Effective Interfacial Area of Packing using Onda's Method Formula

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Effective Interfacial Area represents the total interfacial area per unit volume within a multiphase system. Check FAQs

a W = a \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{σ c}{σ L})}^{0.75}) \cdot {(\frac{L W}{a \cdot μ L})}^{0.1}) \cdot {(\frac{{(L W)}^{2} \cdot a}{{(ρ L)}^{2} \cdot [g]})}^{- 0.05}) \cdot {(\frac{{L W}^{2}}{ρ L \cdot a \cdot σ L})}^{0.2})

a_W - Effective Interfacial Area?a - Interfacial Area per Volume?σ_c - Critical Surface Tension?σ_L - Liquid Surface Tension?L_W - Liquid Mass Flux?μ_L - Fluid Viscosity in Packed Column?ρ_L - Liquid Density?[g] - Gravitational acceleration on Earth?

Effective Interfacial Area of Packing using Onda's Method Example

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Here is how the Effective Interfacial Area of Packing using Onda's Method equation looks like with Values.

Here is how the Effective Interfacial Area of Packing using Onda's Method equation looks like with Units.

Here is how the Effective Interfacial Area of Packing using Onda's Method equation looks like.

0.1758 = 0.1788 \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061}{0.0712})}^{0.75}) \cdot {(\frac{1.4785}{0.1788 \cdot 1.005})}^{0.1}) \cdot {(\frac{{(1.4785)}^{2} \cdot 0.1788}{{(995)}^{2} \cdot 9.8066})}^{- 0.05}) \cdot {(\frac{{1.4785}^{2}}{995 \cdot 0.1788 \cdot 0.0712})}^{0.2})

0.1758m² = 0.1788m² \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061N/m}{0.0712N/m})}^{0.75}) \cdot {(\frac{1.4785kg/s/m²}{0.1788m² \cdot 1.005Pa*s})}^{0.1}) \cdot {(\frac{{(1.4785kg/s/m²)}^{2} \cdot 0.1788m²}{{(995kg/m³)}^{2} \cdot 9.8066m/s²})}^{- 0.05}) \cdot {(\frac{{1.4785kg/s/m²}^{2}}{995kg/m³ \cdot 0.1788m² \cdot 0.0712N/m})}^{0.2})

0.1758 = 0.1788 \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061}{0.0712})}^{0.75}) \cdot {(\frac{1.4785}{0.1788 \cdot 1.005})}^{0.1}) \cdot {(\frac{{(1.4785)}^{2} \cdot 0.1788}{{(995)}^{2} \cdot 9.8066})}^{- 0.05}) \cdot {(\frac{{1.4785}^{2}}{995 \cdot 0.1788 \cdot 0.0712})}^{0.2})

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Effective Interfacial Area of Packing using Onda's Method Solution

Follow our step by step solution on how to calculate Effective Interfacial Area of Packing using Onda's Method?

FIRST Step Consider the formula

a W = a \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{σ c}{σ L})}^{0.75}) \cdot {(\frac{L W}{a \cdot μ L})}^{0.1}) \cdot {(\frac{{(L W)}^{2} \cdot a}{{(ρ L)}^{2} \cdot [g]})}^{- 0.05}) \cdot {(\frac{{L W}^{2}}{ρ L \cdot a \cdot σ L})}^{0.2})

Next Step Substitute values of Variables

∴

a W = 0.1788m² \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061N/m}{0.0712N/m})}^{0.75}) \cdot {(\frac{1.4785kg/s/m²}{0.1788m² \cdot 1.005Pa*s})}^{0.1}) \cdot {(\frac{{(1.4785kg/s/m²)}^{2} \cdot 0.1788m²}{{(995kg/m³)}^{2} \cdot [g]})}^{- 0.05}) \cdot {(\frac{{1.4785kg/s/m²}^{2}}{995kg/m³ \cdot 0.1788m² \cdot 0.0712N/m})}^{0.2})

Next Step Substitute values of Constants

∴

a W = 0.1788m² \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061N/m}{0.0712N/m})}^{0.75}) \cdot {(\frac{1.4785kg/s/m²}{0.1788m² \cdot 1.005Pa*s})}^{0.1}) \cdot {(\frac{{(1.4785kg/s/m²)}^{2} \cdot 0.1788m²}{{(995kg/m³)}^{2} \cdot 9.8066m/s²})}^{- 0.05}) \cdot {(\frac{{1.4785kg/s/m²}^{2}}{995kg/m³ \cdot 0.1788m² \cdot 0.0712N/m})}^{0.2})

Next Step Prepare to Evaluate

∴

a W = 0.1788 \cdot (1 - \exp ((- 1.45 \cdot ({(\frac{0.061}{0.0712})}^{0.75}) \cdot {(\frac{1.4785}{0.1788 \cdot 1.005})}^{0.1}) \cdot {(\frac{{(1.4785)}^{2} \cdot 0.1788}{{(995)}^{2} \cdot 9.8066})}^{- 0.05}) \cdot {(\frac{{1.4785}^{2}}{995 \cdot 0.1788 \cdot 0.0712})}^{0.2})

Next Step Evaluate

∴

a W = 0.175804925321227m²

LAST Step Rounding Answer

∴

a W = 0.1758m²

Effective Interfacial Area of Packing using Onda's Method Formula Elements

Variables

Constants

Functions

Effective Interfacial Area

Effective Interfacial Area represents the total interfacial area per unit volume within a multiphase system.

Symbol: a_W

Measurement: AreaUnit: m²