Your Equation

Library / Heat Transfer / HTR-04

Heat Transfer

Lumped transient cooling

Temperature of a small body cooling in a fluid, with uniform body temperature.

Ref
HTR-04
Rev
v1
Field
Heat Transfer
Result
T [°C]

Uniform initial temperature, constant properties, h and ambient temperature. Check Bi = h(V/As)/ks < 0.1 as a rule of thumb for small internal gradients. Here V/As is entered as the characteristic length Lc.

A small hot object cools quickly at first and then more slowly, approaching the temperature of its surroundings.

Calculator

Solve for
°C
°C
501,000
°C
−40100
kg/m³
50020,000
J/(kg·K)
1005,000
mm
0.1100
W/(m²·K)
52,000
s
13,000
  • Time constant τ400 s
  • Fraction of the temperature gap remaining74.08 %

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.

Temperature at time t (°C) current value 154.6 °C
TiInitial temperatureKey driver
43.5747
tTime
25.1200
LcVolume/area ratio
25197
hHeat transfer coefficient
25195
ρDensity
26.4180
cpSpecific heat
64195
T∞Ambient temperature
138174
  • After one time constant the temperature gap falls to 37%; after three it is about 5%.
  • A large h or a small V/As makes the body respond quickly.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (°C)Local stepChange in result
Ti Initial temperature50 to 1,000 °C43.52747.3703.8+10%+14.8 °C
t Time1 to 3,000 s25.1199.6174.5+10%−3.83 °C
Lc Volume/area ratio0.1 to 100 mm25197.4172.4+10%+3.58 °C
h Heat transfer coefficient5 to 2,000 W/(m²·K)25194.8169.8+10%−3.83 °C
ρ Density500 to 20,000 kg/m³26.44180.2153.8+10%+3.58 °C
cp Specific heat100 to 5,000 J/(kg·K)64.05194.8130.8+10%+3.58 °C
T∞ Ambient temperature−40 to 100 °C137.8174.136.29+10%+0.648 °C

Sweep of Time

Time (s)Temperature at time t (°C)
1199.6
375.993.38
750.851.79
1,12635.49
1,50129.11
1,87526.61
2,25025.63
2,62525.25
3,00025.1

Sweep one variable

Temperature at time t (°C)Hover, touch, or use the arrow keys on the chart
50100150200 50010001500200025003000 Time (s) now 154.6

Variables and how each one contributes

T

Temperature at time t

Result · °C

What the equation gives you. Everything else feeds it.

Ti

Initial temperature

°C · typical 50 to 1,000

Starting temperature of the body. Raises the temperature.

T∞

Ambient temperature

°C · typical −40 to 100

Temperature of the surrounding fluid. Raises the final temperature.

ρ

Density

kg/m³ · typical 500 to 20,000

Density of the body. Raises the time constant.

cp

Specific heat

J/(kg·K) · typical 100 to 5,000

Specific heat of the body. Raises the time constant.

Lc

Volume/area ratio

mm · typical 0.1 to 100

V/As (for a sphere, diameter/6). Raises the time constant.

h

Heat transfer coefficient

W/(m²·K) · typical 5 to 2,000

Surface convection coefficient. Shortens the time constant.

t

Time

s · typical 1 to 3,000

Time since the start. More time brings the body closer to ambient.

Worked example with the values above

  1. τ = ρcpLc/h = 8,000 × 500 × 0.005 / 50 = 400 s.
  2. Decay: e−t/τ = e−0.3 = 0.7408.
  3. T = 25 + (200 − 25) × 0.7408 = 154.6 °C.

Assumptions and limits

  • Biot number below about 0.1.
  • Constant properties and h.

Common mistakes

  • Using the lumped model for thick or low-conductivity bodies.
  • Using the total volume without dividing by surface area.

Statistics and uncertainty

The time constant depends on h to within ±20–30%, so the cooling time is only as good as that estimate.

Where it is used

  • Sensor and thermocouple response.
  • Cooling of small components and shells.

Maths you need

  • Exponential decay: e^(−t/τ)
  • Convection: h

Related formulas

Practise this formula