Library / Heat transfer and thermal systems / HTS-23
Heat transfer and thermal systems
Lumped body with constant internal power
Temperature of a lumped body of capacitance C with constant generated power P and conductance G to ambient Ta.
- Ref
- HTS-23
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- T [K]
A lumped body with constant internal power P, capacitance C and conductance G to ambient follows T = Ti + (Ta + P/G − Ti)[1 − exp(−Gt/C)]. It heads for the steady value Ta + P/G with time constant C/G. About 63 % of the change is complete after one time constant and 99 % after about five. It needs small internal gradients.
Temperature of a lumped body of capacitance C with constant generated power P and conductance G to ambient Ta. It relaxes exponentially towards Ta + P/G with time constant C/G.
Calculator
- Steady-state temperature, Ta + P/G310 K
- Time constant, C/G100 s
- Fraction of the way to steady state, 1 − exp(−Gt/C)0.6321
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.
- After 100 s the body is at 306.3 K, 63.2 % of the way to its steady value of 310 K.
- The time constant is 100 s, so the body needs about 500 s to settle within 1 % of the steady state.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| P Constant internal power | 0 to 100,000 W | 293.7 | 12,940 | 12,640 | +10% | +0.413 % |
| Ta Ambient temperature | 250 to 1,000 K | 281 | 755.1 | 474.1 | +10% | +5.98 % |
| Ti Initial temperature | 250 to 1,000 K | 287.9 | 563.8 | 275.9 | +10% | +3.6 % |
| G Conductance to ambient | 0.01 to 1,000 W/K | 290.1 | 320 | 29.86 | +10% | −0.282 % |
| t Time | 0 to 100,000 s | 300 | 310 | 10 | +10% | +0.114 % |
| C Thermal capacitance | 1 to 1,000,000 J/K | 300 | 310 | 9.995 | +10% | −0.114 % |
Sweep of Time
| Time (s) | Body temperature (K) |
|---|---|
| 0 | 300 |
| 12,500 | 310 |
| 25,000 | 310 |
| 37,500 | 310 |
| 50,000 | 310 |
| 62,500 | 310 |
| 75,000 | 310 |
| 87,500 | 310 |
| 100,000 | 310 |
Sweep one variable
Variables and how each one contributes
Body temperature
What the equation gives you. Everything else feeds it.
Initial temperature
The initial temperature, in K. A higher start raises the temperature early on, but the final value is unchanged. Raising it by 10 % (to 330) moves the result by +3.6 %, from 306.3 to 317.4 K.
Ambient temperature
The ambient temperature, in K. Raises the steady value and the path towards it. Raising it by 10 % (to 319) moves the result by +5.98 %, from 306.3 to 324.7 K.
Constant internal power
The constant internal power, in W. Raises the steady-state rise in direct proportion. Raising it by 10 % (to 110) moves the result by +0.413 %, from 306.3 to 307.6 K.
Conductance to ambient
The conductance to ambient, in W/K. A higher conductance lowers the steady value and shortens the time constant. Raising it by 10 % (to 5.5) moves the result by −0.282 %, from 306.3 to 305.5 K.
Thermal capacitance
The thermal capacitance, in J/K. A larger capacitance slows the response. Raising it by 10 % (to 550) moves the result by −0.114 %, from 306.3 to 306 K.
Time
The time, in s. More time moves the temperature towards the steady value. Raising it by 10 % (to 110) moves the result by +0.114 %, from 306.3 to 306.7 K.
Worked example with the values above
- Steady value: Ta + P/G = 290 + 100/5 = 310 K.
- Approach: 1 − exp(−Gt/C) = 1 − exp(−1) = 0.63212.
- T = 300 + (10) × 0.63212 = 306.32 K.
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