This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Two Dimensional Irrotational Flow”.
1. What is the value of derivative of flow field along characteristic line?
a) Zero
b) Indeterminate
c) One
d) 0.5
View Answer
Explanation: If we consider the direction along the flow field and take the derivatives of these flow properties, the value takes the form of \(\frac {numerator}{denominator} = \frac {0}{0}\) or indeterminate form along the characteristic lines.
2. Which of these equations defines the characteristic curve in an x – y plane?
a) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± (\frac {u^2 + v^{2}}{a^{2}}) – 1}{1 – \frac {u^{2}}{a^{2}}}\)
b) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^{2} + v^{2}}{a^{2}})}}{\frac {u^{2}}{a^{2}}}\)
c) \(\frac {dy}{dx_{char}} = – \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}\)
d) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)
View Answer
Explanation: The approach of method of characteristics entails determining special curves, referred to as characteristics curves, along which the PDE transforms into a family of ordinary differential equations (ODE). If the ODEs have been identified, they can be solved along the characteristic curves to obtain ODE solutions, which can then be compared to the original PDE solution. The characteristic curve is given by:
\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)
3. How many characteristics pass through a point in the flow field if the Mach number is greater than 1?
a) 1
b) 2
c) 4
d) 0
View Answer
Explanation: The characteristic curve is given by the relation
\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)
The term inside the square root \(\frac {u^2 + v^{2}}{a^{2}}\) – 1 can be written in the form of Mach number as follows:
\(\frac {u^2 + v^{2}}{a^{2}}\) – 1 = \(\frac {V^{2}}{a^{2}}\) – 1 = M2 – 1
When the flow is supersonic i.e. Mach number is greater than 1, two characteristics pass through a point in a flow field. Hyperbolic partial differential equation is used to define it.
4. Which partial differential equation is used to define the characteristic at sonic condition?
a) Hyperbolic partial differential equation
b) Parabolic partial differential equation
c) Elliptic partial differential equation
d) Circular partial differential equation
View Answer
Explanation: At sonic condition, when Mach number is equal to 1, there is only one characteristic that passes through the flow field point. This is defined by the parabolic partial differential equation.
\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {M^2 – 1}}{1 – \frac {u^{2}}{a^{2}}}\)
5. For subsonic flows, the characteristics through a flow field point are imaginary.
a) True
b) False
View Answer
Explanation: When the Mach number is less than 1 i.e. the flow is subsonic, the numerator in the characteristic curve is imaginary as the terms inside the square root sign is negative. Thus, methods of characteristics is not used for subsonic flows.
\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(-ve \, term)}}{1 – \frac {u^{2}}{a^{2}}}\)
6. How many characteristic lines pass through the flow field in supersonic flow?
a) 1
b) 2
c) 0
d) 4
View Answer
Explanation: The equation of characteristic curve is given by
\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(-ve \, term)}}{1 – \frac {u^{2}}{a^{2}}}\)
Where u and v are the x and y – component of velocity V. u = Vcosθ and v = Vsinθ. On substituting these values we get
\(\frac {dy}{dx_{char}} = \frac {- \frac {V^2 cosθsinθ}{a^{2}} ± \sqrt {\frac {V^{2}}{a^{2}} (cos^2 θ + sin^2 θ) – 1}}{1 – \frac {V^{2}}{a^{2}} cos^2 θ}\)
The Mach angle is given by μ = sin-1\(\frac {1}{M}\) which can be rearranged as follows sinμ = \(\frac {1}{M}\). But Mach number is given by M2 = \( \frac {V^{2}}{a^{2}} = \frac {1}{sin^2 μ}\). Substituting these we get,
\(\frac {dy}{dx_{char}} = \frac {- \frac {cosθsinθ}{sin^2 μ}±\sqrt {\frac {(cos^2 θ + sin^2 θ)}{sin^2 μ} – 1}}{1 – \frac {cos^2 θ}{sin^2 μ}}\)
Using trigonometry relations, cos2 θ + sin2 θ = 1, thus the square root term in the numerator becomes
\( \sqrt {\frac {(cos^2 θ + sin^2 θ)}{sin^2 μ} – 1} = \sqrt {\frac {1}{sin^2 μ} – 1} = \sqrt {cosec^2 μ – 1} = \sqrt {cot^2 μ – 1} = \frac {1}{tanμ} \)
The characteristic curve thus becomes
\(\frac {dy}{dx_{char}} = \frac {- \frac {cosθsinθ}{sin^2 μ}±\frac {1}{tanμ}}{1 – \frac {cos^2 θ}{sin^2 μ}}\) = tan(θ ± μ)
This results in two characteristic lines through a point in the flow field. One with an angle tan(θ + μ) and the other tan(θ – μ).
7. Characteristic curves passing through a point in a flow field are straight lines.
a) True
b) False
View Answer
Explanation: The right and left running characteristic curves are generally curved as it is given by tan(θ ± μ). Since the value of the Mach angle and the angle θ vary at each point in the flow field, the characteristics are curved instead of being a straight line.
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