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Finite Element Methods · Chapter 8 of 10 · Intermediate

Finite Elements for Heat Transfer

The finite element method is not only for stress. Swap displacement for temperature and stiffness for conductance and the very same machinery solves conduction, sharing matrices with the structural problem term for term.

Readiness check

This chapter applies FEM to a field problem. Tick only what you can do closed-notes.

Recall Fourier's law and steady conduction.
Recall thermal resistance L/kA.
Use the spring assembly from Chapter 2.
Solve a small linear system.
Recall heat rate as conductance times temperature difference.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit conduction in Heat Transfer, Chapter 2.

3 or more weak itemsReview the direct stiffness method in Chapter 2.

The core idea

Conduction is the same finite element problem as a structure, with temperature in place of displacement and conductance kA/L in place of stiffness. The element matrix, assembly, and solve are identical.

K_tT = Q (conductance · temperature = heat)element: (kA/L)[1 −1; −1 1]Q = (kA/L)(T_i − T_j)

The steady conduction equation has the same mathematical form as elastic equilibrium, so the finite element method transfers directly. Each conduction element relates the heat flows at its two nodes to their temperatures through the conductance kA/L, giving the element matrix (kA/L)[1, −1; −1, 1], identical in shape to the spring. Assembling these into a global conductance matrix, applying fixed temperatures as boundary conditions, and solving K_tT = Q gives the nodal temperatures. From them, the heat flow in each element follows as conductance times the temperature difference, and these flows must balance at every node, the thermal version of equilibrium. Convection and heat sources add their own terms, but the backbone is the structural method reused.

The skill works when: you map temperature to displacement and conductance to stiffness, then reuse the structural recipe.

The skill breaks down when: the analogy is forgotten and heat flows do not balance at a node.

The concept. A wall split into two conduction elements with three temperature nodes. With the hot and cold faces fixed, solving for the middle temperature is exactly the spring problem in thermal clothing.

The skills, taught in order

Five skills cover field problems, the conductance matrix, the analogy, the temperature solution, and heat-flow recovery.

8.1 FEM for field problems

Beyond structures, the finite element method solves any field governed by a similar differential equation: heat conduction, seepage, electrostatics, and more. The unknown becomes a scalar field at the nodes, and the same assemble-constrain-solve recipe applies.

8.2 The conduction element

A one-dimensional conduction element relates the nodal heat flows to the nodal temperatures through the conductance kA/L, giving the element matrix (kA/L)[1, −1; −1, 1]. For two and three dimensions the conductance matrix is built from the same B and material steps as the structural stiffness.

8.3 The structural-thermal analogy

The two problems share a structure term for term, which is why one solver handles both.

Structural	Thermal
displacement u	temperature T
force F	heat rate Q
stiffness AE/L	conductance kA/L
Ku = F	K_tT = Q

8.4 Solving for temperatures

Fixed temperatures are essential boundary conditions, applied just like fixed displacements; specified heat flows and sources enter the load vector Q. Solving the reduced K_tT = Q gives the unknown nodal temperatures.

8.5 Recovering heat flow

Once temperatures are known, the heat flow in each element is its conductance times the temperature drop, Q = (kA/L)(T_i − T_j). In steady state with no source, the flow is the same through every element in series, the check that validates the solution.

Engineering connection: because structural and thermal share the method, a thermal-stress analysis chains the two: temperatures from this chapter drive thermal strains in the structural model.

Worked example 1: temperature in a wall

A wall (thermal conductivity k = 15 W/m·K, area A = 0.1 m²) is modelled with two conduction elements, each 0.05 m long. The hot face is held at 200 °C and the cold face at 20 °C, with no internal source. Find the element conductance and the middle temperature.

Figure 1. With equal elements and no source, the middle node sits at the average of the two face temperatures, the linear conduction profile.

ProblemFind the element conductance and the middle temperature for the wall in Figure 1.
Given / findk = 15 W/m·K, A = 0.1 m², L = 0.05 m per element, T₁ = 200 °C, T₃ = 20 °C. Find kA/L and T₂.
AssumptionsSteady conduction, no source; two equal elements.
ModelElement conductance kA/L; the free middle node balances heat from both sides, giving 2(kA/L)T₂ = (kA/L)(T₁ + T₃).
EquationskA/L = 15(0.1)/0.05 = 30 W/KT₂ = (T₁ + T₃)/2
SolvekA/L = 30 W/K. T₂ = (200 + 20)/2 = 110 °C.
CheckThe middle temperature is the average of the faces, the exact linear conduction profile for a uniform wall, which two equal elements reproduce. The conductance cancels from the temperature solution, as it should with no source.
ConclusionThe thermal problem solves exactly like the two-spring structural problem. Temperature has replaced displacement and conductance has replaced stiffness.

Result. kA/L = 30 W/K; T₂ = 110 °C.

Worked example 2: recovering the heat flow

Using the temperatures from Worked Example 1 (T₁ = 200 °C, T₂ = 110 °C, T₃ = 20 °C) and the element conductance 30 W/K, find the heat flow through each element and confirm it is consistent.

Figure 2. The heat flow is conductance times temperature drop, and in steady state it is identical through every element in series, the validation check.

ProblemFind the heat flow through each element in Figure 2 and confirm consistency.
Given / findT₁ = 200, T₂ = 110, T₃ = 20 °C; conductance 30 W/K. Find Q₁, Q₂.
AssumptionsSteady conduction, no source, so the same heat passes through each element.
ModelElement heat flow is conductance times the temperature drop across it.
EquationsQ₁ = (kA/L)(T₁ − T₂)Q₂ = (kA/L)(T₂ − T₃)
SolveQ₁ = 30 × (200 − 110) = 2700 W. Q₂ = 30 × (110 − 20) = 2700 W.
CheckBoth elements carry the same 2700 W, as a series steady-state path with no source must, confirming energy balance at the middle node. The overall conductance kA/L_total = 15(0.1)/0.1 = 15 W/K gives 15 × 180 = 2700 W directly.
ConclusionRecovering and balancing the heat flow is the thermal equilibrium check, exactly the role reactions play in a structural model.

Result. Q = 2700 W through both elements (consistent).

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Forgetting fixed-temperature BCs	Singular conductance matrix	"Is at least one temperature fixed?"	A fixed temperature is the thermal version of a support.
Heat flows not balancing	Different Q in series elements	"Is there a source between them?"	With no source, series elements carry equal heat.
Wrong conductance	Temperatures or heat off	"Did I use kA/L?"	Conductance is k times area over length.
Mixing analogies	Force used where heat belongs	"Is the load a force or a heat rate?"	Map force to heat rate, stiffness to conductance.

Practice ladder

Level 1 · Direct skill

Find the conductance of an element with k = 50 W/m·K, A = 0.02 m², L = 0.1 m.

Show answer

kA/L = 50 × 0.02/0.1 = 10 W/K.

Level 2 · Mixed concept

For the Worked Example 1 wall, what is the middle temperature if the cold face is instead held at 60 °C?

Show answer

T₂ = (200 + 60)/2 = 130 °C, still the average of the two faces for equal elements.

Level 3 · Independent problem

A two-element wall has conductance 20 W/K per element, faces at 150 °C and 30 °C. Find the middle temperature and the heat flow.

Show answer

T₂ = (150 + 30)/2 = 90 °C. Q = 20 × (150 − 90) = 1200 W, the same through both elements.

Level 4 · Transfer to real engineering

For a real conduction problem (a heat sink fin, a composite wall), describe how you would set up the thermal FEM and validate the result.

What good work looks like

Conduction elements with kA/L conductances, fixed temperatures or fluxes as boundary conditions, and a heat-balance or analytic check on the solution.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my conductance matrix and fixed temperatures are right."

"Give me three thermal cases; I will map them to the structural analogy."

"Solve this conduction problem." Assembling and solving yourself is the skill.

"What is the heat flow?" Recovering it and balancing nodes is the point.

Portfolio task

Set up a thermal FEM for a real conduction problem: build the conductance matrix, apply fixed temperatures, solve for temperatures, and recover and balance the heat flow.

Must include: a conductance matrix, a temperature solution, and a balanced heat-flow check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What replaces displacement and stiffness in the thermal problem?

Temperature replaces displacement; conductance kA/L replaces stiffness.

2. Write the conduction element matrix.

(kA/L)[1, −1; −1, 1].

3. What is the thermal system equation?

K_tT = Q, conductance times temperature equals heat rate.

4. What is the thermal boundary condition like a support?

A fixed (prescribed) temperature.

5. How is element heat flow found?

Q = (kA/L)(T_i − T_j), equal through series elements with no source.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the wall temperature and heat flow from a blank page.

+3 daysSolve two new conduction problems.

+7 daysCarry the method into dynamics, Chapter 9.

+30 daysChain temperatures into a thermal-stress analysis.

Textbook mapping

Item	Mapping
Primary source	Logan, A First Course in the Finite Element Method, Chapter 11 (Heat Transfer and Mass Transport)
Cross-reference	Reddy, Ch. 4 · Heat Transfer, Ch. 2
Core topics	8.1 FEM for field problems · 8.2 Conduction element · 8.3 Structural-thermal analogy · 8.4 Solving for temperatures · 8.5 Heat-flow recovery
Engineering connection	Temperatures from here drive thermal strains in a coupled thermal-stress model.
Read next	Chapter 9: Dynamics and Modal Analysis.