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<br />Assuming no mechanical ,.;:)r~ is done on the system, the force (F) <br /> <br />can result only from pressure <br />the points in question and is <br /> <br />energy due to the columns of fluid above <br />, <br />equal to the product of pressure and <br /> <br />strealiltube area. Equation 2..4 can be rel~ritten as: <br /> <br />I/ORK = P2A . dl - PIA . dl <br /> <br />where: <br /> <br />Pl and P2 = pressures at points 1 and 2, respectively. <br /> <br />Substituting equations 2-2, 2-3 and 2-5 into equation 2-1: <br /> <br />2 . <br />W . V2/2g + 12 W + Y02 W + (P2A . dl - PIA . dl) (2-6) <br /> <br /> <br />= W . V~/29 + 11 . W + YOl . W + 6[ <br /> <br />uividing both sides of equation 2-6 by W results in the following: <br /> <br />(V2/29)2 + 12 + Y02 + { P2A . <br /> <br /> <br />= (V2/29)1 + 11 + YOl + 6E/W <br /> <br />dl - PIA <br />W <br /> <br />. dl ) <br /> <br />(2-7) <br /> <br />The total weight W can be expressed as the product of unit weight, <br />y, (in lb/ft3 or kg/m3) times the volume of water flowing where volume <br />is A . dl, (in ft3 or m3). A is the cross sectional area in ft2 or m2 <br />and dl is the unit length in feet or meters. Replacing W in equation <br />2 -7 wi th y . A . d 1 , defi n i ng the change in energy, 6E/W, as 1\ and <br />neglecting the internal energy component, r. yields the Bernoulli <br />equation: <br /> <br />2.04 <br />