Oil flow between fixed plates problem

Given data

$\mu = 0.12\ \text{Pa s}$, $\rho = 900\ \text{kg/m}^3$, $h = 0.01\ \text{m}$, $L = 5\ \text{m}$, $\Delta p = 20\,000\ \text{Pa}$, $b = 1\ \text{m}$.

Pressure gradient magnitude: $\left|\dfrac{dp}{dx}\right| = \dfrac{20\,000}{5} = 4000\ \text{Pa/m}$

(a) Maximum velocity

$$u_{max} = \frac{h^2}{8\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^2 \times 4000}{8 \times 0.12} = 0.4167\ \text{m/s}$$

(b) Discharge

Per unit width:

$$q = \frac{h^3}{12\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^3 \times 4000}{12 \times 0.12} = 2.778\times10^{-3}\ \text{m}^2/\text{s}$$

For $b = 1\ \text{m}$: $Q = 2.778\times10^{-3}\ \text{m}^3/\text{s} = 2.78\ \text{L/s}$.

(c) Mean velocity

$$\bar u = \frac{q}{h} = \frac{2.778\times10^{-3}}{0.01} = 0.2778\ \text{m/s} \quad (= \tfrac23 u_{max})$$

(d) Wall shear stress

$$\tau_w = \frac{h}{2}\left|\frac{dp}{dx}\right| = 0.005 \times 4000 = 20\ \text{Pa}$$

(e) Velocity at $y = 2.5\ \text{mm}$

$$u = \frac{1}{2\mu}\left|\frac{dp}{dx}\right|(hy - y^2) = \frac{4000}{0.24}\left(0.01 \times 0.0025 - 0.0025^2\right) = 0.3125\ \text{m/s}$$

(f) Head loss

$$h_f = \frac{\Delta p}{\rho g} = \frac{20\,000}{900 \times 9.81} = 2.27\ \text{m of oil}$$

Check for laminar flow

Taking the plate gap as the characteristic length:

$$Re = \frac{\rho \bar u h}{\mu} = \frac{900 \times 0.2778 \times 0.01}{0.12} = 20.8 \ll 2000$$

So the flow is laminar.

Answer: $u_{max} = 0.417\ \text{m/s}$, $Q = 2.78\ \text{L/s}$, $\bar u = 0.278\ \text{m/s}$, $\tau_w = 20\ \text{Pa}$, $u(2.5\ \text{mm}) = 0.3125\ \text{m/s}$, $h_f = 2.27\ \text{m}$