Laminar Flow Questions and Answers

Answer: C

Laminar flow is the flow in which the fluid particles move in parallel layers with little mixing or lateral crossover. Turbulent flow has random mixing and eddies, transitional flow lies between the two, and compressible flow concerns the change of density, not the pattern of layers, so laminar flow is the answer.

Answer: B

The average velocity is constant and equal to the average velocity in the fully developed region.

For incompressible flow the volume flow rate is the same at every cross-section, and the pipe area is constant, so the average velocity Q/A cannot change anywhere along the pipe.

What changes in the entrance region is the velocity profile: the boundary layer grows, the fluid near the wall slows down and the core speeds up, until the parabolic profile of fully developed laminar flow is reached. The centreline velocity rises, but the average stays fixed. So the options saying the average increases, decreases or is lower all contradict continuity.

Answer: C

The shear stress at the centerline is zero.

In fully developed pipe flow the shear stress varies linearly with radius, from zero at the axis to its maximum at the wall:

\[ \tau(r) = \frac{r}{2}\left(-\frac{dp}{dx}\right) \]

This follows from a force balance on a cylinder of fluid of radius r, and holds for laminar and turbulent flow alike. At r = 0 the velocity profile has its maximum, so its slope, and hence the shear stress, is zero by symmetry.

It does not vary with axial distance, since the flow is fully developed; it does not depend on centerline velocity; and it is certainly not infinite.

Answer: C

The pressure loss is doubled.

For fully developed laminar flow in a pipe, the Hagen-Poiseuille equation gives the pressure drop over length L:

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

The loss is directly proportional to the average velocity, so doubling V doubles \(\Delta p\). Equivalently, the friction factor f = 64/Re halves as Re doubles while \(V^{2}\) quadruples.

'Halved' and 'unaltered' are wrong. A squared dependence on velocity applies to fully turbulent rough-pipe flow, which would give four times the loss; 'trebled' fits no regime.

Answer: A

The match is P - 2; Q - 3; R - 1.

Between a fixed lower plate and a moving upper plate the velocity profile is the linear Couette profile plus a parabola driven by the pressure gradient:

\[ \begin{aligned} \frac{u}{U} &= \frac{y}{a} \\ &\quad - \frac{a^{2}}{2\mu U}\frac{\partial p}{\partial x}\,\frac{y}{a}\left(1 - \frac{y}{a}\right) \end{aligned} \]

Profile II is the straight line, so \(\partial p/\partial x = 0\). Profile I bulges forward beyond the line, which needs the parabola to add velocity, that is \(\partial p/\partial x \lt 0\), a favourable gradient. Profile III bulges backward, with reversed flow near the lower plate, which needs an adverse gradient, \(\partial p/\partial x \gt 0\). The other options swap these pairings.

Answer: A

The ratio is 0.5.

For laminar pipe flow the Darcy friction factor is \(f = 64/Re\), with \(Re = \rho V D/\mu\). At the same velocity and fluid, doubling the diameter doubles Re, so f halves:

\[ \frac{f_{2D}}{f_{D}} = \frac{Re_{D}}{Re_{2D}} = \frac{1}{2} \]

A ratio of 1.0 would mean f does not depend on Re, which is true only in fully rough turbulent flow, not laminar. 2.0 and 4.0 invert the dependence.

Answer: A

The constant is \(k = \frac{\rho g \sin\theta}{2\mu}\).

For steady, fully developed film flow the momentum balance along the incline is between viscous stress and the component of gravity along the plate, \(\rho g\sin\theta\):

\[ \mu\frac{d^{2}u}{dy^{2}} = -\rho g\sin\theta \]

The given profile \(u = k(2hy - y^{2})\) has \(d^{2}u/dy^{2} = -2k\), so \(-2\mu k = -\rho g\sin\theta\) and \(k = \rho g\sin\theta/(2\mu)\). It also satisfies no slip at the wall (\(u = 0\) at \(y = 0\)) and zero shear at the free surface (\(du/dy = 0\) at \(y = h\)).

The cosine options use the gravity component normal to the plate, which only sets the pressure across the film, and the options without \(\mu\) do not even have the units of k, which must be 1/(m s) for \(ky(2h - y)\) to be a velocity.

Answer: A

The ratio is 1.

With no pressure gradient and no body force, the x-momentum equation for fully developed flow reduces to

\[ \mu\frac{d^{2}u}{dy^{2}} = 0 \]

so the velocity is linear in \(y\) (Couette flow). With \(u(0) = -U_{2}\) and \(u(H) = U_{1}\), \(u = -U_{2} + (U_{1} + U_{2})\frac{y}{H}\), and the shear stress

\[ \tau = \mu\frac{du}{dy} = \frac{\mu(U_{1} + U_{2})}{H} \]

is the same at every \(y\), including both walls. Their ratio is therefore 1, whatever the plate speeds.

The options built from \(U_{1}\) and \(U_{2}\) would require a stress that varies across the gap, which happens only when a pressure gradient adds a parabolic part to the profile.

Answer: D

The relation is \(f = 4C_f\).

The Darcy factor appears in the head-loss equation \(h_f = f\frac{L}{D}\frac{V^{2}}{2g}\), while the Fanning factor is the wall shear stress scaled by dynamic pressure, \(C_f = \tau_w/(\tfrac{1}{2}\rho V^{2})\). A force balance on the pipe gives \(\tau_w = \frac{D}{4}\frac{\Delta p}{L}\), from which \(f = 4C_f\).

For laminar flow this reads \(f = 64/Re\) and \(C_f = 16/Re\). The other ratios (0.25, 0.5, 2) do not match these definitions.

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.