1/29/16 Oregon State University PH 212, Class #121 Other implications of buoyancy: ･ If an object floats in a fluid,  object ≤  fluid. ･ If the object sinks,  object >  fluid. ･ The fraction of a floating object’s volume that is immersed in the fluid is equal to the ratio of its density to that of the fluid: V obj.immersed / V obj.total =  object /  fluid ･ Even when an object does not float (i.e. when it is denser than the fluid—it sinks), the buoyant force is still present: Its weight (as would be indicated by a scale) is reduced by F B.

1/29/16 Oregon State University PH 212, Class #122 Incompressible Fluid Behavior Conservation of mass: The Equation of Continuity “Steady flow” means the same mass per second flows past any given point. (And in an incompressible fluid, the particles don’t “bunch together” or “spread out” within the flow.) Conservation of energy: Bernoulli’s Equation Just as with any other object, unless external work is done on a given mass of fluid, its total mechanical energy remains constant.

1/29/16 Oregon State University PH 212, Class #123 The Speed of a Fluid What you put into a pipe has to come out of the pipe. Steady flow means that the same amounts of mass must both enter and exit the pipe each second; the mass flow rate, dm/dt is constant—at every point in the pipe. But notice: dm/dt = d(  V)dt, or  (dV)dt (because the fluid is incompressible). And dV = Adx, where A is the cross-sectional area of the flow at that point. So the mass flow rate is:  Adx/dt =  Av And this rate is the same at any two points in a steady flow of incompressible fluid:  1 A 1 v 1 =  2 A 2 v 2

1/29/16 Oregon State University PH 212, Class #124 Continuity: Constant Volume Flow Rate (Incompressible Fluids) If a fluid is incompressible, its density  does not change—it’s the same everywhere in the flow—so we can write :  1 A 1 v 1 =  2 A 2 v 2 more simply: A 1 v 1 = A 2 v 2 This equates volume flow rates, Q = Av at any two points. Consider: If the mass flow rate is the same everywhere, what must be true where the pipe is narrower? Either: (a) The fluid particles pack closer together so the same number of particles can still move at the same speed. But this doesn’t happen in an incompressible fluid. So… (b) The particles move faster where the pipe is smaller.

1/29/16 Oregon State University PH 212, Class #125 Which of these units could represent energy density? 1.(N·m)/m 3 2.Pa 3.J/m 3 4.All of the above. 5.None of the above.

1/29/16 Oregon State University PH 212, Class #126 Conservation of Energy: Bernoulli’s Equation Consider some of the forms of mechanical energy that a given mass of (incompressible) fluid may have: ･ Its speed determines its K T : (1/2)mv 2 or (1/2)  Vv 2 ･ Its elevation determines its U G : mgh or  Vgh ･ Its pressure determines its U FL, another form of potential energy (the fluid version of elastic energy): PV After all, the pent-up pressure in a corked bottle can convert to kinetic energy just like a coiled spring.

1/29/16 Oregon State University PH 212, Class #127 So the total mechanical energy in a (non-rotating) incompressible fluid is: PV + (1/2)  Vv 2 +  Vgh More simply, that energy sum per unit volume is: P + (1/2)  v 2 +  gh And unless some work is done on the fluid, this sum is constant everywhere in the flow. For any two points: P 1 + (1/2)  v  gh 1 = P 2 + (1/2)  v  gh 2 The energy forms may interchange (just as with a solid object), but they must still add up to the same total. Bernoulli’s Equation is simply a re-statement of the work- energy theorem as applied to incompressible fluid flow.

1/29/16 Oregon State University PH 212, Class #128 To apply Bernoulli’s Equation, just identify the pressure, speed and height of the fluid at the desired point(s) in the fluid flow. And even more good news: Bernoulli’s Equation is a good starting point (close approximation) even for situations and fluids that aren’t ideal. ･ Air flow ･ Pumps ･ Very slow fluid flow in a large container

1/29/16 Oregon State University PH 212, Class #129 Pressure at Depth in a Static Fluid P 1 + (1/2)  v  gh 1 = P 2 + (1/2)  v  gh 2 Notice what Bernoulli’s Equation says about a static fluid: P 1 +  gh 1 = P 2 +  gh 2 If h 2 > h 1, point 1 is deeper in the fluid than point 2, so: P deep = P shallow +  g  h Example: At what depth in a freshwater lake would the pressure be equal to twice atmospheric pressure? (1 atm = 1.01 x 10 5 Pa, and  water = 1000 kg/m 3.)

1/29/16 Oregon State University PH 212, Class #1210 Tools that Use Effects of Static Fluid Pressure Archimedes’ Principle: ･ All flotation devices Pressure at depth: ･ Pressure gauges ･ Drinking straws and barometers Pascal’s Principle: ･ Hydraulics