Laminar Flow Questions and Answers

Answer: A

The maximum velocity is \(\frac{1}{2\mu}\rho gh^{2}\).

Take y from the plate. For steady, fully developed film flow, gravity balances viscous shear: \(\mu\,d^{2}u/dy^{2} = -\rho g\). With no slip at the plate (\(u = 0\) at \(y = 0\)) and zero shear at the free surface (\(du/dy = 0\) at \(y = h\)):

\[ \begin{aligned} u(y) &= \frac{\rho g}{\mu}\left(hy - \frac{y^{2}}{2}\right) \\ u_{max} &= u(h) = \frac{\rho g h^{2}}{2\mu} \end{aligned} \]

The maximum is at the free surface, where the shear vanishes. \(\frac{1}{8\mu}\rho gh^{2}\) is the centreline value for a channel of width h with walls on both sides, and the other factors do not satisfy these boundary conditions.

Answer: A

The exponent is \(n = 1\).

Fully developed laminar flow in a circular pipe obeys the Hagen-Poiseuille equation:

\[ \Delta P = \frac{32\mu L V}{D^{2}} \]

The pressure drop is directly proportional to the average velocity, so \(K = 32\mu L/D^{2}\) and \(n = 1\). Equivalently, the laminar friction factor \(f = 64/Re\) falls as \(1/V\), cancelling one power of \(V\) in the Darcy-Weisbach \(V^{2}\).

\(n = 2\) is the fully rough turbulent limit, \(n = 1.75\) is smooth-pipe turbulent flow (Blasius friction factor), and 0.5 has no physical basis here.

Answer: B

The ratio is 2.

Fully developed laminar pipe flow (Hagen-Poiseuille flow) has a parabolic profile:

\[ u(r) = u_{max}\left(1-\frac{r^{2}}{R^{2}}\right) \]

Averaging over the circular cross-section:

\[ \begin{aligned} \bar{u} &= \frac{1}{\pi R^{2}}\int_{0}^{R}u_{max}\left(1-\frac{r^{2}}{R^{2}}\right)2\pi r\,dr \\ &= \frac{u_{max}}{2} \end{aligned} \]

So \(u_{max}/\bar{u} = 2\). The value 1.5 belongs to laminar flow between parallel plates, where the profile is parabolic across a slot rather than a circle. 3 and 4 have no basis for laminar flow.

Answer: D

The local velocity equals the average at \(y = \pm\frac{h}{\sqrt{3}}\).

Write the profile as \(u = u_{max}(1-y^{2}/h^{2})\). Its average across the gap is:

\[ \begin{aligned} \bar{u} &= \frac{1}{2h}\int_{-h}^{h}u_{max}\left(1-\frac{y^{2}}{h^{2}}\right)dy \\ &= \frac{2}{3}u_{max} \end{aligned} \]

Setting the local velocity equal to it:

\[ \begin{aligned} 1-\frac{y^{2}}{h^{2}} &= \frac{2}{3} \\ y &= \pm\frac{h}{\sqrt{3}} \approx \pm 0.577h \end{aligned} \]

\(\pm h/\sqrt{2}\) is where the velocity is half the maximum, and \(\pm h/2\) and \(\pm h/3\) give 3/4 and 8/9 of the maximum.

Answer: A

The difference is \(\frac{UL}{2}\).

In plane Couette flow with no pressure gradient the profile is linear, \(u = Uy/L\). The difference in stream function between two streamlines equals the volume flow rate per unit width between them:

\[ \begin{aligned} \psi_{T}-\psi_{B} &= \int_{0}^{L}u\,dy \\ &= \int_{0}^{L}\frac{Uy}{L}\,dy = \frac{UL}{2} \end{aligned} \]

Equivalently, the average velocity is U/2 across a gap L. UL would need the whole gap to move at U, and 2UL and 4UL exceed even that.

Answer: A

The equation of motion for laminar flow of a real fluid is the Navier-Stokes Equation.

The Navier-Stokes equations are Newton's second law for a fluid element including the viscous forces, so they describe real (viscous) fluids. Laminar flow is where they can be solved most readily.

Euler's equation is the same law with viscosity left out, so it applies to ideal fluids, and Bernoulli's equation is its integral along a streamline for steady, ideal, incompressible flow. The Hagen-Poiseuille equation is one particular solution of Navier-Stokes, for laminar flow in a circular pipe, not the general equation of motion.