Gas Dynamics Questions and Answers – Propagating Shock Wave

This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Propagating Shock Wave”.

1. The change in flow properties across the sound wave can be assumed as _______
a) Isentropic
b) Nonisentropic
c) Irreversible
d) Non-adiabatic
View Answer

Answer: a
Explanation: A sound wave is an infinitesimally small pressure wave. Hence the changes across the wave are very small and the speed of the process corresponding to these changes is very fast. Hence if there is no heat transfer in the system, the change in flow properties across the sound wave can be assumed as isentropic.

2. Incompressible fluid medium, the speed of sound is higher than the incompressible fluid medium.
a) True
b) False
View Answer

Answer: b
Explanation: As the speed of the sound is given by;
a = \(\sqrt {\frac {\partial p}{\partial \rho }}\)
The speed of sound is inversely proportional to density change. Now since incompressible fluid medium, the change in density for given pressure change is high, the speed of sound in such medium is lower compared to an incompressible medium where the change in density is lower for a given change in pressure.

3. Which of the following expression is the continuity equation for a moving shock?
a) ρ2Cs = ρ1(Cs – Vp)
b) ρ1Vp = ρ2(Cs – Vp)
c) ρ2Vp = ρ1(Cs – Vp)
d) ρ1Cs = ρ2(Cs – Vp)
View Answer

Answer: d
Explanation: The continuity equation for a stationary shock is given by;
ρ1V1 = ρ2V2
However, for moving shock the velocity of fluid upstream of shock is denoted by Cs relative to wave, hence the velocity of fluid downstream of shock, relative to wave is (Cs – Vp), thus the continuity equation for moving shock is expressed as;
ρ1Cs = ρ2(Cs – Vp)

4. The temperature ratio across the moving shock is a function of which of the following?
a) Pressure ratio
b) Density ratio
c) Entropy ratio
d) Enthalpy ratio
View Answer

Answer: d
Explanation: The entropy change across the moving shock given by Hugoniot relation is;
e2-e1 = \(\frac {p_1+p_2}{2}\)(v1-v2)
Now for a calorically perfect gas e = cvT and v = RT/p,
Therefore substituting these values in Hugoniot relation, we get
\(\frac {T_2}{T_1}=\frac {p_2}{p_1} \big [ \frac { (\frac {γ+1}{γ-1})+(\frac {p_2}{p_1})}{1+(\frac {γ+1}{γ-1})(\frac {p_2}{p_1})} \big ] \)

5. A shock of strength 0.528 is generated by piston traveling in air at 288 K inside the shock tube. Determine the shock speed. Assume perfect gas.
a) 343 m/s
b) 379 m/s
c) 410 m/s
d) 543 m/s
View Answer

Answer: c
Explanation: The shock strength is Δp/P1 = 0.528
P2/P1–1 = 0.528
P2/P1 = 1.528
And speed of sound a = \(\sqrt {γRT } = \sqrt {1.4*287*288}\) = 340.17 m/s
Now the Shock velocity for perfect gas is related to pressure ratio as
cs = a1\(\sqrt {\frac {γ+1}{2γ}( \frac {p_2}{p_1} – 1) + 1}\)
Thus for given pressure ratio = 1.528, Cs = 410 m/s

6. For a very small pressure jump across the shock, the shock is termed as _______
a) Strong shock
b) Weak shock
c) Curved shock
d) Jump shock
View Answer

Answer: b
Explanation: For a very small pressure jump across the shock, P2 is just slightly higher than P1, and according to the relation of the strength of the shock
(P2-P1)/P1 << 1
Hence since there is a minor change in flow properties across the shock, such shock is called the weak shock.

7. A diatomic calorically perfect gas having upstream pressure of 2.4 bar is compressed to 5.1 bar across the shock. If the temperature behind the shock is 700 K, calculate the temperature upstream of shock.
a) 432.26 K
b) 233.93 K
c) 654.37 K
d) 557.46 K
View Answer

Answer: d
Explanation: As for the calorically perfect gas temperature ratio is a function of pressure ratio across moving shock,
\(\frac {T_2}{T_1}=\frac {p_2}{p_1} \big [ \frac { (\frac {γ+1}{γ-1})+(\frac {p_2}{p_1})}{1+(\frac {γ+1}{γ-1})(\frac {p_2}{p_1})} \big ] \)
\(\frac {700}{T_1}\) = 2.125\(\big [ \frac {6+2.125}{1+6(2.125)} \big ] \)
T1 = 557.46 K

8. What will be the density ratio across the strong shock moving in the fluid medium for a monoatomic gas?
a) 6
b) 4
c) 8
d) 0
View Answer

Answer: b
Explanation: For a strong shock moving in medium, according to Hugoniot curve the pressure ratio and density ratio are related by;
\(\frac {\rho_2}{\rho_1}=\frac {1+(\frac {γ+1}{γ-1})(\frac {p_2}{p_1} )}{( \frac {γ+1}{γ-1})+(\frac {p_2}{p_1})}\)
And since P2 >> P1, \(\frac {\rho_2}{\rho_1} = \frac {γ+1}{γ-1}\) = 4 (As for monoatomic gas γ = 1.66)

9. If the piston moving inside the shock tube is having a velocity of 100 m/s with a shock wave being created having a density ratio 1.7 bar across it. Calculate the shock speed.
a) 100 m/s
b) 165.39 m/s
c) 214.59 m/s
d) 242.85 m/s
View Answer

Answer: d
Explanation: The piston speed in terms of density ratio is;
Vp = cs(1-\(\frac {\rho_1}{\rho_2}\))
100 = cs(1-\(\frac {1}{1.7}\))
Cs = 242.85 m/s

10. The speed of very weak shocks is almost equal to the speed of sound ahead of the shock.
a) True
b) False
View Answer

Answer: a
Explanation: Since for very weak shocks the rise in pressure across the shock is very small, P2 ≈ P1.
Thus it results in zero pressure difference, reducing the below expression to,
cs = a1\(\sqrt {\frac {γ+1}{2γ} \big ( \frac {p_2}{p_1} – 1\big )+1}\)
Cs = a1
Therefore, for very weak shocks, the shock speed is equal to the speed of sound ahead of the shock.

Sanfoundry Global Education & Learning Series – Gas Dynamics.

To practice all areas of Gas Dynamics, here is complete set of Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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