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The correct option is (a) (1 – M\(_∞^2\))ϕxx + ϕyy + ϕzz = 0

To explain I would say: When the ASSUMPTIONS of small PERTURBATIONS and transonic, hypersonic conditions are excluded, the linearized velocity potential equation is FOUND out as follows:

(1 – M\(_∞^2\))ϕxx + ϕyy + ϕzz = 0

Where, ϕxx = \(\FRAC {∂^2 ϕ}{∂x^2}\), ϕyy = \(\frac {∂^2 ϕ}{∂y^2}\), ϕzz = \(\frac {∂^2 ϕ}{∂z^2}\)

Correct choice is (b) θc > θcmax

The best explanation: For a given freestream Mach number M∞, there exists a maximum CONE ANGLE for which if we go BEYOND that value, the SHOCK wave originating at the cone’s vertex gets detached. Thus, the condition for detached shock wave is θc > θcmax.

The CORRECT choice is (a) (Cp)CRIT = \(\FRAC {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}\) – 1

Explanation: The formula for the coefficient of PRESSURE for an isentropic flow is given by:

Cp = \(\frac {2}{γM_∞^{2}} \bigg ( \frac {p}{p_∞}– 1 \bigg )\)

For an isentropic flow, the ratio of pressure at a point to the freestream pressure is given by:

\(\frac {p}{p_∞}= \bigg [ \frac {1 + \frac {(γ – 1)}{2} M_∞^{2}}{1 + \frac {(γ – 1)}{2} M^{2}} \bigg ]^{\frac {γ}{γ – 1}} \)

Substituting this in the above equation we get

Cp = \(\frac {2}{γM_∞^{2}} \bigg [ \bigg ( \frac {1 + \frac {(γ – 1)}{2} M_∞^{2}}{1 + \frac {(γ – 1)}{2} M^{2}}\bigg ) ^{\frac {γ}{γ – 1}} – 1 \bigg ] \)

At critical Mach number, local Mach number M = 1 and freestream Mach number is equal to the critical Mach number. Substituting these we finally arrive at the relation:

(Cp)crit = \(\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}\) – 1

Correct CHOICE is (b) False

The best I can explain: The coefficient of pressure over a slender body in a supersonic flow, provided the inclination is small is GIVEN by

Cp = \(\frac {2θ}{\sqrt {M_∞^2 – 1}}\)

The angle θ is positive when measured above the freestream HORIZONTAL and negative when measured below the freestream horizontal. Due to this reason, the coefficient of pressure in the forward portion of the HUMP is positive.

Right choice is (b) False

The explanation is: The linearized perturbation velocity POTENTIAL is DERIVED after taking making assumptions to CONVERT the non – linear equation into linear equation. One of the assumption made is that for the flow between Mach number 0 and 0.8 (transonic flow), the term M\(_∞^2 \big [ \)(γ – 1) \(\frac {U^{‘}}{V_∞} + (\frac {γ + 1}{2}) \frac {u^{‘^2}}{V_{∞}^{2}} + (\frac {γ – 1}{2})(\frac {v^{‘{^2}} + w^{‘^{2}}}{V_∞^2}) \big ] \frac {∂u^{‘}}{∂x}\) in the non – linear perturbation velocity potential equation is ignored because of its negligible value. Due to this assumption, the linearized equation is not applicable for transonic flows.

The correct OPTION is (c) Shock wave is attached to the cone

For explanation: For the oblique shock wave to be attached to the vertex of the cone, the half cone ANGLE θc must be less than the MAXIMUM cone angle θcmax. As soon as θc exceeds θcmax, the oblique shock wave gets DETACHED from the cone.

Correct option is (C) Z – axis

For explanation I would say: The conical flow is obtained by KEEPING a wedge in a y – z PLANE which is rotated about z – axis. This results in an axisymmetric flow in which the flow properties remain constant along the ray in a CONE and DEPEND only on radius r and the axis.

Right option is (a) RAY from a vertex

The EXPLANATION: Flow PROPERTIES such as the pressure and density REMAIN constant along the ray originating from the vertex of the cone INCLUDING on the surface of the cone. Although, the flow properties vary from one ray to another.

The correct option is (a) OBLIQUE shock PROPERTIES

The BEST I can explain: Shock polar is graphical representation of all the properties of the oblique shock waves. It is the locus of all the possible velocities behind the shock wave. The DOWNSTREAM velocities are plotted on the y – axis and the UPSTREAM velocities are plotted on the x – axis.

Right option is (a) True

For explanation: One of the consequences of 3 – dimensional RELIEVING effect in a CONE is that the shock WAVES FORMED are weaker in COMPARISON to the waves formed over the wedge with same half angle kept at the same incoming flow velocity.

The correct option is (a) Axisymmetric flow

For EXPLANATION: The CONICAL flow is symmetric about a PARTICULAR axis hence its flow properties such as pressure, density remain the same in plane which passes through a symmetric line. The flow properties DEPEND only on radius r and the axis hence it is known as axisymmetric flow.

Right option is (b) Mdrag – divergence > Mcrit

For EXPLANATION I would say: Drag –divergence Mach NUMBER is greater than critical Mach number. At this Mach speed, there is a region of LOCAL supersonic flow followed by a shock wave. This results in pressure drag eventually CAUSING boundary layer SEPARATION.

The correct answer is (b) False

For explanation: According to the PRANDTL – Glauert rule, as the limit of Mach number is increased to one, the aerodynamic FORCES – lift and drag becomes infinity which is PRACTICALLY impossible. Thus, this rule is only APPLICABLE for subsonic and supersonic regimes.

Cl = \(\frac {C_{l0}}{\sqrt {1 – M_∞^{2}}}\), Cd = \(\frac {C_{d0}}{\sqrt {1 – M_∞^{2}}}\)

Right choice is (c) 1.62

To EXPLAIN I would say: GIVEN, M\(_∞^2\) = 2, θ = 1.4 deg

For the supersonic FLOW, the LINEARIZED coefficient of pressure for small inclinations is given by:

Cp = \(\frac {2θ}{\sqrt {M_∞^2 – 1}}\)

Substituting the VALUES, we get

Cp = \(\frac {2 × 1.4}{\sqrt {4 – 1}}\) = 1.62

The correct choice is (B) False

The best I can explain: Linearized velocity potential equation is not applicable for hypersonic flows with Mach number greater than 5 as in the non – linear velocity potential equation, there is an assumption made that the magnitude of M\(_∞^2 \big [ \)(γ – 1) \(\frac {u^{‘}}{V_∞} + (\frac {γ + 1}{2}) \frac {u^{‘^2}}{V_{∞}^{2}} + (\frac {γ – 1}{2})(\frac {W^{‘{^2}} + u^{‘^{2}}}{V_∞^2}) \big ] \frac {∂V^{‘}}{∂x}\) is small compared to the left HAND side THUS is neglected to arrive at the linearized velocity potential equation.

Right option is (a) Critical Mach NUMBER

Easiest explanation: Coefficient of pressure at minimum pressure point which is present at the upper surface of the airfoil is given by the relation:

(Cp)crit = \(\FRAC {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}\) – 1

In this the value of gamma is constant depending on the medium. The only varying quantity is critical Mach number. Thus, the coefficient of pressure at minimum pressure point is a FUNCTION of only critical Mach number.

Correct ANSWER is (b) Velocity potential EQUATION

Best explanation: The velocity potential equation is given by:

(1 – \(\FRAC {Φ_x^2}{a^2}\)) Φxx + (1 – \(\frac {Φ_y^2}{a^2}\)) Φyy + (1 – \(\frac {Φ_z^2}{a^2}\)) Φzz – (\(\frac {2Φ_x Φ_y}{a^2}\)) Φxy – (\(\frac {2Φ_x Φ_z}{a^2}\)) Φxz – (\(\frac {2Φ_y Φ_z}{a^2}\)) Φyz

The total velocity potential is related to the perturbation velocity potential by:

Φx = V∞ + Φx, Φy = ϕy, Φz = ϕz

And its double derivative is given by

Φxx = ϕxx, Φyy = ϕyy, Φzz = ϕzz

Substituting these values in the velocity potential equation and multiplying it with a^2 we get

(a^2 – (V∞ + ϕx)^2)ϕxx + (a^2 – ϕy^2) ϕyy + (a^2 – ϕz^2) ϕzz – (2(V∞ + ϕx)ϕy)ϕxy – (2(V∞ + ϕx)ϕz)ϕxz – (2ϕyϕz) ϕyz

The above equation is KNOWN as the perturbation velocity potential equation which is a non linear equation.

The correct option is (a) Vx = V∞ + u^‘

To EXPLAIN I would say: When a SLENDER body is in a uniform laminar flow having small perturbations u^‘, v^‘, w^‘ in x, y, z DIRECTION. The x, y, z components of the velocity is given by:

Vx = V∞ + u^‘

VY = v^‘

Vz = w^‘