Engineering & Science
Nusselt Laminar Film Condensation Coefficient on a Vertical Wall (h̄)
Calculate the mean condensing heat-transfer coefficient h̄ = 0.943 * ((ρ_l * (ρ_l - ρ_v) * g * h_fg * k_l^3) / (μ_l * L * ΔT))^0.25 (W/(m^2·K)) for vertical condenser tubes.
Counterintuitive condenser physics: why DOUBLING ΔT actually REDUCES the heat-transfer coefficient h̄!
In 1916, Wilhelm Nusselt solved the coupled gravity-drainage momentum and conduction equations for a falling liquid condensate film on a vertical isothermal plate or tube of height L at subcooling ΔT = T_sat - T_wall: h̄ = 0.943 * ((ρ_l * (ρ_l - ρ_v) * g * h_fg * k_l^3) / (μ_l * L * ΔT))^0.25! Notice the famous ΔT^-0.25 and L^-0.25 inverse fourth-root scaling: cooling the wall further increases condensation rate, which THICKENS the insulating liquid water film and lowers h̄!
For saturated steam at 100 °C condensing on a vertical tube of length L = 1.20 m held at T_wall = 90 °C (ΔT = 10 K; film properties at 95 °C: ρ_l = 961.9 kg/m^3, ρ_v = 0.598 kg/m^3, h_fg = 2,257,000 J/kg, k_l = 0.677 W/(m·K), μ_l = 0.000297 Pa·s), h̄ reaches an extraordinary 6,126.7 W/(m^2·K); doubling subcooling to ΔT = 20 K thickens the condensate film and trims h̄ by 15.9% to 5,152.0 W/(m^2·K) (while total heat flux q″ = h̄ * ΔT still rises by 2^0.75 = 1.68×)!
| Vertical Steam Condenser Case (L = 1.20 m) | Wall Subcooling ΔT (K) | Mean Condensing h̄ (W/m^2·K) | Heat Flux q″ = h̄ * ΔT (kW/m^2) |
|---|---|---|---|
| Moderate Subcooling (Twall = 90 °C) | 10.0 K | 6,126.7 W/(m^2·K) | 61.27 kW/m^2 |
| Deep Subcooling (Twall = 80 °C) | 20.0 K | 5,152.0 W/(m^2·K) (-15.9%) | 103.04 kW/m^2 (+68.2%) |
| Invalid ΔT <= 0 or ρ_l <= ρ_v | No Condensation | "Invalid Condenser Inputs" | Subcooling Guard |
Nusselt Mean Vertical-Wall Film Condensation Coefficient h̄ (W/(m^2·K))
Evaluates 0.943 * ((ρ_l * (ρ_l - ρ_v) * 9.80665 * h_fg * k_l^3) / (μ_l * L * ΔT))^0.25 across fluid properties B2:B6, tube height B7 (m), and wall subcooling B8 (K).
=IF(OR(B2 <= B3, B3 < 0, B4 <= 0, B5 <= 0, B6 <= 0, B7 <= 0, B8 <= 0), "Invalid Condenser Inputs", ROUND(0.943 * (((B2 * (B2 - B3) * 9.80665 * B4 * (B5 ^ 3)) / (B6 * B7 * B8)) ^ 0.25), 1))Condenser Wall Heat Flux q″ = h̄ * ΔT / 1000 (kW/m^2)
Multiplies h̄ by ΔT and converts W/m^2 to kW/m^2.
=IF(OR(B2 <= B3, B3 < 0, B4 <= 0, B5 <= 0, B6 <= 0, B7 <= 0, B8 <= 0), "Invalid Condenser Inputs", ROUND(0.000943 * (((B2 * (B2 - B3) * 9.80665 * B4 * (B5 ^ 3)) / (B6 * B7 * B8)) ^ 0.25) * B8, 2))Try the example
Use the original values or change them to test your understanding.