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).

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Core concept

Steady Conduction

Steady conduction is heat transfer through a solid (or stagnant fluid) with temperature field independent…

33 min15 Interview15 GATEGATEInterview

Level

Advanced

Topic tags

Steady conduction, Heat Transfer, Sachdeva Heat

Useful for

GATE favourite, Interview ask

Source

Fundamentals of Engineering Heat & Mass Transfer

Learning outcomes

What you'll learn in this topic

1
Fourier law (1-D): — heat flows from hot to cold; $k$ in $\mathrm{W/(m\cdot K)}$ .
2
Plane wall (constant $k$ ): , .
3
Cylinder (radial): , .
4
Sphere (radial): , .

Key Formulas

Important equations & their meanings

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

Worked Examples

Step-by-step solved problems

1Plane-wall conduction rate
2Composite wall with convection
3Critical radius check

Common Mistakes

Avoid these errors in exams & interviews

Using Celsius in formulas that need absolute temperature — **not** required for conduction T, but **required** for radiation T4.
Mixing mm and m in L, A, r (factor-of-10³ errors).
Applying plane-wall formula to a thick cylindrical pipe.

Practice Questions

Strengthen your concepts with questions

30+Practice Questions

Interview Questions

Most asked interview problems

15Interview Questions

GATE Questions

Previous year GATE questions

15GATE Questions

Applications in real world

Heat exchangersLMTD and UA sizing
ElectronicsCooling and heat sinks
InsulationCritical radius & losses
BoilersWall conduction and fins

Visual concept

Hot Source

Resistance

Heat Flow

Cool Sink

Heat flows from hot to cold through thermal resistances.

Recommended Book

Fundamentals of Engineering Heat & Mass TransferRC Sachdeva

Read: Syllabus unit

View on Amazon

University exams

Important Topic

Industry relevance

High

Concept difficulty

Hard

Average time

33 min

Core assumptions (state these in exams)

Before applying textbook formulas, write the assumptions explicitly:

1. Steady state —

\partial \frac{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=\frac{kA\Delta T}{L}

form no longer holds without integrating the ODE.

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)}{\frac{1}{r_1}-\frac{1}{r_2}}

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

Exam trap: Never use

q=\frac{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}}{\frac{1}{h_1A}+\frac{L}{kA}+\frac{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}}=\frac{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=\frac{\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}}=\frac{k}{h}

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

q

should be sensible for the area and

\Delta T

Advanced problem-solving framework

Use this sequence for long-form mastery and repeatable scoring:
1. Identify objective, system boundary, and required output.
2. Write all givens in SI units and classify each as measured, assumed, or estimated.
3. Choose the governing model and relation (Fourier law (1-D):

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

— heat flows from hot to cold;

k

\mathrm{W/(m\cdot K)}

.) with one-line justification.
4. Solve symbolically first to catch structural mistakes early.
5. Substitute values with careful unit tracking.
6. Cross-check by sign, order of magnitude, and limiting case.
7. Write a short engineering conclusion tied to safety, performance, reliability, or cost.

Next, solve one "variant version" of the same problem by changing one assumption (loading type, losses, property constancy, boundary condition, or uncertainty level). This builds transfer ability — essential for difficult exams where numbers and wording are changed deliberately.

For viva/interviews, convert your written method into a 45-second explanation format:
"Objective -> model selected -> key assumption -> result -> practical implication."
This makes your answers concise and technically credible.

Exam, interview, and note-making strategy

To make this topic genuinely reusable, maintain notes in four blocks: concept summary, assumptions checklist, solved template, and common error-correction logic. This transforms passive reading into active revision material for class tests, semester exams, GATE-style practice, and interviews.

A practical weekly cycle:
- Day 1: read and annotate the topic.
- Day 3: solve one moderate numerical from memory.
- Day 5: give a 60-second oral explanation.
- Day 7: solve one mixed problem integrating this topic with a prerequisite.
- Day 14: do a timed review to test retention.

For interview readiness, prepare concise answers to:
1. Where is this used in real engineering?
2. Which assumption is most risky if wrong?
3. How do you sanity-check the result quickly?
4. What trade-off does this result influence?

These four questions are asked repeatedly in technical panels, and practicing them creates confidence.

Use this page as a living notebook: append class doubts, lab observations, previous-year tricks, and personal mnemonics. That personalization is what turns a study page into a repeat-visit resource students trust.

Industry scenarios and decision context

Engineering decisions are made under constraints: deadline, budget, material availability, process capability, safety requirements, and maintenance realities. So while solving steady conduction, do not treat the answer as "final truth" without context. The numerical output is a decision input, not the decision itself.

Ask these context questions after every solved example:
- If load uncertainty increases, does design margin remain acceptable?
- If manufacturing tolerance drifts, will performance degrade critically?
- If operating temperature/humidity changes, are properties still valid?
- If maintenance is delayed, what failure mode appears first?

Students who practice contextual questioning develop judgment faster and perform better in internships, design tasks, and technical interviews. This context-first style is a major retention driver because learners see immediate real-world value.

Long-form revision worksheet

Use this worksheet when preparing notes:
A) One-paragraph concept explanation in your own words.
B) Symbol and units table for key variables.
C) Validity limits and assumptions list.
D) One standard solved pattern with all steps.
E) One variant problem where an assumption changes.
F) One industry-use explanation with failure consequence.
G) Three common mistakes and their correction rules.

If you can fill all seven blocks without external help, your topic depth is strong enough for repeat use and long retention. If not, revisit the corresponding section and strengthen the missing block.

This structured worksheet approach is intentionally longer than quick revision notes because it is designed for durable mastery. It supports exactly the product goal you mentioned: students should keep coming back because the page is complete enough to build serious notes.

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 \frac{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}

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=\frac{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\,\frac{\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)}

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)}{\frac{1}{r_1}-\frac{1}{r_2}}

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

Exam trap: Never use

q=\frac{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}}{\frac{1}{h_1A}+\frac{L}{kA}+\frac{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}}=\frac{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=\frac{\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}}=\frac{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.

Detailed conceptual understanding

Steady Conduction should be studied as a complete reasoning chain: definition, governing assumptions, physical interpretation, boundary conditions, and limits of validity. In heat transfer, strong students do not stop at "what is the formula"; they explain why the model applies, which simplifications are being used, and what error appears when those simplifications break. This is the key difference between memorized learning and engineering understanding.

A high-quality conceptual pass should answer these questions in writing:
1. Which quantity is being predicted or controlled?
2. Which variables dominate sensitivity and why?
3. Which assumptions are explicit, and which are hidden?
4. What real-world effects are neglected in first-pass analysis?
5. Which engineering decision depends on this output?

When revising, rewrite the concept in your own words and attach one real scenario from lab, workshop, project, internship, or industry case. This habit transforms abstract theory into retrievable memory. If a topic cannot be explained without reading the page, it is not yet mastered.

Use this page as a note source: create a "concept map" with cause-effect arrows and keep updating it whenever you solve new problems. Students who maintain evolving concept maps typically retain topics longer and return less to emergency cramming.

Advanced problem-solving framework

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

— heat flows from hot to cold;

k

\mathrm{W/(m\cdot K)}

Exam, interview, and note-making strategy

These four questions are asked repeatedly in technical panels, and practicing them creates confidence.

Industry scenarios and decision context

Common misconceptions and correction patterns

Most weak performance comes from repeated misconception patterns, not from lack of intelligence. Typical patterns include unit inconsistency, wrong model selection, assumption mismatch, and skipping interpretation after substitution.

Correction pattern to practice:
1. Detect: identify exactly where logic diverged.
2. Diagnose: state why that step is invalid.
3. Repair: rewrite with correct model/assumption.
4. Verify: run a sanity check and compare trends.

Maintain a personal "mistake log" with three columns: mistake, reason, correction rule. Reviewing this log before exams has a larger performance impact than reading new theory repeatedly.

Use the same correction discipline in interviews: acknowledge the slip, state corrected logic, and proceed. This demonstrates professional maturity and keeps the discussion positive even when you initially miss a step.

Ask about this topic

What you'll learn in this topic

Key Formulas

Worked Examples

Common Mistakes

Practice Questions

Interview Questions

GATE Questions

Applications in real world

Visual concept

Recommended Book

What should I learn next?

Core assumptions (state these in exams)

Cylindrical and spherical walls

Composite walls and the resistance network

Critical radius of insulation

Step-by-step problem approach

Advanced problem-solving framework

Exam, interview, and note-making strategy

Industry scenarios and decision context

Long-form revision worksheet

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

Detailed conceptual understanding

Advanced problem-solving framework

Exam, interview, and note-making strategy

Industry scenarios and decision context

Common misconceptions and correction patterns

Long-form revision worksheet