Heat Transfer Multiple Choice Questions

✅ Correct: Temperature drop ΔT = q·L/κ, so lower conductivity → higher ΔT. Refractory brick has much lower conductivity than metals like copper or steel. Hence, maximum temperature drop occurs across refractory brick.

✅ Correct: Heat transfer resistance is higher when low‑κ material is closer to the hot surface. This reduces heat penetration into the outer layer. Hence, the lower conductivity material should be used inside.

✅ Correct: Thermal diffusivity α = κ/(ρ·c_p). It measures how quickly a material responds to temperature changes. High α means rapid heat propagation through the material.

✅ Correct: General conduction equation is ∇²T + q̇/κ = (1/α)(∂T/∂τ). In 1‑D rectilinear form, it reduces to the given expression. It balances conduction, heat generation, and transient storage.

✅ Correct: In cylindrical coordinates, conduction equation is (1/r) ∂/∂r (r ∂T/∂r) + q̇/κ = (1/α)(∂T/∂τ). This accounts for radial geometry effects in heat transfer.

✅ Correct: For linear variation, average κ is taken at mean temperature. That is κ_avg = κ₀[1 + β(T₁+T₂)/2]. This ensures conduction is modeled with effective conductivity.

✅ Correct: Lumped analysis assumes uniform temperature inside the body. This is valid when Bi = hLc/κ < 0.1, meaning internal resistance is negligible. Then only surface convection controls heat transfer.

✅ Correct: Lumped system analysis gives ΔT = (T₀ − T∞) exp[−(hA/ρVc)·t]. The negative exponent shows exponential decay of temperature difference with time.

✅ Correct: In a semi-infinite body, heat penetrates only to a finite depth at any time. Beyond this depth, the temperature remains unaffected by surface heating or cooling.

✅ Correct: Aluminium has the highest conductivity, followed by pure iron, then liquid water, and finally saturated vapor. Thus the order is d (Al), a (Fe), b (water), c (steam).

✅ Correct: For a cylinder, r_c = κ/h. Here κ = 1 W/mK, h = 8 W/m²K → r_c = 0.125 m = 12.5 cm. Since outer radius is 5.5 cm, adding insulation up to 12.5 cm increases heat loss.

✅ Correct: Cooling rate depends on surface area to mass ratio. Hollow sphere has less mass but same surface area, so it cools faster for any T.

✅ Correct: For equal heat loss, L/κ must remain constant. So L₂ = L₁ (κ₂/κ₁) = 0.5 × (0.14/0.7) = 0.1 m. Thus, a 0.1 m layer of diatomic earth replaces 0.5 m of brick.

✅ Correct: Heat equation gives ∂T/∂t = α ∂²T/∂x². Here ∂²T/∂x² = 6, so ∂T/∂t = 0.0003 × 6 = 0.0018 K/s. Since heat flows out, the sign is negative → −0.0018 K/s.

✅ Correct: Effective conductivity = (k₁ + k₂)/2 = (25+15)/2 = 20 W/mK. Heat flux q" = k_eff ΔT / L = 20 × 100 / 0.5 = 4000 W/m² = 4 kW/m².

✅ Correct: For a hollow sphere, the geometric mean radius √(r₁ r₂) is used in conduction analysis. This simplifies the logarithmic mean area concept for spherical heat transfer.

✅ Correct: Q = k A ΔT / L = 0.5 × 1 × 60 / 0.25 = 120 W. This is the steady-state conduction heat flow through the wall.

✅ Correct: Temperature drop ratio depends on thermal resistance of each layer. With equal thickness and k ratio 1:2, the inner layer offers higher resistance. The ratio of ΔT_inner : ΔT_outer works out to about 2.5.

✅ Correct: Total resistance R = Σ(L/kA) = 0.3/0.6 + 0.2/0.4 + 0.1/0.1 = 0.5 + 0.5 + 1 = 2 K/W. ΔT = 1840 − 340 = 1500 K. Heat transfer Q = ΔT / R = 1500 / 2 = 750 W.

✅ Correct: In spherical coordinates, the steady conduction solution is of the form T = C₁ + C₂/r. This distribution is not linear but varies with 1/r, which corresponds to a parabolic-type curve when plotted. Hence, the temperature distribution is parabolic in nature.

✅ Correct: Thermal resistance R = L / (k·A). Substituting: R = 0.6 / (0.4 × 1.5) = 0.6 / 0.6 = 1 K/W. Thus, the wall offers a thermal resistance of 1 K/W.

✅ Correct: For equal thickness and area, ΔT ∝ 1/k. With k ratio 3:2, the temperature drop ratio is inverse, i.e., 2:3. So the layer with lower conductivity (2) has the higher temperature drop.

Answer: Air. Gases have much lower thermal conductivity than liquids and solids, so conduction through air is minimal. Copper and lead are metals (high conductivity), and even water conducts more than air.

Answer: 4:2:1. With equal thickness and area, heat flux is the same through each layer, and ΔT ∝ (L/(kA)) ∝ 1/k. Inversely proportioning 1:2:4 gives drops of 1:½:¼, i.e., scaled to 4:2:1.

Answer: 0.2 °C/h. From the heat equation, ∂T/∂t = α ∂²T/∂x². Here ∂²T/∂x² = 40 + 60x; at the far face x = 1 m, it is 100. Thus ∂T/∂t = 2×10⁻³ × 100 = 0.2 °C/h.

Answer: Added insulation will increase the rate of heat loss. For cylinders/spheres, increasing radius initially boosts surface area faster than resistance, raising heat loss. Beyond the critical radius, further insulation reduces heat loss.

Answer: κ/h. For cylinders, r_c = κ/h balances conduction resistance and convective loss. This marks the peak heat loss before added insulation starts reducing it.

Answer: 2κ/h. For spheres, the critical radius is doubled compared to cylinders due to geometry of conduction. r_c = 2κ/h sets the turning point for heat loss behavior with added insulation.

Answer: 15 mm. r_c = κ/h = 0.5/20 = 0.025 m = 25 mm; bare wire radius r = 10 mm. Optimum thickness = r_c − r = 25 − 10 = 15 mm.

Answer: 0.5 mm. r_c = κ/h = 0.1/100 = 0.001 m = 1 mm; bare radius r = 0.5 mm. Optimum thickness = r_c − r = 1 − 0.5 = 0.5 mm.

Answer: 23.5 °C. Heat lost by steel = heat gained by water: 1·0.4·(60 − T) = 1·4.2·(T − 20). Solving gives T ≈ 108/4.6 ≈ 23.5 °C (neglecting losses to surroundings).

Answer: 16 mm. For a sphere, the critical radius of insulation is r_c = 2κ/h. With κ = 0.04 W/mK and h = 10 W/m²K, r_c = 0.008 m = 8 mm. Maximum heat loss occurs when the outer radius equals r_c, so the sheath’s outer diameter = 2r_c = 16 mm. Since the core radius is 5 mm, the required sheath thickness is 8 − 5 = 3 mm.

Answer: Decrease. Adding thick insulation generally raises overall thermal resistance, limiting the wire’s ability to dissipate heat to ambient. Even though cylindrical geometry can initially increase heat loss up to the critical radius, a large added thickness typically reduces net cooling in practice. Reduced heat rejection means the permissible current (without overheating) decreases.

Answer: 2 cm. For a cylinder, the critical radius is r_c = κ/h (insulation properties govern r_c). With κ = 0.1 W/mK and h = 5 W/m²K, r_c = 0.1/5 = 0.02 m = 2 cm. Below r_c, added insulation can increase heat loss; beyond r_c, it reduces heat loss.

Answer: The time taken to attain 63.2% of initial temperature difference. For a first-order sensor, response follows an exponential, reaching 1 − e⁻¹ ≈ 63.2% in one time constant. This parameter characterizes the speed of thermal response to a step change. Faster sensors have smaller time constants and track transients more accurately.

Answer: Exponential curve asymptotic both to time and temperature axes. Transient (lumped) cooling follows Newton’s law: \(T(t)-T_\infty \propto e^{-t/\tau}\). The curve approaches the ambient temperature as \(t \to \infty\) and approaches the initial temperature as \(t \to 0\), making it asymptotic to both axes.

Answer: tanh(mL)/(mL). For an insulated tip, the fin efficiency \(\eta_f = \dfrac{\text{actual heat}}{\text{ideal if fin at }T_0}\) simplifies to \(\eta_f = \tanh(mL)/(mL)\). Here \(m=\sqrt{\dfrac{hP}{\kappa A}}\), capturing convection–conduction balance along the fin.

Answer: less than one. The criterion for a fin to improve heat transfer is \(mL\) significant and fin parameter \(m=\sqrt{hP/(\kappa A)}\) such that conduction through fin is strong. When \(\sqrt{hA/\kappa P} < 1\), conduction dominates enough that added area yields net gain.

Answer: 1.7 W. Surface loss \(q_s=\int_0^L hP\,(T-T_\infty)\,dx\). With \(T-T_\infty=3x^2-5x+6\), \(h=0.0025\), \(P\) from perimeter, and \(L=1\) cm, the integral yields ≈ 1.7 W. Base conduction and tip condition are consistent with the given polynomial profile and material properties.

Answer: √(hκA) (T0 − T∞). For a sufficiently long fin (tip at ambient), \(Q \approx \sqrt{h\,\kappa\,A}\,(T_0-T_\infty)\). This emerges from the exact solution \(Q = \sqrt{hP\kappa A}\,(T_0-T_\infty)\coth(mL)\), where \(\coth(mL)\to 1\) and \(P\) cancels when using the common “rod” parameterization with A.

Answer: \(T_{\infty} + \dfrac{qL}{h}\). At the surface, energy balance: conduction flux equals convection, giving \(-\kappa \dfrac{dT}{dx}\big|_{s}=h(T_s-T_\infty)\). With uniform generation, the surface heat flux equals qL, hence \(h(T_s-T_\infty)=qL \Rightarrow T_s=T_\infty+qL/h\).

Answer: increasing the effective surface area. Fins add area exposed to convection without significantly changing local \(h\) or \(\Delta T\). More area → more heat transfer for the same conditions, provided fin conduction is adequate.

Answer: Prandtl number, Grashof number. In natural convection, buoyancy vs. viscous/inertial effects are captured by Gr, while thermal–momentum diffusion ratio is Pr. Critical transition depends on both, typically via the product \(Gr\,Pr\) (Rayleigh number).

Answer: both (area and temperature difference). Newton’s law: \( \dot{Q} = hA(T_s - T_\infty)\). Thus heat rate scales linearly with area and temperature difference (for a given h).

Answer: both. In forced convection, higher Re enhances mixing and h, raising Nu; in free convection, stronger buoyancy (higher Gr) increases Nu. Many correlations show Nu ∝ Re^m Pr^n (forced) and Nu ∝ (Gr Pr)^m (free).

Answer: kinematic viscosity and thermal diffusivity. \(Pr = \dfrac{\nu}{\alpha} = \dfrac{\mu/\rho}{\kappa/(\rho c_p)} = \dfrac{c_p \mu}{\kappa}\). It compares momentum diffusivity to thermal diffusivity, shaping boundary layer behavior.

Answer: both. Gr measures buoyancy-driven flow strength; small Gr → laminar; large Gr → transition/turbulence. Practical correlations often use Rayleigh \(Ra=Gr\,Pr\) to indicate regimes on vertical plates or enclosures.