Please provide a response to the following hypothetical scenario: An engineer asks for EHS assistance in the planning of a pneumatic proof test. The test article is $1,000$ cubic feet and will be pressed with gaseous nitrogen to $100$ psia.

In your answer, please provide the distance of the personnel keep out zone that you recommend.

For U$.$S$.$ Customary units using air or nitrogen as the

test medium $(k=1\mathrm{.}4),$ this equation becomes

$E=360P_{at}V[1-(\frac{P_a}{P_{at}}{)}^{0\mathrm{.}286}]$

and

$TNT=\frac{E}{1488617}$

See also paras. $6\mathrm{.}2($e$)$ and $6\mathrm{.}2($f$)$ of Article $5\mathrm{.}1.$

The minimum distance between all personnel and the

equipment being tested shall be the greater of

$($a$)$ the following:

$(1)R=30m$ for $E135500000J$

$(2)R=60m$ for $R=100$ftE stored energy $as$ calculated $byeq.(II-1)$

$or(II-2)$

$R=$ actual distance from equipment

$R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}=$ scaled consequence factor; value for

$eq.(III-1)$ shall $be20\frac{m}{k}{g}^{\frac{1}{3}}(50f\frac{t}{l}{b}^{\frac{1}{3}})$

$or$ greater

$TNT=$ energy measured $inTNT,kg(lb),$ deter$-$

mined from $eq.(II-3)or(II-5)E>271000000J(200,000,000ft-lb)R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}{R}_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}6\frac{m}{k}{g}^{\frac{1}{3}}(15f\frac{t}{l}{b}^{\frac{1}{3}})100,000,000$

$(b)$ the following equation:

$R=R_{\text{s}\text{c}\text{a}\text{l}\text{e}\text{d}}(TNT{)}^{\frac{1}{3}}$

where

$E=$ stored energy $as$ calculated $byeq.(II-1)$

$or(II-2)$

$R=$ actual distance from equipment

$R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}=$ scaled consequence factor; value for

$eq.(III-1)$ shall $be20\frac{m}{k}{g}^{\frac{1}{3}}(50f\frac{t}{l}{b}^{\frac{1}{3}})$

$or$ greater

$TNT=$ energy measured $inTNT,kg(lb),$ deter$-$

mined from $eq.(II-3)or(II-5)$

Table III$-1$

For systems where $E>271000000J(200,000,000ft-lb),$

the required distance shall $be$ calculated $byeq.(III-1).$

$If$ the minimum calculated distance cannot $be$

obtained, $an$ alternative value for $R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}$ may $be$ chosen

based $on$ Table III$-1$ for use $ineq.(III-1).$ See also

para. $6\mathrm{.}2(g).$

For example, $to$ prevent lung damage, the distance $a$

person $is$ from the equipment should result $inan{R}_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}$

value $of$ more than $6\frac{m}{k}{g}^{\frac{1}{3}}(15f\frac{t}{l}{b}^{\frac{1}{3}}).$ Note the

structural damage that can occur, which shall $be$

considered$\mathrm{.}135500000J$

$(3)R=100ft$ for

$(4)R=200ft$ for $100,000,000$

$(b)$ the following equation:

$R=R_{\text{s}\text{c}\text{a}\text{l}\text{e}\text{d}}(TNT{)}^{\frac{1}{3}}$

where

$E=$ stored energy $as$ calculated $byeq.(II-1)$

$or(II-2)$

$R=$ actual distance from equipment

$R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}=$ scaled consequence factor; value for

$eq.(III-1)$ shall $be20\frac{m}{k}{g}^{\frac{1}{3}}(50f\frac{t}{l}{b}^{\frac{1}{3}})$

$or$ greater

$TNT=$ energy measured $inTNT,kg(lb),$ deter$-$

mined from $eq.(II-3)or(II-5)$

Table III$-1$

For systems where $E>271000000J(200,000,000ft-lb),$

the required distance shall $be$ calculated $byeq.(III-1).$

$If$ the minimum calculated distance cannot $be$

obtained, $an$ alternative value for $R_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}$ may $be$ chosen

based $on$ Table III$-1$ for use $ineq.(III-1).$ See also

para. $6\mathrm{.}2(g).$

For example, $to$ prevent lung damage, the distance $a$

person $is$ from the equipment should result $inan{R}_{\text{S}\text{C}\text{A}\text{L}\text{E}\text{D}}$

value $of$ more than $6\frac{m}{k}{g}^{\frac{1}{3}}(15f\frac{t}{l}{b}^{\frac{1}{3}}).$ Note the

structural damage that can occur, which shall $be$

considered.