Heat Transfer

7% of exam

Conduction through walls and cylinders, convection and Newton's law of cooling, radiation, transient (lumped) processes, and heat exchangers.

3 concepts

Exam areas

A. Conduction

Conduction and Thermal Resistance Networks

Fourier's law for plane walls and cylinders, the series/parallel thermal-resistance analogy, contact resistance, and the critical insulation radius.

Conduction is where most FE Mechanical heat-transfer points are won or lost, because nearly every problem — a furnace wall, an insulated pipe, a chip on a heat sink — reduces to the same move: turn the geometry into a chain of resistances and divide a temperature difference by their sum. The physics is Fourier's law,

\dot{Q} = -kA\,dT/dx

, but the working tool is the electrical analogy, where temperature plays the role of voltage, heat rate the role of current, and

R = \Delta T/\dot{Q}

the role of resistance. The FE Reference Handbook (Heat Transfer) lays out the resistance formulas for plane walls, cylinders, and convection films; this concept welds them into a procedure you can run cold under exam pressure.

Fourier's law and the plane wall

Fourier's law states that conductive heat flux is proportional to the negative temperature gradient: heat flows down the temperature hill, and the thermal conductivity

k

sets how steeply. For one-dimensional, steady conduction through a plane wall of area

A

and thickness

L

with faces at

T_1

and

T_2

, the gradient is linear and the law integrates to a clean algebraic result. Note that

\dot{Q}

is constant through every layer at steady state — the same watts pass through each slice — which is exactly why the series-resistance idea works.

\dot{Q} = kA\,\frac{T_1 - T_2}{L}

The thermal-resistance analogy

Rewrite each transport step as

\dot{Q} = \Delta T / R

and the whole problem becomes a circuit. A plane conduction layer has

R = L/(kA)

; a convection film has

R = 1/(hA)

. Resistances in series simply add, so the heat rate is the overall temperature drop divided by the total resistance. This single idea — sum the resistances, then divide once — is the backbone of conduction problems, and it lets you find intermediate temperatures by walking the chain one resistor at a time.

\dot{Q} = \frac{\Delta T_{\text{overall}}}{R_{\text{total}}}, \qquad R_{\text{total}} = \sum_i R_i

Composite walls: series and parallel

A multilayer wall bounded by moving fluids is a series chain: inside film, each solid layer, outside film. The overall coefficient

U

collapses that chain into one number through

1/(UA) = R_{\text{total}}

, which is why

\dot{Q} = UA\,\Delta T

. When heat can take two side-by-side paths through a wall (a studded panel, a finned region), those branches are in parallel and combine by reciprocal addition,

1/R = 1/R_1 + 1/R_2

. Recognizing series-versus-parallel topology before plugging numbers is the discriminator on harder items.

\frac{1}{UA} = R_{\text{total}} = \frac{1}{h_1 A} + \sum_j \frac{L_j}{k_j A} + \frac{1}{h_2 A}

Radial conduction through a cylinder

Pipes and tubes conduct radially, and because the area

A = 2\pi r L

grows with radius, the temperature profile is logarithmic rather than linear. The conduction resistance of a cylindrical shell from inner radius

r_1

to outer radius

r_2

therefore carries a

\ln(r_2/r_1)

term. Use the surface area at the relevant radius for each convection film — inside uses

2\pi r_1 L

, outside uses

2\pi r_2 L

— a distinction that trips up examinees who reuse one area everywhere.

R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L}, \qquad \dot{Q} = \frac{2\pi k L\,(T_1 - T_2)}{\ln(r_2/r_1)}

Critical radius of insulation

Adding insulation to a pipe raises the conduction resistance but also enlarges the outer surface, which lowers the convection resistance — so on small-diameter tubes, the first millimeters of insulation can paradoxically increase heat loss. The two effects balance at the critical radius

r_{cr} = k_{\text{ins}}/h

. Below

r_{cr}

, insulation helps the wire shed heat (useful for current-carrying conductors); above it, insulation does its expected job. Always compare

r_{cr}

to the bare-pipe outer radius before assuming insulation reduces loss.

r_{cr} = \frac{k_{\text{ins}}}{h}

Thermal contact resistance

Two solids pressed together touch only at scattered asperities, so heat squeezes through tiny real contact spots and across trapped interstitial gas. This produces a measurable temperature jump at the joint, modeled by a contact resistance. Tabulated values are usually an area-specific resistance

R''_{tc}

\text{m}^2\!\cdot\!\text{K/W}

, converted to a network resistance by dividing by area:

R_{tc} = R''_{tc}/A

. The interface temperature drop is then simply heat flux times

R''_{tc}

, and it can dominate when the surrounding materials are good conductors.

\Delta T_{\text{interface}} = \dot{Q}''\,R''_{tc} = \frac{\dot{Q}}{A}\,R''_{tc}

Exam strategy

Build the resistance network first, on paper, before any number goes in: label each film and each layer, decide series versus parallel, and only then sum. Compute

R_{\text{total}}

, get

\dot{Q}

once, and recover any interior temperature by stepping from a known boundary,

T_{\text{next}} = T_{\text{known}} - \dot{Q}\,R_{\text{step}}

. Keep area straight: per-unit-area work uses

R'' = L/k

and

1/h

with units

\text{m}^2\!\cdot\!\text{K/W}

, while full-network work multiplies by

A

. For cylinders, never use

L/(kA)

— the radial formula has the logarithm. Watch the units of

k

(W/m·K) and confirm your answer's magnitude is physically sane.

Key equations

Fourier's law (1-D)

\dot{Q} = -kA\,\frac{dT}{dx}

Conductive heat rate (W). $k$ = thermal conductivity [W/(m·K)], $A$ = area normal to flow (m²); minus sign gives heat flow down the gradient.

Plane-wall conduction

\dot{Q} = \frac{kA\,(T_1 - T_2)}{L}

Steady 1-D conduction through a slab of thickness $L$ ; faces at $T_1>T_2$ .

Resistance form

\dot{Q} = \frac{\Delta T}{R_{\text{total}}}, \quad R_{\text{total}} = \sum R_i

Electrical analogy; series resistances add. $R$ in K/W, $\Delta T$ in K.

Plane conduction resistance

R_{\text{cond}} = \frac{L}{kA}

Resistance (K/W) of a flat layer; per unit area $R'' = L/k$ .

Convection (film) resistance

R_{\text{conv}} = \frac{1}{hA}

Surface film resistance (K/W); $h$ = convection coefficient [W/(m²·K)].

Cylindrical conduction resistance

R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L}

Radial shell of length $L$ , inner/outer radii $r_1,r_2$ ; logarithmic, not linear.

Overall coefficient

\dot{Q} = UA\,\Delta T, \quad \frac{1}{UA} = R_{\text{total}}

$U$ [W/(m²·K)] lumps the whole series chain; $UA$ is the conductance.

Parallel resistances

\frac{1}{R_{\parallel}} = \frac{1}{R_1} + \frac{1}{R_2}

Side-by-side heat paths (studs, finned regions) combine reciprocally.

Critical insulation radius

r_{cr} = \frac{k_{\text{ins}}}{h}

Outer radius of maximum heat loss for a cylinder; below it, insulation increases loss.

Contact resistance drop

\Delta T_{tc} = \dot{Q}''\,R''_{tc}

Interface temperature jump; $R''_{tc}$ = area-specific contact resistance [m²·K/W].

Worked examples

Composite wall with films on both sides

Problem. A wall is brick ( $L_A = 0.10\,\text{m}$ , $k_A = 0.70\,\text{W/(m·K)}$ ) backed by insulation ( $L_B = 0.05\,\text{m}$ , $k_B = 0.040\,\text{W/(m·K)}$ ). Inside air is at $20\,^\circ\text{C}$ with $h_i = 10\,\text{W/(m}^2\text{·K)}$ ; outside air is at $-10\,^\circ\text{C}$ with $h_o = 40\,\text{W/(m}^2\text{·K)}$ . Find the heat flux and the brick-insulation interface temperature.

Solution. Work per unit area (

A = 1\,\text{m}^2

). Resistances (m²·K/W): inside film

R_i'' = 1/h_i = 0.100

; brick

R_A'' = L_A/k_A = 0.10/0.70 = 0.1429

; insulation

R_B'' = L_B/k_B = 0.05/0.040 = 1.250

; outside film

R_o'' = 1/h_o = 0.025

. Sum:

R_{\text{total}}'' = 0.100 + 0.1429 + 1.250 + 0.025 = 1.518\,\text{m}^2\text{·K/W}

. Flux:

\dot{Q}'' = \Delta T / R_{\text{total}}'' = [20-(-10)]/1.518 = 30/1.518 = 19.8\,\text{W/m}^2

. Walk the chain to the A–B interface: inner surface

T_{s1} = 20 - 19.8(0.100) = 18.0\,^\circ\text{C}

; interface

T_{AB} = 18.0 - 19.8(0.1429) = 15.2\,^\circ\text{C}

. Answer:

\dot{Q}'' = 19.8\,\text{W/m}^2

, interface

T_{AB} = 15.2\,^\circ\text{C}

. Sanity check: continue to the outer surface

15.2 - 19.8(1.250) = -9.5\,^\circ\text{C}

, then across the outer film

-9.5 - 19.8(0.025) = -10.0\,^\circ\text{C}

, recovering the outside air temperature exactly.

\dot{Q}'' = \frac{30\,\text{K}}{1.518\,\text{m}^2\text{·K/W}} = 19.8\,\text{W/m}^2

Insulated steam pipe — loss per meter

Problem. A steel pipe ( $r_1 = 50\,\text{mm}$ , $r_2 = 55\,\text{mm}$ , $k_s = 50\,\text{W/(m·K)}$ ) carries fluid that holds the inner surface at $200\,^\circ\text{C}$ . It is wrapped with $50\,\text{mm}$ of insulation ( $k_{\text{ins}} = 0.050\,\text{W/(m·K)}$ , so $r_3 = 105\,\text{mm}$ ), and the outer surface convects to $25\,^\circ\text{C}$ air with $h_o = 15\,\text{W/(m}^2\text{·K)}$ . Find the heat loss per meter of pipe.

Solution. Use

L = 1\,\text{m}

. Steel shell:

R_s = \ln(55/50)/(2\pi\cdot 50\cdot 1) = 0.09531/314.16 = 3.03\times10^{-4}\,\text{K/W}

(negligible). Insulation:

R_{\text{ins}} = \ln(105/55)/(2\pi\cdot 0.050\cdot 1) = 0.6466/0.3142 = 2.058\,\text{K/W}

. Outer film:

R_o = 1/(h_o\,2\pi r_3 L) = 1/(15\cdot 2\pi\cdot 0.105\cdot 1) = 1/9.896 = 0.1011\,\text{K/W}

. Sum:

R_{\text{total}} = 0.000303 + 2.058 + 0.1011 = 2.160\,\text{K/W}

\dot{Q} = (200-25)/2.160 = 175/2.160 = 81.0\,\text{W per meter}

. Answer:

\dot{Q} = 81.0\,\text{W/m}

. Sanity check: the insulation supplies

95\%

of the resistance, so the steel wall is rightly ignored; loss order-of-magnitude (tens of W/m) is typical for a well-insulated line.

\dot{Q} = \frac{200-25}{R_s + R_{\text{ins}} + R_o} = \frac{175}{2.160} = 81.0\ \text{W/m}

Contact resistance at a bolted joint

Problem. An aluminum bar ( $k_1 = 200\,\text{W/(m·K)}$ , $L_1 = 20\,\text{mm}$ ) is pressed end-to-end against a steel bar ( $k_2 = 15\,\text{W/(m·K)}$ , $L_2 = 20\,\text{mm}$ ). The joint has an area-specific contact resistance $R''_{tc} = 3.0\times10^{-4}\,\text{m}^2\text{·K/W}$ . The free ends are held at $100\,^\circ\text{C}$ (aluminum) and $20\,^\circ\text{C}$ (steel). Find the heat flux and the temperature drop across the contact.

Solution. Per-area resistances (m²·K/W): aluminum

R_1'' = L_1/k_1 = 0.020/200 = 1.00\times10^{-4}

; contact

R_{tc}'' = 3.00\times10^{-4}

; steel

R_2'' = L_2/k_2 = 0.020/15 = 1.333\times10^{-3}

. Sum:

R_{\text{total}}'' = (1.00 + 3.00 + 13.33)\times10^{-4} = 1.733\times10^{-3}\,\text{m}^2\text{·K/W}

. Flux:

\dot{Q}'' = 80/1.733\times10^{-3} = 4.62\times10^{4}\,\text{W/m}^2

. Contact drop:

\Delta T_{tc} = \dot{Q}''\,R_{tc}'' = 4.62\times10^{4}\cdot 3.00\times10^{-4} = 13.8\,^\circ\text{C}

. Answer:

\dot{Q}'' = 4.62\times10^{4}\,\text{W/m}^2

, contact drop

\Delta T_{tc} = 13.8\,^\circ\text{C}

. Sanity check: the contact alone accounts for

3/17.33 \approx 17\%

of the

80\,^\circ\text{C}

span (

\approx 13.8\,^\circ\text{C}

), showing why interfaces matter when the metals conduct well.

\Delta T_{tc} = \dot{Q}''\,R''_{tc} = (4.62\times10^{4})(3.0\times10^{-4}) = 13.8\ \text{K}

Common pitfalls

•Using the plane-wall resistance $L/(kA)$ for a pipe. Radial conduction is logarithmic — use $\ln(r_2/r_1)/(2\pi kL)$ , and remember $r$ is radius, not diameter.
•Reusing one area for every term in a cylindrical network. Inside convection uses $2\pi r_1 L$ , outside convection uses the outer radius — they differ.
•Mixing per-area and full-network resistances. $R''=L/k$ has units m²·K/W; $R=L/(kA)$ has K/W. Keep an entire calculation on one footing.
•Assuming insulation always reduces loss. On small tubes below the critical radius $r_{cr}=k_{\text{ins}}/h$ , the first layer of insulation increases heat loss.
•Forgetting the convection films. The bounding fluid films are resistors in the series chain; dropping them overpredicts $\dot{Q}$ , sometimes badly.
•Treating side-by-side paths as series. Parallel branches (studs through insulation) add reciprocally — summing them as series overestimates resistance.
•Adding contact resistance without dividing by area. Tabulated $R''_{tc}$ is per unit area; the network value is $R''_{tc}/A$ .

References

NCEES FE Reference Handbook — Heat Transfer (Conduction, Thermal Resistance)
Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — Resistance networks, cylindrical conduction, contact resistance
Çengel, Heat and Mass Transfer: A Practical Approach — Critical radius of insulation, composite walls

B. Convection

Convection, Radiation and Lumped Transient Response

Newton's law of cooling and the overall U, the Stefan-Boltzmann radiation law and emissivity, fin heat rate, and lumped-capacitance cooling via the Biot number.

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E. Heat exchangers

Heat Exchangers: LMTD and NTU-Effectiveness

Parallel vs counterflow, the log-mean temperature difference (LMTD) sizing method, the effectiveness-NTU rating method, and the overall coefficient U.

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