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The correct ANSWER is (a) True

Easiest explanation: Finite waves which propagate in some DIRECTION x have properties such as temperature, pressure, density, velocity a FUNCTION of DISTANCE x for time t at an instant. The flow is supposed to be isentropic.

Correct choice is (c) APPEARS stationary

Explanation: Usually shock waves propagate with SUBSONIC or supersonic speed but when the shockwave propagates in a FLOW that moves as well, and it is in the opposite direction with the same velocity as the flow, then it appears to be stationary known as ‘STANDING shock WAVE’.

Right choice is (a) Constant

Explanation: For a stationary SHOCK WAVE, the total enthalpy remains constant across the shock wave which means that h02 = h01. Although, in CASE of moving shock wave, this is not the case. The enthalpy does not remain constant across the shock.

Correct choice is (c) Pressure ratio

The best I can explain: In a stationary WAVE, changes across the shock wave is governed by the Mach number. But, in a moving shock wave it is DEPENDENT majorly on the pressure ratio as seen in the formula below which shows the EQUATION for density and pressure ratio across a moving shock wave.

\( \FRAC {T_2}{T_1} = \frac {p_2}{p_1} \Bigg ( \frac {\frac {γ + 1}{γ – 1} + \frac {p_2}{p_1}}{1 + \frac {γ + 1}{γ – 1} \frac {p_2}{p_1}} \Bigg )\)

\( \frac {\rho _2}{\rho _1} = \frac {1 + \frac {γ + 1}{γ – 1} \frac {p_2}{p_1}}{\frac {γ + 1}{γ – 1} + \frac {p_2}{p_1}}\)

Correct option is (a) \(\FRAC {J_++J_-}{2}\)

To elaborate: On solving the two Riemann invariants J+ and J– for the two CHARACTERISTIC lines passing through a point x, y in an x – y plane, we get the solution for MASS – motion velocity u. It is given by:

u = \(\frac {J_++J_-}{2}\)

Right ANSWER is (a) PLOT of wave motion between t and x

The explanation: When unsteady wave motion i.e. MOVING shock wave is studied, wave diagram is often constructed. It is a sketch of the wave diagram plot on a graph with x – AXIS representing distance and y – axis representing the time. The diagram shows where the incident, reflected shock occurs after applying the boundary conditions.

The correct choice is (a) J+ = u + ∫\(\FRAC {dp}{ρa}\) = constant

Explanation: RIEMANN invariant for a compatibility equation of the C+ characteristic LINE is obtained by integrating the compatibility equation.

The compatibility equation along C+ characteristic line is GIVEN by:

du + \(\frac {dp}{ρa}\) = 0

On integrating the above equation, we get:

J+ = u + ∫\(\frac {dp}{ρa}\) = constant

Where, J+ is the Riemann invariant

The CORRECT ANSWER is (b) False

Easy explanation: Shock tube makes USE of unsteady wave motion. It is a closed tube at both ends having a diaphragm which separates high pressure region known as the driver SECTION and low pressure region known as the DRIVEN section.

Right choice is (b) 2.375

For EXPLANATION: Given, WR = 600 \(\frac {m}{s}\), up = 350 \(\frac {m}{s}\), a2 = 400 m/s

The incident shock wave propagates into the gas ahead of it TOWARDS the WALL with Mach number Ms which is reflected into the gas ahead of it with Mach number MR. This is given by the formula:

MR = \(\frac {W_R + u_p}{a_2}\)

Substituting the VALUES, we get:

MR = \(\frac {600 + 350}{400}\) = 2.375

Correct choice is (b) False

Easy explanation: Hugoniot equation is given by:

e2 – e1 = \( \frac {p_1 + p_2}{2}\)(v1 – v2)

This is DERIVED from the three governing equations – Continuity, momentum and ENERGY equations for a moving SHOCK wave. This equation gives idea of how the thermodynamic VARIABLES such as energy, pressure changes across the normal shock wave. It is not dependent on whether the shock wave is stationary or not.

Right choice is (c) a = \(\FRAC {γ – 1}{4}\)(J+ – J–)

For explanation: The two Riemann equations along characteristic lines are given by:

J+ = u + \(\frac {2a}{γ – 1}\) = constant

J– = u – \(\frac {2a}{γ – 1}\) = constant

Where J+ is along C+ characteristic line and J– is along C– characteristic line. When we SOLVE these two equations, we obtain the local velocity of sound which is given by:

a = \(\frac {γ – 1}{4}\)(J+ – J–)

The correct answer is (c) Linear equations are USED to govern the flow variables

For EXPLANATION: Finite waves unlike the sound waves have high perturbation of density, velocity, temperature etc. They are known to propagate at speed which is an addition of local speed of mass velocity and speed of sound. The shape of the finite WAVE does not remain CONSTANT like the sound waves and the flow variables are governed by the NONLINEAR equations.

Right choice is (a) Waves with large perturbations

Easiest explanation: The TRAVELING waves have SMALL perturbations in ambient conditions having small changes in PRESSURE, density ETC. Such waves are weak waves but FINITE waves have large perturbations at ambient conditions.

Correct answer is (b) WR < W

Best explanation: On observing the wave diagram it is often seen that for a good reflected shock wave CHARACTERISTICS, the VELOCITY of the reflected shock wave is smaller than the velocity of the wave (WR < W). THUS, in the wave diagram, the slope of the reflected shock wave PATH is steeper.

The correct choice is (b) Zero

For EXPLANATION: The intensity of this reflected shock (WR) is such that with velocity up, the originally induced mass motion is stopped dead at the WALL. The mass motion behind the shock wave that is reflected must be zero. Therefore, the reflected shock wave maintains the zero – velocity BOUNDARY state.

The correct choice is (B) Induces gas behind it

The explanation: When the normal SHOCK wave is not opposed to the flow velocity, it propagates with some velocity W into the LABORATORY facility. This leads to the induction of gas behind it to MOVE in the wave direction.

The correct answer is (b) 2.37

The explanation: GIVEN, \(\frac {p_2}{p_1}\) = 8.5, γ = 1.4 (ambient air)

The TEMPERATURE ratio ACROSS a moving shock wave is given by:

\( \frac {T_2}{T_1} = \frac {p_2}{p_1} \Bigg ( \frac {\frac {γ + 1}{γ – 1} + \frac {p_2}{p_1}}{1 + \frac {γ + 1}{γ – 1} \frac {p_2}{p_1}} \Bigg )\)

Substituting the values, we get:

\( \frac {T_2}{T_1}\) = 8.5\( \Bigg ( \frac {\frac {1.4 + 1}{1.4 – 1} + 8.5}{1 + \frac {1.4 + 1}{1.4 – 1} \times 8.5} \Bigg )\) = 8.5\( \Bigg ( \frac {\frac {2.4}{0.4} + 8.5}{1 + \frac {2.4}{0.4} \times 8.5} \Bigg ) \)

\( \frac {T_2}{T_1}\) = 8.5\(\BIG ( \frac {14.5}{52} \big ) \) = 2.37

Right choice is (B) 360 m/s

The explanation is: Given, W = 900 \( \frac {m}{s}\), \( \frac {\rho _2}{\rho _1}\) = 0.6

For a moving SHOCK wave that propagates into a stagnant gas, it induces a mas motion of velocity up.

This velocity is given by the relation:

up = W\(\big (\)1 – \( \frac {\rho _2}{\rho _1} \big ) \)

Substituting the VALUES:

up = 900(1 – 0.6) = 360 m/s