16C (Fluid Dynamics)

For the steady, fully developed ansatz

u = w (r) e_{z}

, both

\partial_{t} u

and

(u \cdot \nabla) u

vanish. The axial Navier-Stokes equation reduces to

0 = - \frac{\partial p}{\partial z} + μ \frac{1}{r} \frac{d}{d r} (r \frac{d w}{d r}) .

Two integrations give

w (r) = \frac{1}{4 μ} \frac{\partial p}{\partial z} r^{2} + C_{1} lo g r + C_{2} .

Regularity at the axis forces

C_{1} = 0

w (R) = 0

w (r) = - \frac{1}{4 μ} \frac{\partial p}{\partial z} (R^{2} - r^{2}) .

For flow in the positive

z

direction,

\partial p / \partial z < 0

b

L

fluid cylinder is

F_{p} = π R^{2} [p (z_{0}) - p (z_{0} + L)] = - π R^{2} L \frac{\partial p}{\partial z} .

From part (a),

μ w^{'} (R) = \frac{R}{2} \frac{\partial p}{\partial z} .

Multiplying this wall shear stress by the lateral area

2 π R L

gives

F_{wall} = 2 π R L μ w^{'} (R) = π R^{2} L \frac{\partial p}{\partial z} .

Consequently

F_{p} + F_{wall} = 0,

which is the required steady force balance.

\dot{M} = ρ \int_{0}^{2 π} \int_{0}^{R} w (r) r d r d θ = - \frac{ρ π R ^{4}}{8 μ} \frac{\partial p}{\partial z} .

Thus

\dot{M} = - \frac{ρ π R ^{4}}{8 μ} \frac{\partial p}{\partial z} .

There are no discussions about this article yet.

There are currently no matching articles.