Heat Transfer, Electronics Thermal Management & Passive Natural Convection
Natural Convection Rayleigh & Churchill-Chu Nusselt Formula
Calculate Vertical Heatsink / Enclosure Plate Laminar Rayleigh Number Ra_L, Exact Churchill-Chu Laminar Nusselt Number Nu_L in Air (Pr = 0.71), Natural-Convection Heat-Transfer Coefficient h (W/m²·K), and Passive Heat Dissipation Q (W).
Size fanless electronics heatsinks with Rayleigh buoyancy and Churchill-Chu Nusselt numbers!
Passive fanless inverters, telecom enclosures, and LED heatsinks rely on buoyancy-driven natural convection along a vertical plate of height L (m) at excess temperature ΔT = T_s - T_inf (K)! Multiplying Air’s Buoyancy Property Modulus M_air = (g * β) / (ν * α) (~1.0 × 10^8 m^-3·K^-1 at mean film temperature ~50°C) by ΔT * L^3 gives the dimensionless Rayleigh Number Ra_L = M_air * ΔT * L^3! For laminar natural convection (Ra_L <= 10^9) in air (Pr = 0.71), Churchill and Chu’s correlation gives Average Nusselt Number Nu_L = 0.68 + (0.67 * Ra_L^0.25) / (1 + (0.492 / 0.71)^(9/16))^(4/9), yielding convective coefficient h = (Nu_L * k_air) / L and superlinear heat rejection Q = h * A * ΔT (~ΔT^1.25)!
Verify on a Fanless Industrial Drive Chassis—Air_Buoyancy_Modulus = 100,000,000 m^-3·K^-1 (B2), Surface_Excess_ΔT = 64.0 K (B3), Vertical_Chassis_Height_L = 0.25 m (B4 -> Ra_L = 10^8 * 64 * 0.25^3 = 100,000,000 in E2, laminar!), Air_Conductivity_k = 0.028 W/(m·K) (B5), and Exposed_Surface_Area_A = 0.40 m² (B6): Exact Churchill-Chu Nusselt Number is 52.10 (E3), Average Natural-Convection h is (52.102 * 0.028) / 0.25 = 5.84 W/(m²·K) (E4), and Passive Convective Heat Dissipated is 5.8354 * 0.40 * 64 = 149.39 W (E5)!
| Natural Convection Metric (M = 10^8, ΔT = 64 K, L = 0.25 m, k = 0.028, A = 0.40 m²) | Formula | Verified Output |
|---|---|---|
| Rayleigh Number Ra_L | Buoyancy_Modulus * ΔT * (L^3) | 100,000,000 (6,250,000 @ ΔT=4 K) |
| Churchill-Chu Nusselt Nu_L | 0.68 + 0.67*(Ra_L^0.25) / (1+(0.492/0.71)^(9/16))^(4/9) | 52.10 (26.39 @ ΔT=4 K) |
| Convective Coeff h (W/m²·K) | (Nu_L * k_air) / L_m | 5.84 W/m²·K (2.96 @ ΔT=4 K) |
| Passive Heat Rejected Q (W) | h * Area_m2 * ΔT_K | 149.39 W (4.73 W @ ΔT=4 K) |
Churchill-Chu Laminar Natural Convection Heat Rejection Formula
Validate Buoyancy_Modulus > 0, ΔT > 0, Height_L > 0, k_air > 0, Area > 0, and laminar bound Ra_L <= 10^9!
=LET(mb_a, B2, dt_k, B3, ht_l, B4, ka_w, B5, ar_m, B6, ra_l, mb_a * dt_k * (ht_l ^ 3), IF(OR(mb_a <= 0, dt_k <= 0, ht_l <= 0, ka_w <= 0, ar_m <= 0, ra_l > 1000000000), "Invalid Laminar Ra_L Input", LET(nu_l, 0.68 + (0.67 * (ra_l ^ 0.25)) / ((1 + ((0.492 / 0.71) ^ (9 / 16))) ^ (4 / 9)), hc_v, (nu_l * ka_w) / ht_l, ROUND(hc_v * ar_m * dt_k, 2))))Standby Low-ΔT Buoyancy Weakening Check (Set ΔT = 4.0 K)
Notice why natural convection is nonlinear! Cut ΔT B3 by 16x from 64 K down to 4 K: Ra_L E2 drops 16x to 6,250,000, cutting Nu_L E3 from 52.10 to 26.39 and h E4 by ~half from 5.84 to 2.96 W/(m²·K)—so Heat Rejected E5 plunges ~31.6x to 4.73 W!
=IF(ISTEXT(E5), E5, ROUND(((0.68 + (0.67 * ((B2 * B3 * (B4 ^ 3)) ^ 0.25)) / ((1 + ((0.492 / 0.71) ^ (9 / 16))) ^ (4 / 9))) * B5) / B4, 2))Try the example
Use the original values or change them to test your understanding.