Hagedorn and Brown correlation

Brief

Hagedorn and Brown is an empirical two-phase flow correlation published in 1965.

It doesn't distinguish between the flow regimes.

The heart of the Hagedorn and Brown method is a correlation for the liquid holdup :$H_L$.

Math & Physics

Following the law of conservation of energy the basic steady state flow equation is:

$144 \frac{\Delta p}{\Delta h} = \bar \rho_m + \frac{f q_L^2 M^2}{2.9652 \times 10^{11} D^5 \bar \rho_m} + \bar \rho_m \frac{\Delta{(\frac{v_m^2}{2g_c}})}{\Delta h}$

where

$\bar \rho_m = \bar \rho_L H_L + \bar \rho_g (1 - H_L)$

Colebrook–White equation for the Darcy's friction factor:

$\frac{1}{\sqrt{f}}= -2 \log \left( \frac { \varepsilon} {3.7 D} + \frac {2.51} {\mathrm{Re} \sqrt{f}} \right)$

Reynolds two phase number:

$Re = 2.2 \times 10^{-2} \frac {q_L M}{D \mu_L^{H_L} \mu_g^{(1-H_L)}}$

Workflow

$M =SG_o\ 350.52\ \frac{1}{1+WOR}+SG_w\ 350.52\ \frac{WOR}{1+WOR}+SG_g\ 0.0764\ GLR$

$\bar \rho_L= \frac{62.4\ SG_o + \frac{Rs\ 0.0764\ SG_g}{5.614}}{B_o} \frac{1}{1+WOR} + 62.4\ SG_w\ \frac{WOR}{1 + WOR}$

$\bar \rho_g = \frac{28.967\ SG_g\ p}{z\ 10.732\ T_R}$

$\mu_L = \mu_o \frac{1}{1 + WOR} + \mu_w \frac{WOR}{1 + WOR}$

$\sigma = \sigma_o \frac{1}{1 + WOR} + \sigma_w \frac{WOR}{1 + WOR}$

$N_L = 0.15726\ \mu_L \sqrt[4]{\frac{1}{\rho_L \sigma_L^3}}$

$CN_L = 0.061\ N_L^3 - 0.0929\ N_L^2 + 0.0505\ N_L + 0.0019$

$v_{SL}$

$v_{SG}$

$N_{LV} = 1.938\ v_{SL}\ \sqrt[4]{\frac{\rho_L}{\sigma}}$

$N_{GV} = 1.938\ v_{SG}\ \sqrt[4]{\frac{\rho_L}{\sigma}}$

$N_{D} = 120.872\ D \sqrt{\frac{\rho_L}{\sigma}}$

$H = \frac{N_{LV}}{N_{GV}^{0.575}}\ \left ( \frac{p}{14.7} \right )^{0.1} \frac{CN_L}{N_D}$

$\frac{H_L}{\psi} = \sqrt{\frac{0.0047+1123.32 H + 729489.64H^2}{1+1097.1566 H + 722153.97 H^2}}$

$B = \frac{N_{GV} N_{LV}^{0.38}}{N_{D}^{2.14}}$

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \psi = \begin{cases} 27170 B^3 - 317.52 B^2 + 0.5472 B + 0.9999, & \mbox{if }B\mbox{ 0.025} \\ -533.33 B^2 + 58.524 B + 0.1171, & \mbox{if }B\mbox{ 0.025} 2.5714 B +1.5962, & \mbox{if }B\mbox{ 0.055} \end{cases}

$H_L$

Nomenclature

$\psi = \begin{cases} 27170 B^3 - 317.52 B^2 + 0.5472 B + 0.9999, & \mbox{if }B\mbox{ = 0.025} \\ -533.33 B^2 + 58.524 B + 0.1171, & \mbox{if }B\mbox{ 0.025} \\ 2.5714 B +1.5962, & \mbox{if }B\mbox{ 0.055} \end{cases}$