Library / Heat transfer and thermal systems / HTS-50
Heat transfer and thermal systems
Lumped radiative cooling towards a cold sink
Time for a lumped body to cool from Ti to Tf by radiation alone to a sink so cold that its irradiation is negligible.
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- HTS-50
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- Heat transfer and thermal systems
- Result
- t [s]
A body cooling only by radiation to a very cold sink has C dT/dt = −εσAT⁴. Integrating from Ti to Tf gives the cooling time t ≈ C(Tf⁻³ − Ti⁻³)/(3εσA). It neglects irradiation from the sink, so the sink must be far colder than Tf, and needs a uniform body temperature. Doubling the absolute temperature range does not simply scale the time, because of the cube.
Time for a lumped body to cool from Ti to Tf by radiation alone to a sink so cold that its irradiation is negligible. Needs a uniform body temperature, constant properties and Tf above the sink temperature.
Calculator
- Initial radiated power, εσAT_i⁴283.5 W
- Final radiated power, εσAT_f⁴116.1 W
- Mean cooling power, C(Ti − Tf)/t178.5 W
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 body takes 56.03 s to cool from 500 K to 400 K, radiating 283.5 W at first and 116.1 W at the end.
- Cooling gets slower as it proceeds: most of the time is spent at the low end, where radiation is weakest.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C Thermal capacitance | 1 to 100,000 J/K | 0.5603 | 56,030 | 56,030 | +10% | +10 % |
| Tf Final temperature | 100 to 499 K | 0.3541 | 7,289 | 7,289 | +10% | −51 % |
| A Radiating area | 0.001 to 10 m² | 0.5603 | 5,603 | 5,602 | +10% | −9.09 % |
| ε Emissivity | 0.05 to 1 | 44.82 | 896.5 | 851.6 | +10% | −9.09 % |
| Ti Initial temperature | 500 to 2,000 K | 56.03 | 113.9 | 57.87 | +10% | +26.1 % |
Sweep of Final temperature
| Final temperature (K) | Cooling time (s) |
|---|---|
| 100 | 7,289 |
| 149.9 | 2,124 |
| 199.8 | 863.2 |
| 249.6 | 413.6 |
| 299.5 | 214.7 |
| 349.4 | 113.5 |
| 399.3 | 56.68 |
| 449.1 | 22.33 |
| 499 | 0.3541 |
Sweep one variable
Variables and how each one contributes
Cooling time
What the equation gives you. Everything else feeds it.
Thermal capacitance
The thermal capacitance of the body, in J/K. Raises the cooling time in direct proportion. Raising it by 10 % (to 110) moves the result by +10 %, from 56.03 to 61.63 s.
Radiating area
The radiating area, in m². Lowers the cooling time in inverse proportion. Raising it by 10 % (to 0.11) moves the result by −9.09 %, from 56.03 to 50.94 s.
Emissivity
The emissivity of the surface. Lowers the cooling time in inverse proportion. Raising it by 10 % (to 0.88) moves the result by −9.09 %, from 56.03 to 50.94 s.
Initial temperature
The initial temperature, in K. A hotter start raises the time slightly, since the hottest period passes quickly. Raising it by 10 % (to 550) moves the result by +26.1 %, from 56.03 to 70.65 s.
Final temperature
The final temperature, in K. A lower final temperature raises the time greatly, since radiation falls as T⁴. Raising it by 10 % (to 440) moves the result by −51 %, from 56.03 to 27.48 s.
Worked example with the values above
- Temperature terms: Tf⁻³ = 1.5625×10−8, Ti⁻³ = 8×10−9; difference = 7.625×10−9 K⁻³.
- Denominator: 3εσA = 3 × 0.8 × 5.67037×10−8 × 0.1 = 1.3609×10−8.
- t = 100 × 7.625×10−9 / 1.3609×10−8 = 56.03 s.
Assumptions and limits
- Constant properties in each phase, a single melting temperature and no losses unless stated.
- Steady or quasi-steady balances, with absolute temperatures where logarithms or powers are used.
Common mistakes
- Forgetting the latent heat term, or counting it twice.
- Using °C in an entropy or radiation formula.
- Using these simple formulas when the Stefan number is not small.
Statistics and uncertainty
Latent heats and specific heats are tabulated to a few per cent, but real storage materials melt over a range and lose heat, so real systems typically deliver 5 % to 20 % less than the ideal values.
Where it is used
- Sizing thermal storage, ice banks and phase-change units.
- Checking exchanger data, mixing streams and cooling times.
Maths you need
- Latent heat: energy per kilogram to change phase
- Heat-capacity rate: C = ṁcp
- Second law: entropy generation cannot be negative