INDUSTRIAL FIRE SAFETY

INDUSTRIAL FIRE SAFETY

DEPARTMENTS

LAST MONTH we presented several formulas for use when dealing with so-called perfect gases. In this issue, some additional formulas are presented, and practical examples worked out. In order to solve for the unknown quantity it may be necessary in some instances to transpose or rearrange the equation. This may make it necessary to brush up on our high school algebra.

It will be noted that many of the formulas are approximate only; but in most practical problems an approximation will suffice. For instance, calculations show the density or unit weight of ethane to be 0.0779 pounds per cubic foot, whereas the handbook tables show 0.079. Such slight errors are not important in most of our problems.

Let us consider a closed vessel containing a volatile liquid with air or other gas in the space above it. It has been standing long enough for evaporation to saturate the air and come to an equilibrium condition. Then as more liquid evaporates, a like amount of vapor will condense on the walls of the container. Consequently, the pressure in the vessel will be the sum of the original atmospheric air pressure (14.7 psia) and the vapor pressure of the liquid. This condition holds true only if the vessel is sealed immediately after the liquid has been placed in it, and before evaporation has produced appreciable vapor which pushes some of the air out through the filling hole.

As an example, an empty steel drum is partially filled with summer-grade gasoline having a vapor pressure of say 9.5 psia, and immediately sealed. When equilibrium is reached, the vapor pressure of the gasoline (9.5 psia) has been added to the atmospheric pressure (14.7 psi) to produce an internal pressure of 24.2 psia, or 9.5 psig. This will explain why there may be a whistle when the drum plug is later removed, and the internal pressure released. However, if the drum was not sealed immediately after the gasoline was put into it, the vapor produced would have pushed some of the air out of the vent, and less pressure would have been built up.

In some chemical plants it may be the practice to use compressed air to deliver liquids from the storage tank, through the piping to the process area. (A safety valve must always be installed on such a tank.) In our hypothetical case we will assume that benzine is in the tank. When the benzine has evaporated and saturated the atmospheric air in the tank, a too-rich vapor-air mixture will be created, so there is little or no explosive hazard.

But when the compressed air is injected into the tank we will have a real problem, for this will dilute the too-rich mixture so that it may be within the explosive range. The explosive limits for benzine are 1.4 percent and 8.0 percent respectively. The vapor pressure of benzine at 70° is about 1.5 psia. Refer to the following:

Formula 4: Total pressure of vapor and compressed air in a vessel at either explo-

sive limit = 100 x vapor pressure in psia. / explosive limit in percent

Total pressure at lower explosive limit = 100 x 1.5 / 1.4

= 107 psia at the LEL Total pressure at upper explosive limit = 100 x 1.5 / 8.0

= 18.7 psia at the UEL

These absolute pressures are converted into 92.3 psig and 4.0 psig respectively. Therefore, it is safe to use compressed air to build up an internal pressure in the tank provided the pressure is a safe amount below 4.0 psig or a safe amount above 92.3 psig; but between these pressures, look out!

Sometimes we have occasion to learn the relative vapor pressure of a flammable liquid such as acetone. (This is the density or weight of a gas relative to that of air, which is 1.000.) Acetone has the chemical formula CH3COCH3 and its molecular weight is 58.

Formula 5: Relative vapor density

(when air is 1.0) = molecular weight / 29

Relative vapor density of acetone

= 58 / 29

= 2.0 (tabular value 2.0)

The true vapor density, or unit weight, is given in pounds per cubic foot. To obtain this figure from the relative vapor density, use:

Formula 6: Vapor density, or unit weight = relative vapor density x 0.075 pound per cubic foot. As an example we will use acetone. We know that the relative vapor density is 2.0, therefore:

Vapor density = 2.0 x 0.075

= 0.150 pound per cubic foot

The volume of vapor produced by 1 gallon of flammable liquid may be obtained by either of the following:

Formula 7: Volume of vapor per gallon

at 70° F = 111 x specific gravity of liquid / relative vapor density

Let us consider Toluol which has a specific gravity of 0.866 and a relative vapor density of 3.14.

Volume of vapor = 111 x 0.866 / 3.14

= 30.6 cubic feet per gallon (tabular value 31.1)

Formula 8: Volume of vapor per gallon

at 70°F = 3225 x specific gravity of liquid / molecular weight

As an example, methanol CH3OH has a specific gravity of 0.796 and a molecular weight of 32.

Volume of vapor = 3225 x 0.796 / 32

= 80.2 cubic feet per gallon (tabular value 79.0)

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