Heat Transfer · Chapter 4
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Heat Transfer · Chapter Four

Transient Heat Conduction

Transient conduction predicts not only how much energy moves, but when a component reaches a temperature that affects safety, quality, or performance.

The decisive modeling question is whether internal gradients matter. This chapter builds a disciplined path from the lumped model to eigenfunction, semi-infinite, and multidimensional solutions, with explicit property-temperature and validity audits.

Learning objectives

  • Interpret Biot and Fourier numbers as resistance and diffusion-time comparisons.
  • Select lumped, finite-body, semi-infinite, or product solutions from geometry and time-scale cues.
  • Calculate temperatures, response times, and stored-energy changes with consistent properties.
  • Apply one-term solutions only within their geometry and Fourier-number limits.
  • Audit signs, units, limiting behavior, and the sensitivity of a thermal design.
4–1

Lumped System Analysis

If internal conduction equalizes temperature much faster than the surface exchanges energy, the body has one meaningful temperature \(T(t)\). An energy balance, \(\rho Vc_p\,dT/dt=-hA_s(T-T_\infty)\), produces exponential approach to the ambient. Radiation may be included only through a justified effective coefficient or a separate nonlinear balance.

Lumped-Capacitance Criterion

A body can be treated as having a single, uniform temperature when internal conduction resistance is negligible compared with surface convection resistance — that is, when the Biot number is small (\(Bi \le 0.1\)). The temperature then decays exponentially toward the ambient value.

\[ Bi = \frac{h L_c}{k} \le 0.1, \qquad L_c = \frac{V}{A_s} \]

Lumped transient temperature
\[ \frac{T(t) - T_\infty}{T_i - T_\infty} = e^{-bt}, \qquad b = \frac{h A_s}{\rho V c_p} \]
\(b\) is the time constant's reciprocal; \(1/b = \rho V c_p/(hA_s)\) is the time to reach about 63% of the total change.

Physical meaning

\(Bi=hL_c/k\) compares internal conduction resistance with external convection resistance. Small \(Bi\) means the surface bottleneck dominates. The thermal time constant \(\tau_c=\rho Vc_p/(hA_s)\) grows with thermal mass and falls with exposed area or convection strength.

Problem-solving strategy

  1. Identify the body, exposed area, initial condition, and imposed ambient history.
  2. Evaluate \(k,\rho,c_p\) at a representative mean body temperature and \(h\) by a correlation valid for the flow.
  3. Compute \(L_c=V/A_s\) and \(Bi\); use lumped analysis only when \(Bi\le0.1\).
  4. Solve the dimensionless exponential relation and then obtain energy from \(mc_p\Delta T\).
  5. Audit the temperature bounds, energy sign, and constant-property assumption.
EXAMPLE 1Cooldown of an Aluminum Battery Busbar
Problem. A \(0.20\times0.040\times0.006\) m aluminum busbar, initially at \(90^\circ\text{C}\), is exposed on all faces to \(25^\circ\text{C}\) forced air with \(h=35\ \text{W/m}^2\text{K}\). Given \(k=180\ \text{W/mK}\), \(\rho=2700\ \text{kg/m}^3\), and \(c_p=900\ \text{J/kgK}\) at the mean temperature, verify the model and find the time to \(40^\circ\text{C}\).

Strategy

A small, highly conductive solid cooling in a fixed ambient suggests lumped capacitance, but the Biot audit must precede its use.

Givens

  • \(T_i=90^\circ\text{C}\), \(T_\infty=25^\circ\text{C}\), target \(40^\circ\text{C}\).
  • Rectangular dimensions and constant supplied properties; all six faces exposed.

Physical/model interpretation

The busbar is the closed system, convection heat leaving is positive, contact leads and radiation are neglected, and uniform temperature is conditional on \(Bi\).

Step-by-step solution
1. Geometry. \(V=4.80\times10^{-5}\ \text{m}^3\), \(A_s=2(ab+ac+bc)=0.01888\ \text{m}^2\), so \(L_c=0.002542\ \text{m}\).
2. Validate. \(Bi=hL_c/k=35(0.002542)/180=4.94\times10^{-4}\ll0.1\); internal gradients are negligible.
3. Formulate. \(\theta/\theta_i=(40-25)/(90-25)=0.23077=\exp[-t/\tau_c]\), where \(\tau_c=\rho c_pL_c/h\).
4. Calculate. \(\tau_c=2700(900)(0.002542)/35=176.5\ \text{s}\), hence \(t=-176.5\ln(0.23077)=258.7\ \text{s}\).

Unit/sign/limit checks

Units: \((\text{kg/m}^3)(\text{J/kgK})(\text{m})/(\text{W/m}^2\text{K})=\text{s}\).

Signs: \(T>T_\infty\), so the body cools and \(t>0\).

Limits: \(t=0\) gives \(90^\circ\text{C}\), while \(t\to\infty\) approaches \(25^\circ\text{C}\); the target lies between them.

Final answer: \(\boxed{t=259\ \text{s}\approx4.31\ \text{min}}\).

Engineering insight

The result scales inversely with \(h\); uncertainty in airflow directly affects cooldown time. Leads, joints, and radiation should be included for release-to-service decisions.

4–2

Transient Conduction with Spatial Effects

When \(Bi>0.1\), temperature depends on position and time. Separation of variables produces an eigenfunction series. For a plane wall, use half-thickness \(L\), \(Bi=hL/k\), and \(Fo=\alpha t/L^2\); this length is not the lumped \(V/A_s\). The first term is generally adequate only for \(Fo\gtrsim0.2\).

Fourier Number

The dimensionless time \(Fo=\alpha t/L^2\) compares elapsed time with the internal diffusion time \(L^2/\alpha\). Thermal diffusivity \(\alpha=k/(\rho c_p)\) controls response, while conductivity alone does not.

One-term approximation (centre temperature)
\[ \theta_0 = \frac{T_0 - T_\infty}{T_i - T_\infty} = A_1\,e^{-\lambda_1^2 \tau}, \qquad \tau = \frac{\alpha t}{L_c^2} > 0.2 \]
Constants \(A_1\) and \(\lambda_1\) depend on geometry and on \(Bi\), read from standard tables.
Total heat transferred up to time \(t\)
\[ \frac{Q}{Q_{\max}} = 1 - \theta_0\,\frac{\sin\lambda_1}{\lambda_1}\ \text{(wall-type relation)}, \qquad Q_{\max} = m c_p (T_i - T_\infty) \]
Key Idea \(Q_{\max}\) is the heat exchanged if the body reaches ambient everywhere. The transient charts (Heisler charts) plot the one-term solution for center temperature, position, and accumulated heat against \(\tau\) and \(Bi\).

Physical meaning

The center is usually the last point to respond. Increasing \(Bi\) steepens internal gradients; increasing \(Fo\) erases them. Tabulated eigenvalues belong to a particular geometry and Biot definition and are not interchangeable.

EXAMPLE 2Center Temperature of a Polymer Panel
Problem. A 40-mm-thick polymer panel initially at \(20^\circ\text{C}\) is heated symmetrically by \(120^\circ\text{C}\) air with \(h=45\ \text{W/m}^2\text{K}\). At the representative mean temperature, \(k=0.20\ \text{W/mK}\) and \(\alpha=1.20\times10^{-7}\ \text{m}^2/\text{s}\). For \(Bi=4.5\), a plane-wall table gives \(\lambda_1=1.283\), \(A_1=1.247\). Find the center temperature after 30 min.

Strategy

The slab is finite across its thickness, both faces see the same ambient, and \(Bi\) is not small; use the plane-wall one-term center solution after checking \(Fo\).

Givens

  • \(2L=0.040\ \text{m}\), \(t=1800\ \text{s}\), \(T_i=20^\circ\text{C}\), \(T_\infty=120^\circ\text{C}\).
  • Supplied mean-temperature properties and matching tabulated constants.

Physical/model interpretation

One-dimensional conduction, constant properties, negligible edge effects, uniform initial temperature, and prescribed constant \(h,T_\infty\) are assumed.

Step-by-step solution
1. Select length. For the wall eigen-solution, \(L=0.020\ \text{m}\), not the full thickness.
2. Audit model. \(Bi=hL/k=45(0.020)/0.20=4.5\), confirming spatial gradients; \(Fo=\alpha t/L^2=1.20\times10^{-7}(1800)/(0.020)^2=0.540>0.2\).
3. Evaluate ratio. \(\theta_0/\theta_i=A_1e^{-\lambda_1^2Fo}=1.247e^{-(1.283)^2(0.540)}=0.5125\).
4. Recover temperature. \(T_0=T_\infty+0.5125(T_i-T_\infty)=120+0.5125(-100)=68.8^\circ\text{C}\).

Unit/sign/limit checks

Units: \(Bi\), \(Fo\), and the temperature ratio are dimensionless.

Signs: \(T_i-T_\infty<0\), so the center remains below ambient while warming.

Limits: The answer lies between \(20\) and \(120^\circ\text{C}\); the center lag is expected at large \(Bi\).

Final answer: \(\boxed{T_0(30\ \text{min})\approx68.8^\circ\text{C}}\).

Engineering insight

A surface measurement would overstate the cold center temperature. Manufacturing cure or food-safety decisions should therefore use the slowest-heating location and temperature-dependent data when ranges are large.

4–3

Transient Conduction in Semi-Infinite Solids

For short times, the penetration depth is small compared with body thickness, so the remote boundary is irrelevant. A practical audit is that the far boundary remains near \(T_i\), often achieved when thickness exceeds roughly \(4\sqrt{\alpha t}\). The solution below requires an abruptly imposed, constant surface temperature and constant properties.

Sudden surface temperature change (error-function solution)
\[ \frac{T(x,t) - T_s}{T_i - T_s} = \operatorname{erf}\!\left(\frac{x}{2\sqrt{\alpha t}}\right) \]
\(\operatorname{erf}\) is the Gauss error function; \(T_s\) is the imposed surface temperature, \(T_i\) the initial uniform temperature.
EXAMPLE 3Thermal Penetration During Die Quenching
Problem. A thick steel die initially at \(220^\circ\text{C}\) has its surface suddenly held at \(30^\circ\text{C}\). With \(\alpha=1.25\times10^{-5}\ \text{m}^2/\text{s}\), find the depth at which the temperature is \(150^\circ\text{C}\) after 45 s. The die is 0.20 m thick; assess the semi-infinite approximation.

Strategy

A short surface-temperature transient in a thick body matches the semi-infinite error-function model; invert the dimensionless temperature and audit penetration against thickness.

Givens

  • \(T_i=220^\circ\text{C}\), \(T_s=30^\circ\text{C}\), \(T=150^\circ\text{C}\), \(t=45\ \text{s}\).
  • \(\alpha=1.25\times10^{-5}\ \text{m}^2/\text{s}\); supplied \(\operatorname{erf}^{-1}(0.6316)\approx0.635\).

Physical/model interpretation

Coordinate \(x\) points inward from the quenched face. Properties and surface temperature are constant; lateral heat flow and phase transformation are neglected.

Step-by-step solution
1. Normalize. \((T-T_s)/(T_i-T_s)=(150-30)/(220-30)=0.6316\).
2. Invert. \(\operatorname{erf}(\eta)=0.6316\) gives \(\eta=x/(2\sqrt{\alpha t})\approx0.635\).
3. Calculate depth. \(2\sqrt{\alpha t}=2\sqrt{(1.25\times10^{-5})(45)}=0.04743\ \text{m}\), so \(x=0.635(0.04743)=0.0301\ \text{m}\).
4. Audit finite thickness. \(4\sqrt{\alpha t}=0.0949\ \text{m}<0.20\ \text{m}\), so the remote face is well beyond the principal penetration region.

Unit/sign/limit checks

Units: \(\sqrt{\alpha t}\) has units of metres.

Signs: \(x>0\) is inward, and \(T_s

Limits: At \(x=0\), the model gives \(T_s\); as \(x\to\infty\), it gives \(T_i\).

Final answer: \(\boxed{x\approx30.1\ \text{mm}}\), with the semi-infinite approximation reasonable.

Engineering insight

The \(\sqrt{t}\) scaling means doubling affected depth requires roughly four times the quench duration. Real steel treatment also requires transformation kinetics, stress, and cracking review.

4–4

Transient Conduction in Multidimensional Systems

A finite cylinder or rectangular block can be viewed as the intersection of infinite one-dimensional bodies. The multidimensional solution is then simply the product of the corresponding 1-D solutions.

Product Solution

For a body formed by the intersection of two or three infinite geometries (e.g. a short cylinder = infinite cylinder ∩ infinite plane wall), the dimensionless transient temperature equals the product of the dimensionless solutions of those geometries, evaluated at the same time.

\[ \theta_{\text{short cyl}}(r,x,t) = \theta_{\text{cyl}}(r,t)\times \theta_{\text{wall}}(x,t) \]

Combined heat transfer for two-body product
\[ \left(\frac{Q}{Q_{\max}}\right)_{\text{total}} = \left(\frac{Q}{Q_{\max}}\right)_1 + \left(\frac{Q}{Q_{\max}}\right)_2\!\left[1 - \left(\frac{Q}{Q_{\max}}\right)_1\right] \]
Course Note

The product rule only works for the dimensionless temperatures, and only when the body is the geometric intersection of standard shapes with the same ambient. Always reduce each direction to its own \(Bi\) and \(\tau\) first, then multiply — never multiply actual temperatures.

Physical meaning

Separable rectangular and cylindrical coordinates allow independent diffusion factors. Every factor lies between zero and one, so adding another finite direction accelerates center response relative to a one-dimensional body.

Chapter recap

  • Use \(Bi\) to decide whether spatial gradients may be neglected and \(Fo\) to measure progress of internal diffusion.
  • Match every finite-body solution to its geometry, characteristic length, boundary conditions, and correlation-table convention.
  • Use semi-infinite solutions only while the remote boundary remains unaffected; multiply dimensionless solutions for separable multidimensional bodies.

Before moving on

  • Can you state the property reference temperature and justify a transient model?
  • Can you verify units, temperature bounds, direction, and limiting behavior?
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