Engineering Questions with Answers - Multiple Choice Questions

MCQs on Finite Waves

1 - Question

What are finite waves?
a) Waves with large perturbations
b) Sound waves
c) Weak waves
d) Waves with small temperature fluctuations

View Answer

a
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.




2 - Question

Finite wave’s properties such as density, velocity are a function of distance.
a) True
b) False

View Answer

a
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.




3 - Question

What is the region in a finite wave called where density increases?
a) Compression region
b) Expansion region
c) Linearized region
d) Propagating region

View Answer

a
Explanation: The density of the finite wave varies along the distance of its propagation. The region where the density increases is known as the finite compression region, and the portion where the density decreases is known as the finite expansion region.




4 - Question

Which of these is not a property of finite wave?
a) The perturbations are large
b) Wave shape changes
c) Linear equations are used to govern the flow variables
d) Wave propagates with added local speed and speed of sound

View Answer

c
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.




5 - Question

Which of these is the compatibility equation along C+ characteristic line?
a) du + dpρa = 0
b) du – dpρa = 0
c) dpuρa = 0
d) duρa = 0

View Answer

a
Explanation: C+ characteristic line is the path along which the governing partial equation can be reduced to the ordinary differential equation. The compatibility equation along C+ line is:
du + dpρa = 0
The compatibility equation along C– line is:
du – dpρa = 0




6 - Question

What is the slope for C– characteristic line?
a) u + a
b) u – a
c) ua
d) ua

View Answer

b
Explanation: At a point (x,y) in an x – y plane there exists paths through this point known as C+ and C– characteristic lines. They represent the path of sound waves which are left and right running. The slope for a C– characteristic line is given by u – a and for C+ characteristic is u + a.




7 - Question

Riemann invariants are obtained by differentiating the compatibility equations for the characteristic lines.
a) True
b) False

View Answer

b
Explanation: When the two compatibility equations for C+ and C– characteristic lines are integrated, we obtain the Riemann invariants. J+ is the Riemann invariant for C+ characteristic line and J– is the Riemann invariant for C– characteristic line.




8 - Question

What is the Riemann invariant for C+ characteristic line?
a) J+ = u + ∫dpρa = constant
b) J+ = u – ∫dpρa = constant
c) J+ = u + dpρa
d) J+ = u – dpρa

View Answer

a
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 + dpρa = 0
On integrating the above equation, we get:
J+ = u + ∫dpρa = constant
Where, J+ is the Riemann invariant




9 - Question

If the Riemann variants are known, how can we compute the mass – motion velocity u?
a) J++J−2
b) J+−J−2
c) J+×J−3
d) J+×J−6

View Answer

a
Explanation: 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 = J++J−2




10 - Question

How is local velocity of sound related to the Riemann invariants?
a) a = γ–14(J+ + J–)
b) u = γ–14(J+ J–)
c) a = γ–14(J+ – J–)
d) a = γ – 1(J+ – J–)

View Answer

c
Explanation: The two Riemann equations along characteristic lines are given by:
J+ = u + 2aγ–1 = constant
J– = u – 2aγ–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 = γ–14(J+ – J–)

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