Worked Example: Gas Well Deliverability & Inflow Performance Relationship (IPR) Modeling

The Inflow Performance Relationship (IPR) defines the mathematical link between flowing bottomhole pressure ($p_{wf}$) and the volumetric production rate ($Q$) delivered from the porous rock reservoir into the wellbore. Accurately modeling this relationship is fundamental to production engineering: it dictates artificial lift design, tubing size selection, surface compression schedules, and field life forecasting.

In natural gas reservoirs, high fluid velocities near the wellbore cause inertial pressure losses (the Forchheimer effect or non-Darcy turbulence). The classic standard for capturing this non-linear behavior is the Rawlins and Schellhardt Backpressure Equation (US Bureau of Mines Monograph 7).

1. Wellhead & Reservoir Test Data

A production well drilled into a deep sandstone gas reservoir underwent a 4-point modified isochronal backpressure test: * Average Static Reservoir Pressure ($\bar{p}_R$): $280.0\text{ barg}$ ($281.01\text{ bar(a)} = 4,075.8\text{ psia}$) * Stabilized Test Flow Rate ($Q_{test}$): $250,000\text{ Sm}^3/\text{day}$ ($8.829\text{ MMSCFD}$) * Flowing Bottomhole Pressure ($p_{wf}$): $240.0\text{ barg}$ ($241.01\text{ bar(a)} = 3,495.3\text{ psia}$) * Deliverability Exponent ($n$): $0.82$ (fitted from multi-rate flow periods) * Gas Specific Gravity ($G$): $0.65$ * Reservoir Temperature: $95.0^\circ\text{C}$ ($203.0^\circ\text{F}$)

2. Mathematical Foundation: Rawlins-Schellhardt Formulation

The empirical backpressure equation states:

$$Q = C \left( \bar{p}_R^2 - p_{wf}^2 \right)^n$$

Where: * $Q$ is gas flow rate (in $\text{MMSCFD}$ or $\text{Sm}^3/\text{day}$) * $\bar{p}R$ is static average reservoir pressure ($\text{psia}$ or $\text{bar}$) * $p$) * $C$ is the stabilized deliverability performance coefficient * $n$ is the backpressure turbulence exponent ($0.5 \le n \le 1.0$) * $n = 1.0$: Completely laminar Darcy flow * $n = 0.5$: Completely turbulent (non-Darcy) flow dominated by inertial resistance}$ is flowing bottomhole pressure ($\text{psia}$ or $\text{bar

3. Step 1: Calculate Pressure-Squared Drawdown ($\Delta p^2$)

Convert pressures to field units ($\text{psia}$): $$\bar{p}_R = 4,075.8\text{ psia} \implies \bar{p}_R^2 = (4075.8)^2 = 1.6612 \times 10^7\text{ psia}^2$$

$$p_{wf} = 3,495.3\text{ psia} \implies p_{wf}^2 = (3495.3)^2 = 1.2217 \times 10^7\text{ psia}^2$$

The drawdown pressure-squared driving force is: $$\Delta(p^2) = \bar{p}_R^2 - p_{wf}^2 = 1.6612 \times 10^7 - 1.2217 \times 10^7 = \mathbf{4.3951 \times 10^6\text{ psia}^2}$$

4. Step 2: Solve for Performance Coefficient ($C$)

Rearranging the backpressure formula:

$$C = \frac{Q_{test}}{\left( \bar{p}_R^2 - p_{wf}^2 \right)^n}$$

Substitute the test rate ($Q_{test} = 8.829\text{ MMSCFD}$) and $n = 0.82$:

$$\left( \Delta(p^2) \right)^n = \left( 4.3951 \times 10^6 \right)^{0.82} = 279.92$$

$$C = \frac{8.829}{279.92} = \mathbf{0.031541\text{ MMSCFD/(psia}^2)^{0.82}}$$

In metric units (using $\text{bar}$ and $\text{Sm}^3/\text{day}$): $$C_{metric} = \frac{250,000}{(281.01^2 - 241.01^2)^{0.82}} = \frac{250,000}{(78,966 - 58,085)^{0.82}} = \frac{250,000}{20,881^{0.82}} = \mathbf{72.06\text{ Sm}^3/\text{day}\cdot(\text{bar}^2)^{0.82}}$$

5. Step 3: Determine Absolute Open Flow Potential (AOFP)

The Absolute Open Flow Potential (AOFP) is the theoretical maximum rate the reservoir could deliver if the flowing bottomhole pressure were reduced to absolute atmospheric pressure ($p_{wf} = 0\text{ psia}$):

$$Q_{AOFP} = C \left( \bar{p}_R^2 - 0 \right)^n = C \left( \bar{p}_R^2 \right)^n$$

$$\left( \bar{p}_R^2 \right)^{0.82} = \left( 1.6612 \times 10^7 \right)^{0.82} = 833.17$$

$$Q_{AOFP} = 0.031541 \times 833.17 = \mathbf{26.279\text{ MMSCFD}}$$

In standard metric units: $$Q_{AOFP} = 26.279 \times 28,316.85 = \mathbf{744,132\text{ Sm}^3/\text{day}}$$

6. Step 4: Construct Full Inflow Performance Curve

Using our fitted parameters ($C = 0.031541, n = 0.82$), we calculate flow rates across bottomhole operating pressures:

7. Engineering Discussion: Deliverability Diagnostics

graph TD
    A[AOFP = 26.28 MMSCFD] --> B[Typical Operating Window: 40% - 65% of AOFP]
    B --> C[Recommended Production Rate: 10.5 - 17.0 MMSCFD]
    C --> D[Pwf = 180 to 230 barg]
    D --> E[Prevents Sand Production & Water Coning]

1. Operating Range vs AOFP:
2. Wells are rarely operated at $>70\%$ of AOFP. Operating at extreme drawdowns triggers wellbore shear failure, sand mobilization, premature water coning, and severe near-wellbore skin damage.

3. For this well, a sustainable production target is $10.5$ to $17.0\text{ MMSCFD}$ ($p_{wf} \approx 180 - 230\text{ barg}$).

4. Significance of $n = 0.82$:

5. An exponent of $0.82$ indicates moderate turbulence in the completion gravel pack or perforation tunnels. If stimulation (hydraulic fracturing or acidizing) is performed, non-Darcy pressure loss is minimized, shifting $n \to 0.95$ and significantly expanding deliverability.