Engineering Questions with Answers - Multiple Choice Questions
Computational Fluid Dynamics – Classification of PDE – 1
1 - Question
Which of these is not a type of flows based on their mathematical behaviour?
a) Circular
b) Elliptic
c) Parabolic
d) Hyperbolic
View Answer
Explanation: The three types of flows based on the mathematical behaviour are Elliptic, Parabolic and Hyperbolic. These behaviours decide the method used to solve the mathematical model of the flows.
2 - Question
Which type of flow does the Laplace’s equation (∂2Φ∂x2+∂2Φ∂y2=0) belong to?
a) Hyperbolic/ Parabolic
b) Hyperbolic
c) Parabolic
d) Elliptic
View Answer
Explanation: The general equation is in this form. A∂2Φ∂x2+B∂2Φ∂x∂y+C∂2Φ∂y2+D∂Φ∂x+E∂Φ∂y+FΦ+G=0 Comparing ∂2Φ∂x2+∂2Φ∂y2=0 with the above equation, A=1 B=0 C=1 To find the type, d=B2-4AC d=-4 As d is negative, Laplace’s equation is elliptical.
3 - Question
The lines along which the derivatives of the dependent variables are indeterminate are called ___________
a) parabolic lines
b) characteristic lines
c) hyperbolic lines
d) transition lines
View Answer
Explanation: The characteristic lines determine the type of the equation. These are defined as the lines along which the derivatives of the dependent variables do not exist.
4 - Question
Find the nature of the second-order wave equation.
a) Hyperbolic/elliptic
b) Parabolic
c) Hyperbolic
d) Elliptic
View Answer
Explanation: The second-order wave equation is ∂2u∂t2=c2∂2u∂x2 The general equation is in this form. A ∂2Φ∂x2+B∂2Φ∂x∂y+C∂2Φ∂y2+D∂Φ∂x+E∂Φ∂y+FΦ+G=0 Comparing ∂2u∂t2−c2∂2u∂x2=0 with the above equation, (let ‘y’ be ‘t’). A=-c2 B=0 C=1 To find the type, d=B2-4AC d=4c2 As d is positive, the second order wave equation is hyperbolic.
5 - Question
The classification of PDEs are governed by ________
a) Their highest order derivatives
b) Their least order derivatives
c) The number of terms
d) The constants
View Answer
Explanation: The highest order derivatives of a PDE determines its type. Only the coefficients A, B and C are used to find the type of A∂2Φ∂x2+B∂2Φ∂x∂y+C∂2Φ∂y2+D∂Φ∂x+E∂Φ∂y+FΦ+G=0.
6 - Question
Which of these equations are used to classify PDEs?
a) b(dydx)−c=0
b) a(dydx)2−b(dydx)=0
c) (dydx)2−(dydx)+1=0
d) a(dydx)2−b(dydx)+c=0
View Answer
Explanation: a(dydx)2−b(dydx)+c=0 is the characteristic equation for searching simple wave solutions. This is used to find the type of PDEs by substituting a, b and c by the coefficients of the second order derivatives of the given PDE.
7 - Question
Find the nature of the one-dimensional heat equation.
a) Circular
b) Elliptic
c) Hyperbolic
d) Parabolic
View Answer
Explanation: The one-dimensional heat equation is ∂T∂t=α∂2T∂x2 The general equation is in this form. A∂2Φ∂x2+B∂2Φ∂x∂y+C∂2Φ∂y2+D∂Φ∂x+E∂Φ∂y+FΦ+G=0 Comparing α∂2T∂x2−∂T∂t=0 with the above equation, (let ‘y’ be ‘t’). A=α B=0 C=0 To find the type, d=B2-4AC d=0 As d is zero, the one-dimensional heat equation is parabolic.
8 - Question
The mathematical classification of inviscid flow equations are different from that of the viscous flow equations because of __________
a) absence of viscosity coefficients
b) absence of higher order terms
c) absence of convective terms
d) absence of diffusive terms
View Answer
Explanation: The type of PDE is determined by the higher order terms. In the inviscid flow equations, the higher order terms are not present. So, the mathematical classification of inviscid and viscous flows is not the same.
9 - Question
Type of compressible flows depend upon _________
a) free stream pressure
b) free stream density
c) free stream velocity
d) free stream temperature
View Answer
Explanation: Classification of compressible flows depend on M∞ (the free stream Mach number). This, in turn, depends on the free stream velocity. So, the type is determined by free stream velocity.
10 - Question
Find the nature of this system.
(1−M2∞)∂u∂x+∂v∂y=0,∂u∂y−∂v∂x=0.
a) Hyperbolic/elliptic
b) Elliptic
c) Hyperbolic
d) Parabolic
View Answer
Explanation: For a system of equations, the general form is a1∂u∂x+b1∂u∂y+c1∂v∂x+d1∂v∂y=0 a2∂u∂x+b2∂u∂y+c2∂v∂x+d2∂v∂y=0 Comparing the given equations with these two equations, a1=1-M∞2, b1=0, c1=0, d1=1 a2=0, b2=1, c2=-1, d2=0 Here, A=a1 c2-a2 c1=M∞2-1 B=-a1 d2+a2 d1-b1 c2+b2 c1=0 C=b1 d2-b2 d1=-1 Now, d=B2-4AC = 4(M∞2-1) Here, When the flow is subsonic (M∞<1), d1), d>1 and the equations are hyperbolic.