1. **State the problem:** Calculate the Energy Fluence $E$ in HNS-1 when a 0.1 mm thick 304L flyer impacts it with an impact pressure of 22.5 GPa and shock velocity in the flyer is 5.36 km/s. Also determine if it will detonate. 2. **Given data:** - Thickness of flyer $d = 0.1$ mm $= 0.0001$ m - Impact pressure $p = 22.5$ GPa $= 22.5 \times 10^9$ Pa - Shock velocity in flyer $U = 5.36$ km/s $= 5360$ m/s - Density of 304L $\rho_0 = 7.903$ g/cm$^3 = 7903$ kg/m$^3$ - $c_0 = 4.57$ km/s $= 4570$ m/s - $s = 1.48$ 3. **Calculate time $t$ for shock wave to travel both directions in flyer:** The shock wave travels through the flyer thickness twice (forward and backward), so $$t = \frac{2d}{U} = \frac{2 \times 0.0001}{5360} = 3.731 \times 10^{-8} \text{ seconds}$$ 4. **Calculate particle velocity $u$ using pressure formula:** Given $$P = \rho_0 c_0 u + \rho_0 s u^2$$ Rearranged as quadratic in $u$: $$\rho_0 s u^2 + \rho_0 c_0 u - P = 0$$ Substitute values: $$7903 \times 1.48 u^2 + 7903 \times 4570 u - 22.5 \times 10^9 = 0$$ Simplify coefficients: $$11692.44 u^2 + 36120710 u - 22500000000 = 0$$ 5. **Solve quadratic for $u$:** Using quadratic formula: $$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=11692.44$, $b=36120710$, $c=-22500000000$ Calculate discriminant: $$\Delta = b^2 - 4ac = (36120710)^2 - 4 \times 11692.44 \times (-22500000000)$$ $$= 1.3057 \times 10^{15} + 1.0523 \times 10^{15} = 2.358 \times 10^{15}$$ Calculate $u$: $$u = \frac{-36120710 \pm \sqrt{2.358 \times 10^{15}}}{2 \times 11692.44}$$ $$= \frac{-36120710 \pm 48566200}{23384.88}$$ Positive root: $$u = \frac{12445490}{23384.88} = 532.1 \text{ m/s}$$ 6. **Calculate Energy Fluence $E$ using formula:** $$E = \frac{p^2 t}{\rho_0 U}$$ Substitute values: $$E = \frac{(22.5 \times 10^9)^2 \times 3.731 \times 10^{-8}}{7903 \times 5360}$$ Calculate numerator: $$ (22.5 \times 10^9)^2 = 5.0625 \times 10^{20}$$ $$ 5.0625 \times 10^{20} \times 3.731 \times 10^{-8} = 1.889 \times 10^{13}$$ Calculate denominator: $$7903 \times 5360 = 4.237 \times 10^{7}$$ Calculate $E$: $$E = \frac{1.889 \times 10^{13}}{4.237 \times 10^{7}} = 445,800 \text{ J/m}^2$$ 7. **Interpretation:** Typical detonation threshold energy fluence for HNS-1 is about 34 cal/cm$^2$. Convert $E$ to cal/cm$^2$: $$1 \text{ J/m}^2 = 0.0239 \text{ cal/cm}^2$$ $$E = 445,800 \times 0.0239 = 10,650 \text{ cal/cm}^2$$ Since $10,650$ cal/cm$^2$ $>>$ 34 cal/cm$^2$, the energy fluence is sufficient to detonate HNS-1. **Final answer:** $$\boxed{E = 4.46 \times 10^5 \text{ J/m}^2 \approx 10,650 \text{ cal/cm}^2, \text{HNS-1 will detonate}}$$