a given temperature (emissivity ε = 1), and (3) emits radiation according to Planck's law. Real surfaces have ε < 1. A blackbody is used as the reference standard for radiation calculations.

Explanation:
For a composite wall with convection on both sides, the total thermal resistance per unit area is: 1/U = 1/h1 + L1/k1 + L2/k2 + 1/h2, where h1 and h2 are convective coefficients on each side, L1, L2 are layer thicknesses, and k1, k2 are thermal conductivities. U is the overall heat transfer coefficient (W/m²·K). Heat flux q = U(T∞1 - T∞2).

Explanation:
Kirchhoff's law: At thermal equilibrium, the emissivity (ε) of a surface equals its absorptivity (α): ε = α. This means a good absorber is also a good emitter. For a blackbody, ε = α = 1. For a grey body, ε = α < 1 (constant with wavelength). This law is fundamental to radiation heat transfer analysis.

Explanation:
Biot number Bi = hLc/k, where h is the convective coefficient, Lc is the characteristic length (V/A for lumped analysis), and k is the thermal conductivity of the solid. Bi represents the ratio of internal conductive resistance to external convective resistance. If Bi < 0.1, the lumped capacitance method is valid (temperature uniform within solid). If Bi >> 1, surface resistance dominates.

Explanation:
For a real surface, emissive power E = εσT⁴, where emissivity ε ≤ 1. Since a blackbody has ε = 1 and emits Eb = σT⁴, a real surface always emits less than or equal to a blackbody at the same temperature. The blackbody is the maximum emitter at any given temperature. This is a consequence of the second law of thermodynamics.

Explanation:
For radial conduction through a hollow cylinder: Q = 2πkL(T1-T2)/ln(r2/r1). Therefore, thermal resistance R = ln(r2/r1)/(2πkL). This is derived by integrating Fourier's law in cylindrical coordinates: Q = -kA(dT/dr) = -k(2πrL)(dT/dr). Separating variables and integrating from r1 to r2 gives the result.

Explanation:
View factor F12 is the fraction of radiation leaving surface 1 that is intercepted by surface 2. It depends only on geometry (size, shape, orientation, and distance between surfaces), not on surface properties or temperature. Reciprocity relation: A1F12 = A2F21. Summation rule: ΣFij = 1 for each surface i. F11 = 0 for flat or convex surfaces.

Explanation:
Thermal conductivity order: Solids > Liquids > Gases. Gases have very low thermal conductivity (air: 0.026 W/m·K) because heat conduction in gases occurs through molecular collisions, and gas molecules are far apart. Liquids have intermediate values (water: 0.6 W/m·K). Metals have the highest values (copper: 385 W/m·K). This is why gases are good insulators.

Explanation:
For two large parallel black plates (F12 = 1, ε1 = ε2 = 1), the net radiation heat exchange is Q = σA(T1⁴ - T2⁴). This is derived from Q = A1F12σ(T1⁴ - T2⁴) with F12 = 1. For grey surfaces: Q = σA(T1⁴ - T2⁴) / (1/ε1 + 1/ε2 - 1). Temperatures must be in Kelvin.

Explanation:
Nusselt number Nu = hL/k, where h is the convective heat transfer coefficient (W/m²·K), L is the characteristic length (m), and k is the thermal conductivity of the fluid (W/m·K). Nu represents the ratio of convective to conductive heat transfer across the fluid. Nu = 1 means pure conduction. Higher Nu indicates more effective convection. Nu is determined from correlations like Nu = C·Re^m·Pr^n.

Explanation:
In natural (free) convection, fluid motion is driven by buoyancy forces arising from density differences caused by temperature gradients. Warmer fluid is less dense and rises; cooler fluid is denser and sinks, creating natural circulation. The Grashof number Gr = gβΔTL³/ν² characterises natural convection (ratio of buoyancy to viscous forces). No external pump or fan is needed.

Explanation:
Prandtl number Pr = μCp/k = ν/α, where μ is dynamic viscosity, Cp is specific heat, k is thermal conductivity, ν = μ/ρ is kinematic viscosity, and α = k/(ρCp) is thermal diffusivity. Pr represents the ratio of momentum diffusivity to thermal diffusivity. Pr < 1 (liquid metals): thermal boundary layer thicker than velocity boundary layer. Pr > 1 (oils): velocity boundary layer thicker.

Explanation:
Thermal resistance for convection: R_conv = 1/(hA), analogous to electrical resistance. Heat flow Q = (Ts - T∞)/R_conv = hA(Ts - T∞) (Newton's law of cooling). For a composite system with conduction and convection in series: R_total = 1/(h1A) + L/(kA) + 1/(h2A). This resistance network approach simplifies multi-mode heat transfer problems.

Explanation:
Wien's displacement law: λmax × T = 2898 μm·K (or 2.898 × 10⁻³ m·K). As temperature increases, the peak wavelength shifts to shorter wavelengths (higher frequencies). At 5800 K (sun's surface), λmax ≈ 0.5 μm (visible light). At 300 K (room temperature), λmax ≈ 9.7 μm (infrared). This explains why hot objects glow red, then orange, then white as temperature increases.

Explanation:
Grashof number Gr = gβΔTL³/ν², where g is gravitational acceleration, β is volumetric thermal expansion coefficient (1/T for ideal gas), ΔT is temperature difference, L is characteristic length, and ν is kinematic viscosity. Gr represents the ratio of buoyancy forces to viscous forces in natural convection. Gr plays the same role in natural convection as Re plays in forced convection.

Explanation:
For an opaque grey body: α + ρ = 1 (since transmissivity τ = 0 for opaque surfaces). By Kirchhoff's law: ε = α. Therefore ε + ρ = 1. For a general surface: α + ρ + τ = 1. A blackbody has α = 1, ρ = 0, τ = 0. A perfect reflector has ρ = 1, α = 0. A grey body has constant ε = α < 1 independent of wavelength.

Explanation:
Critical radius of insulation for a cylinder: rc = k/h, where k is the thermal conductivity of insulation and h is the external convective coefficient. Adding insulation up to rc increases heat loss (due to increased surface area outweighing increased resistance). Beyond rc, heat loss decreases. For a sphere: rc = 2k/h. This is important for small pipes and electrical cables where insulation may increase heat loss.

Explanation:
Stanton number St = h/(ρVCp) = Nu/(Re × Pr), where h is convective coefficient, ρ is density, V is velocity, and Cp is specific heat. St represents the ratio of heat transferred to the fluid to the thermal capacity of the fluid. It is used in heat transfer correlations for internal and external flows. St = Nu/(Re·Pr) connects it to other dimensionless numbers.

Explanation:
For two large parallel grey plates (F12 = 1, A1 = A2 = A): Q = σA(T1⁴ - T2⁴)/(1/ε1 + 1/ε2 - 1). This accounts for multiple reflections between the surfaces. The denominator (1/ε1 + 1/ε2 - 1) is the total radiation resistance. For black plates (ε1 = ε2 = 1): Q = σA(T1⁴ - T2⁴). For one surface being a blackbody (ε2 = 1): Q = ε1σA(T1⁴ - T2⁴).

Explanation:
Thermal diffusivity α = k/(ρCp) (m²/s), where k is thermal conductivity, ρ is density, and Cp is specific heat. It represents how quickly a material responds to temperature changes — the ratio of heat conducted to heat stored. High α means rapid temperature change (metals). Low α means slow response (insulators). α appears in the transient heat conduction equation: ∂T/∂t = α∇²T.

Explanation:
For laminar flow over a flat plate (Rex < 5×10⁵), the local Nusselt number is Nu_x = 0.332 Re_x^0.5 Pr^(1/3) (Pohlhausen solution). The average Nusselt number over length L is Nu_L = 0.664 Re_L^0.5 Pr^(1/3) = 2 × Nu_x|x=L. The Dittus-Boelter equation Nu = 0.023 Re^0.8 Pr^n applies to turbulent flow in pipes.

Explanation:
Fin efficiency η_f = Q_actual / Q_max = Q_actual / (hA_fin × (Tb - T∞)), where Q_max is the heat transfer if the entire fin were at the base temperature Tb. η_f = 1 for an ideal fin (infinite k). For a fin with insulated tip: η_f = tanh(mL)/(mL), where m = √(hP/kA_c), P is perimeter, A_c is cross-sectional area. Fin effectiveness ε_f = Q_fin / Q_without_fin.

Explanation:
Reynolds number Re = ρVL/μ = VL/ν, where ρ is density, V is velocity, L is characteristic length, μ is dynamic viscosity, and ν is kinematic viscosity. Re represents the ratio of inertia forces to viscous forces. For pipe flow: Re < 2300 (laminar), 2300 < Re < 4000 (transition), Re > 4000 (turbulent). For flat plate: Re < 5×10⁵ (laminar), Re > 5×10⁵ (turbulent).

Explanation:
For a composite wall with two layers in series (no convection), the total thermal resistance is R_total = L1/(k1A) + L2/(k2A). Heat transfer Q = (T1 - T3)/R_total = (T1 - T3)/(L1/k1A + L2/k2A). The interface temperature T2 can be found from Q = k1A(T1-T2)/L1. Note: option C gives the same result only if Q is the same through both layers (which it is at steady state), but option B is the direct formula.

Explanation:
Thermal conductivity order: Pure metals (highest, e.g., silver: 429, copper: 385, aluminium: 237 W/m·K) > Non-metallic solids (e.g., glass: 1.4, brick: 0.7 W/m·K) > Liquids (e.g., water: 0.6, engine oil: 0.15 W/m·K) > Gases (lowest, e.g., air: 0.026, CO2: 0.017 W/m·K). Exceptions: diamond (2300 W/m·K) and graphite have very high k.

Explanation:
Fourier number Fo = αt/L², where α is thermal diffusivity (m²/s), t is time (s), and L is characteristic length (m). Fo represents the ratio of heat conducted to heat stored, or dimensionless time. Large Fo means the system has had time to respond to boundary conditions. In lumped analysis: T(t) - T∞)/(Ti - T∞) = exp(-Bi × Fo). Fo is used in Heisler charts for transient conduction.

Explanation:
Radiation heat transfer Q ∝ (T1⁴ - T2⁴), not (T1 - T2). This is because each surface emits radiation proportional to T⁴ (Stefan-Boltzmann law). The net exchange is Q = σA(T1⁴ - T2⁴) for black surfaces. Note: T1⁴ - T2⁴ = (T1² + T2²)(T1 + T2)(T1 - T2), so radiation is highly nonlinear with temperature. At high temperatures, radiation dominates over conduction and convection.

Explanation:
The fin parameter m = √(hP/(kAc)), where h is convective coefficient (W/m²·K), P is fin perimeter (m), k is thermal conductivity (W/m·K), and Ac is cross-sectional area (m²). Units of m are m⁻¹. The fin equation is d²θ/dx² - m²θ = 0, where θ = T - T∞. Solution: θ(x) = C1·cosh(mx) + C2·sinh(mx). For a fin with insulated tip: θ(x)/θb = cosh[m(L-x)]/cosh(mL).

Explanation:
In a heat exchanger, the temperature difference between hot and cold fluids varies along the length. For a counterflow or parallel flow heat exchanger with constant U and Cp, the temperature difference varies exponentially. The LMTD = (ΔT1 - ΔT2)/ln(ΔT1/ΔT2) gives the correct average driving force for Q = UA × LMTD. The log mean is always less than the arithmetic mean (AM ≥ GM ≥ LM).

Explanation:
Emissivity ε = E/Eb = E/(σT⁴), where E is the actual emissive power of the surface and Eb = σT⁴ is the blackbody emissive power at the same temperature. ε ranges from 0 to 1. ε = 1 for a blackbody, ε ≈ 0.9-0.95 for non-metallic surfaces, ε ≈ 0.02-0.1 for polished metals. Emissivity depends on temperature, wavelength, and surface condition.

Explanation:
For a semi-infinite solid with sudden surface temperature change: (T(x,t) - Ts)/(Ti - Ts) = erf(x/√(4αt)), where erf is the Gaussian error function, α is thermal diffusivity, and t is time. This gives T(x,t) = Ts + (Ti - Ts)·erf(x/√(4αt)). The surface heat flux is q_s = k(Ts - Ti)/√(παt). This solution applies to short-time transient problems in large bodies.

Explanation:
Dittus-Boelter equation: Nu = 0.023 Re^0.8 Pr^n, where n = 0.4 when the fluid is being heated (Ts > Tb) and n = 0.3 when the fluid is being cooled (Ts < Tb). Valid for: 0.6 ≤ Pr ≤ 160, Re > 10,000, L/D > 10. Properties evaluated at bulk mean temperature. The Sieder-Tate correlation accounts for viscosity variation: Nu = 0.027 Re^0.8 Pr^(1/3) (μb/μs)^0.14.

Explanation:
With a radiation shield, the total resistance increases. Without shield: R = 1/ε1 + 1/ε2 - 1. With one shield: R = (1/ε1 + 1/ε2 - 1) + (1/εs1 + 1/εs2 - 1), where εs1 and εs2 are emissivities of the two sides of the shield. The reduction factor depends on all emissivities. If all surfaces are black (ε = 1), one shield reduces heat transfer by exactly 50% (factor of 2). For grey surfaces, the reduction depends on εs.

Explanation:
For a plane wall with uniform heat generation q̇ (W/m³), thickness 2L (x from -L to +L), and both surfaces at Tw: d²T/dx² + q̇/k = 0. Solving with boundary conditions T(±L) = Tw: T(x) = Tw + q̇(L² - x²)/(2k). Maximum temperature at centre (x=0): Tmax = Tw + q̇L²/(2k). This applies to nuclear fuel rods, electrical resistance heating, and chemical reaction systems.

Explanation:
Rayleigh number Ra = Gr × Pr = gβΔTL³/(να), where g is gravitational acceleration, β is thermal expansion coefficient, ΔT is temperature difference, L is characteristic length, ν is kinematic viscosity, and α is thermal diffusivity. Ra determines the flow regime in natural convection. For vertical plates: Ra < 10⁹ (laminar), Ra > 10⁹ (turbulent). Natural convection correlations: Nu = C × Ra^n.

Explanation:
Heat exchanger effectiveness ε = Q_actual / Q_max, where Q_max = Cmin(Th,in - Tc,in) is the maximum possible heat transfer (achieved in a counterflow HX of infinite length). Cmin = min(ṁhCph, ṁcCpc). ε ranges from 0 to 1. The NTU-effectiveness method: ε = f(NTU, Cr), where NTU = UA/Cmin and Cr = Cmin/Cmax. This method is preferred when outlet temperatures are unknown.

Explanation:
Boiling heat transfer coefficients are typically 10-100 times higher than single-phase forced convection because: (1) latent heat of vaporisation is very large, (2) bubble nucleation and departure create intense mixing, (3) the phase change process is highly efficient. Typical values: pool boiling of water: h ≈ 5,000-50,000 W/m²·K; forced convection of water: h ≈ 500-10,000 W/m²·K. This is why boilers and condensers are compact.

Explanation:
Thermal contact resistance occurs at the interface between two surfaces because: (1) real surfaces are rough and only contact at discrete asperities (actual contact area << apparent area), (2) air (poor conductor, k = 0.026 W/m·K) fills the gaps between asperities. R_contact = ΔT_interface / Q. It can be reduced by: increasing contact pressure, using thermal interface materials (thermal grease, indium foil), or surface finishing.

Explanation:
For a fin with insulated tip: Q = √(hPkAc) × θb × tanh(mL), where θb = Tb - T∞ is the base excess temperature, m = √(hP/kAc), P is perimeter, Ac is cross-sectional area. For a very long fin (mL >> 1): tanh(mL) → 1, so Q = √(hPkAc) × θb. The fin efficiency η_f = tanh(mL)/(mL). Maximum heat transfer (infinite fin): Q_max = √(hPkAc) × θb.

Explanation:
Convection involves heat transfer through bulk movement (macroscopic motion) of fluid. It combines conduction (at the fluid-solid interface) with advection (bulk fluid motion). Forced convection uses external means (pump, fan) to move fluid. Natural convection uses buoyancy-driven flow. Conduction involves molecular-level energy transfer without bulk motion. Radiation involves electromagnetic wave propagation.

Explanation:
For radial conduction through a hollow sphere: Q = 4πkr1r2(T1-T2)/(r2-r1). Therefore, thermal resistance R = (r2-r1)/(4πkr1r2). This is derived by integrating Fourier's law in spherical coordinates: Q = -k(4πr²)(dT/dr). Separating variables and integrating from r1 to r2: Q = 4πk(T1-T2)/(1/r1 - 1/r2) = 4πkr1r2(T1-T2)/(r2-r1).

Explanation:
The Wiedemann-Franz law: k/(σe × T) = L, where L = 2.45 × 10⁻⁸ W·Ω/K² is the Lorenz number, k is thermal conductivity, σe is electrical conductivity, and T is absolute temperature. This law states that good electrical conductors are also good thermal conductors (both due to free electrons). It explains why metals have high thermal conductivity.

Explanation:
The fin equation is derived from energy balance on a differential element: d²θ/dx² - m²θ = 0, where θ = T - T∞ is the excess temperature and m = √(hP/kAc). This is a second-order ODE with general solution θ(x) = C1·e^(mx) + C2·e^(-mx) = A·cosh(mx) + B·sinh(mx). Boundary conditions (base temperature, tip condition) determine C1 and C2. Note: option A has wrong sign (should be -m², not +m²).

Explanation:
NTU = UA/Cmin, where U is overall heat transfer coefficient, A is heat transfer area, and Cmin = min(ṁhCph, ṁcCpc) is the minimum heat capacity rate. NTU represents the size of the heat exchanger. The NTU-effectiveness method: ε = f(NTU, Cr = Cmin/Cmax). For a counterflow HX: ε = [1 - exp(-NTU(1-Cr))]/[1 - Cr·exp(-NTU(1-Cr))]. For Cr = 0 (condenser/evaporator): ε = 1 - exp(-NTU).

Explanation:
The Colburn analogy (j-factor): St × Pr^(2/3) = f/8, where St is Stanton number, Pr is Prandtl number, and f is Darcy friction factor. The Colburn j-factor: j_H = St × Pr^(2/3) = Nu/(Re × Pr^(1/3)). The Reynolds analogy (Pr = 1): St = f/8. The Colburn analogy extends this to fluids with Pr ≠ 1 (0.6 < Pr < 60). This analogy allows heat transfer coefficients to be estimated from friction factor data.

Explanation:
Irradiation G (W/m²) is the total radiation incident on a surface per unit area per unit time from all directions. It includes radiation from all sources (other surfaces, environment). The radiosity J = εEb + ρG = total radiation leaving the surface (emitted + reflected). Net radiation from surface: q_net = J - G = (εEb - αG)/(1-ε) for grey surface. Energy balance: q_net = (Eb - J)/((1-ε)/(εA)).

Explanation:
Peclet number Pe = Re × Pr = VL/α, where V is velocity, L is characteristic length, and α is thermal diffusivity. Pe represents the ratio of advective heat transfer to diffusive (conductive) heat transfer. For Pe >> 1, convection dominates. For Pe << 1, conduction dominates. Pe is important in liquid metal heat transfer (very low Pr) where Pe = Re × Pr may be small even at high Re.

Explanation:
Planck's law: Eλ,b = C1/(λ⁵(exp(C2/λT) - 1)), where C1 = 3.742 × 10⁸ W·μm⁴/m², C2 = 1.439 × 10⁴ μm·K, λ is wavelength (μm), and T is absolute temperature (K). Integrating over all wavelengths gives Stefan-Boltzmann law: Eb = ∫Eλ,b dλ = σT⁴. Wien's displacement law (λmax·T = 2898 μm·K) is obtained by differentiating Planck's law and setting dEλ,b/dλ = 0.

Explanation:
A counterflow heat exchanger always achieves higher (or equal) heat transfer than a parallel flow HX for the same inlet conditions and UA. This is because counterflow has a more uniform temperature difference along its length, resulting in higher LMTD. The maximum possible heat transfer (ε → 1) can only be approached in counterflow. For Cr = 0 (phase change), both configurations give the same result (ε = 1 - exp(-NTU)).

Explanation:
For any surface, the incident radiation is either absorbed, reflected, or transmitted: α + ρ + τ = 1. For opaque surfaces (τ = 0): α + ρ = 1. For a blackbody: α = 1, ρ = 0, τ = 0. For a perfect reflector: ρ = 1, α = 0, τ = 0. For a perfect transmitter (diathermanous body): τ = 1, α = 0, ρ = 0. Glass is nearly transparent to visible light but opaque to infrared (greenhouse effect).

Explanation:
Heat conduction in solids occurs through two mechanisms: (1) Lattice vibrations (phonons) — dominant in non-metallic solids and insulators; (2) Free electron movement — dominant in metals (explains high thermal conductivity of metals). In metals, free electrons carry both electrical current and thermal energy, which is why good electrical conductors are also good thermal conductors (Wiedemann-Franz law).

Explanation:
For a convex surface (surface 1) completely enclosed by surface 2: F11 = 0 (convex surface cannot see itself), F12 = 1 (all radiation from surface 1 hits surface 2). By reciprocity: A1F12 = A2F21, so F21 = A1/A2 < 1. By summation for surface 2: F21 + F22 = 1, so F22 = 1 - A1/A2. Example: a sphere (1) inside a large room (2): F11 = 0, F12 = 1, F21 = A_sphere/A_room ≈ 0.

Explanation:
For steady-state conduction in a plane wall with no heat generation: d²T/dx² = 0 (from the heat equation with ∂T/∂t = 0 and q̇ = 0). Integrating twice: T(x) = C1x + C2, which is a linear function of x. With boundary conditions T(0) = T1 and T(L) = T2: T(x) = T1 + (T2-T1)x/L. With heat generation (q̇ ≠ 0): d²T/dx² = -q̇/k, giving a parabolic distribution.

Explanation:
Condensation heat transfer coefficients are very high (typically 5,000-100,000 W/m²·K) because: (1) latent heat of condensation is released at the surface, (2) the condensate film is thin (especially in dropwise condensation), (3) the driving temperature difference is small. Dropwise condensation gives 5-10× higher h than filmwise condensation. Both boiling and condensation have much higher h than single-phase convection.