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Correct choice is (b) False

For explanation I would say: According to the Prandtl –Glauert rule, the incompressible pressure coefficient is given by:

Cp = \(\frac {C_{p_0}}{\SQRT {1 – M_∞^{2}}}\)

For AIRFOILS which are THIN, the flow has less expansion resulting in less magnitude of Cp0. This RESULTS in a higher critical Mach number value. On the other hand thick airfoils have higher magnitude of Cp0 because of strong flow expansion. This results in lower critical Mach number.

Correct OPTION is (b) Hyperbolic

To explain: The linearized perturbation VELOCITY POTENTIAL equation for supersonic flow differs with the subsonic flow in the type of PARTIAL differential equation formed. The equation for supersonic flow given as below which is in the form of hyperbolic partial differential equation.

λ^2ϕxx + ϕyy = 0

The correct choice is (a) \(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{DX}\)

The explanation: For an airfoil with x – COMPONENT of velocity as V∞ + u^‘ and y – component of the velocity as v^‘, the SURFACE boundary condition is

\(\frac {df}{dx} = \frac {v^{‘}}{V_∞ + u^{‘}}\) = tanθ

Since it is a thin airfoil, the perturbation vector u^‘ is very SMALL in comparison to the freestream velocity V∞, RESULTING in \(\frac {df}{dx} = \frac {v^{‘}}{V_∞}\) = θ (Where tanθ ~ θ for small angles). Expressing the perturbation v^‘ in terms of velocity potential we get

v^‘ = \(\frac {∂ϕ}{∂x}\)

Substituting this in the above equation:

\(\frac {df}{dx} = \frac {\frac {∂ϕ}{∂x}}{V_∞}\) = θ

\(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{dx}\)

The CORRECT option is (c) \(\FRAC {W^{‘}}{V_∞}\) >> 1

Easy explanation: The perturbation velocity potential equation is given by:

(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 – (2ϕy ϕz) ϕyz

This relation is non – linear in nature and in ORDER to reduce to the linear form, an assumption is made i.e. the perturbations in uniform flow is very small. This results in u^‘, v^‘, w^‘ << V∞.

Correct ANSWER is (c) 1.648

Explanation: Given, M1 = 3.5,θ = 30.2°

From the θ – β – M curve,

For M1 = 3.5 and θ = 30.2°, the value of β is 48°

Normal component of M1 is Mn1 = M1 sinβ = 3.5 × sin48 = 2.60

From the normal SHOCK table (gas table), for Mn1 = 2.6, we get

\(\FRAC {P_{02}}{P_{01}}\) = 0.4601 and Mn2 = 0.5039

Thus the Mach number downstream of the shock wave is

M2 = \(\frac {M_{n2}}{sin⁡(β – θ)} = \frac {0.5039}{sin⁡(48 – 30.2)}\) = 1.648

The correct option is (a) Between the oblique shock and the sonic line

The explanation: The incoming FLOW over a cone is mostly supersonic between the surface of the cone and the oblique shock wave. Except in some cases when the half cone angle is large, flow sometimes becomes SUBSONIC and one of the RAYS ORIGINATING the cone’s vertex ACTS at the sonic line. The flow between the sonic line and the oblique shock thus remains supersonic.

The correct answer is (B) 2

To explain: For a conical flow placed in a freestream Mach number M∞ with a particular cone angle θc, there exists two OBLIQUE shock waves which is a result of one weak and one strong solution. It is similar to the CASE of WEDGE where using θ – β – M graph we OBTAIN two shock waves.

Right choice is (a) Vθ = \(\frac {∂(V_r )}{∂θ}\)

The explanation is: If we apply Crocco’s theorem in spherical coordinates we get,

∇ × V = \(\frac {1}{r^2 sinθ} \begin {vmatrix} e_r & re_θ & (rsinθ) e_ϕ \\

\frac {∂}{∂r} & \frac {∂}{∂θ} & \frac {∂}{∂ϕ} \\

V_r & rV_θ & (rsinθ) V_ϕ \\

\END {vmatrix}\) = 0

On expanding this we get,

∇ × V = \(\frac {1}{r^2 sinθ} \bigg [ \)ER (\(\frac {∂}{∂θ}\)(rsinθ)Vϕ – \(\frac {∂(rV_θ)}{∂ϕ}\)) – reθ(\(\frac {∂}{∂r}\)(rsinθ)Vϕ – \(\frac {∂(V_r)}{∂ϕ}\)) + (rsinθ)eϕ\( \bigg( \frac {∂(rV_θ)}{∂r} – \frac {∂(V_r)}{∂θ} \bigg )\bigg ] \) = 0

For this equation to be valid, the terms inside the bracket are ZERO. Taking the last bracket term,

\( \frac {∂(rV_θ)}{∂r} – \frac {∂(V_r)}{∂θ}\) = 0

Using chain rule to EXPAND this, we get

r\( \frac {∂(V_θ )}{∂r}\) + Vθ\( \frac {∂(r)}{∂r} – \frac {∂(V_r )}{∂θ}\) = 0

Based on the conical flow assumptions, \( \frac {∂}{∂r}\) = 0 and \( \frac {∂}{∂ϕ}\) = 0. Applying this the equation reduced to

\(\frac {∂(V_r )}{∂θ}\) = 0

Which results in the irrotationally CONDITION for a conical flow as Vθ = \(\frac {∂(V_r )}{∂θ}\).

Correct option is (c) Decreases

Easy explanation: USING the shock polar, we notice that as the deflection ANGLE is INCREASED from θA to θB, the downstream velocity magnitude decreases due to the formations of STRONGER shock WAVE.

The correct answer is (b) False

The explanation: Unlike the flow over a wedge which is a two – dimensional flow, the streamline is not parallel to the surface behind the shock wave. The flow field between the shock wave and the CONICAL surface is not uniform causing the STREAMLINES to becomes slightly curved as the PRESSURE is not CONSTANT the conical surface.

Right option is (b) 0.12

Easy explanation: Given, M∞ = 2, ∝ = 3 = 0.052 rad

According to the LINEARIZED theory, the coefficient of lift over a FLAT plate is given by:

cl = \(\frac {4∝}{\sqrt {M_∞^2 – 1}}\)

Substituting the values we get

cl = \(\frac {4 × 0.052}{\sqrt {4 – 1}}\) = 0.12

The correct option is (a) Increases

The best I can explain: In the supersonic REGIME, ACCORDING to the linearized COEFFICIENT of pressure is directly proportional to the local inclination and inversely proportional to\(\sqrt {M_∞^2 – 1}\). Due to this relation, as the Mach number decreases, the coefficient of pressure increases to a point where Mach number reaches transonic regime (M = 1) where the coefficient of pressure becomes INFINITY.

The correct choice is (a) Increases

The explanation: The linearized COEFFICIENT of pressure for a subsonic flow varies with RESPECT to the Mach number as follows:

CP ∝ \(\frac {1}{\SQRT {1 – M_∞^2}}\)

According to this proportionality, as the Mach number increases, the coefficient of pressure also increases.

Right choice is (b) False

Explanation: When the incoming flow PASSES through an OBLIQUE SHOCK in a two – dimensional wedge, the streamline becomes parallel to the surface of the wedge. But, in case of a CONE, the STREAMLINES between the shock wave and the cone are not parallel.

Correct answer is (a) Increases

For EXPLANATION: For a subsonic flow, the linearized coefficient of pressure is given by the equation below according to which when the Mach number is increased, the coefficient of pressure increases. Although, one thing to note is that as this Mach number is increased to UNITY, the coefficient of pressure reaches INFINITY and thus, for transonic regions, this equation fails.

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

Right choice is (b) Curved at the beginning, PARALLEL as surface tends to infinity

To explain I would say: For a conical flow at a given Mach number, the SHOCK formation occurs at the vertex of the CONE. The streamlines behind the shock wave are initially curved due to 3 – dimensional space unlike streamlines behind the shockwave on a wedge which is parallel. These streamlines eventually become parallel at the cone’s surface tends to infinity.

The CORRECT answer is (b) Cp = \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\)– 1)

EASIEST explanation: The RELATION between coefficient of pressure with gamma and MACH number are derived as follows:

Coefficient of pressure is given by Cp = \(\frac {p – p_∞}{\frac {1}{2} ρ_∞ V_∞^{2}}\). Where, p∞, ρ∞, V∞ are freestream pressure, density and velocity. We can manipulate the denominator by multiplying and diving it by γp∞. We get,

\(\frac {1}{2}\)ρ∞\(V_∞^2\) = \(\frac {1}{2}\frac {γp_∞}{γp_∞}\) ρ∞V∞^2 = \(\frac {γ}{2}\) p∞\(\frac {ρ_∞ V_∞^2}{γp_∞}\)

Since \(\frac {γp_∞}{ρ_∞}\) = a^2 the denominator becomes \(\frac {γ}{2}\)p∞\(\frac {V_∞^2}{a_∞^2} = \frac {γ}{2}\) p∞ M\(_∞^2\) Since M = V/a.

Therefore the coefficient of pressure in terms of gamma and Mach number is:

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

Right answer is (a) Φ(x, y, z) = V∞ x + ϕ(x, y, z)

EASY EXPLANATION: When the body is placed in a uniform flow, the y and z components of the local VELOCITY are zero. Since the velocity potential is given by V = ∇Φ and the local velocity is given by V = (V∞ + u^‘)i + v^‘j + w^‘k, we can use PERTURBATION velocity potential to derive the relation.

Perturbation velocity potential is related to the perturbations in x, y, z components as follows:

\(\FRAC {∂ϕ}{∂x}\) = u^‘, \(\frac {∂ϕ}{∂y}\) = v^‘, \(\frac {∂ϕ}{∂z}\) = w^‘

Substituting this in the equation V = ∇Φ = (V∞ + u^‘)i + v^‘j + w^‘k we get,

Φ(x, y, z) = V∞ x + ϕ(x, y, z)

The correct choice is (a) True

To explain: The energy equation is given by

ρ\(\frac {Dh_0}{Dt} = \frac {∂p}{∂t}\) + p\( \DOT {q}\) + ρ(f.V)

Where \(\frac {Dh_0}{Dt}\) is the total derivative of total enthalpy

p is the pressure

\( \dot {q}\) is per rate of heat added/removed

f.V is body FORCES

According to Taylor – Maccoll’s assumptions, the FLOW is said to be steady (\(\frac {∂}{∂t}\) = 0), adiabatic (\( \dot {q}\) = 0), INVISCID and has no external body forces (f.V = 0). Thus, the equation reduces to

ρ\(\frac {Dh_0}{Dt}\) = 0

Which on integrating gives US h0 = const.

The CORRECT choice is (b) False

The best I can explain: Based on Crocco’s theorem, for a steady floe the relation is given by

T∇s = ∇h0 – V × (∇ × V)

Therefore the VORTICITY is related to total ENTHALPY and gradient as (on REARRANGING the terms):

V × (∇ × V) = ∇h0 – T∇s

In a flow having a change in enthalpy or entropy would result in a rotational flow, but since for conical flow, the assumptions result in both change in entropy and enthalpy being zero, we get

V × (∇ × V) = 0

Since curl of velocity i.e. vorticity is zero, thus the flow is irrotational.

The correct choice is (B) 1.66

Easiest explanation: Given, MCRIT = 0.6, γ = 1.4 (air)

The critical MACH number is calculated USING 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

On substituting the values, we get

(Cp)crit = \(\frac {2}{1.4×0.6^{2}} \bigg [ \frac {1 + \frac {(1.4 – 1)}{2} 0.6^{2}}{1 + \frac {(1.4 – 1)}{2}} \bigg ]^{\frac {1.4}{1.4 – 1}}\) – 1

(Cp)crit = 3.97\(\big [ \frac {1.07}{1.20} \big ] \)^3.5 – 1 = 1.66

The correct choice is (c) Shock wave is curved

To elaborate: There are certain assumptions made to determine the CONICAL flow. As per Taylor – Maccoll, the flow is assumed to be axisymmetric about z – axis \(\frac {∂}{∂ϕ}\) = 0, and the cone is assumed to be at zero ANGLE of attack. If it was kept at any other angle then there will be 3 – dimensional effects that will be hard to ACCOUNT for. Also, the flow properties along the ray of the cone is assumed to be CONSTANT \(\frac {∂}{∂r}\) = 0 and the shock wave is straight.

Right answer is (b) False

The best EXPLANATION: The coefficient of pressure as obtained from the linearized supersonic theory and the exact oblique shock theory match only until small value of SEMI angle θ. As the Mach NUMBER increases beyond 4, the linearized theory results are not as ACCURATE at the exact oblique shock theory.

Correct answer is (c) β^2ϕxx + ϕyy = 0

Easy explanation: For a compressible SUBSONIC flow over a thin airfoil, the TWO dimensional linearized perturbation velocity potential EQUATION is given by

β^2ϕxx + ϕyy = 0

In this equation the perturbations are ASSUMED to be small with the value of β = \(\sqrt {1 – M_∞^{2}} \).

Correct OPTION is (d) SUBSONIC flow is excluded

Easiest explanation: In order to obtain the linearized perturbation velocity potential equation, there are few assumptions made. The PERTURBATIONS U^‘, v^‘, w^‘ are assumed to be small in COMPARISON to the free stream velocity. Apart from this the equation is not applicable for transonic flow with Mach number between 0.8 and 1.2 and for hypersonic flow with Mach number greater than 5.

Right option is (a) True

The explanation is: For a PARTICULAR CONE ANGLE, the shock wave formed at the cone’s vertex is USUALLY weaker in strength compared to the shock wave formation on a wedge. This phenomenon is due to the three – dimensional RELIEVING effect.

Right ANSWER is (a) One

The best I can EXPLAIN: Taylor – Maccoll is a one – dimensional EQUATION in which it is dependent on only one UNKNOWN. This term is the radial velocity. THUS the radial velocity is a function of angle θ.

Vr = f(θ)

Right ANSWER is (b) Conical shock

Explanation: When a cone having semi VERTEX angle θc is KEPT in an incoming SUPERSONIC flow, then there is a formation of shock wave. This is an oblique shock wave which has the shape of the cone and the streamlines downstream of the shock are not immediately PARALLEL to the surface.

The correct answer is (a) 0.0987

For explanation I WOULD say: Given, M∞ = 3, ∝ = 4 = 0.0698 rad

According to the linearized theory, the COEFFICIENT of lift over a flat plate is given by:

cl = \(\frac {4∝}{\SQRT {M_∞^2 – 1}}\)

Substituting the values we get

cl = \(\frac { 4× 0.0698}{\sqrt {9 – 1}} = \frac {0.2792}{2.8284}\)0.0987

The correct answer is (b) (θmax)wedge > (θmax)CONE

The EXPLANATION: Due to the three-dimensional relieving effect, for a particular FREESTREAM Mach number, the MAXIMUM cone angle allowed is larger than the maximum wedge angle before the shock wave BECOMES detached.

The correct choice is (b) 2ρVr + ρVθcot⁡θ + ρ\(\frac {∂(V_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0

To elaborate: The general continuity EQUATION is given by

\(\frac {∂}{∂t}\) + ∇.(ρV) = 0.

Since the flow is assumbed to be steady, \(\frac {∂}{∂t}\) = 0.

For SPHERICAL coordinated of a cone, the del operator is expanded as

∇.(ρV) = \(\frac {1}{r{^2}} \frac {∂}{∂r}\)(r^2ρVr) +\(\frac {1}{r sinθ} \frac {∂}{∂θ}\) (ρVθsin⁡θ) + \(\frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

Solving the partial derivatives, we GET

\(\frac {1}{r{^2}}\bigg [ \)r^2\(\frac {∂}{∂r}\)(ρVr) + ρVr\(\frac {∂(r^2)}{∂r} \bigg ] + \frac {1}{r sinθ} \bigg [ \)ρVθ\(\frac {∂}{∂θ}\)(sin⁡θ ) + sin⁡θ\(\frac {∂(ρV_θ)}{∂θ} \bigg ] + \frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

This is equal to

\(\frac {1}{r{^2}}\bigg [ \)r^2\(\frac {∂}{∂r}\)(ρVr) + ρVr(2r)\( \bigg ] + \frac {1}{r sinθ} \bigg [ \)ρVθ(cos⁡θ) + sin⁡θ\(\frac {∂(ρV_θ)}{∂θ} \bigg ] + \frac {1}{r sinθ} \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

Since the flow PROPERTIES are constant along a ray, \(\frac {∂}{∂r}\)(ρVr) = 0 and \( \frac {∂(ρV_ϕ)}{∂ϕ}\) = 0

The equations becomes \(\frac {1}{r{^2}}\)[ρVr(2r)] + \(\frac {1}{r sinθ} \bigg [ \)ρVθ(cos⁡θ) + sin⁡θ \(\frac {∂(ρV_θ)}{∂θ} \bigg ] \) + 0

Multiplying the final equation with r: 2ρVr + ρVθcot⁡θ + ρ\(\frac {∂(ρV_θ)}{∂θ}\) + Vθ\(\frac {∂(ρ)}{∂θ}\) = 0

Right choice is (a) Cp = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

Easy EXPLANATION: The non – linear coefficient of pressure is in the form of the relation below as obtained after binomial expansion.

Cp =–\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {U^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

For TWO dimensional TERMS, the second order terms can be neglected. But for three – dimensional the term \(\frac {v^{‘{^2}}+w^{‘^{2}}}{V_∞^{2}}\)has to be retained since it is not negligible. Apart from that \(\frac {u^{‘^2}}{V_∞^{2}}\)is neglected because the perturbed velocities have magnitudes that are very less than unity. This results in equation for linearized coefficient of pressure for three – dimensional bodies as Cp = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\).

Right answer is (c) 0.8

To elaborate: For an increasing MACH number in a subsonic flow over a body, the coefficient of pressure also increases as a result of Prandtl – Glauert RULE. But, after Mach number 0.8 the EQUATION fails because the flow enters transonic regime where coefficient of pressure TENDS to INFINITY as Mach number tends to unity.

The CORRECT option is (b) The flow around airfoil becomes SUPERSONIC

Explanation: CRITICAL Mach number is of TWO kinds –lower and upper critical Mach number. When the flow around the airfoil is at the upper critical Mach number, then the flow around the ENTIRE airfoil reaches supersonic speed.

Correct choice is (a) Re-entry shuttle

The EXPLANATION: Any OBJECT which has blunt conical nose at the front end TRAVELLING at HIGH supersonic speeds such as re-entry vehicle, missiles have FORMATION of shock waves. Thus, studying conical flow is of great significance.

The correct choice is (C) Is INVALID

For explanation I would say: The linearized velocity potential equation becomes invalid for flows at higher Mach number with LIMIT tends to infinity. It is ALSO not valid for flows over airfoil at higher angle of attack and high thickness to chard ratio.

The correct option is (a) Laplace equation

The best explanation: The linearized velocity potential equation is given by (1 – M\(_∞^2\))ϕxx + ϕyy + ϕzz = 0. When the Mach number approaches to ZERO, the equation TAKES the form ϕxx + ϕyy + ϕzz = 0 which is a form of Laplace equation ∇^2ϕ = 0 and the FLOW BECOMES INCOMPRESSIBLE.

Correct choice is (c) Pressure drag

The explanation: When the flow EXCEEDS critical Mach number, there is a FORMATION of supersonic REGION which is followed by a SHOCK wave. There is a pressure loss across the shock wave which results in large pressure drag.

The correct OPTION is (a) ∇s = 0

For EXPLANATION: The shock wave in the conical flow is assumed to be straight resulting in same INCREASE in the entropy for all streamlines passing through the shock (∇s = 0). This property implies that the conical flow is irrotational as per Crocco’s equation.

Right choice is (b) Incompressible FLOW to the compressible flow for same airfoil

The explanation: The Prandtl – Glauert equation is given by:

Cp = \(\frac {C_{p0}}{\sqrt {1 – M_∞^{2}}}\), Cl = \(\frac {C_{L0}}{\sqrt {1 – M_∞^{2}}}\), CD = \(\frac {C_{d0}}{\sqrt {1 – M_∞^{2}}}\)

This equation relates the pressure/lift/drag coefficient in incompressible flow Cp0 to the pressure/lift/drag coefficient in compressible flow (Cp) for a two – dimensional airfoil with the same profile.

The correct choice is (c) Flow over a cone

Explanation: Since the pressure over the conical SURFACE is not constant unlike the flow over the wedge, the streamlines tend to curve behind the SHOCK wave. The 3 – dimensional nature of conical flow provides the streamlines with an EXTRA space relieving any obstruction from the surface of the body. This is KNOWN as 3D relieving effect.

Correct choice is (b) SWEPT wing

The best I can explain: There are two ways employed to increase the critical Mach number thereby INCREASING drag – divergence Mach number. First is to reduce the thickness of the AIRFOIL, and SECOND is to add sweep to the wing.

The CORRECT answer is (a) Laplace’s equation

The explanation: The COMPRESSIBLE linearized perturbation velocity POTENTIAL equation is transformed into incompressible using a transformed COORDINATE system (ξ, η). The equation is GIVEN by

ϕξξ + ϕηη = 0

This is the Laplace equation representing the incompressible flow in a linearized form.