# MIT2_00AJs09_read04 - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 2.00AJ / 16.00AJ Exploring Sea, Space, & Earth: Fundamentals of Engineering Design Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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2.00AJ/16.00AJ Reading for Lecture #4 2.00AJ/16.00AJ Reading: Hydrostatic Pressure Effects and Archimedes’ Principle Prof. A.H. Techet Pressure Effects I. Hydrostatic Pressure Fluid forces can arise due to flow stresses (pressure and viscous shear), gravity forces, fluid acceleration, or other body forces. For now, let us consider a fluid in static equilibrium – with no velocity gradients (thus no viscous stresses). Forces are then due only to: 1. Pressure acting on the fluid volume 2. Gravity acting on the mass of the fluid 3. External body forces Using standard conventions we will consider pressure to be positive for compression. Recall that we said pressure is isotropic: Consider a triangular volume of fluid with height, dz , length, dx, and unit width, b , into the page: Figure 2.1: Elemental fluid volume version 5.0 updated 2/21/2008 -1- © 2008 A. Techet
2.00AJ/16.00AJ Reading for Lecture #4 The fluid element can support no shear while at rest (by our definition of a fluid). Thus the sum of the forces on the triangle, in the x- and z- directions, MUST equal zero: F x = p x b ( ! z ) " p n b ( ! s )sin # \$ = 0 (2.1) F z = p z b ( ! x ) " p n b ( ! s )cos \$ " 1 2 % b ( ! x ! z ) = 0 (2.2) We can simplify the above equations using simple geometry: ! z = ( ! s )sin " and ! x = ( ! s )cos , (2.3) such that and F x = p x ! p n ( ) " = 0 F z = p z ! p n ! 1/ 2 ( # z ) \$ = 0 (2.4) . (2.5) Taking the limit as x, z goes to zero (i.e. the triangle goes to a point), we see that p x = p z = p n = p . (2.6) Since θ is arbitrary, pressure at a point in a fluid is independent of orientation and is thus isotropic. Pressure (or any stress for that matter) causes NO net force on a fluid element unless it varies spatially! version 5.0 updated 2/21/2008 -2- © 2008 A. Techet

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Reading for Lecture #4 Take for example a small fluid element, δ x, δ y, δ z: Figure 2.2: Fluid Element Volume (z is positive upwards by convention). We will continue to assume that the only forces are due to gravity and pressure gradients. Let’s start by looking at the pressure acting on the top and bottom of the element volume. Using Taylor’s series expansion: p top = po + ! z 2 dp dz 0 + 1 2! z 2 " # \$ % 2 d 2 p dz 2 0 ( ... (2.7) Second term is positive b/c we are taking z positive upwards from the center of the element. p bot = p o ! " z 2 dp dz 0 + 1 2! z 2 # \$ % ( 2 d 2 p dz 2 0 + ... (2.8) Taking the limit as δ z goes to zero we are able to ignore the higher order terms and keep only up to second order. The resultant force due to pressure on the face is F press . © 2008 A. Techet
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