Library / Heat transfer and thermal systems / HTS-24
Heat transfer and thermal systems
Equilibration time constant of two finite bodies
Time constant for two lumped bodies of heat capacity C1 and C2, isolated from their surroundings, to exchange heat through a conductance G.
- Ref
- HTS-24
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- τ [s]
Two lumped bodies with capacitances C1 and C2 joined by a conductance G and isolated from the surroundings equalise their temperatures exponentially. The difference decays with the time constant τ = C₁C₂/[G(C₁ + C₂)]. It is the series combination of the two capacitances divided by G, and is shorter than the single-body time constant of either body.
Time constant for two lumped bodies of heat capacity C1 and C2, isolated from their surroundings, to exchange heat through a conductance G. Their temperature difference falls by a factor e in time τ.
Calculator
- Series capacitance, C₁C₂/(C₁ + C₂)120 J/K
- Time to equalise within 1 %, 4.605τ138.2 s
- Weight of body 1 in the final temperature, C₁/(C₁ + C₂)0.4
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The temperature difference between the bodies falls by a factor of e every 30 s and is within 1 % after 138.2 s.
- The final common temperature is the capacitance-weighted average, with 40 % weight on body 1.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| G Conductance between the bodies | 0.01 to 1,000 W/K | 0.12 | 12,000 | 12,000 | +10% | −9.09 % |
| C1 Capacitance of body 1 | 1 to 1,000,000 J/K | 0.2492 | 74.98 | 74.73 | +10% | +5.77 % |
| C2 Capacitance of body 2 | 1 to 1,000,000 J/K | 0.2488 | 49.99 | 49.74 | +10% | +3.77 % |
Sweep of Conductance between the bodies
| Conductance between the bodies (W/K) | Time constant (s) |
|---|---|
| 0.01 | 12,000 |
| 0.04217 | 2,846 |
| 0.1778 | 674.8 |
| 0.7499 | 160 |
| 3.162 | 37.95 |
| 13.34 | 8.999 |
| 56.23 | 2.134 |
| 237.1 | 0.506 |
| 1,000 | 0.12 |
Sweep one variable
Variables and how each one contributes
Time constant
What the equation gives you. Everything else feeds it.
Capacitance of body 1
The thermal capacitance of body 1, in J/K. A larger capacitance slows equilibration. Raising it by 10 % (to 220) moves the result by +5.77 %, from 30 to 31.73 s.
Capacitance of body 2
The thermal capacitance of body 2, in J/K. A larger capacitance slows equilibration. Raising it by 10 % (to 330) moves the result by +3.77 %, from 30 to 31.13 s.
Conductance between the bodies
The conductance between the bodies, in W/K. A larger conductance shortens the time constant in inverse proportion. Raising it by 10 % (to 4.4) moves the result by −9.09 %, from 30 to 27.27 s.
Worked example with the values above
- Product: C₁C₂ = 200 × 300 = 60,000 J²/K²; sum: C₁ + C₂ = 500 J/K.
- Series capacitance: 120 J/K.
- τ = 120 / 4 = 30 s.
Assumptions and limits
- Constant properties, a uniform initial temperature and convection to a fixed ambient on both sides for the wall solutions.
- For lumped models, a body with small internal temperature differences and constant conductances.
Common mistakes
- Using the full thickness instead of the half-thickness in Bi and Fo.
- Using the first-term form at small Fo, where more terms are needed.
- Applying a lumped model when the Biot number is not small.
Statistics and uncertainty
First-term wall solutions are accurate to within about 2 % for Fo above 0.2, and lumped formulas need a small Biot number, below about 0.1, to stay within a few per cent.
Where it is used
- Cooling and heating of slabs, plates and electronics enclosures.
- Estimating warm-up and settling times.
Maths you need
- Biot number: Bi = hL/k
- Fourier number: Fo = αt/L²
- Time constant: τ = C/G, capacitance over conductance