HEAT CONDUCTION
EQUATION

Differential equations, boundary conditions, and analytical solutions for steady and transient heat conduction problems.

01 // DIMENSIONALITY

One-Dimensional

Temperature varies in one direction only. Heat transfer in other directions is negligible.

Two-Dimensional

Temperature varies mainly in two directions. Third direction variation is negligible.

Three-Dimensional

Most general case. Temperature varies in all three directions.

Steady vs. Transient

Steady

No change with time at any point

Transient

Variation with time (time-dependent)

Coordinate Systems

Rectangular:T(x, y, z, t)

Cylindrical:T(r, φ, z, t)

Spherical:T(r, θ, φ, t)

02 // HEAT GENERATION

Conversion of electrical, nuclear, or chemical energy into thermal energy within a medium. Heat generation is a volumetric phenomenon.

ELECTRICAL

I²R Heating

NUCLEAR

Fuel Rods

CHEMICAL

Exothermic

Rate of heat generation per unit volume [W/m³]

\dot{e}_{\text{gen}} = \frac{I^2 R_e}{V}

Total heat generation rate

\dot{E}_{\text{gen}} = \int_V \dot{e}_{\text{gen}} \, dV

For uniform generation: Ė_gen = ė_gen · V

Surface Temperature with Generation

Under steady conditions, all heat generated must be conducted away:

\dot{E}_{\text{gen}} = \dot{Q}_{\text{conv}} \quad \Rightarrow \quad \dot{e}_{\text{gen}} V = h A_s (T_s - T_\infty)

03 // 1D HEAT CONDUCTION EQUATIONS

Plane Wall (Rectangular)

General transient equation with generation:

\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

Steady, constant k, no generation:

\frac{d^2 T}{dx^2} = 0

Long Cylinder (Cylindrical)

General transient equation with generation:

\frac{1}{r}\frac{\partial}{\partial r}\left(kr \frac{\partial T}{\partial r}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

Steady, constant k, no generation:

\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) = 0

Sphere (Spherical)

General transient equation with generation:

\frac{1}{r^2}\frac{\partial}{\partial r}\left(kr^2 \frac{\partial T}{\partial r}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

Steady, constant k, no generation:

\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right) = 0

Unified Form

\frac{1}{r^n}\frac{\partial}{\partial r}\left(kr^n \frac{\partial T}{\partial r}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

n = 0 (plane wall)n = 1 (cylinder)n = 2 (sphere)

04 // GENERAL 3D EQUATIONS

Rectangular Coordinates

Variable conductivity:

\frac{\partial}{\partial x}\left(k\frac{\partial T}{\partial x}\right) + \frac{\partial}{\partial y}\left(k\frac{\partial T}{\partial y}\right) + \frac{\partial}{\partial z}\left(k\frac{\partial T}{\partial z}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

Constant conductivity (Fourier-Biot equation):

\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{e}_{\text{gen}}}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}

where α = k/(ρc) is thermal diffusivity

Cylindrical Coordinates

\frac{1}{r}\frac{\partial}{\partial r}\left(kr\frac{\partial T}{\partial r}\right) + \frac{1}{r^2}\frac{\partial}{\partial \phi}\left(k\frac{\partial T}{\partial \phi}\right) + \frac{\partial}{\partial z}\left(k\frac{\partial T}{\partial z}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

Spherical Coordinates

\frac{1}{r^2}\frac{\partial}{\partial r}\left(kr^2\frac{\partial T}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(k\sin\theta\frac{\partial T}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial}{\partial \phi}\left(k\frac{\partial T}{\partial \phi}\right) + \dot{e}_{\text{gen}} = \rho c \frac{\partial T}{\partial t}

05 // BOUNDARY CONDITIONS

Specified Temperature

Temperature at surface is known

T(0) = T_1 \quad \text{and} \quad T(L) = T_2

Specified Heat Flux

Heat flux at surface is known

-k\frac{dT}{dx}\bigg|_{x=0} = \dot{q}_0

Convection

Surface exposed to fluid at T∞

-k\frac{dT}{dx}\bigg|_{x=L} = h[T(L) - T_\infty]

Radiation

Surface exchanges radiation

-k\frac{dT}{dx}\bigg|_{x=L} = \varepsilon\sigma[T(L)^4 - T_{\text{surr}}^4]

Insulated Surface (Adiabatic)

No heat transfer across the surface

\frac{dT}{dx}\bigg|_{x=0} = 0

Thermal Symmetry

Plane of symmetry acts as insulated surface

\frac{dT}{dx}\bigg|_{x=L/2} = 0

Interface Boundary

Perfect contact between two materials

T_A(x_0) = T_B(x_0) \quad \text{(temperature continuity)}

-k_A\frac{dT_A}{dx}\bigg|_{x=x_0} = -k_B\frac{dT_B}{dx}\bigg|_{x=x_0} \quad \text{(flux continuity)}

06 // SOLUTION PROCEDURE

Formulate the Problem

Obtain the applicable differential equation in its simplest form and specify all boundary conditions (and initial condition for transient problems).

Obtain General Solution

Solve the differential equation to get a general solution with arbitrary constants (e.g., C₁, C₂).

Apply Boundary Conditions

Use the boundary (and initial) conditions to determine the arbitrary constants and obtain the particular solution.

Key Insight

Under steady-state conditions, the heat conduction rate through a medium is constant. For transient problems, use energy balance principles combined with Fourier's law.

07 // STEADY 1D SOLUTIONS

Plane Wall (No Generation)

Temperature distribution:

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

Heat transfer rate:

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

Long Cylinder (No Generation)

Temperature distribution:

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

Heat transfer rate:

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

Sphere (No Generation)

Temperature distribution:

T(r) = T_1 + \frac{T_2 - T_1}{(1/r_1) - (1/r_2)}\left(\frac{1}{r_1} - \frac{1}{r}\right)

Heat transfer rate:

\dot{Q} = \frac{4\pi k(T_1 - T_2)}{(1/r_1) - (1/r_2)}

EXAMPLE 2

Double-Pane Window

Same window but now with two 3-mm glass layers separated by 12-mm air gap (k_air = 0.026 W/m·°C). Compare with single pane.

KEY INSIGHT

The air gap acts as excellent insulation, significantly increasing the thermal resistance compared to a single pane.

Thermal Resistances

Glass Layers

R_{glass} = 2 \times 0.0016 = 0.0032

Air Gap

R_{air} = 0.1923

Convection

R_{conv} = 0.0417 + 0.0167 = 0.0584

Total Resistance

R_{total} = 0.254 \, °C/W

Results & Comparison

Heat Transfer Rate

\dot{Q} = 114.2 \, W

76% Reduction

Compared to single pane (470.8 W), the double pane window reduces heat loss by nearly a factor of 4.

EXAMPLE 3

Contact Resistance

Two 5-cm diameter, 15-cm long aluminum bars (k = 176 W/m·°C) pressed together at 20 atm. Top at 150°C, bottom at 20°C. h_c = 11,400 W/m²·°C.

GIVEN DATA

A = 0.00196 m²

L = 0.15 m (each)

k = 176 W/m·°C

h_c = 11,400 W/m²·°C

Thermal Network

Bar 1 & 2

R_{bar} = 0.435

Contact

R_c = 0.0448

Total

R_{tot} = 0.915

Analysis

Heat Transfer Rate

\dot{Q} = 142 \, W

Interface Temp Drop

\Delta T_c = 6.4°C

The contact resistance causes a 6.4°C drop, which is about 5% of the total temperature difference.

EXAMPLE 4

Insulated Steam Pipe

Steam at 320°C flows in a stainless steel pipe (k = 15 W/m·°C) with ID 5 cm and OD 5.5 cm. Covered with 3-cm glass wool (k = 0.038). Surroundings at 5°C.

GEOMETRY

r₁ = 0.025 m (inner)

r₂ = 0.0275 m (outer pipe)

r₃ = 0.0575 m (outer insul)

Thermal Resistances (per m)

Inner Conv.

R_i = 0.0796

Pipe Wall

R_{pipe} = 0.0011

Insulation (Dominant)

R_{ins} = 3.149

Outer Conv.

R_o = 0.1846

Total Resistance: R_total = 3.414 °C/W

Results

Heat Loss (per m)

\dot{Q} = 92.3 \, W/m

Drop across Pipe

ΔT = 0.1°C

Drop across Insulation

ΔT = 290.7°C

EXAMPLE 5

Nuclear Rod Heat Generation

Uranium rods (D = 5 cm, L = 1 m) generate heat uniformly at 7 × 10⁷ W/m³. Determine total heat generation rate.

Calculation

Volume of Rod

V = \pi r^2 L = \pi (0.025)^2 (1) = 0.00196 \, m^3

Total Generation Rate

\dot{E}_{gen} = \dot{e}_{gen} V = (7 \times 10^7)(0.00196) = 137.2 \, kW

EXAMPLE 6

Heat Flux in Plate

Large 3-cm-thick stainless steel plate. Heat generated uniformly at 5 × 10⁶ W/m³. Losing heat from both sides. Determine surface heat flux.

ANALYSIS

For steady state, the heat generated within the plate must be dissipated from the surfaces. By symmetry, half the heat goes out each side.

Calculation (per unit area)

Total Heat Generation (per m²)

\dot{Q}_{gen} = \dot{e}_{gen} \times L = (5 \times 10^6)(0.03) = 150,000 \, W/m^2

Surface Heat Flux (each side)

\dot{q}_s = \frac{\dot{Q}_{gen}}{2} = 75,000 \, W/m^2 = 75 \, kW/m^2

EXAMPLE 7

Boundary Conditions

Spherical container (r₁, r₂). Express BCs at inner surface (r = r₁) for:

Specified Temperature (50°C)
Specified Heat Flux (30 W/m² inward)
Convection (h, T∞)

(a) Specified Temp

T(r_1) = 50°C

(b) Heat Flux (Inward)

-k\frac{dT}{dr}\bigg|_{r=r_1} = -30

-k\frac{dT}{dr}\bigg|_{r=r_1} = h[T(r_1) - T_\infty]

EXAMPLE 8

Steel Pan Formulation

Steel pan bottom (L = 0.5 cm, D = 20 cm). Heating unit 1000 W, 85% transferred to pan. Express formulation.

HEAT FLUX

\dot{q}_0 = \frac{0.85 \times 1000}{\pi(0.1)^2} = 27,070 \, W/m^2

Mathematical Formulation

Differential Equation

\frac{d^2T}{dx^2} = 0 \quad \text{for } 0 < x < L

BC at x = 0 (Flux)

-k\frac{dT}{dx}\bigg|_{x=0} = 27,070

BC at x = L (Convection)

-k\frac{dT}{dx}\bigg|_{x=L} = h[T(L) - T_\infty]

EXAMPLE 9

Plane Wall Solution

Wall (L = 0.4 m, k = 2.3, A = 20 m²). Left T₁ = 80°C. Right convection T∞ = 15°C, h = 24.

GOAL

Find temperature distribution T(x) and heat transfer rate.

Solution Process

General Solution

T(x) = C_1 x + C_2

Applying BCs

C_2 = 80, \quad C_1 = -146.2

Final Distribution

T(x) = 80 - 146.2x

Heat Transfer Rate

Rate (Constant through wall)

\dot{Q} = -6,725 \, W

EXAMPLE 10

Resistance Heater Wire

2-kW wire (k = 20, D = 5 mm, L = 0.7 m). Surface Tₛ = 110°C. Find center temperature.

APPROACH

Calculate volumetric heat generation, then use the cylinder max temperature rise formula.

Calculation

Volume

V = 1.374 \times 10^{-5} \, m^3

Generation Rate

\dot{e}_{gen} = 1.456 \times 10^8 \, W/m^3

Max Temp Rise (Center to Surface)

\Delta T_{max} = \frac{\dot{e}_{gen} r_0^2}{4k} = 11.4°C

Center Temperature

T_0 = 110 + 11.4 = 121.4°C

EXAMPLE 11

Fin Efficiency Relations

Relation for efficiency of a long fin (constant cross-section).

Efficiency Formula (Long Fin)

\eta_{fin} = \frac{\text{Actual Heat Transfer}}{\text{Max Possible Heat Transfer}}

\eta_{long\_fin} = \frac{1}{L}\sqrt{\frac{kA_c}{hp}}

Circular Fin (Diameter D)

\eta \approx \frac{1}{2L}\sqrt{\frac{kD}{h}}

Rectangular Fin (Thickness t)

\eta \approx \frac{1}{L}\sqrt{\frac{kt}{2h}}

EXAMPLE 12

Finned Tube Performance

Steam tube (5 cm) at 180°C. Aluminum fins added (OD 6 cm, t=1 mm, 250 fins/m). Air 25°C, h=40.

IMPACT

Fins dramatically increase surface area, enhancing heat transfer despite the added conduction resistance.

Comparison

Bare Tube Rate

\dot{Q} = 974 \, W/m

Finned Tube Rate

\dot{Q} \approx 2650 \, W/m

2.7x Increase

The addition of fins nearly triples the heat transfer rate.

EXAMPLE 13

Circuit Board Cooling

Board (12x18 cm) with 80 chips (0.04 W each). Back cooled by air 40°C, h=50.

Temperature Analysis

Without Fins

T_s = 42.96°C

Rise of ~3°C

With Pin Fins

T_s = 40.2°C

Rise of ~0.2°C

EXAMPLE 14

Thermocouple Response

1-mm sphere junction. k=35, ρ=8500, cₚ=320. h=210. Time to read 99%?

CHECK BIOT

Bi = 0.001 < 0.1

Lumped Analysis Valid

Transient Calculation

Time Constant

b = \frac{h}{\rho c_p L_c} = 0.462 \, s^{-1}

Time to 99%

t = 10 \, s

NOTES_VER: 1.0.0

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