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

1. Choose the correct statement.

a) A shock tube is a thick-walled tube with square or circular or rectangular cross-section, which is divided by a diaphragm

b) A shock tube is a thick-walled tube, which is divided by a membrane with equal pressures on both sides

c) A shock tube is a thick-walled tube with any cross-section, which is has a moving piston to generate pressure difference

d) Shock tubes only produce normal shock waves

View Answer

Explanation: Shock tube is a thick-walled tube with square or circular or rectangular cross-section, which is divided by a diaphragm into two chambers with different pressures. When the diaphragm is removed suddenly, wave motion is set giving rise to shock waves and expansion waves.

2. Which relation is of prime importance to find for shock tube operation?

a) The relation between shock strength and diaphragm pressure ratio

b) The relation between normal shock strength and expansion wave pressure ratio

c) Normal shock relations and Isentropic relations

d) The relation between diaphragm pressure ratio and reservoir conditions

View Answer

Explanation: For shock tube operation, it is of prime importance to develop a relation between shock strength \(\frac {p_2}{p_1}\) and diaphragm pressure ratio \(\frac {p_4}{p_1}\). Once the shock strength is known, all other glow quantities can be found from the normal shock relations.

3. A shock tube may be used as a short-duration wind tunnel by utilizing the flow behind the shock wave. If shock speed is M_{s} = \(\frac {C_S}{a_1}\), density ratio η = \(\frac {\rho_2}{\rho_1}\), where conditions of expansion chamber are denoted with subscript 1 and the conditions in the flow region behind the shock are denoted by subscript 2. Find \(\frac {V_p}{a_1}\), where V_{p} is the piston speed.

a) \(\frac {V_p}{a_1}\) = M_{s} (1 – \(\frac {1}{\eta }\))

b) \(\frac {V_p}{a_1}\) = M_{s} (2 – \(\frac {1}{\eta }\))

c) \(\frac {V_p}{a_1}\) = M_{s} (1 – \(\frac {2}{\eta }\))

d) \(\frac {V_p}{a_1}\) = M_{s} (1 + \(\frac {1}{\eta }\))

View Answer

Explanation: By continuity, ρ

_{1}V

_{1}= ρ

_{2}V

_{2}, which gives V

_{2}= \(\frac {\rho_1}{\rho_2}\)V

_{1}. Piston speed is given by, V

_{p}= V

_{1}– V

_{2}= (1 – \(\frac {\rho_1}{\rho_2}\))V

_{1}. Hence, \(\frac {V_p}{a_1}\) = M

_{s}(1 – \(\frac {1}{\eta }\)).

4. Which non-equilibrium conditions are created due to severe gradients of flow properties across a shock?

a) Viscous stress and vibrational stress

b) Viscous stress and heat transfer

c) Vibrational stress and heat transfer

d) Only viscous stress as the flow is adiabatic

View Answer

Explanation: Viscous stress and heat transfer are two non-equilibrium conditions observed across the flow due to high-temperature gradient and pressure gradient.

5. If \(\frac {p_2}{p_1}\) = 29.65 and \(\frac {p_4}{p_1}\) = 35.92, then what is the value of expansion strength inside a shock tube?

a) 1.2114

b) 2.4229

c) 0.8254

d) 0.2357

View Answer

Explanation: We know that p

_{2}= p

_{3}in a shock tube, so expansion strength is given by, \(\frac {p_3}{p_4} = \frac {p_2}{p_1} \times \frac {p_1} {p_4} = \frac {29.65}{35.92}\) = 0.8254.

6. When are gases said to be ideal?

a) When gases are thermally perfect

b) When gases are calorically perfect

c) When gases are thermally and calorically perfect

d) When gases are thermally and physically perfect

View Answer

Explanation: Gases are said to be ideal when they are thermally and calorically perfect. When the specific heat capacity of the gas does not change and is constant, it is considered as calorically perfect. If internal energy and enthalpy is a function of temperature only then gas is said to be thermally perfect.

7. When is flow considered to be compressible?

a) If the flow is less than Mach number 0.5 or the percentage change in density is more than 10%

b) If the percentage change in the density of the flow is more than 15%

c) If the flow is more than Mach number 0.3 or the percentage change in density is more than 5%

d) If the flow Mach number is greater than one

View Answer

Explanation: Compressible effects are observed in all flow regimes but its effects are only significant from Mach number 0.3 or if the percentage change in density of the flow is more than 5%.

8. Which of the following is true about the regions on the either sides of the contact surface inside a shock tube?

a) Both regions have same temperature and density but may have different velocity and pressure

b) Both regions have same velocity and pressure but may have different density and temperature

c) Both regions have same temperature and pressure but different velocity

d) Both regions have same density and pressure but different temperature

View Answer

Explanation: On the either side of the contact surface, the temperature and density may be different but it is necessary that the pressure and fluid velocity be the same, since contact surface is an imaginary boundary which separates high pressure gases from mixing with low pressure gases.

9. Choose the correct statement from the following.

a) To increase the Mach number, we need to either decrease the internal energy or increase the kinetic energy

b) To increase the Mach number, we need to increase both the internal energy and kinetic energy

c) Mach number doesn’t depend upon the internal energy

d) To increase Mach number, we need to either increase the internal energy or decrease the kinetic energy.

View Answer

Explanation: According to relation, \(\frac {KE}{Internal \, Energy} = \frac {γ(γ-1) M^2}{2}\); it clearly shows that to increase Mach number we need to either decrease internal energy or increase kinetic energy.

10. Flow properties are given for upstream and downstream of a shock wave as T_{1} = 288 K and T_{2} = 690 K. What is the change in enthalpy if specific heat ratio is 1.4 and specific heat capacity at constant volume is 718 \(\frac {J}{Kg*K}\)?

a) 2.88 * 10^{5} \(\frac {J}{Kg}\)

b) 4.038 * 10^{5} \(\frac {J}{Kg}\)

c) 258.2 \(\frac {J}{Kg}\)

d) 2114 \(\frac {J}{Kg}\)

View Answer

Explanation: Change in enthalpy is given by, h

_{2}– h

_{1}= c

_{p}(t

_{2}– t

_{1}) = 1005(690 – 288) = 4.038 * 10

^{5}\(\frac {J}{Kg}\)

.

**Sanfoundry Global Education & Learning Series – Gas Dynamics.**

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