Transient Conduction: Lumped Capacitance
Every chapter so far has assumed steady state: temperatures that do not change with time. Reality is rarely that patient. Polymer parts cool from melt to solid in seconds, pellets quench in water baths, and DSC samples heat at controlled ramps. This chapter develops the lumped capacitance method, the simplest and most widely used tool for transient heat transfer, and identifies precisely when it applies.
Close your notes. Try each question from memory, then reveal the answer. The ones you cannot recall are exactly what this chapter will reinforce.
Write the definition of thermal diffusivity α. Give the formula in terms of other properties, its units, and explain in one sentence what it physically represents.
Write Newton’s law of cooling. Define $h$ in words: what physical resistance does it represent? What are its units?
What does the Biot number compare? Write the formula for a slab of thickness $2L$ and state what $\mathrm{Bi} \ll 1$ means physically.
Why Transient Matters
Picture a polypropylene pellet dropping from the extruder die into a cold-water trough. At the moment it enters the water it is at 220°C. Five seconds later it is cool enough to convey without sticking. But how do you know when five seconds is enough, and when you need ten? The answer requires thinking about how temperature changes over time, not just where it ends up.
In every processing step you have encountered, some period of time was spent reaching steady state. The steady-state chapters taught you the final temperatures once nothing is changing. The transient chapters explain the journey. That journey determines cycle time in injection molding, residence time in a quench bath, and the cooling rate that sets the crystallinity of a semicrystalline polymer. Those three variables directly control part quality.
The simplest transient model is the lumped capacitance method. It treats the entire object as a single mass at a uniform temperature, like a coffee mug that is uniformly hot throughout. This simplification is physically justified when internal conduction is fast compared to external convection, which the Biot number tells you. When it holds, the problem reduces to a first-order ODE with an elegant exponential solution.
After each layer is deposited, the bead must cool enough before the next layer arrives or the newly deposited polymer cannot support its weight. The cooling rate of a 0.4 mm bead in the part-cooling fan airstream is a lumped capacitance problem: the bead is so thin (Bi ≈ 0.02 in 2 m/s airflow) that its temperature is uniform across its cross-section at every moment. The thermal time constant for a PLA bead in a 40°C enclosure is roughly 2–5 seconds, setting a firm ceiling on how fast you can print reliably.
Before reading on — the governing equation for this chapter is $d\theta/dt = -\text{const}\times\theta$. Does that shape remind you of anything? It is exactly the same equation as radioactive decay, an RC circuit discharging, and the growth of bacteria in a limited medium. They all say: the rate of change is proportional to the current value. That always produces an exponential. You already know what the solution looks like. The only new question is what sets the timescale — and that is where the physics lives.
The Lumped Capacitance Idea
Consider a small object initially at temperature $T_i$, suddenly plunged into a fluid at temperature $T_\infty$ (where $T_\infty \neq T_i$). The fluid removes heat from the surface at rate $\dot{Q} = h A_s (T - T_\infty)$. That heat comes from the internal thermal energy of the object.
If we assume that the object has a single, uniform temperature $T(t)$ at all times, we can write an energy balance on the whole object. The rate of change of internal energy equals the net heat loss:
This is a first-order linear ODE in $T(t)$. Introducing the temperature excess $\theta = T - T_\infty$ (which vanishes at steady state), the equation becomes:
Equation (8.2) is separable. Rearrange:
$$\frac{d\theta}{\theta} = -\frac{h A_s}{\rho V c_p}\,dt$$Integrate both sides from the initial condition $\theta = \theta_i$ at $t = 0$ to $\theta$ at time $t$:
$$\int_{\theta_i}^{\theta} \frac{d\theta'}{\theta'} = -\frac{h A_s}{\rho V c_p}\int_0^t dt'$$ $$\ln\!\left(\frac{\theta}{\theta_i}\right) = -\frac{h A_s}{\rho V c_p}\,t$$Exponentiate both sides:
$$\frac{\theta}{\theta_i} = \exp\!\left(-\frac{h A_s}{\rho V c_p}\,t\right)$$Substituting back $\theta = T - T_\infty$ and $\theta_i = T_i - T_\infty$:
$$\frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\!\left(-\frac{h A_s}{\rho V c_p}\,t\right)$$The Biot Number: When Can We Lump?
The lumped capacitance assumption says the temperature is uniform throughout the solid at every instant. That is only true if heat is transported through the interior so quickly that no gradient can build up. The Biot number measures whether this condition is satisfied.
Think of two resistances in series: the resistance to conduction through the solid ($L_c / k$) and the resistance to convection at the surface ($1/h$). If the conduction resistance is tiny compared to the convection resistance, the solid equilibrates internally almost instantly and its temperature is nearly uniform. The ratio of these two resistances is the Biot number:
Lumped capacitance is valid when $\mathrm{Bi} \leq 0.1$. At Bi = 0.1 the error in assuming uniform temperature is less than 5%. Beyond Bi = 0.1 the temperature gradient inside the solid becomes significant and the single-temperature model fails.
Polymers have low thermal conductivity (PLA 0.13, HDPE 0.44, nylon 0.25 W/mK). A polymer pellet (r = 5 mm) in an agitated water bath (h = 200 W/m²K) gives Bi = 200 × (0.005/3) / 0.25 = 1.33. Lumped capacitance is not valid here. The method does apply to thin polymer films in air (Bi < 0.05 is typical), extruded beads cooled by a fan, and fine fibres in a fluid stream.
The Biot number $\mathrm{Bi} = hL_c/k$ uses the solid's conductivity $k$. The Nusselt number $\mathrm{Nu} = hL/k_f$ uses the fluid's conductivity $k_f$. They look identical but measure completely different things: Bi is about the solid's internal resistance, Nu is about the fluid's convective enhancement.
The Exponential Decay Solution
When Bi ≤ 0.1, the lumped capacitance model is valid and the solution is:
This is a pure exponential decay: the classic signature of a first-order system. Every radioactive decay, every RC circuit discharge, and every lumped thermal body cool by the same mathematical law. The rate is set by the single parameter $\tau$.
The Thermal Time Constant τ
The time constant has a clean physical interpretation. The numerator $\rho V c_p$ is the thermal capacitance of the object: how much energy it can store per degree. The denominator $h A_s$ is the thermal conductance of the surface: how much power it loses per degree of temperature excess. Their ratio gives the time scale over which the object loses a significant fraction of its excess heat.
Think of it like a water tank with a drain. A large tank (high $\rho V c_p$) takes long to empty. A wide drain (high $h A_s$) empties it quickly. $\tau$ is the tank volume divided by the drain flow rate, per unit of water level. Doubling $h$ halves $\tau$ and speeds up cooling by exactly the same factor.
After one time constant ($t = \tau$), the temperature excess has dropped to 36.8% of its initial value. After two time constants it is at 13.5%. After five it is at 0.7%, effectively at $T_\infty$. Engineers often say a system is "thermally settled" after 3–5 time constants.
For a flat slab of thickness $2L$ (cooled on both faces, $L_c = L$): $\tau = \rho c_p L / h$. Doubling the thickness doubles $\tau$ (more mass to cool); doubling $h$ halves $\tau$ (faster surface stripping). This direct proportionality is one reason thin polymer films cool so much faster than thick ones.
The Fourier Number: Dimensionless Time
The solution in (8.4) can be rewritten in a fully dimensionless form using two dimensionless groups. Substituting $\tau = \rho c_p L_c / h$ into the exponent:
$$\frac{t}{\tau} = \frac{h t}{\rho c_p L_c} = \frac{h}{k} \cdot \frac{k t}{\rho c_p L_c^2} = \mathrm{Bi} \cdot \frac{\alpha t}{L_c^2} = \mathrm{Bi} \cdot \mathrm{Fo}$$This lets us write the lumped capacitance solution as:
Dimensional form
$$\frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau}$$Two parameters: τ (seconds), which depends on material, geometry, and h.
Dimensionless form
$$\theta^* = e^{-\mathrm{Bi}\cdot\mathrm{Fo}}$$One product: Bi·Fo. A single curve describes all lumped capacitance problems, regardless of material or size.
Start with the dimensional energy balance (Eq. 8.1):
$$\rho c_p V \frac{dT}{dt} = -h A_s (T - T_\infty)$$Divide through by volume $V$ and rearrange:
$$\rho c_p \frac{dT}{dt} = -\frac{h A_s}{V} (T - T_\infty)$$Identify the natural scales. The natural temperature scale is $\Delta T = T_i - T_\infty$ (the initial excess). The natural length scale is $L_c = V/A_s$. We need a time scale $\tau$ — the equation itself will tell us what it should be. Introduce dimensionless variables:
$$\theta^* = \frac{T - T_\infty}{T_i - T_\infty}, \qquad t^* = \frac{t}{\tau}$$Substituting $T = T_\infty + \theta^*(T_i - T_\infty)$ and $dt = \tau\, dt^*$:
$$\rho c_p \frac{(T_i - T_\infty)}{\tau}\frac{d\theta^*}{dt^*} = -\frac{h}{L_c}(T_i - T_\infty)\theta^*$$Cancel $(T_i - T_\infty)$ and rearrange:
$$\frac{d\theta^*}{dt^*} = -\underbrace{\frac{h\tau}{\rho c_p L_c}}_{\displaystyle = 1 \text{ if we choose } \tau = \rho c_p L_c / h}\theta^*$$Choosing $\tau = \rho c_p L_c / h$ makes the coefficient equal to 1 and gives the cleanest form $d\theta^*/dt^* = -\theta^*$. Now express $t^*$ in terms of physical groups by writing $\tau = (\rho c_p L_c / h)$ and introducing $\alpha = k/(\rho c_p)$:
$$t^* = \frac{t}{\tau} = \frac{ht}{\rho c_p L_c} = \underbrace{\frac{hL_c}{k}}_{\mathrm{Bi}} \cdot \underbrace{\frac{k t}{\rho c_p L_c^2}}_{\alpha t/L_c^2 = \mathrm{Fo}} = \mathrm{Bi} \cdot \mathrm{Fo}$$Conclusion: Bi and Fo are not arbitrary names — they emerge directly from non-dimensionalising the energy balance. Bi measures how hard the surface is working relative to the material's internal conduction. Fo measures how far the thermal "wave" has travelled through the material relative to its size. Their product $\mathrm{Bi}\cdot\mathrm{Fo}$ is the natural dimensionless time for the lumped system.
The Fourier number is most useful when comparing two materials or geometries. A high-diffusivity material reaches the same Fo in less time: it "processes" the temperature disturbance faster. Steel (α = 1.2×10−5 m²/s) reaches Fo = 1 roughly 60 times faster than HDPE (α = 2×10−7 m²/s) for the same geometry. This is why steel parts cool in a water bath in seconds while thick HDPE parts take minutes.
When Lumped Capacitance Fails (Bi > 0.1)
When Bi exceeds 0.1, the temperature is no longer uniform inside the solid at any given instant: the surface has already cooled significantly while the core is still near $T_i$. The single-temperature model breaks down, and you must solve the full partial differential equation with spatial coordinates.
For simple geometries (infinite slab, long cylinder, sphere), exact series solutions exist. For long times (Fo > 0.2) only the first term of the series matters, giving a solution that still looks like an exponential but with a correction coefficient and modified time constant. These are tabulated in Heisler charts, which you will encounter if you take a dedicated transport phenomena course.
Lumped capacitance is the primary transient tool at this level. When you calculate Bi > 0.1, you have three options: (1) note that the method fails and leave the calculation there, (2) use a look-up Heisler chart (outside this text), or (3) recognise that the surface and the core cool at different rates and frame a bounding estimate. Chapter 9 will revisit the semi-infinite approximation, which is often more useful for thick polymer bodies at short times.
Interactive: Cooling Curve Explorer
Use the sliders to set the thermal time constant and initial temperature. Watch the temperature decay in real time. The dashed vertical line marks one time constant τ; the grey dot shows the temperature at that moment.
Worked Examples
Example 8.1: HDPE Film Cooling in a Packaging Line
An HDPE film exits a die at 220°C and is cooled by an air knife (h = 25 W/m²K, T∞ = 20°C). The film is 3 mm thick (Lc = 1.5 mm = 0.0015 m) with k = 0.44 W/mK, ρ = 950 kg/m³, cp = 2,000 J/kgK. (a) Verify that lumped capacitance applies. (b) Find τ. (c) Find the temperature after 2 minutes.
Solution:
(a) Biot number check:
$$\mathrm{Bi} = \frac{h L_c}{k} = \frac{25 \times 0.0015}{0.44} = 0.085 < 0.1 \quad \checkmark$$Lumped capacitance is valid (barely — verify assumption in any real application).
(b) Time constant:
$$\tau = \frac{\rho c_p L_c}{h} = \frac{950 \times 2000 \times 0.0015}{25} = 114 \text{ s}$$(c) Temperature after 120 s:
$$T = 20 + 200 \exp\!\left(-\frac{120}{114}\right) = 20 + 200 \times 0.350 = 20 + 70 = 90°\text{C}$$The film is still at 90°C after two minutes in this air stream. If the line needs the film below 60°C before winding, a much higher h (water cooling, forced convection) or thinner gauge is needed.
Example 8.2: Time to Safe Handling of an Injection-Molded Part
A thin ABS panel (Lc = 2 mm = 0.002 m) is ejected from a mold at 160°C and set on an air-cooled conveyor (h = 15 W/m²K, T∞ = 25°C). k = 0.17 W/mK, ρ = 1050 kg/m³, cp = 1,400 J/kgK. The part can be packaged when T ≤ 50°C. How long must it wait?
Solution:
$$\mathrm{Bi} = \frac{15 \times 0.002}{0.17} = 0.176$$Bi = 0.18 > 0.1, so lumped capacitance is marginal here. In an exam setting, proceed with the method and note the assumption; in practice, use a Heisler chart. For illustration:
$$\tau = \frac{1050 \times 1400 \times 0.002}{15} = 196 \text{ s}$$ $$\frac{T - T_\infty}{T_i - T_\infty} = \frac{50 - 25}{160 - 25} = \frac{25}{135} = 0.185$$ $$t = -\tau \ln(0.185) = -196 \times (-1.687) \approx 330 \text{ s} \approx 5.5 \text{ min}$$Note the caveat: because Bi > 0.1, the surface cools faster than the core. The core will still be above 50°C at 330 s; the true wait time is longer. This illustrates exactly why checking the Biot number matters before applying lumped capacitance.
Concept Questions
An object with a thermal time constant of 200 s is cooling from 180°C to a 20°C environment. After 400 s, its temperature is approximately:
A polymer part has τ = 150 s. After one time constant the dimensionless temperature (T − T∞)/(Ti − T∞) equals:
Key Concept
The lumped capacitance solution is $\theta/\theta_i = e^{-t/\tau}$. At $t = \tau$: $e^{-1} = 0.368$, meaning 36.8% of the initial excess remains. At $t = 2\tau$: $e^{-2} = 0.135$. Note carefully: 0.368 is the remaining fraction, not the lost fraction (which is $1 - e^{-1} = 0.632$).
You double the convection coefficient h while keeping everything else constant. What happens to the thermal time constant τ?
A polymer sheet has τ = 200 s. You replace it with the same sheet twice as thick (double Lc). What is the new τ?
Key Concept
$\tau = \rho c_p L_c / h$. The time constant is linear in Lc (twice as thick → twice the τ) and inversely proportional to h (twice the convection → half the τ). This tells you the fastest way to speed up cooling: increase h (switch from air to water cooling, or add forced flow), not just reduce wall thickness.
Problem Set
Part A — Conceptual Problems
Explain in your own words what the Biot number measures. Why is the threshold value Bi = 0.1 chosen? What physical error does lumped capacitance introduce when Bi is large?
The thermal time constant τ = ρcpLc/h can be interpreted as a ratio. What physical quantity is in the numerator, and what is in the denominator? Why does a higher heat capacity lead to a larger τ?
Two polymer beads of the same material are placed in the same quench bath. Bead A is a sphere of radius 2 mm; Bead B is a sphere of radius 4 mm. Which bead cools faster, and by what factor? Your answer should reference the time constant formula and the characteristic length for a sphere.
A process engineer wants to reduce the time constant of an injection-molded HDPE part from 300 s to 150 s. She proposes two options: (a) use a water-cooled mold instead of an air-cooled one (increases h by a factor of 8), or (b) redesign the part so the wall thickness is halved. Both options achieve τ < 150 s. Without doing a full calculation, explain which option is more straightforward to implement in practice and identify one potential disadvantage of each approach.
Starting from the lumped energy balance $\rho c_p V\, dT/dt = -h A_s (T - T_\infty)$, introduce the dimensionless temperature $\theta^* = (T - T_\infty)/(T_i - T_\infty)$ and dimensionless time $t^* = t / \tau$ (with $\tau$ to be determined). (a) Show by substitution that the equation becomes $d\theta^*/dt^* = -\theta^*$ when $\tau = \rho c_p L_c / h$. (b) Now express $t^* = t/\tau$ in terms of $\alpha$, $h$, $k$, $t$, and $L_c$. Identify the two dimensionless groups Bi and Fo that appear, and write $t^* = \mathrm{Bi}\cdot\mathrm{Fo}$ explicitly. (c) In one sentence, explain why an HDPE part (α ≈ 2×10−7 m²/s) takes roughly 60 times longer to reach the same Fo as a steel part of the same geometry (α ≈ 1.2×10−5 m²/s).
Part B — Calculation Problems
A thin HDPE film (large flat sheet) is cooled on both faces by still air. Characteristic length Lc = 1 mm = 0.001 m. Properties: k = 0.44 W/mK, ρ = 950 kg/m³, cp = 2000 J/kgK. Film coefficient: h = 19 W/m²K. (a) Calculate the Biot number and confirm that lumped capacitance is valid. (b) Find the thermal time constant τ (in seconds).
Use: Bi = hLc/k; τ = ρcpLc/h
Bi = (internal conduction resistance) / (external convection resistance); τ = (thermal energy stored) / (rate of heat removal per degree)
A nylon-6 pellet is approximated as a sphere of radius r = 5 mm (Lc = r/3). It is dropped into an agitated water bath where h = 200 W/m²K. Properties of nylon-6: k = 0.25 W/mK. Calculate the Biot number. Does lumped capacitance apply?
Use: Bi = hLc/k; for a sphere, Lc = r/3
Bi > 0.1 means the inside of the object cannot keep up with what the surface is losing — a temperature gradient builds up through the solid
Note: Bi > 0.1 means lumped capacitance does not apply to this pellet.
Using the same HDPE film from P1 (τ = 100 s), the film starts at Ti = 200°C and is cooled in 20°C air. Find its temperature after exactly one time constant (t = 100 s).
Use: T(t) = T∞ + (Ti − T∞)exp(−t/τ); recall exp(−1) = 0.3679
T∞ is the baseline the object decays toward; the exponential gives the fraction of the original temperature excess that remains
A PP sheet (τ = 200 s) is heated to 200°C and then quenched in an ice bath at T∞ = 0°C. Find its temperature after 600 s (= 3τ). Report to three significant figures.
Use: T(t) = T∞ + (Ti − T∞)exp(−t/τ); recall exp(−3) = 0.04979
After 3τ, only ≈5% of the original excess remains — the object is almost completely equilibrated with its surroundings
An injection-molded part is ejected at Ti = 160°C and cooled on a conveyor in T∞ = 20°C air. Its thermal time constant is τ = 80 s. How long (in seconds) must the operator wait before the surface temperature drops to 60°C (the safe handling limit)?
Use: t = −τ ln[(T − T∞)/(Ti − T∞)]; recall ln(7/2) ≈ 1.253
Invert the exponential by taking a natural log — the time to reach a target T is proportional to τ, scaled by how far that target is from T∞
In Junior Lab you measure the cooling curve of a small metal cylinder using a thermocouple embedded at its centre. You expect an exponential decay, but the data shows that the centre temperature lags significantly behind the surface temperature for the first 30 seconds before the two converge. The cylinder is 50 mm in diameter and sits in a 30°C air stream with h ≈ 80 W/m²K. (a) Without calculating, explain whether you would expect lumped capacitance to apply to this object. (b) What does the lag between surface and centre tell you physically? (c) Propose one simple modification to the experimental setup that would make the lumped capacitance model valid, and explain your reasoning using the Biot number.
Chapter Summary
The lumped capacitance method applies when Bi = hLc/k ≤ 0.1, meaning convection at the surface is slow compared to conduction through the interior, so the solid maintains a nearly uniform temperature at all times. When this holds, the energy balance reduces to a first-order ODE whose solution is a pure exponential decay.
The temperature evolves as $(T - T_\infty)/(T_i - T_\infty) = \exp(-t/\tau)$, where the thermal time constant $\tau = \rho c_p L_c / h$ sets the rate. A large capacitance (ρcpLc) slows cooling; a large film coefficient h speeds it up. After five time constants the object is within 1% of equilibrium with its surroundings.
The same solution can be expressed in dimensionless form as $\exp(-\mathrm{Bi} \cdot \mathrm{Fo})$, where Fo = αt/Lc² is the Fourier number. Low thermal diffusivity α in polymers means large Fo is reached slowly, explaining why polymer parts take far longer to cool than metal parts of the same size.
When Bi > 0.1 the single-temperature model fails: the surface cools faster than the core and the temperature distribution inside the solid must be tracked spatially. Heisler charts or series solutions handle this regime. Chapter 9 continues the story by examining how a moving fluid generates the film coefficient h in the first place, and what controls its magnitude.
By the final exam, you will be expected to write these equations from memory.
In the next chapter, we examine forced convection: the physics that determines the film coefficient h, how boundary layers form and grow along a surface, and the Nusselt-number correlations that let you predict h from flow conditions in polymer processing lines.