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Engineering LibreTexts

3.3: The Equivalent Liquid Model

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  • Assuming that the pressure losses in a pipe are proportional to the kinetic energy of the eddies and the kinetic energy of the eddies is proportional to the mixture density and the line speed and assuming that there are no losses due to sliding friction or collisions, the pressure losses can be determined by:

    \[\ \Delta \mathrm{p}_{\mathrm{m}}=\lambda_{\mathrm{l}} \cdot \frac{\Delta \mathrm{L}}{\mathrm{D}_{\mathrm{p}}} \cdot \frac{\mathrm{1}}{\mathrm{2}} \cdot \rho_{\mathrm{m}} \cdot \mathrm{v}_{\mathrm{l} \mathrm{s}}^{\mathrm{2}}\]

    The hydraulic gradient im (for mixture) is now:

    \[\ \mathrm{i}_{\mathrm{m}}=\frac{\Delta \mathrm{p}_{\mathrm{m}}}{\rho_{\mathrm{l}} \cdot \mathrm{g} \cdot \Delta \mathrm{L}}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{l}}} \cdot \frac{\lambda_{\mathrm{l}} \cdot \mathrm{v}_{\mathrm{l}_{\mathrm{s}}}^{\mathrm{2}}}{\mathrm{2} \cdot \mathrm{g} \cdot \mathrm{D}_{\mathrm{p}}}=\frac{\rho_{\mathrm{m}}}{\rho_{\mathrm{l}}} \cdot \mathrm{i}_{\mathrm{l}}\]

    The above assumptions are valid as long as the particles are small enough to be considered part of the eddies. So for larger particles this may not be true anymore.

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