x u =
= 0, are: 
= thermodynamic internal energy, p = pressure, T = temperature,
= density,
= viscosity, KH = heat conduction coefficient, F = external force per unit mass,
, and
.
, temperature T, and three components of the velocity vector u. Together with the equation of state such as the ideal gas law - p V = n R T, the six equations are just enough to determine the six dependent variables. In general, all of the dependent variables are functions of all four independent variables. Usually, the Navier-Stokes equations are too complicated to be solved in a closed form. However, in some special cases the equations can be simplified and may admit analytical solutions (see "Differential Equation" for a very brief introduction).
u = 0. It is an idealization used to simplify analysis. In reality, all fluids are compressible to some extent.
usL)/
, where us is the mean fluid velocity, and L is the characteristic length, e.g., the cross-section of the pipe) can be used to evaluate whether viscous or inviscid equations are appropriate to the problem. High Reynolds numbers indicate that the inertial forces are more significant than the viscous forces. However, even in high Reynolds number regimes certain problems require that viscosity be included. In particular, problems calculating net forces on bodies (such as the wings on aircraft) should use viscous equations. Stokes flow occurs at very low Reynold's numbers, such that inertial forces can be neglected compared to viscous forces. ![]() |
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instead modeled using one of a number of turbulence models and coupled with a flow solver that assumes laminar flow outside a turbulent region. Turbulence usually occurs below a Reynold's numbers of 3000. It causes increased energy loss (as heat), more drag (on the moving body), and generates sound wave (noise). |
Figure 01 Turbulent Flow [view large image] |
Figure 02 Laminar Flow [view large image] |
Modern vehicle and aircraft designs always try to minimize the turbulence by adopting a smooth surface and streamlined contour. |
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0 = - dp / dz + (1/r) d (r d v / dr) / dr ---------- (4)where v denotes the velocity field in the z direction as shown in Figure 02. For a constant pressure drop (per unit length) d p / dz is a negative constant, the solution becomes: v (r) = vm ( 1 - r2 / R2 ) ---------- (5) where vm = (R2 / 4 ) (- dp / dz) ---------- (6) |
Figure 03a Cylindrical Co- ordinates [view large image] |
is the velocity at the centre, and R is the radius of the pipe. The boundary conditions are v = vm at r = 0 and v = 0 at r = R. |
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Q = v 2 r dr = [( R4)/(8 )] (-dp/dz) ---------- (6a)The mass flow rate is just Q x . The flow rate formula in Eq.(6a) is also applicable approximately to the blood flow in the artery. The p and v become temporally varying pulses (Figure 03b). The formula shows that shrinking artery as well as thick blood will |
Figure 03b Blood Pressure [view large image] |
lower the blood flow rate. |
~ infinity.
) (- dp / dz) + v0 ---------- (7) ![]() |
In case when the pressure gradient is absent, i.e., (- dp / dz) = 0 in Eqs.(4) and (7). The above solution is reduced to v(r) = vm = v0, that is, the flow has an uniform velocity similar to the case shown in Figure 04a for the uniform flow of inviscid fluid. But in this example, the viscosity term disappears in Eqs.(4) and (7) because (- dp /dz) = 0; and thus has no influence on the flow. It seems that the flow can last forever without a driving force. In reality, it would finally come to a halt by the dissipative effects and the discharge at the end of the pipe (in this example an infinitely long pipe is assumed). |
Figure 04a Uniform Flow [view large image] |
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(- dp / dz) = g sin(A) ---------- (9)where is the density of the fluid, g = 980 cm/sec2 is the acceleration of gravity, and
|
Figure 04b Flow by Gravity Feed [view large image] |
sin(A) is related to the slant of the pipe. Such method is a convenient way to transport liquid, but it does not work for gas since its density is about 1000 times lower. |
) dp/dz ---------- (10)
[(ux1)2 - (ux2)2]} / 2 = (p2 - p1) ---------- (11)![]() |
where the subscripts 1 and 2 respectively denote the flow velocity ux, and the pressure p to the upper and lower layers of the air flow. It shows that if the structure of the wings are designed to create a higher flow velocity in the upper layer, then a net pressure in the |
Figure 05 Lift for Aeroplane |
upward direction is created to lift the aeroplane up into the air provided such force can overcome the weight [see Figure 05, the length of the arrows is proportional to the magnitude of ux (blue) and p (red)]. The force to lift up the plane is just f = P x Sref , |
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where P = p2 - p1 is the net upward pressure, and Sref (Reference Area) is the effective area acted upon by P. Figure 06 shows the Sref for an aircraft and a helicopter. The actual force F on these objects is determined further by the formula F = f x CL, where CL is the Coefficient of Lift. It depends on the type of aircraft as shown in Figure 06, where the angle of attack is |
Figure 06 Ref. Area & Lift Coeff. [view large image] |
defined as the angle between the wing and the direction of the airstream. Most aircrafts will behave similarly to the Cessna 172 while high-speed planes with short wingspans, like fighters, will more closely resemble the Lightning data. |
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The lift for helicopter can be derived from Eq.(2) of the Navier-Stokes Equations in a similar way (as for the aeroplane) with the role of ux and uz interchanged because the airflow pattern is different (see Figure 07). Thus for uz >> ux , Eq.(10) becomes: uz [d(uz)/dz] = - (1/ ) dp/dz ---------- (12)Integrating Eq.(12) yields: { [(-uz2)2 - (-uz1)2]} / 2 = (p2 - p1) ---------- (13)
|
Figure 07 Helicopter Airflow [view large image] |
where uz is negative as it is pointing toward the negative z direction. It has a form similar to Eq.(11) for the aeroplane. Computation for the lifting force follows exactly the same line as developed previously. |
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In stationary position, the helicopter's engine provides only the lifting force. According to the principle of angular momentum conservation, the body would turn in the opposite direction of the rotating blades. To stabilize the helicopter, a tail rotate is installed to counteract this trend. By applying more or less pitch angle to the tail rotor blades, it can |
Figure 08 Helicopter |
also be used to make the helicopter turn left or right. Forward motion is achieved by tilting the spinning rotor in the direction of the flight (see Figure 08). There are many factors related to the rotor blades to limit the maximum speed of a helicopter at about 400 km/h. |
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By rotating the rotor blades and the z-axis 90o, Eq. (13) is applicable to calculate the forward pressure (or thrust) of a propeller on a ship or airplane (Figure 09). Since the density for water is about 1000 times higher than air, the formula shows that the propeller can generate more thrust with lower flow speed on a ship or boat. It also explains why the prop plane cannot fly high up into the rare atmosphere, where the air density is very low.
|
Figure 09 A Prop Plane [view large image] |
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relatively high, central-pressure zone. As the air diverges from the central region, it is deflected by the Coriolis force in a clockwise circulation (Figures 10 & 12). Thus, most Highs are generally elliptical in shape following their formation. But as they interact with other air masses and topography, and are distorted by forces of the upper atmosphere, high pressure cells often become long and |
Figure 10 High Pressure Ridge |
Figure 11 Low Pressure Cell |
narrow in shape, and is referred to as high pressure ridge in the weather map. Since the air at high altitude is dry, the High is usually associated with fair weather. |
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Mathematically, the sinking and rising air can be explained by Archimedes' principle as discussed in the section of Hydrostatics. Cooler air mass will sink as it is denser than the surrounding air, and vice versa for the warmer air mass. The swirling motion of air on the horizontal plane is determined by the Navier-Stokes Equations in Eq.(2). Since the radial velocity ur is usually much smaller than the circular velocity v = r u in the core region, the radial component of the equation (in cylindrical coordinates) can be reduced to: |
Figure 12 Coriolis Force |
2 v sin = (1/ ) p/ r ---------- (14a) |
is the angular velocity of the Earth's rotation, and
is the angle of the latitude (Figure 12). Eq.(14a) shows that for the low pressure cell, the Coriolis force on the left-hand side is balanced by the pressure gradient force on the right-hand side, and the air circulates in the counter-clock wise direction as depicted in Figure 11. When the radial velocity diverges from the center as for the case of high pressure cell, both forces change sign leaving the equation in exactly the same form with the circular velocity moving in clock wise direction.
, and v
does not depend on z, then the circular component of Eq.(2) (in cylindrical coordinates) can be simplified to:
/r) (
v
/
) = - 2
ur sin
---------- (14b)
= 0 at the equator, the effect of the Coriolis force vanishes there as shown in both Eqs.(14a) and (14b).![]() |
The Great Red Spot (Figure 13) in Jupiter provides a very good example to illustrate the Coriolis force at work. It is a high pressure cell located 22o South of the equator. Thus, the rotational vector is pointing inward to the center (instead of pointing outward as for the case in the Northern hemisphere), and the swirling gas is circulating counter-clock wise. This system of anticyclonic storm has existed for up to 400 years. The long lifetime cannot be attributed entirely to the higher rotational speed (about twice as much as that for the Earth), and hence the stronger Coriolis force. It is suggested that the lack of solid surface to |
Figure 13 Great Red Spot |
provide friction may play a part contrary to the hurricane on Earth, which always break up shortly after landfall. Note that the oval shape is caused by the constriction from the neighboring cloud bands. |
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phenomena such as spiral galaxy, hurricane, and drain in the sink (Figures 14, and 15). By neglecting the thickness of the spiral flow, Eq.(2) of the Navier-Stokes Equations in cylindrical coordinates can be expressed in the form: ur ( ur/ r) + u ( ur/![]() ) = Fr ---------- (15a)ur [ (r u )/ r] + u [ (r u )/![]() ] = 0 ---------- (15b)where we have assumed further that the force acts on the fluid |
Figure 14 Barred Spiral Galaxy |
Figure 15 Hurricane |
only in the radial direction. In addition by assuming constant density for the fluid, Eq.(1) of the continuity equation becomes: (1/r) [ (r ur)/ r] +
(1/r) [ (r u )/![]() ] = 0 ---------- (16) |
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A relationship between the velocity components is obtained by substituting Eq.(16) to Eq.(15b): ur = b u ---------- (17)where b is a constant having the dimension of length. This formula can be integrated once more to yield: r = a + b ---------- (18)
|
Figure 16 Archimedean Spiral |
where a is another constants of integration. Eq.(18) expresses the trajectory of an Archimedean spiral (see curve on the left of Figure 16). So far there is no restriction on whether the movement is to plunge inward or to expand. |
is the relative rate of change between r and
, a small value of b makes the winding very tight and vice versa.
-
or 180o out of phase (see curves on the right of Figure 16).
0, the spiral assumes a broad sweeping pattern much like the hurricane in Figure 15 instead of one slim locus.
2 / r![]() |
with M to be the mass of the central black hole, while the mass on the disk within r is given by M' = 2 ![]() r dr, where = 0 (r0/r) is the surface density in unit of gm/cm2, 0 is its cutoff value at the edge of the galaxy r = r0, and the third term is the centrifugal force. Thus, the radial and rotational components of the velocity can be expressed as:ur2 = GM / r + 2 G 0r0 - r2u 2 ---------- (19b),r2u 2 = r (GM + 2 G 0r0 r) / (b2 + r2) ---------- (19c),
|
Figure 17a Rotation Curve for Milky Way [view large image] |
which has the same profile of the rotation curve for the disk as shown in Figure 17a with a peak at rm (4![]() 0r0b/M) b for 4![]() 0r0b >> M. The observed curve takes into account of the dark matter in the halo. |
r02
0
1011Msun,
3x106Msun,
30 kpc,
0.5 kpc,
0.015 kpc, which implies tightly wound arms, and
0r0b/M)
33 - enough to qualify for a peak in the rotation curve.
122 km/sec, and ur
0.06 km/sec respectively.
2GM/c2 matter plunging into the hole with radial velocity ur
c / 21/2 (where c is the velocity of light), but the rotation velocity reduces to rsu
4 km/sec.
= 0 to 2
as shown in Figure 14 is about 20 kpc. This is obvious a contradiction, and that's why the simple spiral flow in galaxy from fluid dynamics consideration alone is at best a toy model as mentioned earlier.
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dp/dr (which has overwhelmed the Coriolis and centrifugal forces in this case): ur2 / r = (1/ ) dp/dr ---------- (20). The dependence of p on r is shown in Figure 17b. The observational data (for the "Charlie" type hurricane) can be approximated by the empirical formula: p = p0 [5.5 - e-k(r - re)] ---------- (21) for r re, where re is the distance from the center to the wall of the eye (Figure 17b), p0 = 220 mb and k is a constant related to the steepness of the pressure gradient. Thus, the pressure gradient is: dp/dr = kp0e-k(r - re) ---------- (22) for r reor ur2 = (krp0/ ) e-k(r - re) ---------- (23) for r re.At the wall of the eye, r = re, and ur2 = krep0/ .
|
Figure 17b Pressure Gradient of Hurricane |
The rotation curve is given by r2u 2 = (kr3p0/b2 ) e-k(r - re), which reproduces a profile similar to the curve in Figure 17b with a maximum at rm = 3/k. This kind of analysis is also applicable to the drain in the sink although on a much smaller scale. |
40 km,
10 km,
)
2 at re.
0.075 km-1,
)] re
20 km.
turns one cycle. There is no tightly wound arms. This model seems to represent the actual system quite well. Meanwhile, the calculated radial and circular velocities at the wall of the eye are ur
4.6 m/h and reu
2.3 m/h respectively indicating a calm region there.
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The term "hydrostatics" is applied to the study of fluids at rest, including both liquids and gases. For this special case, u = 0 in the Navier-Stokes equations, the external force F = -g (in the negative y direction) and the pressure gradient is just dp. Eq.(2) is reduced to the simple form: dp = - g dy ---------- (24).Taking point 1 and 2 as shown in Figure 18, integration of Eq.(24) yields: p2 - p1 = - g (y2 - y1) ---------- (25)
|
Figure 18 Hydrostatics |
Figure 19 Atmosphere [view large image] |
which shows that the pressure is less in higher elevation as it is well known in the Earth's atmosphere (Figure 19). |
V g ---------- (26)![]() |
Submarine controls the submerging and surfacing by admitting water into a tank or pumping it out. When the submarine is on the surface its ballast tanks are full of air (Figure 20). The air inside the hull brings the average density of the entire ship below the density of saltwater. Because the sub is less dense, it floats. The average density of a submerging submarine is increased by flooding the ballast tanks with water through flood ports on the bottom of the tanks with the release of air out of the top of the tanks (Figure 20). Once the overall density of the submarine is equal to the water around it, it has the neutral buoyancy and will remain at that depth. High pressure air is blown into the ballast tanks to push the water out when the submarine is prepared to surface again. |
Figure 20 Submarine |
Air does not provide a lot of buoyant force because its density is about 1000 times lower. Only balloon filled with low density gas such as hydrogen or helium can rise up to the sky. |
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Solid body does not flow like fluid, the Navier-Stokes Equations doesn't seem to be applicable in such case. However, the equation can be used to calculate the critical mass for the self-gravity of a solid body to overcome its resistant forces so that it assumes a hydrostatic equilibrium (nearly round) shape. Solid |
Figure 21 Deformation of Solid Body |
Figure 22 Geometric Shape [view large image] |
body actually has elastic property, that makes it to deform in response to the external force as shown by the deformation of the moon by the gravitational tug of a planet in Figure 21. |
0 in the Navier-Stokes Equation (2), which then becomes:
p /
---------- (27).
p = dp / dr = Y / r ---------- (29)
r. ---------- (30)
Y /
r, the solid body does not has enough gravitational pull to overcome the resistance, its geometric shape would be irregular as shown by the Itokawa asteroid in Figure 22.
r, the solid body just has enough gravitational pull to overcome the resistance, its geometric shape would be round as shown by the Earth/Moon in Figure 22.
Y /
r, the solid body has more than enough gravitational pull to overcome the resistance. The object would shrink until equilibrium is achieved.
= 4
r3/ 3, thus Eq.(30) can be rewritten as:
/3)Y/G)1/2r2 ---------- (31)
8x109r2) . If the mass M on the left-hand side of Eq.(31) is greater than the value on the right-hand side, the value of Y is then adjusted to maintain equilibrium at the observed radius r.| Object | Type | Mass M (gm) | Radius or Size r (cm) | 8x109r2(gm) | Adjusted Y (1012dynes/cm2) | Geometric Shape |
|---|---|---|---|---|---|---|
| Neutron Star | Star | 0.2 - 6.0x1033 | ~1.0x106 | ~8.0x1021 | Super-hard | Spherical |
| Mercury | Planet | 3.6x1026 | 1.1x108 | 1.0x1026 | 13 | Spherical |
| Venus | Planet | 5.7x1027 | 5.7x108 | 2.6x1027 | 4.8 | Spherical |
| Earth | Planet | 6.0x1027 | 6.0x108 | 2.9x1027 | 4.0 | Spherical |
| Moon | Satellite | 7.2x1025 | 7.8x107 | 4.9x1025 | 2.0 | Spherical |
| Mars | Planet | 6.6x1026 | 3.0x108 | 7.2x1026 | 0.84 | Spherical |
| Pluto | Dwarf Planet | 1.2x1025 | 1.2x108 | 1.2x1026 | 0.01 | Spherical |
| Phobos | Satellite of Mars | 1.0x1019 | 2.7x106 | 5.8x1022 | N/A | Irregular |
| Deimos | Satellite of Mars | 1.8x1018 | 1.5x106 | 1.8x1022 | N/A | Irregular |
| Itokawa | Asteroid | 3x1013 | 5.4x104 | 2.4x1019 | N/A | Irregular |
| Comet | Planetesimal | ~6x1015 | ~6x104 | ~3x1019 | N/A | Irregular |
| Human | Living Organism | ~105 | ~2x102 | ~3x1014 | N/A | Definitely not Spherical |
| Neutron | Composite Particle | 1.67x10-24 | ~10-13 | ~8x10-17 | N/A | Unknown |
| Electron | Elementary Particle | 9.1x10-28 | ~10-16 | ~8x10-23 | N/A | Unknown |