Sphere at rest in a uniform stream
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Consider a solid sphere of radius a at rest with its center being the origin of the system (r, theta, curly-phi).
The sphere is immersed in an infinite stream of an ideal fluid of density rho, with uniform velocity U in the positive z-direction.
5 answers
This is good content; thanks for sharing. However, is there a question here?
I've been watching diogenes' posts, and it seems he is posting both questions AND answers, not expecting any of our math or science tutors to reply. To me this is questionable use of Jiskha.
We consider only the n = 1 mode as described above: the corresponding solution to Laplace's equation is of the form:
phi(r, theta) = (A r + B/r^2) cos(theta)
We can adjust A and B in order to satisfy the boundary conditions, which are:
BC1:
v goes to U e_z as r goes to infinity
BC2:
v dot e_r = (d/dr)(phi) = 0 on r==a
We find (check!):
phi = U (r + a^3/(2 r^2)) cos(theta), r>=a
This function satisfies Laplace's equation and the boundary conditions, and so is the unique solution to the problem.
The corresponding velocity components are:
v_r = (d/dr)(phi) = U(1 - a^3/r^3) cos(theta)
v_theta = (1/r)(d/dr)(phi) = - U(1 + a^3/(2 r^3)) sin(theta)
v_curly-phi = (1/(r sin(theta))) (d/d curly-phi)(phi) = 0
The stagnation points are the points where v = 0, and here they occur for the points where r = a and theta = 0, pi.
Next we may try to draw the streamlines, the curves which have everywhere the same direction as v.
In spherical coordinates, a small displacement is given by
dr_ = dr e_r + r dtheta e_theta + r sin(theta) dcurly-phi e_curly-phi
So, if a streamline is given by the parametric form:
r = r(s)
theta = theta(s)
curly-phi= curly-phi(s)
then the tangent to this curve is given by:
{
(d/ds)(r)
,r (d/ds)(theta)
,r sin(theta) (d/ds)(phi)
}
This tangent must be parallel to the velocity field v at every point and so that streamlines are defined by the equations:
(1/v_r)(d/ds)(r) = (1/v_theta)(d/ds)(theta) = (1/v_curly-phi)(sin(theta))(d/ds)(curly-phi)
Here, they read:
dr / ( (1-a^3/r^3)cos(theta) ) = (- r dtheta) / ( (1 + a^3/ (2 r^3))sin(theta) ) = (r sin(theta) dcurly-phi) / 0,
r>=a
The second equality gives:
dcurly-phi = 0
i.e. curly-phi = constant.
The first equality, multiplied across by 2, gives:
( ( (2+a^3)/r^3 ) / ((r - a^3)/r^2) ) dr = -2 (cos(theta)/sin(theta)) dtheta
Multiply top and bottom by r in the first term gives
((2 r + a^3/r^2) / (r^2 - a^3)/r ) dr = -2 (cos(theta)/ sin(theta) ) dtheta
which we can integrate into:
ln(r^2 - a^3/r) = -2 ln( sin(theta) ) + C'
where C' is a constant of integration.
Re-arranging we find the equation for the streamlines:
(r^2 - a^3/ r ) sin(theta)^2 = C, r >= a
where C >= 0 is a constant.
In particular we notice that corresponding to C = 0 we have the streamline composed of axis:
theta = 0
and the contour of the sphere r = a.
By fixing a as a = 1, and picking several increasing values of C, we can easily generate the streamlines shown in Figure 3.
Applying Bernoulli's theorem in the absence of body forces, we find that the pressure on a streamlines is:
p(r, theta) = p_infinity + (1/2) rho U^2 (1 - (1- a^3/r^3)^2 cos(theta)^2 - (1- a^3/(2 r^3))^2 sin(theta)^2 )
where p_infinity is the pressure at infinity.
In particular, the pressure on the surface of the sphere r = a (corresponding to the C = 0 streamline) is:
p(a, theta) = p_infinity + (1/2) rho U^2 (1 - (9/4) sin(theta)^2)
The maximum and minimum values of the pressure on the sphere are thus:
p_max(a) = p_infinity + (1/2) rho U^2,
p_min(a) = p_infinity - (5/8) rho U^2,
taking place at theta == 0 and theta == pi (this is at the location of the rear and forward stagnation points) for the maxima and at theta == pi/2 (on the equator) for the minima.
Note that in the latter case it is possible to have p_min = 0, when U reaches the critical value of sqrt( (8 p_infinity) / (5 rho) ). This is known as cavitation, and has the effect that the fluid peels away from the surface leaving a vacuum behind, i.e. forming bubbles. It is seen for example in torpedoes or at the tips of propeller blades.
MAPLE-code
---------------------------
> restart: with(plots): a := 1
C := [0.00001, 0.01, 0.05, 0.1, 0.3, 0.6, 1.0, 1.5, 2.0, 2.5, 3.2]:
display(seq(implicitplot((r^2 - a^3/r )* sin(theta)^2 = C[i], r = 0..3*a, theta=0..2*pi, coords=polar, numpoints=90000, color=black, scaling=constrained), i=1..11));
Figure 3: MAPLE code used to draw some streamlines associated with a uniform flow moving around a fixed solid sphere. Note the presence of the stagnation points. The full 3D picture is obtained by rotating about the vertical axis (axial symmetry baby! (we are looking at it from the front)).
The net force F acting on the sphere is given by the effect of the fluid pressure on the surface
F = - int_{r==a} (p n dS) = int_{r==a} (p e_r dS) = int_{curly-phi==0}^{2 pi} int_{theta = 0}^{pi} p(a, theta) e_r a^2 sin(theta) dtheta dcurly-phi
phi(r, theta) = (A r + B/r^2) cos(theta)
We can adjust A and B in order to satisfy the boundary conditions, which are:
BC1:
v goes to U e_z as r goes to infinity
BC2:
v dot e_r = (d/dr)(phi) = 0 on r==a
We find (check!):
phi = U (r + a^3/(2 r^2)) cos(theta), r>=a
This function satisfies Laplace's equation and the boundary conditions, and so is the unique solution to the problem.
The corresponding velocity components are:
v_r = (d/dr)(phi) = U(1 - a^3/r^3) cos(theta)
v_theta = (1/r)(d/dr)(phi) = - U(1 + a^3/(2 r^3)) sin(theta)
v_curly-phi = (1/(r sin(theta))) (d/d curly-phi)(phi) = 0
The stagnation points are the points where v = 0, and here they occur for the points where r = a and theta = 0, pi.
Next we may try to draw the streamlines, the curves which have everywhere the same direction as v.
In spherical coordinates, a small displacement is given by
dr_ = dr e_r + r dtheta e_theta + r sin(theta) dcurly-phi e_curly-phi
So, if a streamline is given by the parametric form:
r = r(s)
theta = theta(s)
curly-phi= curly-phi(s)
then the tangent to this curve is given by:
{
(d/ds)(r)
,r (d/ds)(theta)
,r sin(theta) (d/ds)(phi)
}
This tangent must be parallel to the velocity field v at every point and so that streamlines are defined by the equations:
(1/v_r)(d/ds)(r) = (1/v_theta)(d/ds)(theta) = (1/v_curly-phi)(sin(theta))(d/ds)(curly-phi)
Here, they read:
dr / ( (1-a^3/r^3)cos(theta) ) = (- r dtheta) / ( (1 + a^3/ (2 r^3))sin(theta) ) = (r sin(theta) dcurly-phi) / 0,
r>=a
The second equality gives:
dcurly-phi = 0
i.e. curly-phi = constant.
The first equality, multiplied across by 2, gives:
( ( (2+a^3)/r^3 ) / ((r - a^3)/r^2) ) dr = -2 (cos(theta)/sin(theta)) dtheta
Multiply top and bottom by r in the first term gives
((2 r + a^3/r^2) / (r^2 - a^3)/r ) dr = -2 (cos(theta)/ sin(theta) ) dtheta
which we can integrate into:
ln(r^2 - a^3/r) = -2 ln( sin(theta) ) + C'
where C' is a constant of integration.
Re-arranging we find the equation for the streamlines:
(r^2 - a^3/ r ) sin(theta)^2 = C, r >= a
where C >= 0 is a constant.
In particular we notice that corresponding to C = 0 we have the streamline composed of axis:
theta = 0
and the contour of the sphere r = a.
By fixing a as a = 1, and picking several increasing values of C, we can easily generate the streamlines shown in Figure 3.
Applying Bernoulli's theorem in the absence of body forces, we find that the pressure on a streamlines is:
p(r, theta) = p_infinity + (1/2) rho U^2 (1 - (1- a^3/r^3)^2 cos(theta)^2 - (1- a^3/(2 r^3))^2 sin(theta)^2 )
where p_infinity is the pressure at infinity.
In particular, the pressure on the surface of the sphere r = a (corresponding to the C = 0 streamline) is:
p(a, theta) = p_infinity + (1/2) rho U^2 (1 - (9/4) sin(theta)^2)
The maximum and minimum values of the pressure on the sphere are thus:
p_max(a) = p_infinity + (1/2) rho U^2,
p_min(a) = p_infinity - (5/8) rho U^2,
taking place at theta == 0 and theta == pi (this is at the location of the rear and forward stagnation points) for the maxima and at theta == pi/2 (on the equator) for the minima.
Note that in the latter case it is possible to have p_min = 0, when U reaches the critical value of sqrt( (8 p_infinity) / (5 rho) ). This is known as cavitation, and has the effect that the fluid peels away from the surface leaving a vacuum behind, i.e. forming bubbles. It is seen for example in torpedoes or at the tips of propeller blades.
MAPLE-code
---------------------------
> restart: with(plots): a := 1
C := [0.00001, 0.01, 0.05, 0.1, 0.3, 0.6, 1.0, 1.5, 2.0, 2.5, 3.2]:
display(seq(implicitplot((r^2 - a^3/r )* sin(theta)^2 = C[i], r = 0..3*a, theta=0..2*pi, coords=polar, numpoints=90000, color=black, scaling=constrained), i=1..11));
Figure 3: MAPLE code used to draw some streamlines associated with a uniform flow moving around a fixed solid sphere. Note the presence of the stagnation points. The full 3D picture is obtained by rotating about the vertical axis (axial symmetry baby! (we are looking at it from the front)).
The net force F acting on the sphere is given by the effect of the fluid pressure on the surface
F = - int_{r==a} (p n dS) = int_{r==a} (p e_r dS) = int_{curly-phi==0}^{2 pi} int_{theta = 0}^{pi} p(a, theta) e_r a^2 sin(theta) dtheta dcurly-phi
Ah, unless you have separation or viscosity, the net force is zero.
and that is separation, not cavitation.
Separation is when viscosity has removed enough kinetic energy from the fluid so that it can no longer reach the pressure at the rear stagnation point and separates from the body leaving a wake of lower pressure than at the same angle at the bow of the sphere.
Cavitation is when the lowest pressure reaches the vapor point of the fluid and it turns to gas (cavitation bubble)
Both mean that the maximum pressure at the rear is lower than at the front. That is drag.
Separation is when viscosity has removed enough kinetic energy from the fluid so that it can no longer reach the pressure at the rear stagnation point and separates from the body leaving a wake of lower pressure than at the same angle at the bow of the sphere.
Cavitation is when the lowest pressure reaches the vapor point of the fluid and it turns to gas (cavitation bubble)
Both mean that the maximum pressure at the rear is lower than at the front. That is drag.
2 answers