This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Hugoniot Equation”.
1. In the Hugoniot relation, the changes across the shock are related in terms of ________
a) Chemical properties
b) Electrical properties
c) Thermodynamic properties
d) Mechanical properties
View Answer
Explanation: Since the shock can be visualized as a thermodynamic device that compresses the gas, the Hugoniot relation only relates the thermodynamic quantities across the shock, without any reference to velocity and Mach number.
2. Which of the following relation represents the Hugoniot relation?
a) Δe = -PavΔv
b) Δe = -Pav/Δv
c) Δs = -PavΔv
d) Δs = -Pav/Δv
View Answer
Explanation: The Hugoniot relation implies that “the change in internal energy is equal to the mean pressure across the shock times the change in specific volume.”
i.e. e2-e1 = \(\frac {p_1+p_2}{2}\)(v1-v2)
e2-e1 = –\(\frac {p_1+p_2}{2}\)(v2-v1)
Δe = -PavΔv
3. Which curve is the locus of all possible pressure-volume conditions behind the normal shocks of different strengths for given upstream p1 and v1?
a) Isentropic curve
b) Hugoniot curve
c) Kelvin-Plank curve
d) Prandlt curve
View Answer
Explanation: As in thermodynamic equilibrium state the specific energy can be expressed as a function of pressure and specific volume i.e.
e = e(p, v)
Then from the Hugoniot relation, p2 = f(p1, v1, v2)
This means that for given p1 and v1, it represents p2 as a function of v2. And the plot of this relation in pv diagram is called Hugoniot curve, and such curve is the locus of all possible pressure-volume conditions behind the normal shocks of different strengths for given upstream p1 and v1.
4. A gas having a specified volume of 3.12 m3/Kg passes through normal shock traveling at 530 m/s. If the specific volume behind the shock is 1.51 m3/Kg, then calculate the increase in pressure across the shock.
a) 0.325 bar
b) 0.125 bar
c) 0.589 bar
d) 0.464 bar
View Answer
Explanation: From the Hugoniot curve, the slope of the curve is defined as;
\(\frac {p_2-p_1}{v_2-v_1}=-(\frac {V_1}{v_1})\)2
Now for given values, V1 = 530 m/s, v1 = 3.12 m3/Kg, v2 = 1.51 m3/Kg
Hence, p2-p1 = –\((\frac {V_1}{v_1})\)2v2-v1
p2-p1 = –\((\frac {530}{3.12})\)2(1.51-3.12)
p2-p1 = 46458.85 pa = 0.464 bar
5. Along the Hugoniot curve, the pressure is bounded by which of the following boundaries?
a) ∞ ≤ p ≤ 0
b) 0 ≤ p ≤ ∞
c) 1 ≤ p ≤ 0
d) 0 ≤ p ≤ 1
View Answer
Explanation: The Hugoniot curve represented by pv diagram asymptotically approaches to the line
p = -(γ-1)/(γ+1)
As a result of this, the entire range of pressure ratio occurs between zero and infinite, i.e. the pressure on the Hugoniot curve is bounded by 0 ≤ p ≤ ∞.
6. Across the detonation wave,__________
a) Pressure increases and density decreases
b) Pressure decreases and density increases
c) Both pressure and density decreases
d) Both pressure and density increases
View Answer
Explanation: A supersonic wave is termed as a detonation wave. Therefore in detonation heat and radial diffusion do not control the velocity, but the shock wave structure developed by the supersonic waves causes the flow to slow down, increasing its pressure, density, and temperature across the shock.
7. For deflagrations, the minimum Mach number lies between zero and one.
a) True
b) False
View Answer
Explanation: At the deflagration region in a Hugoniot curve, the velocity of the gas increases while decreasing its pressure and density. Hence the deflagrations correspond to maximum propagation speed for shock waves, we can conclude that all deflagrations propagate at subsonic speeds, hence Mach number lying between zero and one.
8. If the air flow having an upstream pressure 1.89 bar passes through a shock wave, compressing to pressure 2.6 bar across the shock, determine the ratio os specific volume.
a) 0.528
b) 1.254
c) 3.687
d) 4.932
View Answer
Explanation: According to Hugoniot curve the pressure ratio and density ratio are related by;
\(\frac {p_2}{p_1} = \frac {(\frac {γ+1}{γ-1})(\frac {v_1}{v_2})-1}{(\frac {γ+1}{γ-1})-(\frac {v_1}{v_2} )} \)
Therefore for given flow properties \(\frac {2.6}{1.89}=\frac {6(\frac {v_1}{v_2})-1}{6-(\frac {v_1}{v_2}) }\)
i.e. \(\frac {v_2}{v_1}\) = 1.254
Sanfoundry Global Education & Learning Series – Gas Dynamics.
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