San José State University

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Thayer Watkins
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 The Expansion of the Fireball of an Explosion

The fireball of an explosion is the shock wave that emanates from it. It usually takes a spherical shape if the explosion takes place in the air or a hemispherical form if the explosion takes place on the ground. This is an exercise in dimensional analysis to obtain an equation for the radius of the fireball of an explosion as a function of time, the initial energy, the ambient density of the air and its ambient pressure. The dimensions of these variables in terms of mass, length and time are as follows:

• Radius of fireball R: [L]
• Energy E: [ML²/T²]
• Time after the explosion t: [T]
• Ambient air density ρ: = [M/L³]
• Ambient air pressure p: [M/LT²]

The formula for R is of the form

#### R = AEαtβργpδ

where A is a constant.

The dimensions of the variables require that

#### [L] = [ML²/T²]α[T]β[M/L³]γ[M/LT²]δ

This means that the exponents have to satisfy the following three equations to balance the dimensions for mass, length and time on the two sides of the equation:

#### α + γ + δ = 0 2α − 3γ − δ = 1 −2α + β − 2δ = 0

This is a set of three equations in four unknowns. A solution may be obtained by setting the value of one variable to a specified amount and solving for the remaining three variables. Let δ=0. Then the second and third equations may be added to give

#### β −3γ = 1

Likewise the first equation can be multiplied by 2 and added to the third equation to give

#### β + 2γ = 0

These two equations have the solution α = 1/5, β = 2/5 and γ = −1/5.

Thus the formula for the radius of the fireball of an explosion is

## An Application

The above analysis was worked out by the British scientist Goeffrey Taylor in 1941. The United States exploded atomic bombs in 1945, but kept the technical details concerning the bomb, such as the total energy of the explosion, secret. In 1947 a movie of the explosion was released. The total energy of the explosion was still kept secret. Goeffrey Taylor measured the expansion of the fireball as function of time and fitted the following equation to the data. His equation was

#### R = Bt2/5or, in logarithmic form log(R) = log(B) + (2/5)log(t)

where B=A(E/ρ)1/5

From the parameter log(B) of the fitted equation he was able to determine the value of E, the energy of the explosion.

## A More Detailed Analysis

Dimensional Analysis is based on a theorem developed by E. Buckingham, an American engineer. The theorem is usually called the Buckingham π Theorem. It effectively states that there exist dimensionless quantities in a problem such that one dimensionless quantity is a function of the other dimensionless quantities. In the above problem one dimensionless quantity, call it π0, is

#### R[Et²/ρ]−1/5

Another dimensionless variable can be found by solving the equations

#### α + γ + δ = 0 2α − 3γ − δ = 0 −2α + β − 2δ = 0

Again this is three equations in four unknowns. One variable can be specified and the rest solved for. For this case let δ=1. The equations to be solved are then

#### α + γ = −1 2α − 3γ = 2 −2α + β − = −2

The solution is α=−2/5, β=6/5, γ=−3/5 and, of course, δ=1.

This dimensionless variable, call it π1, is

#### π1 = E−2/5t6/5ρ−3/5p which can be expressed as π1 = p[t6/(E2ρ3)]1/5

According to the Buckingham π Theorem

#### π0 = f(π1) or, equivalently R[Et²/ρ]−1/5 = f(π1) and hence R = [Et²/ρ]1/5f(π1)

Source:

G.W. Bluman and S. Kumei, Symmetries and Differential Equations, Spring-Verlag, 1989.