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?
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.
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:
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.
