Boiling & Condensation Sizing

🌀 Visual Phase-Change Engine

📝 Configuration

🔥 Pool Boiling ❄️ Film Condensation

Cylinder Diameter ($D$) [m]:

Cylinder Length [m]:

Temperatures

🔴 Surface Temp ($T_s$) [°C]:

💧 Saturation Temp ($T_{sat}$) [°C]:

Surface Properties

Surface-Fluid Preset:

Surface Emissivity ($\varepsilon$):

💧 Phase Properties Presets

Fluid Setup Preset:

Liquid properties

Density ($\rho_l$) [kg/m³]:

Viscosity ($\mu_l$) [Pa·s]:

Thermal Cond. ($k_l$) [W/m·K]:

Prandtl Number ($Pr_l$):

Specific Heat ($C_{p,l}$) [J/kg·K]:

Vapor properties

Density ($\rho_v$) [kg/m³]:

Viscosity ($\mu_v$) [Pa·s]:

Thermal Cond. ($k_v$) [W/m·K]:

Prandtl Number ($Pr_v$):

Specific Heat ($C_{p,v}$) [J/kg·K]:

Phase Change Constants

Latent Heat ($h_{fg}$) [J/kg]:

Surface Tension ($\sigma$) [N/m]:

📊 Results & Visualization

Configure inputs and click Sizing phase change to view results.

📘 Calculation Methodology

Mathematical Model & Theory

Phase-change heat transfer includes boiling and condensation. Nucleate pool boiling is modeled using Rohsenow's correlation for heat flux:

$$q_s'' = \mu_l h_{fg} \left[\frac{g(\rho_l - \rho_v)}{\sigma}\right]^{1/2} \left[\frac{C_{p,l} (T_s - T_{sat})}{C_{sf} h_{fg} Pr_l^{1.7}}\right]^3$$

Film condensation on a vertical surface uses Nusselt's analysis:

$$h_{cond} = 0.943 \left[\frac{g \rho_l (\rho_l - \rho_v) k_l^3 h_{fg}'}{\mu_l (T_{sat} - T_s) L}\right]^{1/4}$$

Worked Engineering Example

Problem Statement:
Water is boiled at 1 atm ($T_{sat} = 100$°C) on a clean copper surface ($C_{sf} = 0.013$, surface area $A = 0.05$ m²) maintained at 115°C. Water liquid properties: $\rho_l = 958$ kg/m³, $\mu_l = 2.8 \times 10^{-4}$ Pa·s, $C_{p,l} = 4217$ J/kg·K, $Pr_l = 1.75$, $h_{fg} = 2.257 \times 10^6$ J/kg, $\sigma = 0.059$ N/m. Find the heat transfer rate.

Step-by-step Solution:
1. Calculate excess temperature: $\Delta T_e = T_s - T_{sat} = 115 - 100 = 15$°C.
2. Apply Rohsenow's equation:
$$q_s'' = (2.8\times 10^{-4}) (2.257\times 10^6) \left[\frac{9.81 \times (958 - 0.6)}{0.059}\right]^{1/2} \left[\frac{4217 \times 15}{0.013 \times 2.257\times 10^6 \times 1.75^{1.7}}\right]^3$$ $$q_s'' = 631.96 \times [159,484]^{1/2} \times [0.000868]^3 = 631.96 \times 399.35 \times 6.54 \times 10^{-10} = 0.165 \text{ MW/m}^2 = 165,000 \text{ W/m}^2$$ 3. Calculate heat transfer rate:
$$Q = q_s'' \times A = 165,000 \times 0.05 = 8250 \text{ W}$$
Final Result:
Boiling heat transfer rate is 8.25 kW.