Engineering Questions with Answers - Multiple Choice Questions
Aerodynamics – Linearized Velocity Potential Equation
1 - Question
What will be the x – component of velocity for a slender body which is immersed in uniform flow having perturbations?
a) Vx = V∞ + u‘
b) Vx = V∞ + v‘
c) Vx = V∞ + w‘
d) Vx = V∞
View Answer
Explanation: 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‘
2 - Question
What is the velocity potential for a slender body in uniform flow with perturbations?
a) Φ(x, y, z) = V∞ x + ϕ(x, y, z)
b) Φ(x, y, z) = V∞ z + ϕ(x, y, z)
c) Φ(x, y, z) = V∞ y + ϕ(x, y, z)
d) ∇Φ = u‘i + v‘j + (V∞ + w‘)k
View Answer
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: ∂ϕ∂x = u‘, ∂ϕ∂y = v‘, ∂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)
3 - Question
In which equation is total velocity and it double derivative substituted to obtain the perturbation velocity potential equation?
a) Momentum equation
b) Velocity potential equation
c) Perturbation equation
d) Enthalpy equation
View Answer
Explanation: The velocity potential equation is given by: (1 – Φ2xa2) Φxx + (1 – Φ2ya2) Φyy + (1 – Φ2za2) Φzz – (2ΦxΦya2) Φxy – (2ΦxΦza2) Φxz – (2ΦyΦza2) 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 a2 we get (a2 – (V∞ + ϕx)2)ϕxx + (a2 – ϕy2) ϕyy + (a2 – ϕz2) 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.
4 - Question
Linearized perturbation velocity potential equation is applicable for transonic flow.
a) True
b) False
View Answer
Explanation: 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 M2∞[(γ – 1) u‘V∞+(γ+12)u‘2V2∞+(γ–12)(v‘2+w‘2V2∞)]∂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.
5 - Question
Which of these assumptions are not made while obtaining the linearized perturbation velocity potential equation?
a) Small perturbations are there
b) Transonic flow is excluded
c) Hypersonic flow is excluded
d) Subsonic flow is excluded
View Answer
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.
6 - Question
Linearized velocity potential equation is applicable to hypersonic flow.
a) True
b) False
View Answer
Explanation: 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 M2∞[(γ – 1) u‘V∞+(γ+12)u‘2V2∞+(γ–12)(w‘2+u‘2V2∞)]∂v‘∂x is small compared to the left hand side thus is neglected to arrive at the linearized velocity potential equation.
7 - Question
Which of these assumptions is invalid for the linearized velocity potential equation?
a) u‘V∞ << 1
b) v‘V∞ << 1
c) w‘V∞ >> 1
d) w‘V∞ << 1
View Answer
Explanation: The perturbation velocity potential equation is given by: (a2 – (V∞ + ϕx)2) ϕxx + (a2 – ϕy2) ϕyy + (a2 – ϕz2) 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∞.
8 - Question
Which of these is linearized velocity potential equation?
a) (1 – M2∞)ϕxx + ϕyy + ϕzz = 0
b) ϕxx + (1 – M2∞)ϕyy + ϕzz = 0
c) ϕxx + ϕyy + (1 – M2∞)ϕzz = 0
d) (1 – M2∞)[ϕxx + ϕyy + ϕzz] = 0
View Answer
Explanation: When the assumptions of small perturbations and transonic, hypersonic conditions are excluded, the linearized velocity potential equation is found out as follows: (1 – M2∞)ϕxx + ϕyy + ϕzz = 0 Where, ϕxx = ∂2ϕ∂x2, ϕyy = ∂2ϕ∂y2, ϕzz = ∂2ϕ∂z2
9 - Question
Which equation is satisfied when the Mach number approaches to zero in linearized velocity potential equation?
a) Laplace equation
b) Momentum equation
c) Energy equation
d) Euler’s equation
View Answer
Explanation: The linearized velocity potential equation is given by (1 – M2∞)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.
10 - Question
What happens to the linearized velocity potential equation for flow over high thickness – chord ratio?
a) Becomes zero
b) Becomes 1
c) Is invalid
d) Becomes infinity
View Answer
Explanation: 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.