By J. A. Fox (auth.)

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**Additional resources for An Introduction to Engineering Fluid Mechanics**

**Example text**

21 Area of plate = 2 m 2 = 0·5 + 1 cos 60° = 1·00 Force = area x pressure at centroid = 2 x 9810 x 1·00 Depth of centroid = 19·6 kilonewtons This force acts perpendicularly to the plate so it is acting in a direction which is 30° to the vertical. Location of centre of pressure-: X=x +kb/x i = O·S/cos 60° + 1 = 2·0 0·333 2·00 X= 2·00+--= 2·17 m 26 An Introduction to Engineering Fluid Mechanics Questions Questions throughout this book are set in both the British and the SI systems of units. The values quoted in square brackets are in the SI system.

10 Fig. 11 44 An Introduction to Engineering Fluid Mechanics for the element is simply 21Trv. C then The concept of circulation is not limited to small elements. If a large element is drawn and this is split up into small elements as shown in Fig. 12 it will be Fig. 12 seen that circulations around the interior small elements are neutralised by those around adjacent elements except at the boundary of the large element. Thus the circulation of all the small elements when summed must equal the circulation around the large element.

7 is an example of rotation without shearing deformation. In Fig. 5 a case is illustrated in which both shearing deformation and rotation is occurring. 5 The Navier-Stokes equations When the Euler equations were developed it was assumed that there were no frictional forces acting upon the fluid element. In any real fluid such forces must exist and the Euler equations cannot give an accurate description of the behaviour of the fluid especially if its viscosity is large. Shear stresses must be exerted on all faces of the parallelepiped shown in Fig.