Lagrangian Versus Eulerian Approach

-
Lagrangian Versus Eulerian Approach
Lagrangian Approach
Method of description that follows the particle is referred to as the Lagrangian method of description. In Lagrangian approach we analyze a fluid flow by assuming the fluid to be composed of a very large number of particles whose motion must be described.
How to do it?
Identify (or label) a material of the fluid; track (or follow) it as it moves, and monitor change in its properties. The properties may be velocity, temperature, density, mass, or concentration, etc in the flow field.
Eulerian Approach
In control volume analyses, it is convenient to use the field, or Eulerian, method of description, which focuses attention on the properties of a flow at a given point in space as a function of time. In the Eulerian method of description, the properties of a flow field are described as functions of space coordinates and time.
How to do it?
Identify (or label) a certain fixed location in the flow field and follow change in its property, as different materials pass through that location. In such case, the following property, say temperature is recorded by the sensor.
For better understanding watch these videos

Comments

Post a Comment

Popular posts from this blog

MCQs for Machine design

-
1.  Question 1 Material Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi​ Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6AL-4V. All of the bars have identical cross section areas of 1 in^2. An axial stress of 20 ksi is placed on each of the bars. Which bar will yield? AISI 4130 Steel Aluminum 6061 T4 Ti-6AL-4V All of the bars will yield Question 2 1 point 2.  Question 2 Material Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi​ Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6
Read more

Engineering Drawing - Orthographic Projection - General Questions

-
Image
1. The top, front, and bottom views align in this manner: A. Horizontally B. Vertically C. According to the planar views D. Parallel to the frontal plane 2. If a plane is parallel to the plane of projection, it appears: A. True size B. As a line or edge C. Foreshortened D. As an oblique surface 3. This line pattern is composed of three dashes, one long dash on each end with a short dash in the middle: A. Object B. Hidden C. Center D. Phantom 4. the plane upon which the top view is projected: A. Horizontal B. Frontal C. Profile D. Base 5. An advantage of this type of view is that each view shows the object all the way through as if it were transparent: A. Planar B. Horizontal C. Auxiliary D. Orthographic 6. This type of surface is tipped to all principal planes of projection and does not appear true size in any standard view: A. Foreshortened B. Parallel C. Orthographic D. Oblique
Read more

Engineering Drawing - Sectional Views - General Questions

-
Image
1. Drafters should use a ________ in a section view of a mechanical part that includes the cylindrical view of a threaded hole. A. Center line B. Hatch line C. Poly line D. Dimension line 2. The section view drawing in which one fourth of an object has been marked for removal is known as a ________ section. A. full B. half C. quarter D. none of the above  3. In offset sections, offsets or bends in the cutting plane are all: A. 90 degrees B. 180 degrees C. Either 90 or 180 degrees D. 30, 60, or 90 degrees 4. When filling an area with a hatch pattern in AutoCAD the drafter needs to be able to ________. A. see the entire bounding area to hatch B. set Ortho on C. turn ISO grid off D. set the layer to Defpoints
Read more