Calculator Inputs
Layout becomes 3 columns on large screens, 2 on smaller, and 1 on mobile.
Example Data Table
A realistic set of inputs and typical outputs.
| Case | Tᵢ (°C) | Tₐ (°C) | Dᵢ (mm) | tₚ (mm) | tᵢₙₛ (mm) | hᵢ | hₒ | Q (W) | q′ (W/m) |
|---|---|---|---|---|---|---|---|---|---|
| Hot fluid, insulated | 150 | 25 | 50 | 3 | 30 | 800 | 12 | ~70–95 | ~70–95 |
| Hot fluid, no insulation | 150 | 25 | 50 | 3 | 0 | 800 | 12 | ~250–400 | ~250–400 |
Ranges vary with material properties and convection coefficients.
Formula Used
Thermal resistance network for a multilayer cylinder.
1) Convection resistances
Rconv,i = 1 / (hi · 2πr1L)
Rconv,o = 1 / (ho · 2πr3L)
2) Conduction resistances
Rpipe = ln(r2/r1) / (2πkpL)
Rins = ln(r3/r2) / (2πkinsL)
3) Heat transfer rate
Rtotal = Rconv,i + Rpipe + Rins + Rconv,o
Q = (Ti − Ta) / Rtotal
q′ = Q / L
4) Critical insulation radius (optional)
rcrit = kins / ho
If r3 is below rcrit, adding insulation can increase heat loss.
How to Use This Calculator
A quick workflow for practical decisions.
- Enter inside temperature and ambient temperature.
- Set pipe geometry: inner diameter, wall thickness, and length.
- Enter pipe conductivity and convection coefficients.
- Add insulation thickness and conductivity, or choose no insulation.
- Press Calculate Heat Loss to see results above the form.
- Use CSV or PDF buttons to export your latest run.
FAQs
Plain HTML, short answers, no accordions.
1) What does this calculator estimate?
It estimates steady heat transfer through a cylindrical pipe wall and optional insulation, including inside and outside convection. It reports heat rate, per‑meter loss, resistances, and surface temperatures.
2) Can I use Celsius or Kelvin?
Yes. The calculation uses the temperature difference, so °C and K work the same for ΔT. Keep both temperatures in the same unit to avoid mistakes.
3) Why is convection included?
Convection can dominate the total resistance when the pipe is thin or highly conductive. Including hᵢ and hₒ gives a more realistic heat loss than conduction alone.
4) What is “critical insulation radius”?
For cylinders, adding insulation increases area while adding resistance. Below rcrit = kins/hₒ, heat loss may increase with insulation. Above it, insulation reduces heat loss.
5) What values should I use for hₒ in air?
Natural convection in still air is often around 5–25 W/m²·K. Wind increases hₒ significantly. If you are unsure, run a sensitivity check using a low and high estimate.
6) Why do results change a lot with insulation k?
Insulation conductivity strongly affects the logarithmic conduction resistance term. Small changes in kins can shift Rins and the total resistance, especially when insulation thickness is large.
7) Does it handle multiple insulation layers?
This version supports one insulation layer. For multiple layers, you can approximate by summing each layer’s ln(r₂/r₁)/(2πkL) and using the final outer radius for outside convection.
8) What are the main limitations?
It assumes steady state, uniform properties, and one‑dimensional radial heat flow. It ignores fittings, supports, radiation effects, and axial conduction. Use it for quick engineering estimates, not final safety sign‑off.