ME295/346 Gas Dynamics

Assignment #3 - The Shock Tube - Pt. 1 - SOLUTION

Spring 2005 - D.L. Hitt

Due Date: Tuesday, February 22

1. The Shock Speed vs. Pressure Ratio. In class we showed that the shock velocity depends entirelyaccording to the pressure ratio (P2/P1) according to

cs = a1

sγ + 1

2γ

µp2p1− 1¶+ 1 (1)

where a1 is the speed of sound upstream of the shock.

(a) Construct a plot of the �shock Mach number� Ms = cs/a1as a function of the pressure ratio for(P2/P1)∈ [1,∞]. Verify that this plot asymptotes to the behavior discussed in class.Solution. A plot of the behavior appears below.

2 4 6 8 10 12 14 16 18 200.5

1

1.5

2

2.5

3

3.5

4

4.5

P2/P1

M1

exactasymptotic

(b) Using the normal shock relations, determine an expression for the Mach number Mp on the backside of the shock as a function of the pressure ratio.

Solution. The important thing to realize in developing an expression for Mp is that the normalshock relations previously derived were for a stationary shock and therefore all velocities werealready in the laboratory reference frame. In the present case we having a moving shockwave. While we can still use the �standing shock� relations to obtain Mp as a function ofthe pressure ratio, we must recognize that this Mach number is relative to a reference framemoving at the shock speed and not the laboratory frame. To establish what we will need to

ME295/346 - S05 Assignment #3 - Solution

know, we write down the deÞnition of the induced Mach number relative to the lab frame instate �2�

Mp =upa2

where up is the induced velocity (and thus cs−up is the velocity relative to the moving shock).Next, we re-write this as

upa2

=

µupa1

¶µa1a2

¶=

µupa1

¶rT1T2

where we make use of the fact from ideal gas behavior that the sonic speed goes as the squareroot of the static temperature. Thus we will need an expression for the static temperatureratio across the shock. Let�s look at up/a1:

upa1=upcs

csa1=upcs

sγ + 1

2γ

µp2p1− 1¶+ 1

This follows from our previous derivation for the shock Mach number. The ratio up/cs is bestexamined in terms of mass conservation:

ρ1u1 = ρ2u2

ρ1cs = ρ2 (cs − up)

orupcs= 1− ρ1

ρ2

Thus, the density ratio also is needed across the shock. In short, we have

Mp =

µ1− ρ1

ρ2

¶sγ + 1

2γ

µp2p1− 1¶+ 1

rT1T2J

where the temperature and density ratios need to be expressed in terms of the imposedpressure ratio. Actually, we only need the temperature ratio since, for an ideal gas,

ρ1ρ2=

µp2p1

¶µT1T2

¶and so

Mp =

·1−

µp2p1

¶µT1T2

¶¸sγ + 1

2γ

µp2p1− 1¶+ 1

rT1T2J (2)

So we need an expression for the temperature ratio only as a function of the imposed pressureratio. The most direct approach to this is to invoke the Hugoniot relation previously discussedin class. Since the Hugoniot relation is a purely thermodynamic description of the state changeacross a shock it is independent of the reference frame. The Hugoniot relation states that

e2 − e1 = 1

2(p1 + p2)(v1 − v2)

ME295/346 - S05 Assignment #3 - Solution

For an ideal, calorically perfect gas we know that e = CvT and v = RT/p. Substituting thisinto the Hugoniot relation we have

Cv (T2 − T1) =1

2(p1 + p2)

µRT1p1

− RT2p2

¶CvT1

µT2T1− 1¶

=1

2p2

µ1 +

p1p2

¶µRT1p2

¶µp2p1− T2T1

¶CvR

µT2T1− 1¶

=1

2

µ1 +

p1p2

¶µp2p1− T2T1

¶Recalling that Cv/R = (γ − 1)−1,

2

γ − 1µT2T1− 1¶=

µ1 +

p1p2

¶µp2p1− T2T1

¶With a bit of algebraic manipulation, this equation may be solved for the temperature ratiowith the Þnal result

T2T1=

µp2p1

¶" γ+1γ−1 +

p2p1

1 + γ+1γ−1

p2p1

#J (3)

Finally, we substitute Eq. (3) into Eq. (2) to obtain, after some more algebraic manipulation,1

Mp =1

γ

µp2p1− 1¶vuut 2γ

γ+1p2p1+ γ−1

γ+1

vuuuut 1 + γ+1γ−1

³p2p1

´γ+1γ−1

³p2p1

´+³p2p1

´2 J (4)

A plot of this functional behavior is given below:

0 10 20 30 40 50 60 70 80 90 1000

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Pressure Ratio, p 2/p1

Indu

ced

Flow

Mac

h N

umbe

r, M

p

Note that the induced Mach number Mp asymptotes to a Þnite value as the pressure ratiobecomes extremely large.

1non-trivial manipulation that is!

ME295/346 - S05 Assignment #3 - Solution

(c) Show that the limiting value for Mp is

Mp →s

2

γ (γ − 1) ≈ 1.89 for air

Solution. We examine Eq. (4) in the limit of p2/p1 À 1. Doing so, we can make the approxi-mations

Mp =1

γ

µp2p1− 1¶vuut 2γ

γ+1p2p1+ γ−1

γ+1

vuuuut 1 + γ+1γ−1

³p2p1

´γ+1γ−1

³p2p1

´+³p2p1

´2≈ 1

γ

µp2p1

¶vuut 2γγ+1p2p1

vuuut γ+1γ−1³p2p1

´=

1

γ

r2γ

γ + 1

γ + 1

γ − 1

=

s2

γ(γ − 1) ≈ 1.89 for γ = 1.4 J

This is in agreement with the behavior of the preceding Þgure from part (b).

2. Do Problem 7.4 in the Anderson text.

Solution. Here we are looking for the conditions under which the induced ßow behind the movingshock will be accelerated to supersonic conditions (Mp = 1.5). From Problem #2, Eq. (4) givesus an expression to Þnd the necessary pressure ratio to produce these conditions. From Eq. (4),a numerical solution approach 2 yields

p2p1= 15.46 J

For this pressure ratio, Eq. (1) gives as a shock Mach number

Ms = 3.66 J

For the upstream conditions of air at 300 K, the sonic speed (assuming ideal gas behavior) is

a1 =pγRT = 340 m/s

Finally, the shock speed must be

cs =Msa1 = 1270 m/s J

3. D�Alembert�s Solution to the Linear Acoustic Wave Equation. It was derived in class thatthe density condensation obeys the linear wave equation

∂2�s

∂t2− a2 ∂

2�s

∂x2= 0.

It can be derived that the most general solution to the equation is

�s(x, t) = A F (x− at) +B G(x+ at)

where A,B are constants that depend on the initial condition. Verify this is the case.2This requires either a root-Þnding or trial-and-error approach to solve for p2/p1. The plot from Problem 2(b) is helpful in

getting a good initial guess.

ME295/346 - S05 Assignment #3 - Solution

Solution. We can show that any function of the form �s = f(x± at) is a solution by direction substi-tution. By the chain rule

∂2�s

∂t2= (±a) (±a) f 00 = a2f 00

∂2�s

∂x2= f 00

where 0 denotes differentiation with respect to the argument. Substituting into the wave equation,we Þnd

a2f 00 − a2f 00 = 0and thus �s = f(x ± at) is a solution. Since both ± cases satisfy the equation, the most generalsolution is a linear combination of the two:

�s(x, t) = A F (x− at) +B G(x+ at)

Here A,B are constants and the ± solutions are called F,G since they need not be the samefunction.

Graduate Student Problems

4. A Proof of D�Alembert�s Solution. The method of characteristics suggests the existence of �naturalvariables� for the linear wave equation (as opposed to simply x and t). Consider the natural variablesdeÞned by

ξ = x+ at η = x− at.

(a) Show that the wave equation expressed in terms of ξ and η the is given by

∂2�s

∂ξ∂η= 0.

Solution. Using the chain rule

∂

∂ξ=

∂

∂x

∂x

∂ξ+∂

∂t

∂t

∂ξ

=∂

∂x+1

c

∂

∂t∂

∂η=

∂

∂x

∂x

∂η+∂

∂t

∂t

∂η

=∂

∂x− 1c

∂

∂t

Substituting

∂2�s

∂ξ∂η=

∂

∂ξ

·∂�s

∂η

¸=∂

∂ξ

·∂�s

∂x− 1c

∂�s

∂t

¸=

µ∂

∂x+1

c

∂

∂t

¶·∂�s

∂x− 1c

∂�s

∂t

¸=

∂2�s

∂x2− 1c

∂2�s

∂x∂t+1

c

∂2�s

∂t∂x− 1

c2∂2�s

∂t2

=∂2�s

∂x2− 1

c2∂2�s

∂t2

= 0

ME295/346 - S05 Assignment #3 - Solution

(b) Show that by integrating twice, once with respect to η and once with respect to ξ, that the generalform for u is

�s(ξ, η) = F (ξ) +G(η)

where the forms F and G are arbitrary.

Solution. Beginning with∂2�s

∂ξ∂η= 0,

we integrate Þrst with respect to η∂�s

∂ξ= h(ξ)

where h is any function of ξ alone. We integrate again, this time with respect to ξ

�s =

Z ξ

h(s) ds+G(η)

where G is any function of η alone. Recognizing that this integral just represents a functionof ξ, we can call it F (ξ) where F is the antiderivative.

∴ �s = F (ξ) +G(η) as required.

ME295/346 - S05 Assignment #3 - Solution