• DECEMBER 2019

One Dimensional Waves

Application of Shock-Expansion Theory – Lesson 4 Intro Into 1D Wave Motion

• So far, we have examined shock and expansion waves generated by flows over objects. These waves are stationary in the object’s reference frame, and steady-state analysis can be applied to derive their relations. • In this lesson we will discuss propagation of transient shocks and expansion waves in one dimension. • We will also briefly outline propagation of small amplitude acoustic waves. • Finally, we will combine these elements of the transient wave theory in the classical Sod shock tube problem.

2 Moving Normal Shock Waves

Stationary shock • The governing equations of a moving normal shock can be readily derived from those obtained for the stationary motion shock using the following variables: 푢 푢 Gas motion downstream of 2 1 ‐ 푊 – velocity of the gas ahead of the shock relative to the upstream of shock, shock shock 2 1 ‐ 푊 − 푢푝 – velocity of the gas behind the shock relative to the shock

Shock moving with velocity 푊 휌1푊 = 휌2(푊 − 푢푝)

2 Motion induced Stagnant gas 푝 + 휌 푊2 = 푝 + 휌 푊 − 푢 푢 > 0 푊 1 1 2 2 푝 by moving shock 푝 ahead of 2 shock ℎ2 + 1Τ2 푣 = ℎ1 + 1Τ2 푊 − 푢푝 2 1

• The of the shock is given by:

2 푀1 = 푊Τ푎1 where the is: 푎1 = (푑푝Τ푑휌)1

3 Moving Normal Shock Waves

• The “jump” relations developed for stationary normal shocks can be easily rewritten for the moving shock using the transformation equation from the previous slide. For application to practical problems, it is more convenient to use the pressure ratio 푝2Τ푝1 as the basic independent parameter. For an ideal gas, the moving shock relations become:

훾 − 1 훾 + 1 푝 1/2 2 shock velocity 푊 = 푀1푎1 = 푎1 + 2훾 2훾 푝1 푝 + 푝 Hugoniot equation. This is identical in form to the stationary shock 푒 − 푒 = 1 2 (푣 − 푣 ) 2 1 2 1 2 case since it relates thermodynamic variables independent of whether the shock is moving or stationary. • Relations for the density and temperature ratios can be obtained from the Hugoniot equation as:

훾 + 1 푝 훾 + 1 푝 1 + 2 + 2 휌2 훾 − 1 푝 푢1 푇 푝 훾 − 1 푝 = 1 = 2 = 2 1 훾 + 1 푝 훾 + 1 푝 휌1 + 2 푢2 푇1 푝1 1 + 2 훾 − 1 푝1 훾 − 1 푝1

4 Moving Normal Shock Waves (cont.) • The gas velocity behind the wave is:

1/2 2훾 This is the velocity of moving gas behind the shock “experienced” by an 푎 푝 훾 + 1 푢 = 1 2 − 1 observer in the laboratory frame of reference. Its magnitude can best be 푝 푝 훾 − 1 훾 푝1 2 + estimated in terms of the local Mach number: 푝1 훾 + 1

1/2 1/2 1/2 푢푝 푢푝 푎 푢푝 푇 2훾 훾 + 1 푝 = 1 = 1 1 + 2 푢푝 1 푝2 훾 + 1 훾 − 1 푝 푎2 푎1 푎2 푎1 푇2 ⟹ = − 1 1 ⟶ 2 푝 훾 − 1 2 푎2 훾 푝1 2 훾 + 1 푝2 푝2 푝 + 훾 + 1 + 푝2 훾(훾 − 1) 1 훾 − 1 푝1 푝1 → ∞ 푝1

⟹ In the limit of a very strong shock, 푢푝/푎2 → 1.89 for air. Thus gas flow can be accelerated to supersonic speeds by strong shocks.

• Finally, we should note that, unlike the case of stationary shocks, the total enthalpy ℎ0 is no longer constant for moving shocks: 퐷ℎ0 휕푝 휕푝 휌 = , ≠ 0 ⟹ ℎ0 is not constant. 퐷푡 휕푡 휕푡

5 Acoustic Equations

• Acoustic (sound) waves are small-amplitude disturbances propagating in (most notably in air). Given the assumption of smallness of acoustic fluctuations, governing questions of gas dynamics can be reduced to a set of linear acoustic equations. By applying a linearization technique and thermodynamic considerations, the equations reduce to the acoustic wave equation:

휕2휌 휕2푝 = 푎2 ∇휌 ⟺ = 푎2 ∇p 휕푡2 ∞ 휕푡2 ∞

Here 푎∞ is the constant speed of sound. Note the equation can be written in the same form for density or pressure.

6 Acoustic Equations (cont.) • For one-dimensional waves, the wave equation has a solution of the form:

휌(푥, 푡) = 퐹 푥 − 푎∞푡 + 퐺(푥 + 푎∞푡)

i.e., the solution is a superposition of left- and right-traveling simple waves propagating at the sound speed along characteristic lines with the slope 푑푥Τ푑푡 = ±푎∞in the 푥 − 푡 plane.

• The same form solution also holds for the velocity: characteristics 푑푥 푑푥 푢(푥, 푡) = 푓 푥 − 푎∞푡 + 푔(푥 + 푎∞푡) = 푎 = −푎 푑푡 ∞ 푑푡 ∞ 푡 푡 • The relation between velocity, pressure and density is:

푡 푢 = ± 푎∞휌Τ휌∞ = ± 푝Τ(휌∞푎∞) 2

푡 • The wave shape does not change with time in the acoustic 1 approximation. 0 0 푥 푥 퐹 푥 퐺 푥

7 Isentropic Nonlinear Waves of Finite Amplitude

• Previous analysis of small-amplitude acoustic waves is valid only if the wave amplitude is small. If such an assumption cannot be made, the wave can no longer be described by the linearized equations. Such waves are called finite waves. These waves have the following common characteristics: ✓ Fluctuation of density, temperature, velocity, etc., are large ✓ Any local part of the wave propagates at the local velocity 푢 + 푎 relative to the laboratory frame ✓ The wave shape changes with time ✓ Full non-linear governing equations must be used to describe wave motion ✓ The finite wave approach is suitable for all real waves, linear or non-linear

• The complete non-linear governing equations need to be solved to describe finite waves. Such a mathematical solution was obtained by Riemann and Earnshaw almost 160 years ago, and we will summarize its main steps here.

• Our derivation will be limited to 1D for clarity, but 3D generalization is also possible.

8 Isentropic Nonlinear Waves of Finite Amplitude (cont.) • From our discussion of thermodynamics, the density is a function of pressure and entropy, thus:

휕휌 휕휌 휕휌 퐷휌 1 퐷푝 1 휕푝 휕푝 휕푢 푑휌 = 푑푝 + 푑푠 = 푑푝 ⟹ = ⟹ + 푢 + 휌 = 0 휕푝 휕푠 휕푝 퐷푡 푎2 퐷푡 푎2 휕푡 휕푥 휕푥 푠 푑푝 푠

1/푎2 = 0 for isentropic flows continuity equation • By adding and subtracting the new form of the continuity equation from the 1D momentum equation: 휕푢 휕푢 1 휕푝 휕푝 휕푢 휕푢 1 휕푝 휕푝 + (푢 + 푎) + + (푢 + 푎) = 0 + (푢 − 푎) − + (푢 − 푎) = 0 휕푡 휕푥 휌푎 휕푡 휕푥 휕푡 휕푥 휌푎 휕푡 휕푥

• Considering these equations along two specific lines (characteristics) in 푡 퐶 characteristic 푥 − 푡 − 퐶+ characteristic the plane given by: 푑푥 = 푢 + 푎 푑푡 푑푥 = 푢 − 푎 푑푡 푑푥 푑푥 = 푢 − 푎 = 푢 + 푎 we obtain compatibility equations valid along characteristics: 푑푡 푑푡 푡 푑푝 푑푝 1 푑푢 + = 0 푑푢 − = 0 휌푎 휌푎 푥 along 푑푥Τ푑푡 = 푢 + 푎 along 푑푥Τ푑푡 = 푢 − 푎 푥1

9 Isentropic Nonlinear Waves of Finite Amplitude (cont.)

• Integrating the two compatibility equations along respective characteristic lines gives rise to the Riemann invariants, which can be expressed for a calorically perfect gas as:

푑푝 2푎 푑푝 2푎 퐽 = 푢 + න = 푢 + = const 퐽 = 푢 − න = 푢 − = const

+ 휌푎 훾 − 1 − 휌푎 훾 − 1 ⟹ along 퐶+ characteristic along 퐶− characteristic

훾 − 1 1 푎 = (퐽 − 퐽 ) 푢 = (퐽 + 퐽 ) 4 + − 2 + −

• If 퐽+ and 퐽− are known at a point in the 푥 − 푡 plane, the velocity and speed of sound can be determined from the above relations.

10 Centered Expansion Wave • Let’s consider a tube filled with gas. A piston initially at rest is 4 3 suddenly withdrawn to the right with the constant velocity 푢3. This 푢3 will produce an expansion wave propagating to the left. 푢3 piston • As soon as the wave starts propagating, it begins to flatten. The 퐶− 푡 velocity has a linear profile across the expansion wave and 퐶− pressure has a corresponding profile. 퐶− Piston path 3 푥 = −푢3푡 Expansion • From the analysis of Riemann invariants 퐽+ and 퐽−, and by considering: front 푥 = 푎4푡

1. 퐽+ and 퐽− are constant along respective characteristics 퐶+ and 퐶− 4 2. Regions 3 and 4 are regions with constant flow properties 푥 푢 it can be shown 퐽+ = const through the expansion wave ⟹ 푢3 푢4 2푎 2푎4 푎 훾 − 1 푢 푥 푢 + = 푢4 + = const ⟹ 훾 − 1 훾 − 1 = 1 − 푝 푝 푎4 2 푎4 4 푝3 any local point in = 0 Relation between 푢 and a in a 푥 expansion simple expansion wave.

11 Centered Expansion Wave (cont.)

• Relations between the local gas velocity and temperature, pressure and density are obtained from the isentropic relations:

2 푇 훾 − 1 푢 푝 훾 − 1 푢 2훾/(훾−1) 휌 훾 − 1 푢 2/(훾−1) = 1 − = 1 − = 1 − 푇4 2 푎4 푝4 2 푎4 휌4 2 푎4

• Noting that the expansion wave shown in the sketch is the left-running wave expanding along the fan of 퐶− characteristics, and considering characteristics are straight lines through the origin: 푑푥 = 푢 − 푎 푑푡 ⟹ 푥 = 푢 − 푎 푑푡 ⟹ from the relation between 푢 and a ⟹

2 푥 Variation 푢 as a function of 푥 and 푡 valid for the expansion region between 푢 = 푎4 + 훾 + 1 푡 its front and tail, −푎4 ≤ 푥Τ푡 ≤ 푢3 − 푎3

• Similar relations can be obtained for the right-running expansion wave.

12 Sod Shock Tube Problem

• The shock tube problem was first studied by Gary A. Sod in 1978. driver section driven section • Let us consider a one-dimensional tube with a diaphragm initially separating regions of high pressure (left) and low pressure (right). 4 1

The tube is closed on both ends. 푝4, 푇4, 푎4, 훾4, 푅4 푝1, 푇1, 푎1, 훾1, 푅1 diaphragm • The gases on either side of the diaphragm can have different 0 푥 molecular weights and thermodynamic properties. 푝4 푝 • All flow field properties depend on distance, 푥, and time, 푡. 푝1 • The basic parameter of the shock tube is the diaphragm pressure ratio, 푝4Τ푝1. 0 푥

• At 푡 = 0, the diaphragm is suddenly removed which initiates transient dynamics of gases in the tube.

13 Sod Shock Tube Problem (cont.)

4 3 푢푝 2 푊 1

Contact surface separating driver and Expansion propagating to Normal shock propagating driven gas and moving at the velocity of the left to the right gas behind the shock

• At 푡 > 0 the initial pressure jump splits into a propagating into the driven section and an expansion wave propagating into the driver section with the velocity 푊.

• Propagating normal shock increases the pressure behind it and induces a gas motion with velocity 푢푝.

• The interface between gases in the driver and driven section is called the contact surface, which can be thought of as a slip surface. The gases remain separated by this surface and the entropy changes across it are discontinuous.

• The pressure and velocity, however, do not change across the contact surface, 푝3 = 푝2 and 푢3 = 푢2 = 푢푝.

• The expansion wave propagates to the left, smoothly decreasing the pressure in regions 4.

14 Sod Shock Tube Problem (cont.) • The flow field inside the shock tube after the diaphragm is removed can be completely described by the diaphragm pressure ratio, 푝4Τ푝1. All the relations follow from the previous analysis of moving shocks and expansions.

2훾4/(훾4−1) (훾4−1)/2훾4 푝3 훾4 − 1 푢3 ⟹ 2푎4 푝2 = 1 − 푢3 = 1 − 푝4 2 푎4 푝3 = 푝2 훾4 − 1 푝4 Pressure drop across the expansion

1/2 −2훾4/(훾4−1) 2훾 푝4 푝2 (훾4 − 1)(푎1Τ푎4)(푝2Τ푝1 − 1) 푎1 푝2 훾 + 1 ⟹ = 1 − 푢 = 푢 = − 1 푝1 푝1 Τ 푝 2 푝 훾 − 1 푢3 = 푢2 2훾1 2훾1 + (훾1 + 1))(푝2 푝1 − 1 훾 푝1 2 + 푝1 훾 + 1 Implicit relation for the shock strength 푝2Τ푝1 as a Gas velocity induced by the shock function of the diaphragm pressure 푝4Τ푝1

푝3 푝3 푝1 푝2Τ푝1 • The expansion strength is obtained as: = = 푝4 푝1 푝4 푝4Τ푝1 • Once the shock strength is known, all other flow quantities can be found from the shock and expansion relations.

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Sod Shock Tube Problem (cont.) Shock

Pressure Distribution along the shock tube

16 Summary

• In this lesson we discussed unsteady wave motion in one dimension.

• We covered moving normal shock, acoustic and finite nonlinear waves, and centered expansion wave .

• Finally, we combined this theory to describe the flow field inside a shock tube.

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