College of Engineering and Computer Science Mechanical Engineering Department
ME 692 – Computational Fluid Dynamics Spring 2002 Ticket: 57541 Instructor: Larry Caretto
Solutions to Sixth Homework Assignment 1. Review the grid transformation notes leading to equation [20] on page 5. Show that the remaining, unique, off-diagonal terms, g13 and g23, which are not shown in equation [20] are zero. The metric coefficients, gij are given by equation [18] from the grid-transformation notes, where yi are the Cartesian coordinates and xi represent the new coordinate system:
g ij
yk ym y y y y y y km 1 1 2 2 3 3 xi x j xi x j xi x j xi x j
[18]
The necessary partial derivatives for the cylindrical coordinates are shown in equation [19] from the notes.
y1 cos( x2 ) x1 y 2 sin( x2 ) x1 y3 0 x1
y1 x1 sin( x2 ) x2 y 2 x1 cos( x2 ) x2 y3 0 x2
y1 0 x3 y2 0 x3 y3 1 x2
[19]
Using equation [18] and the appropriate derivatives from [19] gives the following results, which shows that the remaining off-diagonal metric coefficients are zero. g13
y1 y1 y 2 y 2 y3 y3 cos( x2 )[0] sin( x 2 )[0] [0][1] 0 x1 x3 x1 x3 x1 x3
g 23
y1 y1 y 2 y 2 y3 y3 x1 sin( x 2 )[0] x1 cos( x 2 )[0] [0][1] 0 x 2 x3 x 2 x3 x 2 x3
2. For the spherical polar system the three coordinates are x1, the distance from the origin to a point on a sphere, x2, the counterclockwise angle on the x-y plane from the x axis to the projection of the r coordinate on the x-y plane, and x3 = the angle from the vertical axis to the line from the origin to the point. (These coordinates are more conventionally called r, θ, and φ.) The transformation equations from Cartesian coordinates (y1, y2, and y3) to spherical polar coordinates are given by the following equations: x1 = y12 + y22 + y32, x2 = tan-1(y2/y1), and x3 = tan-1[ y12 + y22/y3]. The inverse Engineering Building Room 2303 E-mail:
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ME 692, L. S. Caretto, Spring 2002
Homework Six Solutions
transformation to obtain Cartesian coordinates from spherical polar coordinates is: y1 = x1 cos(x2)sin(x3), y2 = x1 sin(x2)sin(x3), and y3 = x1cos(x3). Find all components of the metric tensor for this transformation. Verify that this is an orthogonal coordinate system. What are the three possible differential areas for this system? What is the volume element for this system? The solution to this equation requires the application of equation [18] from the notes, shown in problem one, to the coordinate transformation spherical coordinates. The partial derivatives required for this computation are the same as those shown in equation [19], but applied to the coordinate transformation given in the problem statement. Those partial derivatives are shown below. Note that these derivatives use the transformation equations for transforming the spherical coordinate system back into Cartesian coordinates.
y1 y1 y1 cos( x2 ) sin( x3 ) x1 sin( x2 ) sin( x3 ) x1 cos( x2 ) cos( x3 ) x1 x2 x3 y 2 y 2 y 2 sin( x2 ) sin( x3 ) x1 cos( x2 ) sin( x3 ) x1 sin( x2 ) cos( x3 ) x1 x2 x3 y3 y3 y3 cos( x3 ) 0 x1 sin( x3 ) x1 x2 x2 Using these derivatives in equation [18], gives the following metric coefficients. g11
y1 y1 y 2 y 2 y3 y3 cos 2 ( x2 ) sin 2 ( x3 ) sin 2 ( x2 ) sin 2 ( x3 ) x1 x1 x1 x1 x1 x1
cos 2 ( x3 ) sin 2 ( x3 ) cos 2 ( x2 ) sin 2 ( x2 ) cos 2 ( x3 ) sin 2 ( x3 ) cos 2 ( x3 ) 1 g12
y1 y1 y 2 y 2 y3 y3 x1 sin( x2 ) cos( x2 ) sin 2 ( x3 ) x1 x2 x1 x2 x1 x2
x1 sin( x2 ) cos( x2 ) sin 2 ( x3 ) cos( x3 )[0] 0 g13
y1 y1 y 2 y 2 y3 y3 x1 cos 2 ( x2 ) cos( x3 ) sin( x3 ) x1 x3 x1 x3 x1 x3
x1 sin 2 ( x2 ) cos( x3 ) sin( x3 ) x1 cos( x3 ) sin( x3 )
x1 cos( x3 ) sin( x3 ) cos 2 ( x2 ) sin 2 ( x2 ) 1 0 g 22
y1 y1 y 2 y 2 y3 y3 2 2 x1 sin( x2 ) sin( x3 ) x1 cos( x2 ) sin( x3 ) [0]2 x2 x2 x2 x2 x2 x2
x12 sin 2 ( x2 ) cos 2 ( x2 ) sin 2 ( x3 ) x12 sin 2 ( x3 )
Homework Six Solutions
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ME 692, L. S. Caretto, Spring 2002
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y1 y1 y 2 y 2 y3 y3 x12 sin( x2 ) sin( x3 ) cos( x2 ) cos( x3 ) x2 x3 x2 x3 x2 x3
x12 sin( x2 ) sin( x3 ) cos( x2 ) cos( x3 ) x1 sin( x3 )[0] 0 g 33
y1 y1 y 2 y 2 y3 y3 2 2 x1 cos( x2 ) cos( x3 ) x1 sin( x2 ) cos( x3 ) x3 x3 x3 x3 x3 x3
x1 sin( x3 ) x12 cos ( x3 ) x1 cos 2 ( x2 ) sin 2 ( x2 ) x12 sin 2 ( x3 ) 2
x12 cos 2 ( x3 ) x12 sin 2 ( x3 ) x12 We have the usual symmetry relationship for the off-diagonal coefficients so that g21 = g12 = 0, g31 = g13 = 0, and g32 = g23 = 0. The differential areas are found from equation [26] in the notes, which is copied below.
(dS ) i g jj g kk g 2jk dx j dxk (no summation) i, j , k cyclic i 1, 2, 3
[26]
Using the metric coefficients just found, we obtain the following differential areas. 2 (dS )1 g 22 g 33 g 23 dx2 dx3 x12 sin 2 ( x3 ) x12 0dx2 dx3 x12 sin( x3 )dx2 dx3
(dS ) 2 g 33 g11 g132 dx2 dx3 x12 (1) 0dx3 dx1 x1dx3 dx1 (dS ) 3 g11 g 22 g122 dx1dx2 (1) x12 sin 2 ( x3 ) 0dx1dx2 x1 sin( x3 )dx1dx2
In these equations, (dS)i represents the differential area facing the ith coordinate direction. For example, x1 corresponds to the r coordinate direction. So the equation that we have just derived tells us that the differential area in this direction is r2sin(φ)dφdθ, where φ is the polar angle and θ is the angle in the x-y plane. This differential area represents the differential area on a spherical surface, a distance r from the origin, when the change in the polar angle creates a differential length of rdφ on the surface of the sphere and the angle in the x-y plane creates a differential path length rsin(φ)dθ. The differential volume element for this coordinate system is found from equation [30] in the notes, which is copied below. dV
g dx1dx2 dx3 Det ( g ij ) dx1dx2 dx3
[30]
For an orthogonal coordinate system, such as the one we have here, where all the off-diagonal gij are zero the determinant is simply the product of the diagonal terms. This gives the differential volume element by the following equation.
dV g11 g12 g13 dx1dx2 dx3 (1) x12 sin 2 ( x3 ) x12 dx1dx2 dx3 x12 sin( x3 )dx1dx2 dx3
