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The correct answer is (a) θ + ν(M) = const

For explanation I WOULD SAY: The K+ CONSTANT along the C+ characteristic line is obtained by INTEGRATING the compatibility equation dθ = ∓\(\sqrt {M^2 – 1}\frac {dV}{V}\). Where, ν(M) is the Prandtl – Meyer function and we obtain the K+ constant as θ + ν(M) = const.

The correct option is (b) STEADY, supersonic flow

To explain: METHOD of characteristics is a type of numerical technique used today for computing flow FIELD of steady supersonic FLOWS. This method was developed in the year of 1920s which is a very CLASSICAL numerical approach.

Right answer is (a) θ3 = \(\frac {(K_- )_1 + (K_+ )_2}{2}\)

Easiest explanation: The value of K+ is CONSTANT along left running Mach WAVE and K– is constant along right running Mach wave. Point 3 lies at the intersection of characteristic lines passing through point 1 and 2 THUS, considering the characteristic curve passing through point 1,

(K–)1 = (K–)3

θ1 + ν1 = θ3 + ν3

θ3 + ν3 = (K–)1 (equation 1)

SIMILARLY along the characteric curve through point 2,

(K+)2 = (K+)3

θ2 – ν2 = θ3 – ν3

θ3 – ν3 = (K+)2 (equation 2)

Solving equation 1 and 2 we get

θ3 = \(\frac {(K_- )_1 + (K_+ )_2}{2}\)

Right choice is (B) False

For explanation I would say: 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.

Correct choice is (b) 2

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 M^2 = \( \frac {V^{2}}{a^{2}} = \frac {1}{sin^2 μ}\). 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, cos^2 θ + sin^2 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).

The correct option is (a) True

Explanation: In an IDEAL case, with infinite grid POINTS, the numerical solution is exact. But in real LIFE, the number of grid point for computing the flow field properties is finite and thus, the higher order terms are USUALLY NEGLECTED causing ‘truncation error’.

Right option is (a) Neglecting higher ORDER terms

To explain I would say: In NUMERICAL solution, while expanding flow field PROPERTIES in terms of Taylor series, higher order terms are often neglected. When these terms are neglected, it leads to ERRORS known as truncation error.

The correct option is (b) INDETERMINATE

To EXPLAIN: 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.

Correct answer is (b) False

Easiest EXPLANATION: K+ and K– CONSTANTS ALONG the characteristic lines are given by:

θ + ν(M) = const (along C+ characteristic LINE)

θ – ν(M) = const (along C+ characteristic line)

These equations relate the MAGNITUDE of velocity with the direction of C+ and C– characteristic lines which is known as a hodograph characteristic. These help in giving us the solution for the methods of characteristics.

The CORRECT option is (a) Taylor SERIES

Explanation: The flow field PROPERTIES such as the pressure, density at the discrete points in a grid (xy coordinate plane) are found out by EXPANDING these properties in USING Taylor series. If we know the pressure at some point (i,j), then the pressure at the point (i + 1,j) point can be computed using this series as follows:

ui + 1,j = ui,j + \(\frac {∂P}{∂x_{i,j}}\) ∆x + \(\frac {∂^2 P}{∂x^{2}_{i,j}} \frac {∆x^{2}}{2} + \frac {∂^3 P}{∂x^{3}_{i,j}} \frac {∆x^{3}}{3}\) + ⋯.

Correct answer is (a) Designing supersonic nozzle

Easy explanation: Methods of characteristics are used in designing the supersonic nozzle. For this, a line with known supersonic flow fields is REQUIRED. Upon going downstream, depending on where the point LIES (internal/wall) the method of characteristics is applied to find the flow FIELD.

Correct answer is (d) \(\frac {DY}{dx_{char}} = \frac {- \frac {UV}{a^{2}} ± \sqrt {(\frac {u^2 + V^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)

The explanation is: 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:

\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)

Correct option is (d) Determining temperature

Explanation: Hodograph is a velocity diagram which has its applications in finding the GRAPHICAL solution of the methods of characteristics in high TECH computers as it related velocity to characteristic LINES, obtaining the position of STARS, PLANETS, swinging Artwood’s machine as well as in metrological department.

The correct choice is (a) True

The best 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.

\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(-ve \, term)}}{1 – \frac {u^{2}}{a^{2}}}\)

Right answer is (a) RECTANGULAR grid

The best I can explain: For computing the flow FIELD properties at points in the flow, there is OFTEN grid considered in a plane with known properties at one of the grid point. USING that, properties at other grid points can be computed. In the method of characteristics, non – rectangular grid point is considered and for finite – difference solution, rectangular grid is considered.

The correct CHOICE is (a) Domain of dependence

Explanation: If a point P lies at the intersection of the left and right running waves, the flow FIELD properties at point P is DEPENDENT on the flow field information in the region UPSTREAM i.e. domain of dependence.

Correct choice is (b) False

Easy explanation: Unlike subsonic FLOW where the disturbances are felt in the ENTIRE flow field, this is not the CASE in supersonic flow. In STEADY supersonic linearized flow, the disturbances/perturbations are felt only in LIMITED regions.

Right CHOICE is (c) Area WITHIN the DOWNSTREAM characteristics

The best I can explain: For a point A which is present in a flow field with incoming steady supersonic flow, there will be two CHARACTERISTIC lines PASSING through them. The area formed between these two characteristics in the downstream region is known as region of influence.

Correct option is (b) Parabolic partial differential equation

The best I can explain: 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.

\(\frac {DY}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {M^2 – 1}}{1 – \frac {u^{2}}{a^{2}}}\)

Correct option is (b) 2

The best 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 = M^2 – 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.

The correct option is (c) Designing CONTOUR of NOZZLE

Explanation: Despite the method of characteristics being HIGHLY intensive in terms of the COMPUTATIONAL load, the invention of high speed computers have made the possibility to USE method of characteristics in designing the contour of the supersonic nozzle.

Correct choice is (a) True

Best EXPLANATION: Riemann invariants are obtained by integrating the TWO compatibility equations obtained along the CHARACTERISTIC LINES C+ and C–. These are as follows:

J+ = u + ∫\(\frac {dp}{ρa}\) = const

J– = u – ∫\(\frac {dp}{ρa}\) = const

J+ and J– are hence constant along the characteristic lines.

Correct option is (b) Area within the upstream characteristics

The explanation is: If we consider a point A in the flow FIELD through which two CHARACTERISTIC lines PASS (C+ and C–) then area which is between the two upstream characteristics of both left and right RUNNING waves is known as domain of dependence.

Right choice is (c) dθ = \(\sqrt {M^2 – 1} \frac {DV}{V}\)

Explanation: If we set the numerator of the combination of momentum, continuity and energy equation REPRESENTED by Cramer’s RULE as zero, we get

(1 – \(\frac {u^{2}}{a^{2}}\))dudy + (1 – \(\frac {v^{2}}{a^{2}}\))dxdv = 0

On rearranging the terms we get

(1 – \(\frac {v^{2}}{a^{2}}\))dxdv = – (1 – \(\frac {u^{2}}{a^{2}}\))dudy

\(\frac {dv}{du} = \frac { – (1 – \frac {u^{2}}{a^{2}})dy}{(1 – \frac {v^{2}}{a^{2}})DX}\)

Substituting the value of characteristic curve in the above equation –

\(\frac {dy}{dx_{char}} =\frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}}\)

We get, \(\frac {dv}{du} = \frac{- (1 – \frac {u^{2}}{a^{2}})}{(1 – \frac {v^{2}}{a^{2} })} \big [ \frac { – \frac {uv}{a^{2}} ± \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}} \big ] = \frac {\frac {uv}{a^{2}} ∓ \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {v^{2}}{a^{2}}}\)

u and v are the x and y – component of velocity V. u = Vcosθ and v = Vsinθ. On substituting these VALUES we get

\(\frac {d(Vsinθ)}{d(Vcosθ)} = \frac {M^2 cosθsinθ ∓ \sqrt {M^2 – 1} }{1 – M^2 sin^2 θ}\)

dθ = ∓ \(\sqrt {M^2 – 1}\frac {dV}{V}\)

This the characteristic line along the C+ characteristic line is .dθ = + \(\sqrt {M^2 – 1}\frac {dV}{V}\).

Right option is (b) du – \(\frac {dp}{ρa}\) = 0

To elaborate: 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 – \(\frac {dp}{ρa}\) = 0

The compatibility equation along C+ line is:

du + \(\frac {dp}{ρa}\) = 0

The CORRECT answer is (b) Rounding number to a significant figure

To explain I would say: Numerical solutions are OBTAINED using the computers which often rounds off every number to a certain significant DIGIT. Each of these rounding off results in small ERRORS. If we decrease the truncation errors, the round – off error INCREASES and these errors is a function of step size ∆x.

Right option is (a) u + a

To explain I would say: The right running wave has a CHARACTERISTIC line as C+ which has a positive slope. It’s slope is given by \(\frac {DY}{DX}\) = u + a, where a is the local speed of sound which is equal to \(\sqrt {γRT}\). Its compatibility equation is given by du + \(\frac {dp}{ρa}\) = 0.

The correct choice is (a) INVISCID flow

For explanation I would say: For computing supersonic steady inviscid flows, there are two MAJOR numerical TECHNIQUES that are used. The first being the methods of CHARACTERISTICS which is an older TECHNIQUE limited to inviscid flows. The other method is the finite – difference method which is applicable for both viscous and inviscid flows.