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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
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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. |