Engineering Questions with Answers - Multiple Choice Questions
Aerodynamics – Newtonian Theory
1 - Question
What is the use of Newtonian theory?
a) Get pressure distribution over hypersonic body
b) Get pressure distribution over supersonic body
c) Get coefficient of drag over hypersonic body
d) Get coefficient of lift over hypersonic body
Explanation: We use Newtonian theory for the purposes of Hypersonic flow problems. It gives us the pressure distribution over the surface of the hypersonic body for both straight and curved surfaces. Newtonian theory gives an approximate estimate rather than an exact result.
2 - Question
What are the limiting conditions for Newtonian theory?
a) M∞ → ∞
b) γ → ∞
c) M∞ → ∞, γ → 1
d) M∞ → 1, γ → ∞
Explanation: Newtonian theory is applicable only under certain limitations. The first being the Mach number tends to infinity as we are talking about hypersonic flows and γ tends to 1. Only under these conditions are the results accurate.
3 - Question
Newtonian theory is more accurate for two – dimensional bodies than three – dimensional bodies.
Explanation: One of the conclusions derived from Newtonian theory is that the result of coefficient of pressure is more accurate for three – dimensional bodies such as the cone compared to the two – dimensional bodies such as a wedge.
4 - Question
Which of these equations is Newton’s sine – squared law?
a) Cp = 2sin2 θ
b) Cp = sin2 θ
c) Cp = sin2θ2
d) Cp = sin2θ4
Explanation: According to Newton’s sine – squared law, the pressure coefficient over the surface is proportional to the sine square of the angle between the tangent to the surface and freestream velocity (θ). It is given by: Cp = 2sin2 θ
5 - Question
What is the density behind the shock in hypersonic flow?
Explanation: The oblique shock relation is given by: ρ2ρ∞=(γ+1)M2∞sin2β(γ–1)M2∞sin2β+2 Applying the limit M∞ → 1 ρ2ρ∞=γ+1γ–1 Applying the limit γ → 1 ρ2ρ∞ → ∞ Thus, the density behind the shock wave in a hypersonic flow is infinitely large.
6 - Question
For which angle θ (between tangential pressure and freestream) does maximum coefficient of pressure exist for a blunt body?
a) θ = 0
b) θ = π
c) θ = π/2
d) θ = 2π
Explanation: For a blunt body, the coefficient of pressure is maximum at π/2. This result is derived from Newtonian theory according to which the coefficient of pressure is related to θ by: Cp = 2sin2 θ For sin π/2, which is the stagnation point, we get maximum value of coefficient of pressure.
7 - Question
What is the value of Cp at stagnation point for a hypersonic body?
Explanation: According to the Newtonian theory, Cp = 2sin2 θ. The maximum value for Cp exists at stagnation point where θ = π/2. For this value, the value of Cp is 2. This is contrast to the maximum value of Cp = 1 for an incompressible flow.
8 - Question
What is modified Newtonian pressure law?
a) Cp = Cp max sin2 θ
b) Cp = Cp max sin θ
c) Cp = Cp max cos2 θ
d) Cp = Cp max cos θ
Explanation: Lester Lees proposed modified Newtonian pressure law for blunt surfaces at hypersonic flow. According to the modified formula, coefficient 2 in Newtonian theory is replaced with Cp max. The maximum pressure over a blunt body arises at the stagnation point.
9 - Question
For which of these conditions was Newtonian law devised?
a) Force on a flat plate in hypersonic flow
b) Force on inclined plane in moving fluid
c) Force on curved airfoil in moving fluid
d) Force on flat plate in subsonic flow
Explanation: Isaac Newton worked on a fluid – dynamics theory which would address the force on an inclined plane in a moving fluid applicable to low – speed flow which was later applied to hypersonic flow. According to the law, the force varies as square of sine of the deflection angle.
10 - Question
The fluid flow in Newtonian theory is modelled as which of the following?
a) Stream of particles in rectilinear motion
c) Stream of particles in conic section
d) Particles moving in a random direction
Explanation: Newton formulated the Newtonian theory by modeling the fluid flow as a stream of particles that move in a rectilinear motion. The individual particles of the flow stream do not move in a random direction. The particles lose the momentum normal to the surface once it strikes. Its tangential momentum is retained without any loss.
11 - Question
In which direction does the streamline move downstream the shockwave in a hypersonic flow over a wedge?
a) Straight line
b) Parallel to the surface
c) No fixed direction
d) Perpendicular to the surface
Explanation: In a hypersonic flow, there is a formation of a thin shock layer over a wedge surface. The streamline pattern follows the trend such as upstream of the shock wave, the flow moves in a straight – line. On moving downstream of the shock wave, the flow is parallel to the wedge – surface inclined at the wedge – angle.
12 - Question
What is the coefficient of pressure over the surface which is in shadow (no impact from the flow present)?
Explanation: For each body, the freestream particle impacts the surface in the frontal region. The rear side remains un – impacted by this flow. This portion is known as the ‘shadow’ where no impact pressure is felt. The coefficient of pressure in this region is zero. For example for a flat plate that is kept at 90 degrees to the incoming freestream flow, the frontal region feels the impact, whereas the back surface is in shadow and the coefficient of pressure in this region is thus zero.
13 - Question
What the coefficient of lift for an inviscid flow is over a flat plate as obtained from Newtonian theory?
a) cd = 2sin3α
b) cd = 2sin2α
c) cd = 2sin2αcosα
d) cd = sin2αcosα
Explanation: For a flat plate of chord c kept at an angle α with freestream velocity, the pressure coefficient on the lower surface according to the Newtonian theory is given by: cpl = 2sin2α Coefficient of pressure on the upper surface is: cpu = 0 It is zero because the backward side of the flat plate does not experience the impact from the flow and is in shadow. The normal force coefficient formula is: cn = Nq∞S Where, S is the area of the flat plate = (c)(1). Integrating the coefficient of pressure on both the upper and lower surfaces yields the normal force coefficient: cn = 1c∫c0 (cpl – cpu)dx = 1c∫c0(2sin2 α – 0)dx = 1c(2sin2 α)c cn = 2sin2 α Coefficient of lift is: cl = cncosα = 2sin2αcosα