Chapter 5: Transient Conduction

Source files used: Heat Transfer (13), Heat Transfer (14), related review comments in later combined-mode lectures; textbook Chapter 5 as support. No worked examples are included.

1. Big Picture

Chapter 5 adds time to conduction.

In Chapters 2 and 3, many problems were steady: temperature did not change with time. In Chapter 5, temperature changes as an object heats up or cools down.

The big decision in Biddle’s lecture is the Biot number:

$$\boxed{Bi=\frac{hL_c}{k}}$$

The Biot number compares internal conduction resistance to external convection resistance.

• Small $Bi$: temperature inside the body is nearly uniform.
• Large $Bi$: temperature varies significantly inside the body.

So the first question in a transient conduction problem is not “which equation do I use?” It is:

$$\boxed{\text{Can I treat the object temperature as spatially uniform?}}$$

If yes, use the lumped heat capacity model. If no, use spatial transient conduction methods such as one-term approximations or Heisler/Grober charts.

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2. Core Ideas

2.1 Lumped Heat Capacity Model

The lumped model assumes the whole object has one temperature at a given time:

$$T=T(t)$$

Temperature changes with time but not with position inside the solid.

This is valid when internal conduction is fast compared with surface convection. In that case, the inside of the object does not have enough thermal resistance to develop a major temperature gradient.

Biddle’s decision rule:

$$Bi<0.1 \quad \Rightarrow \quad \text{lumped model is acceptable}$$

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2.2 Spatial Effects

If $Bi>0.1$, the object cannot be treated as isothermal. Then temperature depends on position and time:

$$T=T(x,t) \quad \text{or} \quad T=T(r,t)$$

For a plane wall, centerline and surface temperatures are different. For a cylinder, center and surface temperatures are different. The center usually responds more slowly than the surface.

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2.3 Characteristic Length

For lumped capacitance:

$$L_c=\frac{V}{A_s}$$

where:

• $V$: object volume,
• $A_s$: surface area exposed to convection.

For one-term transient charts/equations, the characteristic length depends on geometry:

Geometry	One-Term $Bi$ and $Fo$ length	$V/A_s$ (lumped Bi check only)
Plane wall of total thickness $2L$	$L$, half-thickness	$L$
Long cylinder	$r_o$	$r_o/2$
Sphere	$r_o$	$r_o/3$

⚠️ Common error: $r_o/2$ (cylinder) and $r_o/3$ (sphere) are the $V/A_s$ values used only when checking lumped capacitance ($Bi = hL_c/k < 0.1$). Once you move to the one-term approximation, use $r_o$ for both $Bi$ and $Fo$. Confirmed in textbook Ch. 5 and Solutions Manual (Problem 5.72 comments).

Do not mix the lumped $V/A_s$ length with the Heisler/one-term geometry length.

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2.4 Fourier Number

The Fourier number measures dimensionless time:

$$Fo=\frac{\alpha t}{L^2}$$

or for cylinders/spheres:

$$Fo=\frac{\alpha t}{r_o^2}$$

Large $Fo$ means more time has passed for heat to diffuse through the object.

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2.5 Energy Transferred

In transient conduction, you may be asked not only for temperature but also for how much energy has been gained or lost.

For lumped systems:

$$Q=\rho V c_p\left(T_i-T(t)\right)$$

for cooling, with sign chosen according to the problem.

Biddle explicitly corrected a lecture note mistake: for energy stored in a cylinder, use volume $\pi r^2L$, not surface area.

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3. Main Governing Equations and Formulas

3.1 Biot Number

$$Bi=\frac{hL_c}{k}$$

where:

• $h$: convection coefficient,
• $L_c$: characteristic length,
• $k$: solid thermal conductivity.

Use to decide whether lumped capacitance is valid.

Decision:

$$Bi<0.1 \Rightarrow \text{lumped heat capacity model}$$ $$Bi>0.1 \Rightarrow \text{spatial transient method}$$

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3.2 Characteristic Length for Lumped Model

$$L_c=\frac{V}{A_s}$$

Use only for the lumped-capacitance Biot number.

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3.3 Lumped Capacitance Temperature Response

$$\frac{T(t)-T_\infty}{T_i-T_\infty}=\exp\left(-\frac{hA_s}{\rho Vc_p}t\right)$$

or

$$\frac{\theta}{\theta_i}=e^{-t/\tau}$$

where:

$$\theta=T-T_\infty$$ $$\theta_i=T_i-T_\infty$$ $$\tau=\frac{\rho Vc_p}{hA_s}$$

$\tau$ is the thermal time constant.

Use when $Bi<0.1$.

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3.4 Lumped Heat Rate

Instantaneous convection heat rate:

$$\dot{q}(t)=hA_s\left[T(t)-T_\infty\right]$$

Heat rate decreases as the object temperature approaches the fluid temperature.

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3.5 Lumped Energy Transferred

Energy transferred to/from the surroundings by convection over time:

$$Q=\int_{0}^{t} hA_s\theta\,dt$$

Using the lumped result $\theta=\theta_i e^{-t/\tau}$:

$$Q=(\rho cV)\theta_i\left[1-\exp\left(-\frac{t}{\tau}\right)\right]$$

For cooling:

$$Q=\rho Vc_p\left(T_i-T(t)\right)$$

Maximum possible energy transfer:

$$Q_0=\rho Vc_p\left(T_i-T_\infty\right)$$

Fraction transferred:

$$\frac{Q}{Q_0}=1-\frac{T(t)-T_\infty}{T_i-T_\infty}$$

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3.6 Fourier Number

Plane wall:

$$Fo=\frac{\alpha t}{L^2}$$

Long cylinder or sphere:

$$Fo=\frac{\alpha t}{r_o^2}$$

Use with one-term approximations or transient charts.

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3.7 One-Term Approximation: Plane Wall

For a plane wall of total thickness $2L$, with convection on both sides:

$$\frac{\theta}{\theta_i}=C_1e^{-\zeta_1^2Fo}\cos\left(\zeta_1\frac{x}{L}\right)$$

where:

$$\theta=T-T_\infty$$ $$Fo=\frac{\alpha t}{L^2}$$ $$x^*=\frac{x}{L}$$

$C_1$ and $\zeta_1$ come from the transient conduction table as functions of $Bi=hL/k$.

At the centerline, $x=0$, so $\cos(0)=1$:

$$\frac{\theta_0}{\theta_i}=C_1e^{-\zeta_1^2Fo}$$

Use when $Bi>0.1$ and the one-term approximation is valid, commonly for sufficiently large $Fo$ such as $Fo>0.2$.

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3.8 One-Term Approximation: Long Cylinder

For a long cylinder:

$$\frac{\theta}{\theta_i}=C_1e^{-\zeta_1^2Fo}J_0\left(\zeta_1\frac{r}{r_o}\right)$$

where:

$$Fo=\frac{\alpha t}{r_o^2}$$

and $J_0$ is the Bessel function of the first kind.

At the centerline, $r=0$, $J_0(0)=1$:

$$\frac{\theta_0}{\theta_i}=C_1e^{-\zeta_1^2Fo}$$

Use table values for $C_1$ and $\zeta_1$.

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3.9 One-Term Approximation: Sphere

For a sphere:

$$\frac{\theta}{\theta_i}=C_1e^{-\zeta_1^2Fo}\frac{\sin(\zeta_1r/r_o)}{\zeta_1r/r_o}$$

At the center:

$$\frac{\theta_0}{\theta_i}=C_1e^{-\zeta_1^2Fo}$$

Use sphere table values for $C_1$ and $\zeta_1$.

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3.10 Multidimensional Product Solution

For shapes that can be built from intersections of one-dimensional shapes, approximate:

$$\left(\frac{\theta}{\theta_i}\right)_{multi-D} =\prod \left(\frac{\theta}{\theta_i}\right)_{1D}$$

Examples:

• finite cylinder = infinite cylinder solution × plane wall solution,
• rectangular solid = plane wall solution in $x$ × plane wall solution in $y$ × plane wall solution in $z$.

Use only when the textbook charts/tables support the geometry.

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4. Problem-Solving Workflow

4.1 Lumped-Capacitance Workflow

1. Identify the body, volume $V$, and exposed surface area $A_s$.
2. Compute $L_c=V/A_s$.
3. Compute $Bi=hL_c/k$.
4. If $Bi<0.1$, use lumped capacitance.
5. Compute $\tau=\rho Vc_p/(hA_s)$.
6. Use $\theta/\theta_i=e^{-t/\tau}$.
7. Solve for unknown time, temperature, or heat transfer.

4.2 Spatial Transient Workflow

1. Compute $Bi$ using the geometry-specific length.
2. Since $Bi>0.1$, do not use lumped capacitance.
3. Identify geometry: plane wall, cylinder, sphere.
4. Compute $Fo=\alpha t/L^2$ or $\alpha t/r_o^2$.
5. Get $C_1$ and $\zeta_1$ from the proper table for the geometry and Bi.
6. Use the one-term expression to find center temperature.
7. Use position factor to find off-center or surface temperature if needed.
8. Use energy charts/equations if asked for total energy transferred.

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5. Decision Rules / Decision Trees

5.1 Main Transient Decision

Compute $Bi=hL_c/k$.

$Bi<0.1$? → lumped heat capacity: $T=T(t)$ only.

$Bi>0.1$? → spatial effects matter: $T=T(x,t)$ or $T=T(r,t)$.

5.2 Which Length?

Checking lumped capacitance? → $L_c=V/A_s$

Using plane-wall transient chart/equation? → L = half-thickness

Using long-cylinder chart/equation? → $L=r_o$

Using sphere chart/equation? → $L=r_o$

5.3 What Is Being Asked?

Temperature after time t? → use temperature ratio equation/chart

Time to reach temperature? → solve temperature ratio for $t$ or $Fo$

Energy transferred? → use $Q/Q_0$ relation or lumped energy balance

Surface temperature? → find center temperature first, then use surface/center relation

5.4 Lumped Heating vs Cooling

Object hotter than fluid? → cooling, $T$ decreases toward $T_\infty$

Object colder than fluid? → heating, $T$ increases toward $T_\infty$

In both cases: $\theta/\theta_i=\exp(-t/\tau)$

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6. Important Tables / Correlations Needed

6.1 Biot Number Rule

Biot Number	Model
$Bi<0.1$	Lumped capacitance acceptable
$Bi>0.1$	Spatial gradients important

6.2 Geometry Lengths

Geometry	Lumped $L_c$	One-Term/Chart Length
Plane wall, total thickness $2L$	$V/A_s$	half-thickness $L$
Long cylinder	$V/A_s$	radius $r_o$
Sphere	$V/A_s$	radius $r_o$

6.3 Table 5.1 Type Data

For one-term approximations, use tables that give:

• $\zeta_1$,
• $C_1$,
• as functions of $Bi$,
• separately for plane wall, cylinder, and sphere.

Do not use plane wall constants for a cylinder or sphere.

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7. Key Takeaways

• Chapter 5 is about temperature changing with time.
• Always start with the Biot number.
• $Bi<0.1$ means the lumped model is valid.
• Lumped capacitance assumes uniform temperature inside the object.
• The lumped response is exponential.
• $\tau=\rho Vc_p/(hA_s)$ controls how fast the object responds.
• $Bi>0.1$ means spatial gradients matter.
• For spatial effects, use plane wall/cylinder/sphere one-term solutions or charts.
• Use the correct characteristic length for the method.
• For energy stored in a cylinder, use cylinder volume $\pi r^2L$, not surface area.