Discussion
A thin film of an incompressible, Newtonian liquid (density \(\rho\), viscosity \(\mu\)) with an uniform thickness (\(h\)) is flowing down on a vertical plate. The flow is driven by gravity (\(g\)) alone. Assume zero shear stress condition at the free surface. The maximum velocity is given by
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.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development