Question

Two vertical parallel plates are spaced 0.01 ft apart. If the pressure decreases at a rate of 60 psf/ft in the vertical z-direction in the fluid between the plates, what is the maximum fluid velocity in the z-direction? The fluid has a viscosity of 10-3 lbf-s/ft2 and a specific gravity of 0.80.

Answer 1

Answer:

umax = 0.1259ft/s

Explanation:

Given:

•Distance between plates, B = 0.01ft

•Pressure difference decrease, $\frac{dp}{dz}=60ps/ft$

•Fluid viscosity, u = 10^-³lbf-s/ft²

Specific gravity, S = 0.80

Max velocity in the z-direction will be:

$u_max= [\frac{B^2y}{8u}]\frac{dh}{ds}$

$But h = \frac{P}{y}+z$

Substituting for h in the first equation, we have:

$\frac{d}{dz}[\frac{p}{y}+z]$

$\frac{dh}{dz}=\frac{1}{y}\frac{dp}{ds}+\frac{dz}{dz}$

$= \frac{1}{0.8*62.4}(-60)+1$

= -0.20192

Substituting dh/dz value in the first equation (umax), we have:

$umax = \frac{0.01^2(0.8*62.4)}{8*10^-^3}(-0.20192)$

umax = 0.1259ft/s