Library / Heat Transfer / HTR-02

Heat Transfer

Heat loss through a pipe wall

Steady radial heat flow through a cylindrical pipe or insulation layer.

Ref: HTR-02
Rev: v1
Field: Heat Transfer
Result: Q̇ [W]

R_{cyl} = \frac{\ln (\frac{r_{o}}{r_{i}})}{2 π k L}

For a cylindrical layer of length L, constant conductivity k and negligible end effects. For each surface film, use its own area 2πrL. Heat rate Q = ΔT/R_cyl.

Heat leaks through a pipe’s insulation more slowly the thicker and less conductive it is, but each extra bit of thickness helps less than the one before.

Calculator

Solve for

Q̇Heat lossComputed

r_iInner radius

5200

r_oOuter radius

10400

kThermal conductivity

W/(m·K)

0.02400

LPipe length

11,000

ΔTTemperature difference

5500

Resistance R_cyl0.2758 K/W
Heat loss per metre29.01 W/m

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.

Heat loss (W) current value 290.1 W

kThermal conductivityKey driver

1452,900,000

LPipe length

2929,000

r_iInner radius

−5,9702,250

r_oOuter radius

−3,6202,980

ΔTTemperature difference

18.11,810

Doubling the outer radius does not halve the loss because the resistance rises only as ln(r_o/r_i).
At very small pipe sizes, thin insulation can increase heat loss because the surface area grows faster than the resistance.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (W)	Local step	Change in result
k Thermal conductivity	0.02 to 400 W/(m·K)	145	2,901,000	2,901,000	+10%	+10 %
L Pipe length	1 to 1,000 m	29.01	29,010	28,980	+10%	+10 %
r_i Inner radius	5 to 200 mm	−5,965	2,253	8,218	+10%	+15.9 %
r_o Outer radius	10 to 400 mm	−3,619	2,983	6,602	+10%	−12.1 %
ΔT Temperature difference	5 to 500 K	18.13	1,813	1,795	+10%	+10 %

Sweep of Outer radius

Outer radius (mm)	Heat loss (W)
10	−219.4
15.86	−441.7
25.15	33,910
39.88	430.5
63.25	216.6
100.3	144.7
159.1	108.7
252.2	86.98
400	72.52

Sweep one variable

Vary

Heat loss (W)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Q̇

Heat loss

Result · W

What the equation gives you. Everything else feeds it.

r_i

Inner radius

mm · typical 5 to 200

Inner radius of the layer. A larger r_i lowers the resistance.

r_o

Outer radius

mm · typical 10 to 400

Outer radius of the layer. A larger r_o raises the resistance.

Thermal conductivity

W/(m·K) · typical 0.02 to 400

Conductivity of the layer. Lowers the resistance.

Pipe length

m · typical 1 to 1,000

Pipe length. Directly proportional.

ΔT

Temperature difference

K · typical 5 to 500

Across the layer. Directly proportional.

Worked example with the values above

ln(r_o/r_i) = ln(50/25) = 0.6931.
R_cyl = 0.6931 / (2π × 0.04 × 10) = 0.2758 K/W.
Q = ΔT/R = 80 / 0.2758 = 290.1 W.

Assumptions and limits

Radial steady conduction with negligible end effects.
Constant k.

Common mistakes

Using diameters in the logarithm of radius ratios (the ratio is the same, but be consistent).
Using a flat-wall formula for a thick pipe wall.

Statistics and uncertainty

Insulation conductivity changes with temperature and moisture by 10–30%, which dominates the uncertainty.

Where it is used

Steam and chilled-water pipe lagging.
Cable and hose heat-loss estimates.

Maths you need

Conduction: Fourier’s law
Natural logarithm: ln

Related formulas

Practise this formula