This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Normal Shock Waves”.
1. Which kind of flow is observed cross the normal shock wave?
a) Isentropic
b) Adiabatic
c) Reversible
d) Isothermal
View Answer
Explanation: The flow across the normal shock wave is adiabatic as there is no heat transfer takes place during the process and the temperature increases due to the conversion of kinetic energy into internal energy across the shock wave.
2. How does the flow velocity changes across the normal shock wave?
a) Remains the same
b) increases
c) Decreases
d) Becomes zero
View Answer
Explanation: According to Prandlt relation, the relation between the critical Mach number upstream and downstream of the shock is given by;
M\(_2^*=\frac {1}{M_1^*}\)
Hence from the above equation, the flow velocity changes from subsonic to supersonic across the normal shock wave and vice versa. However since to produce the normal shock wave the flow has to be supersonic at upstream, the flow velocity always changes from supersonic to subsonic across a normal shock wave.
3. If the shock wave is the infinitely weak normal shock, then what will be the value of downstream Mach number after the shock?
a) M = 0.8
b) M = 1.0
c) M = 1.7
d) M =1.9
View Answer
Explanation: The infinitely weak normal shock wave occurs at upstream Mach number M = 1. Now Since the relation between Local Mach number upstream and downstream of the normal shock is expressed as;
M\(_2^2=\frac {1+[(γ-1)/2]M_1^2}{γM_1^2-[(γ-1)/2]}\)
Therefore at M1 = 1, M2 = 1.
Such infinitely weak normal shock wave is termed as Mach wave.
4. The strength of the shock is defined by _____________
a) ΔP/P1
b) P1/ΔP
c) M1/M2
d) M2/M1
View Answer
Explanation: As the flow passes through shock, its pressure increases. Hence the strength of the shock is governed by the ratio of the difference in pressure across the shock to upstream pressure and is defined as,
(P2-P1)/P1 = ΔP/P1
Therefore the larger the pressure difference across the shock, the stronger the shock is and vice versa.
5. The flow has an upstream Mach number M = 1.3 and pressure of 3.4 bar downstream of the shock. Calculate the pressure upstream of the shock. Assume γ = 1.4.
a) 3.412
b) 4.598
c) 2.765
d) 1.883
View Answer
Explanation: The static pressure ratio across the normal shock wave is given by;
\(\frac {P_2}{P_1}\) = 1 + \(\frac {2γ}{γ+1}\) (M\(_1^2\) – 1)
And for upstream Mach number M1 = 1.3 and P2 = 3.4,
\(\frac {3.4}{P_1}\) = 1 + \(\frac {2*1.4}{1.4+1}\) (1.32 – 1)
P1 = 1.883 bar
6. Calculate the Mach number behind the shock if the upstream Mach number is infinite for an air.
a) 0
b) 1
c) 0.528
d) 0.378
View Answer
Explanation: For an upstream Mach number = infinite, the Mach number relation reduces to;
M\(_2^2=\frac {1+[(γ-1)/2]M_1^2}{γM_1^2-[(γ-1)/2]}\)
M2 = \(\sqrt {(γ-1)/2γ}\)
And assuming the calorically perfect gas with γ = 1.4,
M2 = 0.378
7. The thickness of the shock wave is of the order of __________
a) 10-5 inch
b) 10-5 mm
c) 10-5 cm
d) 10-5 m
View Answer
Explanation: The shock wave is a very thin region across which the flow properties changes drastically. And the thickness of the shock wave is of the order of 10-5 cm.
8. How the flow behaviors across the shock, if the shock wave is infinitely weak, i.e. Mach wave.
a) Non-isentropic
b) Isentropic
c) Reversible
d) Isothermal
View Answer
Explanation: For an infinitely weak shock wave i.e. Mach wave, the upstream Mach number M = 1. Now since the entropy difference across the shock in terms of upstream Mach number is given by;
s2 – s1 = cpln\( \big [ \big (1+ \frac {2γ}{γ+1} (M_1^2-1) \big ) ( \frac {2+(γ-1) M_1^2}{(γ+1) M_1^2} ) \big ]\) – Rln\( [ 1+\frac {2γ}{γ+1}(M_1^2-1) ] \)
For M1 = 1, s2 – s2 = 0, i.e. s2 = s1
Hence for an infinitely weak shock wave, the flow behaves isentropically across the shock.
9. How is the density ratio across the normal shock wave is expressed?
a) \(\frac {ρ_2}{ρ_1} = \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2}\)
b) \(\frac {ρ_1}{ρ_2} = \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2}\)
c) \(\frac {ρ_2}{ρ_1} = \frac {(γ+1)M_2^2}{2+(γ-1)M_2^2}\)
d) \(\frac {ρ_1}{ρ_2} = \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2}\)
View Answer
Explanation: As per the continuity equation, ρ1v1 = ρ2v2
Also from the Prandlt relation, V1V2 = a*2
Hence, \(\frac {ρ_2}{ρ_1} = \frac {V_1}{V_2 } = \frac {V_1^2}{V_2*V_1} = \frac {V_1^2}{a^{*2}}\)
\(\frac {ρ_2}{ρ_1}\) = M\(_1^{*2} = \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2}\)
10. If the air travels at Mach number 1.46, then calculate the temperature ratio across the shock.
a) 3.36
b) 8.54
c) 4.16
d) 5.95
View Answer
Explanation: From the equation of state, P = ρRT
According to this the temperature ratio across the shock,
\(\frac {T_2}{T_1} = ( \frac {p_2}{p_1} )(\frac {ρ_1}{ρ_2} )\)
\(\frac {T_2}{T_1} = \big [ 1+\frac {2γ}{γ+1} (M_1^2-1) \big ] \big [ \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2} \big ] \)
Therefore at M1 = 1, \(\frac {T_2}{T_1}\) = 4.16
11. The total temperature across the stationary normal shock wave is constant.
a) True
b) False
View Answer
Explanation: As per the energy equation, cpT1 + \(\frac {u_1^2}{2}\) = cpT2 + \(\frac {u_2^2}{2}\)
Also, the total temperature is given by; cpT0 = cpT + \(\frac {u^2}{2}\)
Therefore from the above equations; cpT01 = cpT02
T01 = T02
i.e. the total temperature across the stationary normal shock wave remains constant.
12. If the change in entropy across the shock is 425 J/K, calculate the total pressure ratio across the shock for an air.
a) 1.23
b) 2.71
c) 2.98
d) 3.16
View Answer
Explanation: The ratio of total pressure across the shock in terms of entropy change is given by;
\(\frac {p_{0,2}}{p_{0,1}}\) = e-(s2-s1)/R
Therefore for given values, \(\frac {p_{0,2}}{p_{0,1}}\) = e-425/287
\(\frac {p_{0,2}}{p_{0,1}}\) = 2.71
13. What will be the density ratio for air traveling at Mach number M1 = infinite?
a) 3
b) 4
c) 5
d) 6
View Answer
Explanation: As per the normal shock relations for a perfect gas the ratio of density downstream and upstream of the shock is given by;
\(\frac {ρ_2}{ρ_1} = \frac {(γ+1)M_1^2}{2+(γ-1)M_1^2}\)
Hence at M = infinite \(\frac {ρ_2}{ρ_1} = \frac {(γ+1)}{(γ-1)}\)
\(\frac {ρ_2}{ρ_1}\) = 6
14. If the downstream Mach number behind the shock is 0.8, calculate the strength of the Shock.
a) 0.724
b) 0.523
c) 0.692
d) 0.956
View Answer
Explanation: From the relation of upstream and downstream Mach number;
M\(_2^2=\frac {1+[(γ-1)/2]M_1^2}{γM_1^2-[(γ-1)/2]}\)
For M2 = 2.6 bar, M1 = 1.273
And pressure ratio; P2/P1 = \( \big [ 1 + \frac {2γ}{γ+1} (M_1^2-1) \big ] \)
P2/P1 = 1.724
Now the strength of the shock, = (P2 – P1) / P1
= (P2/P1) – 1
= 1.724 – 1
= 0.724
15. At Mach number M = 6.3, the gas can be termed as a perfect gas.
a) True
b) False
View Answer
Explanation: At M = 6.3, the flow is termed as hypersonic. Due to this the additional flow properties related to hypersonic flow such as high temperature and low density effects comes into account. And because of these parameters the flow cannot be treated as perfect gas anymore.
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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