ECH141- Hydrostatics

# Still another way of stating pascals law is that the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: till another way of stating Pascal's law is that the pressure is the same for any two points at the same elevation in a continuous body of fluid. The following figure illustrates this point. 2 Hydrostatics.nb Stress in Hydrostatic Systems The mathematical statement of Cauchy's First Law is a statement for the balance of linear momentum of a fluid body Ρv V t m Ρb V t m t t m n (1) A t We showed in a previous set of notes that the stress vector t n at a point in a fluid can be expressed in terms of the stress tensor T: t nT n (2) where the stress tensor T is defined as a sum of dyadic products T Tij ei ej For a fluid at rest Eqn. (1) becomes Ρb V 0 m t t m n A (3) t And since a fluid at rest cannot support any shear forces, the stress vector becomes t pn n (4) where p is the hydrostatic pressure. To show that the pressure is the same in every direction we consider a small tetrahedron shown in Figure 2 D C A B Our tetrahedron has 3 mutually orthogonal faces (ADC, ABD, ABC) with areas x, y, z , and a slant face (BDC) with area . The outward directed unit normals to these 4 faces are -i, -j, -k, and n. We apply (4) to our tetrahedron to get Ρg V 0 m t t m n A (5) t Since the volume of our tetrahedron is small we express the volume integral in terms of average values: Hydrostatics.nb Ρg V Ρg V (6) t m Let the stress acting on the face with outward directed normal -i be t with normals -j and -k be t t m A n t t and t j t i where we have used the fact that t related to the area of the slant face in jn kn z j t t y j py x , and in a similar way on faces A k t n A z t x i px y A j y t i . Thus the surface integral in (6) can be written as k A i x x 3 y k pz k t z z (7) n n pn p n . The surface areas of the 3 mutually orthogonal faces rae by simple projections nx ny nz (8) Thus our balance of hydrostatic forces becomes Ρg 0 V n pn Now we divide through by 0 n pn i nx px j ny py and take the limit V/ i nx px j ny py k nz pz (9) 0 to get k nz pz (10) Recall that n i nx j ny k nz (11) Substituting for n in (11) we get 0 i nx pn px j ny pn py k nz pn pz (12) If we take the scalar product of (13) with i , j , k respectively we get pn px 0, pn py 0, pn pz 0 (13) Hence the stress vector on any arbitrary surface in a fluid at rest is given by t pn n (14) Note that the stress vector t n acts in a direction opposite to that defined by the unit normal...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online