Consider flow in a blood vessel with cross-sectional area A(x, t). For simplicity we assume that the flow is a plug flow, with velocity that is a scalar quantity u and is a function of axial distance along the vessel only. Poiseuille flow becomes plug flow in the limit of zero viscosity, and so in the following analysis we omit consideration of viscous forces. The volume of the vessel of length L is f^ A(x, t) dx, and thus conservation
of mass requires that
Taking the partial derivative of (15.131) with respect to L and replacing L by x gives
According to Newton's law, the rate of change of momentum is equal to the total force exerted. Thus, if P is defined to be the excess pressure generated by the heart (i.e., the difference between the actual pressure and the resting pressure), then conservation of momentum demands that d / 'L
dt (pj^ A(x,t)u(x,t)dx^ = pA(x, t)u2(x, t)\°x=L + P(x,t)A(x,t)\°x=L. (15.133)
Note that pA(0,t)u(0,t) is the rate at which mass enters the vessel across the surface x = 0, so that pA(0, t)u2(0, t) is the rate at which momentum enters the vessel across this surface. Differentiating (15.133) with respect to L and replacing L by x, we find that p ((Au)t + (Au2)x) = -(PA)x. (15.134)
This second equation can be simplified by expanding the derivatives and using (15.132) to get p (ut + uux) = -Px (15.135)
as the equation for the conservation of momentum.
For simplicity we assume that the vessel is a linear compliance vessel with
In the analysis that follows, one can use a more general relationship between area and pressure, but the basic conclusions remain unchanged. With this expression for the cross-sectional area, the conservation equation becomes c(Pt + uPx) + A(P)ux = 0. (15.137)
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