Heat Transfer

Syllabus

Modes of heat transfer; one-dimensional heat conduction, resistance concept and electrical analogy, heat transfer through fins; unsteady conduction, lumped parameter system, Heisler's charts; thermal boundary layer, dimensionless parameters in free and forced convection, correlations; radiative heat transfer, black and grey surfaces, shape factors, network analysis; heat exchanger performance, LMTD and NTU methods.

Key concepts and formulas

Conduction. Fourier's law q = −kA dT/dx. Thermal resistances (heat flow = ΔT/ΣR):

Critical radius of insulation: r_c = k/h for a cylinder and 2k/h for a sphere. Adding insulation to a pipe of radius below r_c increases heat loss until the outer radius reaches r_c.

Fins. m = √(hP/(kA_c)). A long fin loses q = √(hPkA_c) θ_b; a fin with an insulated tip loses √(hPkA_c) θ_b tanh(mL). Fin efficiency for the insulated tip is tanh(mL)/(mL).

Unsteady conduction. If the Biot number Bi = hL_c/k < 0.1 (L_c = V/A), the lumped model applies: (T − T_∞)/(T_i − T_∞) = e^−t/τ, with time constant τ = ρVc/(hA). For a sphere, L_c = r/3; for a long cylinder, r/2.

Convection. Newton's law q = hA(T_s − T_∞). Nusselt number Nu = hL/k; Prandtl number Pr = ν/α; Grashof number for free convection. Laminar flat plate: Nu_x = 0.332 Re_x^1/2Pr^1/3. Turbulent pipe flow (Dittus-Boelter): Nu = 0.023 Re^0.8Prⁿ (n = 0.4 heating, 0.3 cooling). Fully developed laminar pipe flow: Nu = 3.66 (constant wall temperature) and 4.36 (constant heat flux).

Radiation. Stefan-Boltzmann: E_b = σT⁴ with σ = 5.67 × 10⁻⁸ W/m²K⁴. Wien's law λ_maxT = 2898 μm K. Shape factor reciprocity A₁F₁₂ = A₂F₂₁ and summation ΣF = 1. Two large parallel grey plates: q/A = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1). A radiation shield with the same emissivity halves the exchange between two plates.

Heat exchangers. Q = UA ΔT_lm, with LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). Counterflow gives a larger LMTD than parallel flow for the same end temperatures. NTU = UA/C_min; effectiveness ε = Q/Q_max with Q_max = C_min(T_h,in − T_c,in).

Worked example

A furnace wall has 200 mm of firebrick (k = 1.2 W/mK) and 100 mm of insulating brick (k = 0.15 W/mK), with inside and outside surface temperatures of 900 °C and 50 °C. Find the heat flux. Then find the LMTD of a counterflow exchanger where hot fluid cools from 150 to 90 °C and cold fluid heats from 30 to 70 °C, and the critical radius for k = 0.1 W/mK and h = 10 W/m²K.

import math

R = 0.2 / 1.2 + 0.1 / 0.15
q = (900 - 50) / R
print(f"total resistance = {R:.4f} m2K/W, heat flux = {q:.2f} W/m2")
dT1, dT2 = 150 - 70, 90 - 30
print(f"counterflow LMTD = {(dT1 - dT2) / math.log(dT1 / dT2):.3f} K")
print(f"critical radius = {0.1 / 10 * 1000:.0f} mm")
sigma = 5.67e-8
print(f"black body at 1000 K emits {sigma * 1000 ** 4 / 1000:.2f} kW/m2")

total resistance = 0.8333 m2K/W, heat flux = 1020.00 W/m2
counterflow LMTD = 69.521 K
critical radius = 10 mm
black body at 1000 K emits 56.70 kW/m2

Practice questions