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Chapter 10 Fluids Phases Solid Liquid Gas Fluids Plasma Density p “rho” p = m/v SI Units Sometimes g/ cm3 1 kg/m3 = .001 g/cm3 Ex: Al p= 2.7 g/cm3 =2700 kg/m3 Kg/m3 specific gravity – the ratio of the density of that substance to the density of water at 4°C SG No Units The SG of any substance will be equal numerically to its density in g/cm3 or 10-3 times its density in kg/m3 continued… SG Pb= 11.3 Alcohol = .79 Al= 2.7 SG will tell you if substance floats or not >1 sink <1 Float Atmospheric Pressure Pressure due to the Atmosphere, changes with depth Earth’s Atmosphere is complicated P for air changes No distinct top surface to measure h Atmospheric Pressure is 1.013 x 105 Pa 14.7 psi 10 -3 This is another unit: Atm 1 atm = 1.013 x 105 N/m2 (Pa) = 101.3 kPa another unit is the bar 1 bar= 1 x 105 N/m2 1 bar= 100 kPa Gauges measure pressure. They measure over and above atmospheric pressure. To get absolute pressure, one must add atmospheric pressure to gauge pressure. P= Patm + PG Example car tire gauge reads 220 kPa, Absolute pressure within the tire is 220 kPa + 101 kPa =321 kPa 33 psi + 14.7 psi = 47.7 psi Pressure P= f/A Force/ Area Force applied to area SI unit is N/m2 Pascal, Pa 1 Pa = 1 N/m2 PSI? Feet psi Fluids exert a pressure equal in all directions *overhead picture* car tire, swimming pool In fluids, force always acts to the surface as depth increases within a fluid, so does pressure Formulas P= f/A f= mg =ma m= pv m=pAh P=pA hg A P= pgh Pressure is directly proportional to density of liquid and to depth within liquid This is just for the liquidNOT any external force on the liquid Example Prob. Blaise Pascal, French 1623 - 1662 States that pressure applied to a confined fluid increases the pressure throughout by the same amount Pascal's Principle carries with it hydraulics (pg. 280) Changing the Area changes the force For this to be true, the fluid must not compress (effectively they don’t) Pin = Pout Input Output Fout Fin Aout Ain Fin= Fout(Ain) Aout A small force can be used to exert a larger force by making the area of one piston larger than the area of another Small input area, Large output area greatly multiplies the input force F= 200 in A= 100 F = 2000 out A= 1000 Fout Fin MA Mechanical advantage If area is 20x greater then output force will be 20x greater Homework! Buoyancy 3 types of buoyancy: + rise - sink neutral equilibrium All objects appear to weigh less when submerged in a fluid Why a buoyant force? pg. 283 F1 F2> F1 F2 F2 is greater b/c there is more pressure at the lower depth This is how we derive the formula… FB = F2 – F1 FB = Apf gh2-Apf gh1 = Apfg(h2- h1) Apfgh (Ah = Vol) pfgV Also… pV = mass mfg=fB Recall F= AP p= m/v w= mg Vol= Ah so…. Fbuoyant = pfgV= mfg This is Archimedes Formula (Principle) In English= The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the object Lets Test out Archimedes Idea and see if he was right… 65.95 .646 Find m of object ____g weight ___N 3 3 7 7 x 10-6 Find V of Object ____cm _____m 58.5 g .573 Find m of submerged ____ weight ____N Find buoyant force on object 1. 2. 3. 4. fB = weight (air) – weight (submerged) = _______ .646 N - _______ .573 N .073 fB= _______N Now check w/ formula fB= pfgV 7 x 10 - 6 3 = 1 x 103 kg/m3(9.8m/s2) __________m .069 = ______ What about fB = Mfg % error = 5.7% Why do steel ships float? How can fish suspend themselves in H20 How do submarines work? Restatement : fB= Wf = pf Vg The volume of an object can also be found by: Vf= m pf = w gp Archimedes Principle applies to both submerged and floating objects fB = weight of object objects pfVdispg = poVog Vdisp = po Vo pf --for floating --g cancels Shortcut– sometimes applied (if p is known) Density of floating object = fraction of object that density of fluid floating in is submerged. wood Log 900 kg/m3 water 1000 kg/m3 9/10 of log is underwater = .9 90 % of log is submerged fluid dynamics (hydrodynamics if water) If flow is smooth- streamline or Laminar if flow is erratic, currents present – turbulent Viscosity- internal resistance within fluid Syrup (high viscosity) Water (low viscosity) Flux- term used to describe Volume of fluid passing through a given area each second Velocity D= Vt Area Rate of Flow - assume ideal fluid, frictionless, laminar flow rate R units= m3/s (volume/ time) m/s x m2 (m3/s) R= VA Velocity x Area Equation of continuity states that rate of flow remains constant. Velocity and area will change (inversely), but rate (m3/s) stays the same. Area 1 Velocity 1 Area 2 Velocity 2 because R(m3/t) stays the same… A1V1= A2V2 Large Velocity – Small Area V1= A2V2 A1 Fast Small Velocity- Large Area River Ex. Slow Narrow Wide Blood Flow Example pg. 288 Rate of flow of blood in human body stays same –Big Aorta to Small Capillaries Follow along w/ the book to see how the sample problem is solved. We can also find Power of a moving fluid Power = PR Power = pressure x flow rate N/m2 x m3/s Nm/s J/s Power = work / time Where the velocity of a fluid is high, the pressure is low, where the velocity is low, the pressure is high. 1 2 V slow V high P high P low continued… This makes sense because if the pressure at 2 were high it would slow down the fluid in 1, because the fluid has sped up, this corresponds to a lower pressure Want proof?! Car’s Convertible Top -- jeep Tarp in back of truck carburetor airplane wing chimneys/ outhouses perfume atomizer Ventilation in burrows hanging ping pong balls ping pong ball = funnel Here’s the equation P1 + ½ pV12 + pgy1 = P2 + 1/2pV22 + pgy2 When solving for P2 the formula looks like this: P2 = P1 + pg(y1-y2)+ ½ p(v21-v22) P= pressure p= density of flowing fluid g= gravity y= height V= velocity **Overhead Picture** Bernoulli’s Equation is really an expression of the law of energy conservation. The formula is derived from work energy principle - pg. 290 Go to overhead for equation… h A Vel