Steady conduction is heat transfer through a solid (or stagnant fluid) with temperature field independent of time. Fourier’s law relates heat rate to the temperature gradient. Plane walls, cylinders, and spheres are solved with thermal resistances in series/parallel networks (RC Sachdeva / GATE HT).

What is Fourier’s law of heat conduction? State its assumptions.

Fourier’s law:. Heat flows from high to low temperature. Assumptions for the simple form: steady state, one-dimensional flow, constant, no generation, homogeneous isotropic material.

Define thermal resistance and write for a plane wall.

Thermal resistance relates temperature difference to heat rate:. For a plane wall, with unit.

Why is temperature distribution linear in a plane wall with constant ?

From Fourier’s law with constant and,, so is linear between the two face temperatures.

Write the radial conduction formula for a hollow cylinder.

and. Area grows with radius, so the log form appears.

Explain the electrical analogy for conduction.

Temperature difference ↔ voltage, heat rate ↔ current, thermal resistance ↔ electrical resistance. Series layers add; parallel paths add conductances.

What is critical radius of insulation for a cylinder?

. If bare radius, adding thin insulation can increase heat loss because surface area grows faster than conduction resistance.

When do you include convective film resistance ?

Whenever fluid temperatures (not surface temperatures) are given. Then is placed at the fluid–solid interface in the thermal circuit.

How do you find interface temperature in a composite wall?

First find. Then for each layer. Interface temperature follows by subtracting successive drops from the hot side.

A wire of radius is insulated with the material above. Will thin insulation reduce heat loss?

Bare, so thin insulation increases heat loss until outer radius exceeds.

Unit of thermal conductivity in SI is:

(also since size is identical).

Back to Heat & Mass Transfer

Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Heat & Mass Transfer

Steady Conduction

Steady conduction is heat transfer through a solid (or stagnant fluid) with temperature field independent of time. Fourier’s law

q = -kA\frac{dT}{dx}

relates heat rate to the temperature gradient. Plane walls, cylinders, and spheres are solved with thermal resistances

R_{\mathrm{th}}

in series/parallel networks (RC Sachdeva / GATE HT).

Exam tip: never replace LMTD with an arithmetic mean of end ΔT unless ΔT_a ≈ ΔT_b; keep U and A in consistent SI units.

Key formulas & points

Skim these first — then read the full notes below.

Fourier law (1-D): $q_x = -kA\frac{dT}{dx}$ — heat flows from hot to cold; $k$ in $\mathrm{W/(m\cdot K)}$ .
Plane wall (constant $k$ ): $q = \frac{kA(T_1-T_2)}{L}$ , $R_{\mathrm{wall}}=\frac{L}{kA}$ .
Cylinder (radial): $q = \frac{2\pi k L(T_1-T_2)}{\ln(r_2/r_1)}$ , $R_{\mathrm{cyl}}=\frac{\ln(r_2/r_1)}{2\pi k L}$ .
Sphere (radial): $q = \frac{4\pi k(T_1-T_2)}{1/r_1-1/r_2}$ , $R_{\mathrm{sph}}=\frac{1/r_1-1/r_2}{4\pi k}$ .
Composite series: $q=\frac{\Delta T_{\mathrm{overall}}}{\sum R_i}$ ; $\Delta T_i = q R_i$ .
Surface convection film: $R_{\mathrm{conv}}=\frac{1}{hA}$ .
Critical insulation radius (cylinder): $r_{\mathrm{cr}}=\frac{k_{\mathrm{ins}}}{h}$ ; insulation below $r_{\mathrm{cr}}$ can increase heat loss.

Topic details

Definition and physical meaning

Conduction is the transfer of thermal energy through a medium by molecular interaction (lattice vibration in solids; molecular collisions in fluids) without bulk motion of the medium.

Steady conduction means the temperature field

T(\mathbf{r})

does not change with time:

\frac{\partial T}{\partial t}=0

Consequently the heat flux field is divergence-free in the absence of generation:

\nabla\cdot\mathbf{q}=0\quad(\text{no generation})

Physical parameters

Symbol	Meaning	SI unit
$T$	Temperature	$\mathrm{K}$ (or °C)
$q$	Heat transfer rate	$\mathrm{W}$
$q''$	Heat flux	$\mathrm{W/m^2}$
$k$	Thermal conductivity	$\mathrm{W/(m\cdot K)}$
$A$	Area normal to heat flow	$\mathrm{m^2}$
$L$ , $r$	Thickness / radius	$\mathrm{m}$
$h$	Convective coefficient	$\mathrm{W/(m^2\cdot K)}$
$R_{\mathrm{th}}$	Thermal resistance	$\mathrm{K/W}$

Typical

k

(order of magnitude): metals

10

–

400

, building materials

0.1

–

2

, insulation

0.02

–

0.1

, air

\sim 0.026\,\mathrm{W/(m\cdot K)}

at room temperature.

Core assumptions (state these in exams)

Before applying textbook formulas, write the assumptions explicitly:

1. Steady state —

\partial T/\partial t = 0

.
2. One-dimensional heat flow (plane wall:

x

only; cylinder/sphere:

r

only).
3. Constant thermal conductivity

k

(or use average

k

over the temperature range).
4. No internal heat generation unless the problem states generation (

\dot{q}

q_{g}

).
5. Homogeneous isotropic material in each layer.
6. Perfect thermal contact between layers (contact resistance neglected unless given).
7. Constant cross-section for plane wall; for cylinder/sphere use the correct radial area

A(r)

If generation is present, the governing equation becomes

\frac{d}{dx}\left(kA\frac{dT}{dx}\right)+\dot{E}_{\mathrm{gen}}=0

and the simple

q=kA\Delta T/L

form no longer holds without integrating the ODE.

Fourier’s law — derivation summary

Statement (1-D Cartesian). The heat rate in the

+x

direction is proportional to area and to the negative temperature gradient:

q_x=-kA\frac{dT}{dx}

Why the minus sign? Heat flows spontaneously from high

T

to low

T

. If

\frac{dT}{dx}<0

(temperature decreases in

+x

), then

q_{x}>0

Heat flux form:

q_x''=-k\frac{dT}{dx}\qquad\left[\mathrm{W/m^2}\right]

Vector form (isotropic solid):

\mathbf{q}''=-k\nabla T

Derivation sketch (continuum). Consider a slab of thickness

\Delta x

and area

A

. Net energy into a control volume with no generation and steady state must be zero. Empirically,

q\propto A\,\Delta T/\Delta x

. Taking the continuum limit

\Delta x\to 0

and inserting the minus sign for direction yields Fourier’s law. The proportionality constant is

k(T)

, measured experimentally.

Units check:

[k]=\frac{[q]\,[L]}{[A]\,[\Delta T]}=\frac{\mathrm{W}\cdot\mathrm{m}}{\mathrm{m^2}\cdot\mathrm{K}}=\mathrm{W/(m\cdot K)}

Fig — Plane wall (RC Sachdeva / GATE schematic)

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Plane wall with constant k

Integrate Fourier’s law from

x = 0

(

T = T_{1}

) to

x = L

(

T = T_{2}

) with constant

k

and

A

q\int_0^L dx=-kA\int_{T_1}^{T_2}dT

qL = kA(T_{1}-T_{2})

q=\frac{kA(T_1-T_2)}{L}

Temperature distribution is linear:

T(x)=T_1-\frac{T_1-T_2}{L}x

Thermal resistance of the wall:

R_{\mathrm{wall}}=\frac{L}{kA}\qquad\Rightarrow\qquad q=\frac{T_1-T_2}{R_{\mathrm{wall}}}

This is the conduction analogue of Ohm’s law

I = \frac{V}{R}

Cylindrical and spherical walls

For a hollow cylinder (inner radius

r_{1}

, outer

r_{2}

, length

L

), area grows as

A(r)=2\pi r L

. Integrating Fourier’s law in radial coordinates:

q_r=\frac{2\pi k L(T_1-T_2)}{\ln(r_2/r_1)}

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

Temperature is logarithmic in

r

T(r)=T_1-\frac{T_1-T_2}{\ln(r_2/r_1)}\ln\left(\frac{r}{r_1}\right)

For a hollow sphere:

q_r=\frac{4\pi k(T_1-T_2)}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}

R_{\mathrm{sph}}=\frac{1/r_1-1/r_2}{4\pi k}

Exam trap: Never use

q=kA\Delta T/L

with a single “mean” area for a thick pipe unless the problem explicitly allows a thin-wall approximation (

r_2\approx r_1

Composite walls and the resistance network

Series layers (heat flows through layer 1, then 2, …): the heat rate

q

is the same through each layer; temperature drops add:

q=\frac{T_{\mathrm{hot}}-T_{\mathrm{cold}}}{\sum_i R_i}

\Delta T_i=q R_i

Parallel paths (side-by-side materials spanning the same

\Delta T

): conductances add:

\frac{1}{R_{\mathrm{eq}}}=\sum_i\frac{1}{R_i}

Including convection at a surface exposed to fluid at

T_\infty

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

Example plane wall with convection on both sides:

q=\frac{T_{\infty,1}-T_{\infty,2}}{\dfrac{1}{h_1A}+\dfrac{L}{kA}+\dfrac{1}{h_2A}}

Overall heat-transfer coefficient

U

is defined by

q=UA\Delta T_{\mathrm{overall}}

, so

\frac{1}{UA}=\sum R_i

Critical radius of insulation

For an insulated cylinder (pipe) of outer insulation radius

r

, heat loss to ambient involves

R_{\mathrm{total}}=\frac{\ln(r/r_i)}{2\pi k_{\mathrm{ins}}L}+\frac{1}{h(2\pi r L)}

Differentiating

q(r)

(or

R_{\mathrm{total}}

) with respect to

r

and setting

\frac{dq}{dr} = 0

gives the critical radius

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

Physical meaning
- If bare outer radius

r_o < r_{\mathrm{cr}}

, adding a thin layer of insulation can increase heat loss (area increase dominates).
- If

r_o > r_{\mathrm{cr}}

, adding insulation decreases heat loss.

For a sphere,

r_{\mathrm{cr}}=2k_{\mathrm{ins}}/h

This is a favourite GATE conceptual question — always compare

r_{o}

with

r_{\mathrm{cr}}

before concluding that “more insulation is better.”

Step-by-step problem approach

1. Sketch geometry; mark known temperatures (surface or fluid) and materials.
2. Decide plane / cylinder / sphere; write assumptions (steady, 1-D, constant

k

, …).
3. Draw the thermal resistance circuit including convection films if fluid temperatures are given.
4. Compute each

R_{i}

with consistent SI units (convert mm → m).
5. Find

q=\Delta T_{\mathrm{overall}}/\sum R

.
6. Recover interface temperatures via

\Delta T_i=qR_i

if asked.
7. For insulation problems, compute

r_{\mathrm{cr}}=k/h

and interpret.
8. Check units and order of magnitude (

q

should be sensible for the area and

\Delta T

Common mistakes in exams

• Using Celsius in formulas that need absolute temperature — not required for conduction

\Delta T

, but required for radiation

T^4

.
• Mixing mm and m in

L

A

r

(factor-of-

10^3

errors).
• Applying plane-wall formula to a thick cylindrical pipe.
• Omitting

\frac{1}{(hA)}

when only fluid temperatures are known.
• Adding series resistances incorrectly as parallel (or vice versa).
• Assuming insulation always reduces heat loss without checking

r_{\mathrm{cr}}

.
• Forgetting that

A

for a cylinder wall is

2\pi r L

(varies with

r

), not a constant plane area.

Calculator

Plane-wall conduction

q = \frac{k A \Delta T}{L}

Conductivity kW/m·KArea Am²ΔTKThickness Lm

Result

9,600W

q = kAΔT/L = (40×0.2×60) / 0.05 = 9600 W

Worked examples

Try the problem first — open the solution when you are ready to check.

Plane-wall conduction rate

Problem

A plane wall has

k=40\,\mathrm{W/(m\cdot K)}

A=0.2\,\mathrm{m^2}

L=0.05\,\mathrm{m}

, and face temperatures

T_1=120^\circ\mathrm{C}

T_2=60^\circ\mathrm{C}

. Find the steady heat rate

q

Solution

Formulas (Indian textbook notation)

$\Delta T=T_1-T_2=60\,\mathrm{K}$

Formulas (Indian textbook notation)

$q=\frac{kA\Delta T}{L}=\frac{40\times 0.2\times 60}{0.05}=\frac{480}{0.05}=9600\,\mathrm{W}=9.6\,\mathrm{kW}$

Also

R_{\mathrm{wall}}=L/(kA)=0.05/(40\times 0.2)=0.00625\,\mathrm{K/W}

, so

q=\Delta T/R=60/0.00625=9600\,\mathrm{W}

Composite wall with convection

Problem

A furnace wall: firebrick

L_1=0.2\,\mathrm{m}

k_1=1.0\,\mathrm{W/(m\cdot K)}

; steel shell

L_2=0.01\,\mathrm{m}

k_2=40\,\mathrm{W/(m\cdot K)}

. Inside gas

T_{\infty,i}=800^\circ\mathrm{C}

h_i=50\,\mathrm{W/(m^2\cdot K)}

; outside air

T_{\infty,o}=30^\circ\mathrm{C}

h_o=20\,\mathrm{W/(m^2\cdot K)}

. Area

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

. Find

q

Critical radius check

Problem

A wire of outer radius

r_o=1\,\mathrm{mm}

is to be insulated with material

k=0.05\,\mathrm{W/(m\cdot K)}

. Ambient convection

h=10\,\mathrm{W/(m^2\cdot K)}

. Will a thin insulation layer reduce heat loss?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
What is Fourier’s law of heat conduction? State its assumptions.
Model answer
Fourier’s law: $q_x=-kA\dfrac{dT}{dx}$ . Heat flows from high to low temperature. Assumptions for the simple form: steady state, one-dimensional flow, constant $k$ , no generation, homogeneous isotropic material.
2
Define thermal resistance and write $R_{\mathrm{th}}$ for a plane wall.
Model answer
Thermal resistance relates temperature difference to heat rate: $q=\Delta T/R_{\mathrm{th}}$ . For a plane wall, $R_{\mathrm{wall}}=\dfrac{L}{kA}$ with unit $\mathrm{K/W}$ .
3
Why is temperature distribution linear in a plane wall with constant $k$ ?
Model answer
From Fourier’s law with constant $q$ and $k$ , $\dfrac{dT}{dx}=-\dfrac{q}{kA}=\text{constant}$ , so $T(x)$ is linear between the two face temperatures.
4
Write the radial conduction formula for a hollow cylinder.
Model answer
$q=\dfrac{2\pi k L(T_1-T_2)}{\ln(r_2/r_1)}$ and $R_{\mathrm{cyl}}=\dfrac{\ln(r_2/r_1)}{2\pi k L}$ . Area grows with radius, so the log form appears.
5
Explain the electrical analogy for conduction.
Model answer
Temperature difference ↔ voltage, heat rate $q$ ↔ current, thermal resistance ↔ electrical resistance. Series layers add $R$ ; parallel paths add conductances $\frac{1}{R}$ .
6
What is critical radius of insulation for a cylinder?
Model answer
$r_{\mathrm{cr}}=\dfrac{k_{\mathrm{ins}}}{h}$ . If bare radius $r_o<r_{\mathrm{cr}}$ , adding thin insulation can increase heat loss because surface area grows faster than conduction resistance.
7
When do you include convective film resistance $\frac{1}{(hA)}$ ?
Model answer
Whenever fluid temperatures (not surface temperatures) are given. Then $R_{\mathrm{conv}}=1/(hA)$ is placed at the fluid–solid interface in the thermal circuit.
8
How do you find interface temperature in a composite wall?
Model answer
First find $q=\Delta T_{\mathrm{overall}}/\sum R$ . Then for each layer $\Delta T_i=qR_i$ . Interface temperature follows by subtracting successive drops from the hot side.
9
Differentiate conduction, convection, and radiation in one line each.
Model answer
Conduction: molecular energy transfer in a continuum without bulk motion. Convection: energy transport by fluid motion plus conduction near the wall. Radiation: electromagnetic emission, needs no medium, $\propto T^4$ .
10
What is overall heat transfer coefficient $U$ ?
Model answer
Defined by $q=UA\Delta T_{\mathrm{overall}}$ so $\dfrac{1}{UA}=\sum R_i$ . $U$ lumps wall conduction and surface convection into one coefficient.
11
Why can’t you use $q=kA\Delta T/L$ for a thick pipe?
Model answer
In a cylinder, heat-flow area $A(r)=2\pi r L$ varies with radius. Integration of Fourier’s law gives the logarithmic formula, not the plane-wall form.
12
State SI units of $k$ , $h$ , $q$ , and $R_{\mathrm{th}}$ .
Model answer
$k$ : $\mathrm{W/(m\cdot K)}$ ; $h$ : $\mathrm{W/(m^2\cdot K)}$ ; $q$ : $\mathrm{W}$ ; $R_{\mathrm{th}}$ : $\mathrm{K/W}$ .
13
What happens to heat flux if wall thickness doubles at fixed $\Delta T$ and $k$ ?
Model answer
For a plane wall $q\propto 1/L$ , so heat rate (and average flux) halves when $L$ doubles, all else equal.
14
Explain series vs parallel thermal resistances with an example.
Model answer
Series: multilayer furnace wall — same $q$ , $\Delta T$ splits by $R_{i}$ . Parallel: two materials side-by-side spanning the same $\Delta T$ — conductances add.
15
How would you reduce heat loss from a steam pipe in practice?
Model answer
Use insulation with low $k$ and outer radius well above $r_{\mathrm{cr}}$ ; reduce outer $h$ with cladding/weather shield; minimize bare fittings; ensure good joint sealing to avoid thermal bridges.

Exams & GATE

1
Textbook: RC Sachdeva Ch. 2–3 / Incropera Ch. 3.
2
Always draw the thermal circuit before writing equations.
3
State SI units: $k$ in W/(m·K), $h$ in W/(m²·K), $q$ in W, $T$ in K or °C ( $\Delta T$ identical).
4
GATE favourites: composite walls, insulated pipes, and critical-radius conceptual MCQs.

📖 Standard books (India)

Fundamentals of Engineering Heat & Mass Transfer — RC Sachdeva
Read: Syllabus unit
Heat transfer and heat exchangers

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (state these in exams)

Fourier’s law — derivation summary

Plane wall with constant k

Cylindrical and spherical walls

Composite walls and the resistance network

Critical radius of insulation

Step-by-step problem approach

Common mistakes in exams

Plane-wall conduction

Worked examples

Plane-wall conduction rate

Composite wall with convection

Critical radius check

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE