## Fluid Dynamics and the Navier-Stokes Equations

The steady flow in a pipe is a very simple solvable case. Since the left-hand side of Eq.(2) vanishes for steady flow, and there is no external force, it can be written in cylindrical coordinates (Figure 03a) as: 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.

• Eq.(5) shows that the constant velocity contours are arranged in layers of concentric cylinders with the maximum velocity vm at the center. There is no mixing of layers. This is one way to define laminar flow.
• • The volume flow rate Q is defined as the volume flowing through per unit time:
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.

• The definition of vm in Eq.(6) shows that the flow is driven by the pressure gradient ( - dp / dz). Viscosity is one of the factors in determining the velocity of the flow, which becomes slower for more viscous fluid. The radius of the pipe R also has an influence on the flow, which becomes slower in smaller pipe. The flow comes to a halt with vm = 0 at the limit of either ( - dp / dz) = 0 or ~ infinity.
• The flow comes to a stop at the wall. This is the consequence of an assumption that the frictional force between the fluid and the wall is infinite. The effect is similar to the thin film of liquid sticking to a solid surface. If this is not the case, then vm would take the more general form:

vm = (R2 / 4 ) (- dp / dz) + v0 ---------- (7)

where v0 is the velocity of the flow at the wall. The solution for the velocity field in Eq.(5) becomes:

v (r) = vm ( 1 - r2 / R2 ) + (r / R)2 v0---------- (8)
• 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] • Instead of maintaining the pressure gradient by a pump, the flow can be driven by gravity if it is tilted by an angle A with respect to the horizon (Figure 04b). Then

(- 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.

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