# Power of Shockwaves

My question is about shockwaves and their power when they are created/how do they lose their power?

Let's say that we have ground 0 with 10 grams of TATP on it. The detonation velocity of TATP is 5,300m/s. How can we calculate the pressure it will make and the shockwave's power when it is detonated. Also,how can we calculate how fast the shockwave will lose power?

• While this is fine as a theoretical question, note that we can't make any recommendations about safety here. In particular, you shouldn't rely on a result you get here to determine how close you can stand to an explosion. To make this point to you and especially to others who may read this question, I removed your last paragraph. (It's still accessible in the revision history of course.) – David Z Feb 24 '14 at 20:38
• @DavidZ , your edit was a good move. I've worked around organic peroxides in industry, and they ARE NOT to be trifled with! – David White Aug 3 '16 at 2:22

I know this is unrelated, but I'd recommend avoiding doing anything with TATP if possible. I manufactured it once and didn't get hurt, but there have been a fair number of people who have been maimed or killed while working with it.

The pressure of the front should decrease as $1/r$, as explained on the Wiki page on spherical sound sources.

However, that is irrelevant to your question of calculating safe distances; obviously shrapnel flux decreases as $1/r^2$, but over the relevant distances in question, shrapnel velocity does not significantly decrease. As a result, calculating a "safe" distance from an explosion in some sort of storage container based on shockwave mechanics alone is nonsense.

• I don't want to calculate a safe distance from an explosive which is inside a container.That would be a bomb.I want to calculate the power of an explosive which is not inside anything.Example: let's say that we have 5 grams of TATP.How do we calculate how much powerful de explosion will be?I mean the dangerous and the safe distance.It's something with the amount of gases the substance will produce or what? – user41229 Feb 24 '14 at 13:53
• That seems like a pointless thing to calculate; real experimental evidence and experience always trumps 'theoretical calculations' regarding complicated phenomena like explosive effects. For 5 g of an unconfined high-order explosive, stay about 10 m away, wear serious ear protection and stay behind a plexiglas shield (if you have access to one), and try it. Based on that, you should be able to approximate safe distances for other masses based on the $1/r^2$ loss in power. This is fundamentally an inexact science. – DumpsterDoofus Feb 24 '14 at 14:07
• So,there is no way yo calculate how much powerful or fast the shockwave will be when it's produced if we know the amount of explosive/amount of gases relased and their temperature? – user41229 Feb 24 '14 at 14:19
• I'm sure there is a way and that you can read about it in any reference book on civil or military engineering, I'm just saying that for practical purposes, it's not really worthwhile to spend the effort trying to mathematically model it. If you're in the military and you're blowing something up, you probably use a lookup table based on the centuries of experience people have accumulated while blowing things up. – DumpsterDoofus Feb 24 '14 at 14:26
• By the way, DumpsterDoofus, I tend to advise against making safety recommendations like "stand x meters away from an explosion." It's not against the rules, really, since readers are supposed to know we're not liable for the consequences of someone using information they find here, but it's nice to remove any doubt by not providing any safety-related advice at all. – David Z Feb 24 '14 at 20:41

# Background

What you are talking about is effectively a blast wave. The following is an excerpt from an answer I wrote at https://physics.stackexchange.com/a/242450/59023.

In the case of Taylor-Sedov-von Neumann like solutions [e.g., see pages 192-196 in Whitham, 1999], the relevant parameter is the position of the shock wave at $r = R\left( t \right)$, given by: $$R\left( t \right) = k \left( \frac{ E_{o} }{ \rho_{up} } \right)^{1/5} \ t^{2/5} \tag{1}$$ where $t$ is time from the initial release of energy, $E_{o}$, from a point source, $\rho_{up}$ is the ambient gas mass density, and $k$ is a dimensionless parameter used for scaling. These solutions are founded upon two assumptions given as follows:

1. the explosion resulted from a sudden release of energy $E$ from a point source and $E$ is the only dimensional parameter introduced by the explosion;
2. the disturbance is so strong that the ambient air/gas pressure and speed of sound can be neglected compared to those in the blast wave.

The second assumption implies the strong shock limit, namely that the downstream parameters are given by (in shock frame): \begin{align} U_{dn} & = \left( \frac{ 2 }{ \gamma + 1 } \right) U_{up} \tag{2a} \\ \rho_{dn} & = \left( \frac{ \gamma + 1 }{ \gamma - 1 } \right) \rho_{up} \tag{2b} \\ P_{dn} & = \left( \frac{ 2 }{ \gamma + 1 } \right) \rho_{up} \ U_{up}^{2} \tag{2c} \end{align} where subscript $up$($dn$) corresponds to upstream(downstream) averages, $\gamma$ is the ratio of specific heats, $U_{j}$ is bulk flow speed in region $j$ (i.e., $up$ or $dn$), and $P_{j}$ is the pressure (here just using dynamic or ram pressure) in region $j$.

It can be seen from Equation 1 the only parameter of the ambient air/gas relevant is the density, $\rho_{up}$. Note that the shock speed, $U_{up} \rightarrow U_{shn}$, will then be given by $dR/dt$ (i.e., the time derivative of Equation 1), or: \begin{align} U_{shn}\left( t \right) & = \frac{ 2 \ k }{ 5 } \left( \frac{ E_{o} }{ \rho_{up} } \right)^{1/5} \ t^{-3/5} \tag{3a} \\ & = \frac{ 2 \ k^{5/2} }{ 5 } \sqrt{ \frac{ E_{o} }{ \rho_{up} } } \ R^{-3/2} \tag{3b} \end{align} Since the upstream density and pressure are assumed constant, then the upstream sound speed, $C_{s,up}$, must be constant as well.

...how do they lose their power?

The idea behind a blast wave is that a thin region of compressed gas (or maybe a solid from a piece of an exploding device), which is the piston in this scenario, moves out against the ambient gas faster than the local speed of sound. Because the piston produces a shock wave, it, by definition, produces entropy. Entropy is generated due to the irreversible transformation of the bulk flow kinetic energy density across the shock (in the shock frame the upstream is "fast and cool" while the downstream is "slow and hot"). This is effectively like having a drag and/or friction force on the shock wave causing it to slow down over time, as can be seen from Equation 3b above.

How can we calculate the pressure it will make...

See Equation 2c above.

...and the shockwave's power when it is detonated...

Power is energy per unit time or $dE/dt$, so figure out how much energy is released when combusting 10 grams of TATP (I found an energy release estimate of ~3.6 kJ $cm^{-3}$ at http://www.wydawnictwa.ipo.waw.pl/cejem/cejem-3-4-2009/price.pdf, though I am not sure what the access restrictions might be).

If we take a naive guess that the energy as a function of time goes as: $$E\left( t \right) = \frac{ 1 }{ 2 } \ \rho_{dn} \ U_{shn}^{2}\left( t \right) \tag{4}$$ then we can substitute the estimates from Equations 2b and 3a for $\rho_{dn}$ and $U_{shn}\left( t \right)$, respectively, to find: $$E\left( t \right) = \frac{ 2 \ \left( \gamma + 1 \right) \ k^{2} }{ 25 \ \left( \gamma - 1 \right) } \ \rho_{up}^{3/5} \ E_{o}^{2/5} \ t^{-6/5} \tag{5}$$ Since everything in Equation 5 is constant with respect to time, we can see that $$\frac{ dE }{ dt } \sim -\frac{ 6 }{ 5 \ t } \ E\left( t \right) \tag{6}$$ The initial power would be estimated by determining $E_{o}$ and $\Delta t$, or the amount of time necessary for energy $E_{o}$ to be released. Then one would just take the ratio to get the power, or $P \sim E_{o}/\Delta t$.

Also,how can we calculate how fast the shockwave will lose power?

See Equation 6.

# References

• Whitham, G.B. (1999), Linear and Nonlinear Waves, New York, NY: John Wiley & Sons, Inc.; ISBN:0-471-35942-4.