Purging with Nitrogen ·

Table of Contents

Before any maintenance or hot work (like welding) can begin on equipment containing hydrogen ($\text{H}_2$), the line must be proven safe. This is achieved by purging with an inert gas, typically nitrogen ($\text{N}_2$). This process involves replacing a dangerous gas with a safe one. The core question is: how much $\text{N}_2$ is “enough”?

The Hazard: The Flammability Envelope #

The primary danger is the combination of $\text{H}_2$ (fuel) and $\text{O}_2$ (oxidizer, from air) with an ignition source (welding).

We must stay outside the flammability envelope. The key safety parameters are:

$\text{LEL}_{\text{H}_2}$: The Lower Explosive Limit. This is the minimum fuel concentration to ignite. $\text{LEL}_{\text{H}_2} \approx 4.0\%$ (or $40,000 \ \text{ppm}$)
$\text{UEL}_{\text{H}_2}$: The Upper Explosive Limit. $\text{UEL}_{\text{H}_2} \approx 75.0\%$
$\text{LOC}_{\text{O}_2}$: The Limiting Oxygen Concentration. This is the minimum $\text{O}_2$ needed for a $\text{H}_2$ fire.

$$\text{LOC}_{\text{H}_2} \approx 5.0\% \ \text{O}_2$$

For hot work, the line must be opened to air (which is $\approx 21%\ \text{O}_2$). Therefore, we must ensure the $\text{H}_2$ concentration,

$C_{\text{H}_2}$, is well below its LEL.

A standard industrial target is $C_{\text{target}} < 10%\ \text{LEL}$.

$$ C_{\text{target}} < 0.10 \times \text{LEL}_{\text{H}_2} = 0.10 \times 4.0\% = 0.4\% \ \text{H}_2 $$

Our goal is to purge the line from $C_i = 100%\ \text{H}_2$ (initial state) to $C_f < 0.4%\ \text{H}_2$ (final state).

The Mathematics of Purging #

We must calculate the required purge volume, $V_p$, as a multiple of the line’s internal volume, $V_L$. This ratio is the number of volume changes, $N$.

$$ N = \frac{V_p}{V_L} $$

We can model this purge using the conservative perfect mixing (CSTR) model. The final concentration $C_f$ is an exponential decay of the initial concentration $C_i$:

$$ C_f = C_i \cdot e^{-N} $$

To find the required number of volumes $N$, we rearrange the formula: $$ \frac{C_f}{C_i} = e^{-N} $$ $$ N = -\ln\left(\frac{C_f}{C_i}\right) = \ln\left(\frac{C_i}{C_f}\right) $$

Using our target values:

$C_i = 100\%$
$C_f = 0.4\%$

$$ N = \ln\left(\frac{100}{0.4}\right) = \ln(250) \approx 5.52 $$

This means a minimum of $5.52$ line volumes of $\text{N}_2$ are required to hit the target. Standard practice applies a safety factor, often rounding up to $N=7$ or even $N=10$ volumes.

Estimation Example #

Estimate the $\text{N}_2$ volume for a typical pipeline.

Pipe Diameter: $D = 8 \ \text{in} \approx 0.2032 \ \text{m}$
Pipe Length: $L = 150 \ \text{m}$

1. Find Line Volume ($V_L$): The pipe’s cross-sectional area $A$ is: $$ A = \frac{\pi D^2}{4} = \frac{\pi \times (0.2032 \ \text{m})^2}{4} \approx 0.0324 \ \text{m}^2 $$ The total line volume $V_L$ is: $$ V_L = A \times L = 0.0324 \ \text{m}^2 \times 150 \ \text{m} \approx 4.86 \ \text{m}^3 $$

2. Find Purge Volume ($V_p$): Using a safe $N=10$ volume changes: $$ V_p = V_L \times N = 4.86 \ \text{m}^3 \times 10 = 48.6 \ \text{m}^3 $$

This $48.6 \ \text{m}^3$ is the volume of $\text{N}_2$

required at the line’s operating pressure, $P_{op}$. If $P_{op} = 5 \ \text{barg}$ ($\approx 6 \ \text{bar abs}$), the equivalent volume at standard pressure ($P_{std} \approx 1 \ \text{bar abs}$) is found using the Ideal Gas Law ($P_1V_1 = P_2V_2$):

$$ V_{std} = \frac{V_p \times P_{op}}{P_{std}} = \frac{48.6 \ \text{m}^3 \times 6 \ \text{bar}}{1 \ \text{bar}} = 291.6 \ \text{m}^3 $$

This is a significant volume of nitrogen. Always confirm the final concentration ($C_f < 0.4\%$) using a calibrated $\text{H}_2$ analyzer at the vent before stopping the purge.

Dilution Factor #

The dilution factor (DF) can be calculated as:

$$ \text{DF} = \frac{\text{Volume of N}_2}{\text{Volume of H}_2 + \text{Volume of N}_2} $$

Example Calculation #

Suppose a pipeline has a volume of 1000 m³ and contains 50 m³ of hydrogen. To ensure the hydrogen concentration is below the LEL, we need to add a sufficient volume of nitrogen.

Initial Concentration of H₂:

$$ \text{Initial Concentration of H}_2 = \frac{50 \ \text{m}^3}{1000 \ \text{m}^3} \times 100% = 5% $$

Required Volume of N₂:

To dilute the hydrogen concentration below the LEL of 4%, we need to add enough nitrogen to reduce the hydrogen concentration. Let’s denote the required volume of nitrogen as $V_{\text{N}_2}$.

$$\text{Final Concentration of H}_2 = $$

$$\frac{50 \ \text{m}^3}{1000 \ \text{m}^3 + V_{\text{N}_2}} \times 100\% < 4\%$$

Solving for $V_{\text{N}_2}$ :

$$ \frac{50}{1000 + V_{\text{N}_2}} < 0.04 $$

$$ 50 < 0.04 \times (1000 + V_{\text{N}_2}) $$

$$ 50 < 40 + 0.04 V_{\text{N}_2} $$

$$ 10 < 0.04 V_{\text{N}_2} $$

$$ V_{\text{N}_2} > \frac{10}{0.04} $$

$$ V_{\text{N}_2} > 250 \ \text{m}^3 $$

Therefore, at least 250 m³ of nitrogen is required to dilute the hydrogen concentration below the LEL.