Purging process vessels

# Purging (blanketing) process vessels and piping from oxygen

## Introduction

To reduce oxygen concentration in process vessels and piping is a common task in those industries where flammable chemicals are processed. Before starting up a plant after a major shutdown adequate procedures to reduce the concentration of oxygen shall be adopted, to prevent the potential formation of flammable mixtures. This article presents methods and equations to help defining a successful procedure for purge and inert (blanket) process vessels and piping using an inert gas such as nitrogen. The method is implemented in the Plant Purging solver at www.chesolver.com

We will talk about two methods of purging:
• Method 1: Dilution and venting
• Method 2: Sweep-through

For an extensive treatment of the subject we suggest to read reference [1].

## Defining the maximum acceptable oxygen content

Before starting any calculation it is necessary to fix the maximum acceptable concentration of oxygen. The Lower Oxidant (Oxygen) Concentration (LOC) is the reference parameter and is defined here below:

• LOC: The concentration of oxidant in a fuel-oxidant-diluent mixture below which a deflagration cannot occur under specified conditions.

The LOC depends upon the temperature, pressure, and fuel concentration as well as the type of diluent. It is expressed in volume percent.

Values of LOC are provided in the table here below, where for each category a representative substance with a conservative value of LOC has been selected. The calculation of the LOC for mixtures of substances is treated in reference [1].

Representative values of Lower Oxidant Concentration (LOC)
Gas/Vapour Category Substance LOC (%vol)
Paraffins Ethane 9.5
Cycloparaffins Cyclopropane 10.2
Olefins Ethylene 8.5
Aromatics Diethylbenzene 8.5
Alcohols Methanol 8.5
Ethers Propylene oxide 6.6
Inorganic compounds Carbon disulfide 4.3
Inorganic compounds Carbon monoxide (in air) 5.1
Inorganic compounds Hydrogen (in air) 4.6
Fuels Natural gas 10.5
Fuels Motor Gasolines 10.5
Fuels Kerosene 8.5

## Method 1: purging by dilution and venting

The oxygen contained in the volume to be purged is removed using a series of dilution and venting cycles. After each cycle, a stepwise reduction of concentration is obtained. The procedure is repeated until the concentration of oxygen is less than the maximum allowed, which shall be less than the LOC of the substances that will be processed in the plant. The cycle can be divided in two different phases:

• Dilution: the gas existing in the volume to be purged is diluted with an inert gas, typically nitrogen
• Venting / Extraction: the diluted mixture containing the inert gas and oxygen is vented or extracted from the volume to be purged.

To to properly design and prepare the purging procedure one has to calculate

• the minimum number of cycles required to lower oxygen concentration below the prescribed limit
• the amount of inert gas required

The input data for the problem will be

• an estimation of the volume to be purged
• the initial and final concentration of oxygen
• the operating pressure interval

## Solving equations

At the beginning of the process the volume is at pressure $$P_1$$ and contains an amount $$n$$ of oxygen moles. During the dilution phase, we change the pressure to $$P_2$$ by adding an inert gas. The concentration of oxygen reduces from $$x_1$$ to $$x_2$$, but the number of moles remains $$n$$. We may say that

$P_1\, x_1 \frac{V}{RT} = n = P_2\, x_2 \frac{V}{RT}$

In writing the above equation we used the ideal gas law $$n = p\, \frac{V}{RT}$$ and the partial pressure $$p = P\, x$$. The volume to be purged is constant and we assume to operate at a constant temperature, so the contribution $$\frac{V}{RT}$$ cancels out and we have that the concentration of oxygen after the pressurization (dilution) is

$x_2 = \frac{P_1}{P_2} x_1$

The next step is to vent the system to restore the pressure at the original value $$P_1$$. The concentration of oxygen will not be affected. If we now repeat the cycle at same values of pressure, we would get a lower concentration $$x_3$$

$x_3 = \frac{P_1}{P_2} x_2 = \frac{P_1}{P_2} \frac{P_1}{P_2} x_1 = \left( \frac{P_1}{P_2} \right)^2 x_1$

This leads to the general expression for the mole fraction of oxygen after $$k$$ cycles where we pressurise at pressure $$P_h$$ and depressurise at pressure $$P_l$$

$$x_k = \left( \frac{P_l}{P_h} \right)^k x_0 \label{eq:mole_frac_k}$$

Reworking equation \eqref{eq:mole_frac_k} we have the number of cycles required to reduce the mole fraction of oxygen from $$x_0$$ to a prescribed value $$x$$

$$k = \frac{ \ln\left( \frac{x}{x_0} \right)} {\ln\left( \frac{P_l}{P_h} \right) } \label{eq:k}$$

The consumption of inert gas is always the same at each cycle, because it depends only on the working pressures and temperature. The total number of moles can be easily calculated from the ideal gas law

$$N = k\, \left( P_h - P_l \right) \frac{V}{RT} \label{eq:N}$$

We have more or less implicitly assumed so far that system is pressurised and vented. If vacuum can be developed, for example with a vacuum pump, the procedure can save some inert gas. In this case, the venting step corresponds to applying vacuum to the plant and to extract the oxygen while the dilution step corresponds to releasing vacuum with an inert gas.

## Example 1 - Pressure purging

During a major shutdown a process vessel of $$20\,m^3$$ volume has been opened for maintenance and is left full of air at atmospheric pressure ( $$1\, bar(a)$$ ). Before introducing the process fluids it is decided to lower the oxygen content from $$21%$$ down to $$2%$$. For this purpose, the vessel will be pressurised at 2 bar(g) using nitrogen an then vented to restore atmospheric pressure. The minimum number of pressurization/depressurization cycles is calculated from equation \eqref{eq:k}

$k = \frac{ \ln\left( \frac{21}{2} \right)} {\ln\left( \frac{1}{3} \right) } = 2.14 \rightarrow 3$

so 3 cycles will be used. The final mole fraction of oxygen is calculated from equation \eqref{eq:mole_frac_k}

$x = \left( \frac{1}{3} \right)^3 21 = 0.8\% < 2 \%$

The total amount of nitrogen that will be required for the operation is calculated from equation \eqref{eq:N}, assuming an operating temperature of $$25 ^oC$$

$N = 3\times \left( 3 - 1 \right)\, 10^5 \times \frac{20}{8.314\times 298} = 4.8\,kmol = 109\, Nm^3$

## Example 2 - Vacuum purging

The vessel of example 1 is purged using a vacuum pump that can develop a vacuum pressure of 80 mbar. We have

$k = \frac{ \ln\left( \frac{21}{2} \right)} {\ln\left( \frac{0.08}{1} \right) } = 0.93 \rightarrow 1$
$N = 1\times \left( 1 - 0.08 \right)\, 10^5 \times \frac{20}{8.314\times 298} = 0.74\,kmol = 17\, Nm^3$

the conclusion is that vacuum purging requires less cycles and a lower amount of purge gas.

## Method 2: sweep-through purging

The sweep-through method consist of feeding inert gas into the system from one opening and to vent the resulting mixture from another opening, in a continuous once-through operation. The concentration of oxygen is progressively reduced in time until it reaches the desired level. This method is mostly suitable for scenarios where dead volumes can be minimised and the flow of gas can reach reasonably well all parts of the plant.

## Solving equations

The starting point is a mass balance on the oxygen. At a certain instant of time, we inject a molar flow rate $$\dot{N}$$ of inert gas and correspondly vent the same amount of mixture with a mole fraction $$x$$. The differential change in the moles of oxygen in the system is

$dn = -\dot{N}x(t)dt = -\dot{N} \frac{n(t)}{N} dt$

this expression can be integrated from the start to the end of the process

$\int_{n_0}^n \frac{dn}{n} = \int_{0}^t -\frac{\dot{N}}{N}dt$

to obtain

$\ln \left( \frac{n}{n_0} \right) = -\frac{\dot{N}}{N}t$

or equivalently

$\ln \left( \frac{x}{x_0} \right) = -\frac{\dot{N}}{N}t$

using the ideal gas law applied to the volume to be purged we get the final result

$\dot{N}t = N \ln \left( \frac{x_0}{x_1} \right) = \frac{PV}{RT} \ln \left( \frac{x_0}{x_1} \right)$

which represents the amount of purge gas to be used for a prescribed change in oxygen concentration. The result obtained so far assumes perfect mixing of the inert gas and oxygen, but this is never the case. To account for this inefficiency the mixing coefficient $$K$$ is introduced, so the final equation becomes

$$\dot{N}t = \frac{1}{K}\,\frac{PV}{RT} \ln \left( \frac{x_0}{x_1} \right) \label{eq:Nt}$$

Most authorities recommend a K-value of not greater than 0.25. The table here below shows some K-value from different ventilation scenarios.

Recommended values of K for different ventilation arrangements
Method of air supply Single Exhaust Opening Multiple Exhaust Opening
No Positive Supply
Infiltration through cracks 0.2 0.3
Infiltration through open doors or windows 0.2 0.4
Forced Air Supply
Grilles and registers 0.3 0.5
Diffusers 0.5 0.7
Perforated ceiling 0.8 0.9

## Example 3 - Sweep through

We will estimate the amount of purge gas required to purge the same vessel of example 1, operating at atmospheric pressure ( $$1\,bar(a)$$ ) and $$25^oC$$. Using equation \eqref{eq:Nt} with $$K=0.2$$ we have

$\dot{N}t = \frac{1}{0.2}\,\frac{1\times 10^5 \times 20}{8.314\times 298}\, \ln \left( \frac{21}{2} \right) = 9.5\, kmol = 213\, Nm^3$

## Conclusions

In this paper we addressed the problem of the elimination of oxygen from process vessels and piping before introducing flammable material, to avoid the potential formation of combustible mixtures. For this purpose an inert gas like nitrogen can be used. We developed the equations for two operating methods: dilution and venting, sweep-through. Known the LOC of the flammable material a suitable maximum oxygen concentration can be fixed. With the proposed method it is then possible to calculate the number of dilution cycles and the amount of purge gas (for dilution and venting) or the total amount of purge gas (for sweep through), to reduce the concentration of oxygen below the fixed limit.