Question

Fluid Mechanincs :

Oil with γ-8.53 KN/m 3 and a viscosity μ-0.191 N s/m2 flows through the vertical pipe shown in the diagram at a rate of Q 4 x 10-4 m3/s. Gage fluid has γ-12.74 kN/m 3 Determine the manometer reading, h. Is the flow laminar? Hints. Ap-pi-p2 + γ f and hi-h-h2 + f S: Governing Equation(s):

Answer 1

Reynolds Number:

$Q = vA \Rightarrow v = Q/A = 4 * 10^{-4}/( \pi * 0.01^2 )= 1.273 m/s$

$R = \frac{\gamma v D}{\mu g} = \frac {8.53*10^{3} * 1.273 * 0.02}{0.191 * 9.81} = 115.91$

The flow is laminar

$\Delta P = P_1 -P_2 = \frac{128\mu l Q}{\pi D^4} -\gamma l = 43.7 kN/m^2$

$\Delta P = \gamma_{oil}\;h_1-\gamma_{gf} \;h+\gamma_{oil}\;h_2$

.

$\Delta P =- \gamma_{oil}\;(h_1+h_2)+\gamma_{gf} \;h = -\gamma_{oil}\;(l+h)+\gamma_{gf} \;h = -\gamma_{oil}l+(-\gamma_o+ \gamma_f)h$

$h = \frac{\Delta P - \gamma_{oil}l }{(\gamma_o - \gamma_f)} = \frac{43.7+8.530*4}{12.740-8.530} = 18.48 m$