Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the no-slip condition? What causes it?

Short Answer

Expert verified
Answer: The no-slip condition is a fluid flow boundary condition which states that the fluid velocity at a solid boundary is equal to the velocity of the solid surface, meaning the fluid particles at the interface share the same speed as the solid surface. This condition arises from the interaction between fluid molecules and the solid surface, as fluid particles tend to stick to the solid due to forces like adhesive forces, Van der Waals forces, and electrostatic forces.

Step by step solution

01

Definition of No-Slip Condition

The no-slip condition is a common fluid flow boundary condition which states that, at a solid boundary, the fluid velocity is equal to the velocity of the solid surface. In other words, the fluid particles at the interface between the fluid and a solid surface have the same speed as the solid surface.
02

Cause of No-Slip Condition

The no-slip condition arises from the interaction between fluid molecules and the solid surface. When fluid particles are in contact with a solid surface, they tend to stick to it due to various forces, which include adhesive forces, Van der Waals forces, and electrostatic forces. As the fluid is in contact with the solid surface, the velocity of fluid particles decreases due to friction. Ultimately, fluid particles at the interface share the same velocity with the solid surface forming a layer known as the boundary layer.
03

Example of the No-Slip Condition

To better understand the no-slip condition, let's consider a common example: fluid flow over a flat horizontal plate. When the fluid flows over a stationary plate, the no-slip condition implies that the fluid particles in direct contact with the stationary plate will have zero velocity. As we move away from the plate, the layer of fluid with a non-zero velocity increases, thus forming a boundary layer. The formation of the boundary layer highlights the impact of the no-slip condition on the fluid flow characteristics.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider an airplane cruising at an altitude of \(10 \mathrm{~km}\) where standard atmospheric conditions are \(-50^{\circ} \mathrm{C}\) and $26.5 \mathrm{kPa}\( at a speed of \)800 \mathrm{~km} / \mathrm{h}$. Each wing of the airplane can be modeled as a \(25-\mathrm{m} \times 3-\mathrm{m}\) flat plate, and the friction coefficient of the wings is \(0.0016\). Using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions. Answer: $89.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$

For the same initial conditions, one can expect the laminar thermal and momentum boundary layers on a flat plate to have the same thickness when the Prandtl number of the flowing fluid is (a) Close to zero (b) Small (c) Approximately one (d) Large (e) Very large

During air cooling of a flat plate $(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, the convection heat transfer coefficient is given as a function of air velocity to be \(h=27 \mathrm{~V}^{0.85}\), where \(h\) and \(V\) are in \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. At a given moment, the surface temperature of the plate is \(75^{\circ} \mathrm{C}\) and the air $(k=0.0266 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( temperature is \)5^{\circ} \mathrm{C}$. Using appropriate software, determine the effect of the air velocity \((V)\) on the air temperature gradient at the plate surface. By varying the air velocity from 0 to \(1.2 \mathrm{~m} / \mathrm{s}\) with increments of $0.1 \mathrm{~m} / \mathrm{s}$, plot the air temperature and plate temperature gradients at the plate surface as a function of air velocity.

What is a similarity variable, and what is it used for? For what kinds of functions can we expect a similarity solution for a set of partial differential equations to exist?

The upper surface of a \(1-\mathrm{m} \times 1-\mathrm{m}\) ASTM B 152 copper plate is being cooled by air at \(20^{\circ} \mathrm{C}\). The air is flowing parallel over the plate surface at a velocity of $0.5 \mathrm{~m} / \mathrm{s}$. The local convection heat transfer coefficient along the plate surface is correlated with the correlation \(h_{x}=1.36 x^{-0.5}\), where \(h_{x}\) and \(x\) have the units \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m}\), respectively. The maximum use temperature for the ASTM B152 copper plate is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). Evaluate the average convection heat transfer coefficient \(h\) for the entire plate. If the rate of convection heat transfer from the plate surface is \(700 \mathrm{~W}\), would the use of ASTM B152 plate be in compliance with the ASME Code for Process Piping?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free