Conjugate Heat Transfer Calculator

Compute coupled convection–conduction heat transfer, including fouling and optional contact resistance. Use Design mode to size the wall thickness or outer radius for a target heat rate.

This calculator uses a steady 1‑D thermal resistance model. The heat rate is:

Q = (T∞₁ − T∞₂) / R_total

1. Choose a geometry: slab, cylinder, or sphere.
2. Enter hot and cold bulk fluid temperatures and convection coefficients.
3. Enter wall conductivity and the required dimensions for the chosen geometry.
4. Optionally add fouling and contact factors for more realistic estimates.
5. Select Design mode if you want the thickness or outer radius that meets a target heat rate.
6. Press Compute to see results below the header, then export CSV or PDF.

1) What conjugate heat transfer means

Conjugate heat transfer (CHT) describes heat moving through a solid wall while the adjacent fluids exchange heat by convection. In many practical systems, neither side alone controls performance; the overall heat rate depends on both boundary layers and the wall’s conduction path.

2) Resistance modeling ties the physics together

This calculator uses a steady one‑dimensional resistance network where Q = (T∞1 − T∞2)/R_total. Convective layers contribute 1/(hA), conduction depends on geometry, and optional fouling/contact terms add realistic penalties. The approach is fast for screening designs and sensitivity checks.

3) Typical convection coefficients provide context

Convection coefficients vary by flow regime and fluid. Natural convection air often falls near 5–25 W/m²·K, forced air commonly 20–200 W/m²·K, and forced water can reach 200–10,000 W/m²·K. If your computed heat rate seems small, low h is frequently the cause.

4) Wall conductivity controls the internal drop

Material conductivity spans orders of magnitude. Typical values include insulation foams around 0.03–0.06 W/m·K, polymers near 0.15–0.40 W/m·K, stainless steels about 14–20 W/m·K, and aluminum roughly 150–230 W/m·K. A small k or thick wall quickly increases R_cond.

5) Geometry changes areas and conduction paths

Curved walls have different inner and outer areas, so the calculator reports A1, A2, and a log‑mean area for consistent overall U definitions. For cylinders, R_cond &propto ln(r2/r1), meaning thin coatings add little resistance until the radius ratio grows noticeably.

6) Fouling and contact losses are often underestimated

Fouling factors can rival convection resistance over long operation. Light clean‑water service may be 0.0001–0.0002 m²·K/W, cooling‑water scaling can reach 0.0002–0.001, and heavy deposits may exceed 0.01. Contact resistance can range from 1×10⁻⁵ to 1×10⁻³ m²·K/W for good interfaces, and much higher for rough, dry joints.

7) Interpreting outputs: Q, U, and surface temperatures

The heat rate Q sets each temperature drop: ΔT = Q R. For example, if T∞1 − T∞2 = 70 K and R_total = 0.20 K/W, then |Q| = 350 W. Surface temperatures follow by subtracting the convective and fouling drops on each side.

8) Using design mode responsibly

Design mode solves a thickness (slab) or outer radius (cylinder/sphere) for a target |Q| under the same boundary assumptions. Validate units and sweep h and fouling to bracket uncertainty. The model is steady and 1-D; add radiation or CFD for strong multidimensional effects.

1) What problems is this calculator best for?

It is best for steady, one‑dimensional heat transfer across a wall with convection on both sides, including optional fouling and contact losses. It is ideal for early sizing and quick comparisons.

2) Why can the heat rate appear negative?

The sign follows the temperature difference (T∞1 − T∞2). A negative value indicates heat flows from side 2 toward side 1. Use |Q| when you only need magnitude.

3) Which area reference should I choose for U?

For curved walls, select log‑mean area for a balanced definition, or inner/outer if your standard specifies it. The heat rate is unchanged; only the reported U value depends on the reference area.

4) How do I handle multiple wall layers?

Add each layer’s conduction resistance and sum them in series, then combine with convection and fouling. If you only know an overall effective conductivity, you can approximate the stack as one equivalent layer using that value.

5) Do Celsius and Kelvin change the result?

No. Only temperature differences matter in the resistance model, and a 1 K difference equals a 1 °C difference. Choose the unit that matches your input data.

6) When should I include fouling and contact factors?

Include them when surfaces age, scale, corrode, or when interfaces are pressed together imperfectly. If uncertain, start at zero and run a sensitivity sweep. Small increases can noticeably reduce U and Q.

7) My surface temperatures look unrealistic. What should I check?

Confirm units, geometry dimensions, and that h is not off by 10× or 100×. Extremely low h or very large fouling can create large drops. Also ensure the hot side really has the higher bulk temperature.