Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2018, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here.
Given. Steady gravity-driven discharge from a large tank through a horizontal pipe; the tank surface is atmospheric and effectively at rest, the pipe exit is atmospheric.
Find. (a) the pipe head loss $h_f$; (b) the kinematic viscosity $\nu$; (c) whether the flow is laminar.
Figure 3. Oil drains from a large tank through a horizontal pipe; the depth $h$ is the driving head.
Approach. Apply the energy equation from the (still) oil surface to the pipe exit to get the head loss, assume the pipe flow is laminar (Hagen–Poiseuille) to back out the viscosity, then confirm the assumption with the Reynolds number.
Mean pipe velocity. With $A=\tfrac{\pi}{4}D^2 = 1.327\times10^{-4}\ \text{m}^2$,
$$V=\frac{Q}{A}=\frac{2.78\times10^{-4}}{1.327\times10^{-4}}=2.09\ \text{m/s}.$$
(a) Head loss. Energy from surface (1, at rest, elevation $h$) to exit (2, velocity $V$, datum), both atmospheric:
$$h = \frac{V^{2}}{2g}+h_f \;\Rightarrow\; h_f = h-\frac{V^{2}}{2g}=3-\frac{2.09^{2}}{2(9.81)}.$$
$$\boxed{h_f = 2.78\ \text{m}.}$$
(b) Kinematic viscosity. Assume fully-developed laminar flow, for which $h_f = \dfrac{32\,\mu L V}{\rho g D^{2}}$. Solving for $\mu$ and dividing by $\rho$:
$$\mu = \frac{h_f\,\rho g D^{2}}{32 L V}=0.0309\ \text{Pa}\cdot\text{s}, \qquad \boxed{\nu=\frac{\mu}{\rho}=3.44\times10^{-5}\ \text{m}^2/\text{s}.}$$
(c) Laminar? Check the Reynolds number:
$$Re=\frac{VD}{\nu}=\frac{(2.09)(0.013)}{3.44\times10^{-5}}=792 .$$
Since $Re = 792 \lt 2300$, the flow is $\boxed{\text{laminar}}$, consistent with the Hagen–Poiseuille assumption used in (b).
