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Compute convective heat transfer (Newton's law of cooling): heat flux q = h·ΔT (W/m²), heat flow rate Q = h·A·ΔT (W), and convective thermal resistance R_conv = 1/(h·A) (K/W). h is the convective heat-transfer coefficient (W/(m²·K)); ΔT is the temperature difference between the surface and the fluid (K; a °C difference equals a K difference, a °F difference is converted by ×5/9); A the heat-transfer area. ΔT may be negative (indicating reverse heat flow), but h and A must be positive. Area in m²/cm².
Execution
Fill in the form, run the tool, and review the result in one place.
Samples
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Tool usage guide
The Heat Convection Calculator applies Newton’s law of cooling to calculate heat flux q, heat flow rate Q, and convective thermal resistance from the heat-transfer coefficient, temperature difference, and transfer area.
Background
A thermal engineer needs to estimate heat transfer from a 1 m² surface exposed to air.
Problem
Calculate the heat flux, total heat flow, and convective resistance for h = 25 W/(m²·K) and ΔT = 20 K.
How to use
Enter 25 for h, 20 for ΔT, select K, enter an area of 1, and select m².
h = 25 W/(m²·K); ΔT = 20 K; A = 1 m²Outcome
The calculator returns q = 500 W/m², Q = 500 W, and R_conv = 0.04 K/W.
Background
A designer is evaluating heat transfer to water across a 200 cm² surface.
Problem
Determine the heat flux, heat flow rate, and convective resistance for h = 500 W/(m²·K) and a 50 °C temperature difference.
How to use
Enter 500 for h, 50 for ΔT, select °C, enter an area of 200, and select cm².
It calculates heat flux q in W/m², heat flow rate Q in W, and convective thermal resistance R_conv in K/W.
Heat flux is calculated with q = h·ΔT.
Yes. A Celsius difference equals the same kelvin difference, while a Fahrenheit difference is converted by multiplying by 5/9.
Yes. A negative ΔT indicates heat flow in the reverse direction.
The heat-transfer coefficient h and area A must be positive. The temperature difference may be negative.
h = 500 W/(m²·K); ΔT = 50 °C; A = 200 cm²Outcome
The 50 °C difference is treated as 50 K. Results are q = 25,000 W/m², Q = 500 W, and R_conv = 0.1 K/W.
Background
A student is checking a case where the fluid is hotter than the surface, so the selected temperature difference is negative.
Problem
Calculate the signed heat flux and heat flow for h = 25 W/(m²·K), ΔT = -10 K, and A = 0.5 m².
How to use
Enter 25 for h, -10 for ΔT, select K, enter 0.5 for the area, and select m².
h = 25 W/(m²·K); ΔT = -10 K; A = 0.5 m²Outcome
The negative temperature difference produces q = -250 W/m² and Q = -125 W, indicating reverse heat flow; R_conv = 0.08 K/W.