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The Resistance Analogy
Heat transfer through a solid or across a fluid boundary follows the same mathematical structure as electrical current through a resistor: heat flux is driven by a temperature difference, opposed by a thermal resistance. This analogy — known as the thermal resistance network method — lets engineers decompose complex multi-layer wall problems into simple Ohm's-law-style circuits.
$$ Q = \frac{\Delta T}{R_{total}}, \qquad R_{total} = \sum R_i \text{ (series)} $$where Q is the heat flow rate [W], ΔT is the overall temperature difference [K or °C], and Ri are the individual resistances [K/W].
Conduction Resistance — Flat Wall
For a homogeneous slab of thickness L, cross-sectional area A, and thermal conductivity k:
$$ R_{cond} = \frac{L}{k A} $$Units: [m] / ([W/m·K] × [m²]) = K/W. The resistance increases with thickness (longer path) and decreases with conductivity (better conductor) and area (more parallel paths).
Convection Resistance
At any solid-fluid interface with convective heat-transfer coefficient h and area A:
$$ R_{conv} = \frac{1}{h A} $$A thin, high-conductivity wall flanked by low-h convective resistances (e.g., air on both sides) has its overall performance dominated by convection, not conduction. This is why insulation and surface enhancement (fins) matter far more than switching from aluminium to copper in most building and HVAC applications.
Series Resistance: Composite Flat Wall
For a wall built from n layers in series (e.g., brick + insulation + plasterboard), with convection on both sides:
$$ R_{total} = \frac{1}{h_1 A} + \frac{L_1}{k_1 A} + \frac{L_2}{k_2 A} + \cdots + \frac{L_n}{k_n A} + \frac{1}{h_2 A} $$ $$ Q = \frac{T_{fluid,1} - T_{fluid,2}}{R_{total}} $$Intermediate interface temperatures are found by stepping through the network:
$$ T_{interface,i} = T_{fluid,1} - Q \cdot \left(\frac{1}{h_1 A} + \sum_{j=1}^{i} \frac{L_j}{k_j A}\right) $$Worked Example: Cold Store Wall
A refrigerated warehouse wall consists of: external steel plate (t = 6 mm, k = 50 W/m·K), polyurethane foam insulation (t = 150 mm, k = 0.026 W/m·K), internal steel plate (t = 6 mm). External ambient h = 25 W/m²·K at 35°C; internal h = 10 W/m²·K at −20°C. Per unit area (A = 1 m²):
$$ R_{ext,conv} = \frac{1}{25 \times 1} = 0.0400 \ \text{K/W} $$ $$ R_{steel,1} = \frac{0.006}{50 \times 1} = 0.000120 \ \text{K/W} $$ $$ R_{insul} = \frac{0.150}{0.026 \times 1} = 5.769 \ \text{K/W} $$ $$ R_{steel,2} = 0.000120 \ \text{K/W} $$ $$ R_{int,conv} = \frac{1}{10 \times 1} = 0.1000 \ \text{K/W} $$ $$ R_{total} = 0.0400 + 0.000120 + 5.769 + 0.000120 + 0.1000 = 5.909 \ \text{K/W} $$ $$ Q = \frac{35 - (-20)}{5.909} = \frac{55}{5.909} = 9.31 \ \text{W/m}^2 $$The insulation accounts for 5.769 / 5.909 = 97.6% of total resistance. The steel plates contribute virtually nothing — a classic result confirming that conductivity of the structural layers is irrelevant compared to the insulation thermal performance.
Cylindrical Resistance — Pipe Walls & Tube Bundles
For a hollow cylinder (pipe) of inner radius ri, outer radius ro, length L, and conductivity k:
$$ R_{cyl} = \frac{\ln(r_o / r_i)}{2\pi k L} $$The logarithmic form arises because the cross-sectional area for heat flow is not constant — it increases with radius. For a multi-layer pipe (insulation over pipe over liner):
$$ R_{total} = \frac{1}{h_i \cdot 2\pi r_i L} + \frac{\ln(r_1/r_i)}{2\pi k_1 L} + \frac{\ln(r_2/r_1)}{2\pi k_2 L} + \frac{1}{h_o \cdot 2\pi r_o L} $$Worked Example: Steam Pipe with Insulation
A 4-inch (DN100) schedule 40 steel pipe carries 180°C steam. Pipe: inner radius ri = 52.5 mm, outer radius r1 = 60.3 mm, ksteel = 50 W/m·K. Covered with 75 mm calcium silicate insulation (k = 0.065 W/m·K), outer radius r2 = 135.3 mm. Ambient air: T∞ = 25°C, ho = 12 W/m²·K. Steam hi = 8 000 W/m²·K. Per metre of pipe length (L = 1 m):
$$ R_{steam} = \frac{1}{8000 \times 2\pi \times 0.0525 \times 1} = \frac{1}{2639} = 0.000379 \ \text{K/W} $$ $$ R_{steel} = \frac{\ln(60.3/52.5)}{2\pi \times 50 \times 1} = \frac{\ln(1.149)}{314.2} = \frac{0.1390}{314.2} = 0.000442 \ \text{K/W} $$ $$ R_{insul} = \frac{\ln(135.3/60.3)}{2\pi \times 0.065 \times 1} = \frac{\ln(2.244)}{0.4084} = \frac{0.8074}{0.4084} = 1.977 \ \text{K/W} $$ $$ R_{amb} = \frac{1}{12 \times 2\pi \times 0.1353 \times 1} = \frac{1}{10.20} = 0.0980 \ \text{K/W} $$ $$ R_{total} = 0.000379 + 0.000442 + 1.977 + 0.0980 = 2.076 \ \text{K/W per metre} $$ $$ Q = \frac{180 - 25}{2.076} = \frac{155}{2.076} = 74.7 \ \text{W/m} $$Outer pipe surface temperature:
$$ T_{outer} = T_{amb} + Q \times R_{amb} = 25 + 74.7 \times 0.0980 = 25 + 7.3 = 32.3°C $$The outer insulation surface is only 32.3°C despite the 180°C steam inside — confirming the insulation is working effectively. An uninsulated pipe would radiate and convect approximately 1 200 W/m — about 16× more heat loss.
Parallel Resistance Networks
When heat can flow simultaneously through two different paths (e.g., a composite wall with a metal stud embedded in insulation), the resistances are in parallel:
$$ \frac{1}{R_{parallel}} = \frac{1}{R_A} + \frac{1}{R_B} $$ $$ R_{parallel} = \frac{R_A \cdot R_B}{R_A + R_B} $$For a wall with studs at regular spacing, the effective overall resistance combines a series network (inner/outer convection) with a parallel section (stud + insulation):
$$ R_{eff} = R_{conv,1} + R_{parallel,section} + R_{conv,2} $$Thermal Bridging: Why Studs Matter
In building physics and process insulation, metal conductors (studs, brackets, hangers) that penetrate an insulation layer create thermal bridges. Even if the metal occupies only 5% of the wall cross-section, its much lower resistance can dominate the parallel combination. For a typical steel stud (k = 50 W/m·K) in 150 mm mineral wool (k = 0.040 W/m·K) at 10% area fraction:
$$ R_{stud} = \frac{0.150}{50 \times 0.10} = 0.030 \ \text{K/W} \quad R_{wool} = \frac{0.150}{0.040 \times 0.90} = 4.167 \ \text{K/W} $$ $$ R_{parallel} = \frac{0.030 \times 4.167}{0.030 + 4.167} = \frac{0.125}{4.197} = 0.0298 \ \text{K/W} $$The parallel resistance (0.030 K/W) is virtually identical to the stud alone — the stud short-circuits the insulation almost entirely. This is why process insulation designers use standoff brackets rather than continuous metal contact wherever possible.
Contact Resistance
At the interface between two solid surfaces, microscopic roughness means actual contact exists only at asperities (high points). The rest of the interface is filled with air (a poor conductor). The resulting contact resistance Rc (or thermal contact conductance hc = 1/Rc) depends on:
- Surface roughness (RMS roughness σ)
- Interface pressure (higher pressure → more asperity contact)
- Interfacial material (air, thermal grease, indium foil)
- Material hardness
Representative values for engineering surfaces:
| Interface | hc (W/m²·K) | Rc" (m²·K/W) |
|---|---|---|
| Steel–steel (ground, no filler) | 3 000 – 10 000 | 0.0001 – 0.00033 |
| Aluminium–aluminium (ground) | 4 000 – 40 000 | 0.000025 – 0.00025 |
| Steel–steel + thermal grease | 20 000 – 100 000 | 0.00001 – 0.00005 |
| Fin on tube (interference fit) | 1 000 – 6 000 | 0.00017 – 0.001 |
Contact resistance is especially important in electronics cooling and in finned-tube air coolers where poorly fitted fin collars can double or triple the fin base resistance.
Fin Efficiency and the Extended Surface
Fins increase the effective heat-transfer area on a surface with a low convection coefficient (typically the air side). However, the fin tip is farther from the base, so the temperature gradient along the fin reduces its effectiveness. The fin efficiency ηf compares actual heat transfer to the idealised case where the entire fin is at the base temperature:
$$ \eta_f = \frac{Q_{actual}}{Q_{ideal}} = \frac{\tanh(mL)}{mL} $$ $$ m = \sqrt{\frac{h P}{k A_c}} $$where P is the fin perimeter [m], Ac is the fin cross-sectional area [m²], k is fin material conductivity, h is the convective coefficient on the fin surface, and L is fin height [m]. For a thin rectangular fin of width w and thickness t: P = 2w (≈ 2 for unit width), Ac = wt.
Overall Surface Efficiency
For a surface with Nf fins each of area Af and unfinned base area Ab:
$$ \eta_o = 1 - \frac{N_f A_f}{A_{total}}(1 - \eta_f) $$The effective convective resistance for the finned surface is then:
$$ R_{fin,eff} = \frac{1}{\eta_o h A_{total}} $$ηo typically ranges from 0.70 to 0.95 for well-designed finned surfaces. Values below 0.60 indicate fins are too tall, too thin, or made from a low-conductivity alloy.
Radiation Resistance
Radiation heat transfer can be linearised around a mean temperature to give a radiation resistance analogous to convection:
$$ q_{rad} = h_{rad}(T_s - T_{surr}) $$ $$ h_{rad} = \varepsilon \sigma (T_s + T_{surr})(T_s^2 + T_{surr}^2) $$ $$ R_{rad} = \frac{1}{h_{rad} A} $$where ε is surface emissivity and σ = 5.67 × 10⁻⁸ W/m²·K⁴ (Stefan-Boltzmann constant). Radiation and convection resistances on an external surface are in parallel:
$$ R_{ext} = \frac{1}{(h_{conv} + h_{rad})A} $$At high surface temperatures (> 300°C), radiation can contribute 30–60% of total external heat loss and must not be neglected in bare-surface insulation calculations.
Complete Network: Heat Exchanger Tube Wall
A shell-and-tube exchanger tube wall represents the canonical process engineering thermal resistance network (referenced to outer tube area Ao):
$$ \frac{1}{U_f A_o} = \underbrace{\frac{1}{h_o A_o}}_{\text{shell-side film}} + \underbrace{\frac{R_{fo}}{A_o}}_{\text{shell fouling}} + \underbrace{\frac{\ln(d_o/d_i)}{2\pi k_w L}}_{\text{tube wall}} + \underbrace{\frac{R_{fi}}{A_i}}_{\text{tube fouling}} + \underbrace{\frac{1}{h_i A_i}}_{\text{tube-side film}} $$Each term can be evaluated and compared to identify the controlling resistance — the bottleneck that limits performance. Spending capital to reduce an already-small resistance yields little benefit; improving the dominant resistance delivers the greatest improvement per dollar.
Design Insight: Controlling Resistance
| Dominant Resistance | Typical Service | Improvement Strategy |
|---|---|---|
| Shell-side film | Gas cooling, viscous liquid | Add baffles, reduce baffle pitch, use enhanced tubes |
| Tube-side film | Laminar flow, high viscosity | Increase velocity, use twisted-tape inserts, add tube passes |
| Fouling | Crude preheat, cooling water | Increase velocity, add anti-fouling treatment, reduce cleaning interval |
| Wall conduction | Thick-wall high-pressure service | Switch from low-alloy to higher-conductivity material, thin wall if code allows |
Try the Calculator
Use the Heat Transfer Coefficient Calculator panel on this page to build your own thermal resistance network: enter film coefficients, wall conductance, and fouling resistances to compute the overall U, heat duty, and outlet temperatures instantly.
Heat Exchanger Rating
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