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Material Properties for Sat. Liquid and Sat. Vapor |
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September 17, 2012, 20:11 |
Material Properties for Sat. Liquid and Sat. Vapor
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Bradley J
Join Date: Jul 2012
Location: Cincinnati, OH
Posts: 12
Rep Power: 14 |
Hello everyone,
I am looking to see if my Material Properties are correct for my simulation. I am doing bubble growth due to boiling. I have patched a small hemisphere as vapor for the starting point with UDFs guiding the rest. I want my temperature of vapor and liquid to be 373.15 K. I have the wall at a higher temperature. This post is mainly about the material properties for saturated liquid and saturated vapor at atmospheric pressure. I have the two materials as h2o liquid and h2o vapor. I want to make sure that the changes I have made to the material properties are correct. (All material properties were founded from Fundamentals of Heat and Mass Transfer - 6th Edition Incropera - Appendix A.6 pg 949 - Properties of Saturated Water) This is what I put into Fluent (All properties are CONSTANT): I am unsure of Standard Enthalpy and Reference Temperature. The heat of vaporization that I am using 2257 kj/kg. Properties of Saturated Water at 373.15 K: Material: water-liquid h2o<l> Density: 957.8544 kg/m^3 Cp: 4217 j/(kg*K) Thermal Conductivity: 0.68 W/(m*K) Viscosity: 0.000279 kg/(m*s) Molecular Weight: 18.0152 kg/kgmol Standard State Enthalpy: 419040 j/kg*18.0152 kg/kgmol Reference Temperature: 373.15 K Properties of Saturated Vapor at 373.15 K: Material: water-vapor h2o Density: 0.5956 kg/m^3 Cp: 2029 j/(kg*K) Thermal Conductivity: 0.0248 W/(m*K) Viscosity: 0.000012 kg/(m*s) Molecular Weight: 18.0152 kg/kgmol Standard State Enthalpy: 2676100 j/kg*18.0152 kg/kgmol Reference Temperature: 373.15 K The change in Standard State Enthalpy does come out to the heat of vaporization. I have also changed the interaction surface temperature to 0.0589 N/m. I have also patched the whole domain to the saturation temperature of 373.15 K. Please let me know if this is a correct set up or if my properties are incorrect. Thank you so much for all your help. Bradley. |
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