Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 15, Problem 4CP
To determine

A sketch of unstructured grid using three-sided cell.

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The center of mass for a human body can be determined by a segmental method. Using cadavers, it is possible to determine the mass of individual body segments (as a proportion of total body mass) and the center of mass for each segment (often expressed as a distance from one end of the segment). Finding the overall body center of mass can be a complex calculation, involving more than 10 body segments. Below, we will look at a simplified model that uses just six segments: head, trunk, two arms, and two legs. Search y X As a percentage of total body mass, the head is 10%, the two arms are 10%, the trunk is 56%, and the two legs are 24%. The center of mass for each segment is given as an (x,y) coordinate, both units in cm: head = (0, 165), arms = (0, 115), trunk = (0, 95), and legs = (0, 35). Assume the body mass for the individual is 88 kg and their total height is 180 cm. Determine they and y coord 99+ H of mass
I am trying to find a Direction Cosine Matrix (DCM) for the Euler angle body 1-2-3 sequence. I tried making my own function and using the MATLAB function, but the result is a matrix that is transpose of each other. I mean that transpose(EA123toDCM) = E123toDCM. Why is that? Also, for the E123toDCM line, I am using the sequence 'ZYX'. Is that correct or should it be 'XYZ'? I know that that for a DCM of sequence 1-2-3 = R3(theta1)*R2(theta2)*R1(theta3). Is ZYX sequence the same as a 1-2-3 sequence?   EA = [pi/3; -pi/4; -pi/6];EA123toDCM = EA123DCM(EA) E123toDCM = angle2dcm(EA(1,1), EA(2,1), EA(3,1), 'ZYX')   function [R] = EA123DCM(EA)     theta1 = EA(1,1);    theta2 = EA(2,1);    theta3 = EA(3,1);            R1 = @(a)[1    0      0    ;              0  cos(a) -sin(a);              0  sin(a) cos(a)];              R2 = @(a)[cos(a)  0 sin(a) ;                0     1   0    ;              -sin(a) 0 cos(a)];              R3 = @(a)[ cos(a) -sin(a) 0;               sin(a) cos(a)  0;…
An incompressible CFD code is used to simulate the flow of gasoline through a two-dimensional rectangular channel in which there is a large circular settling chamber. Flow enters from the left and exits to the right as shown. A time-averaged turbulent flow solution is generated using a turbulence model. Top–bottom symmetry is assumed. Inlet velocity V is known, and outlet pressure Pout is also known. Generate the blocking for a structured grid using four-sided blocks, and sketch a coarse grid using four-sided cells, being sure to cluster cells near walls. Also be careful to avoid highly skewed cells. Label the boundary conditions that should be applied to every edge of every block of your computational domain.
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