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Unformatted text preview: Problems on Shocks 1. Assume you are given equations for the jump conditions across a hydrodynamic shock, namely U 1 U = γ- 1 γ + 1 + 2 (1 + γ ) M 2 , (1) P 1 ρ U 2 = 2 γ + 1- γ- 1 γ (1 + γ ) M 2 , (2) and ρ 1 ρ = U U 1 (3) where ρ , U , and P are the mass denisty, shock normal velocity, and pressure, respec- tively, and the subscripts 0 and 1 refer to upstream from and downstream from the shockwave. What are the asymptotic shock values (values for very large Mach num- bers) for the density ratio ( ρ 1 /ρ ), velocity ratio ( U 1 /U ), and normalized pressure ( P 1 / ( ρ U 2 ))? Assume γ = 5 / 3. Solution: ρ 1 /ρ → 4, U 1 /U → 1 / 4, and P 1 / ( ρ U 2 ) → 3 / 4. 2. Again assume you are given equations (1), (2), and (3) from problem 1, and assume that γ = 5 / 3. If the density jump across the shock is 2, what is the corresponding Mach number? Solution: If ρ 1 /ρ = 2, then U 1 /U = 0 . 5. Solving equation (1) for M yields M = √ 3. 3. How do the Rankine Hugoniot (RH) jump conditions depend on the details of the shock structure (i.e., the detailed physics on how the plasma is slowed and heated? Solution: They don’t. The beauty of the RH conditions are that as long as some mechanism exists to maintain the shock, the jump conditions will be valid. Thus, one can study the macroscopic consequences of the shock without ever having to look into the detailed microphysics. 4. How does the Earth’s bow shock differ from the simple hydrodynamical shock? Solution: It differs in at least 3 ways. First, it has finite curvature (it is not a simple one-dimensional object). This allows the shock to stand at a fixed position upstream from the Earth. Second, there are no Coulomb collisions, so the means by which the shock is created and maintained is very different from shock waves in the atmosphere or...
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- Spring '07
- Fluid Dynamics, Mach number, Shock wave, Scramjet