## Brief

Liquid loading is a phenomenon when the gas phase does't provide sufficient transport energy to lift the liquids out of the well.

In 1969 Turner et al. published an empirical correlation defining the Liquid loading gas velocity[1].

## Math & Physics

The minimum gas velocity to remove the liquid equation:

$v_g = 1.593\ \sigma^{1/4}\ \frac{({\rho_L-\rho_g})^{1/4}}{\rho_g^{1/2}}$[2] (1)

Note that original paper[1] uses 20.4 constant which is good if the surface tension units are in 'lbf/ft'. Also 20.4 is calculated by adjusting the theoretical equation constant by 20 percent. For the reference PROSPER uses 2.04. PQplot uses 1.593.

The minimum gas rate to remove the liquid equation:

$q_g = 3.067\ \frac{P\ v_g\ A}{T\ z}$ (2)

## PQplot workflow

For every point in the VLP curve:
1. Critical gas velocity at the wellhead is calculated by eq (1)
2. Wellhead flowing gas velocity is calculated

$v_{g wellhead} = \frac{q_g \times 10^6}{86400 A}\ \frac{14.7}{P}\ \frac{T}{520}\ \frac{z}{1}$

$v_{g wellhead} < v_{g critical}$

4. Critical gas rate is the last VLP curve point at which liquid loading flag is 1.

## Discussion

The higher the gas rate the higher the gas velocity.

The lower the wellhead flowing pressure the higher the gas rate.

The bigger the tubing ID the higher the gas rate.

In case when the gas rate is limited by the Reservoir deliverability smaller tubing ID will increase the gas velocity.

## Example. Calculating critical gas velocity and rate

Example 3-2 from [2]

### Input data

$P = 400$, flowing wellhead pressure, psia
$T =120$, flowing wellhead temperature, °F
$\rho_L =67$, liquid density, lbm/ft3
$\sigma = 60$, water surface tension, dyne/cm
$SG_g = 0.6$, gas specific gravity
$z =0.9$, gas compressibility factor

Production string = 2-3/8 inch tubing with 1.995 in ID, A = .0217 ft2
Water is the produced liquid

### Solution

Calculate the gas density:

$\rho_g = \frac{28.967\ SG_g\ p}{z\ 10.732\ T_R} = \frac{28.967\ 0.6\ 400}{0.9\ 10.732\ (120+460)}=1.24$, lbm/ft2

Calculate the critical gas velocity form (1):

$v_g = 1.593\ \sigma^{1/4}\ \frac{({\rho_L-\rho_g})^{1/4}}{\rho_g^{1/2}} = 1.593\ 60^{1/4}\ \frac{({67-1.24})^{1/4}}{1.24^{1/2}}=11.3$, ft/sec

Calculate the critical gas rate form (2):

$q_g = 3.067\ \frac{P\ v_g\ A}{T\ z} = 3.067\ \frac{400\ 11.3\ 0.0217}{(120+460)\ 0.9}=0.578$, MMscfd

### Solution with PQplot

PQplot calculates the following critical gas rate:

$q_g = 0.56$, MMscfd

## Nomenclature

$A$ = flow area, ft^2
$P$ = flowing wellhead pressure, psia
$q_g$ = gas rate, MMscf/d
$\rho_g$ = gas density, lbm/ft3
$\rho_L$ = liquid density, lbm/ft3
$\sigma$ = surface tension, dyne/cm (ref values: 60 - water, 20 - condensate) [1]
$SG_g$ = gas specific gravity
$T$ = flowing wellhead temperature, °R
$v_g$ = gas velocity, ft/sec
$z$ = gas compressibility factor at flowing P & T, dimensionless