Chapter 10 Fluids Phases Solid Solid Liquid Liquid Gas Gas Fluids Fluids Plasma Plasma Densit Density.

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Presentation transcript:

Chapter 10 Fluids

Phases Solid Solid Liquid Liquid Gas Gas Fluids Fluids Plasma Plasma Densit Density

p “rho” p “rho” p = m/v SI Units Kg/m 3 p = m/v SI Units Kg/m 3 Sometimes g/ cm 3 Sometimes g/ cm 3 1 kg/m 3 =.001 g/cm 3 1 kg/m 3 =.001 g/cm 3 Ex: Al p= 2.7 g/cm 3 Ex: Al p= 2.7 g/cm 3 =2700 kg/m 3

specific gravity – the ratio of the density of that substance to the density of water at 4°C 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/cm 3 or times its density in kg/m 3 The SG of any substance will be equal numerically to its density in g/cm 3 or times its density in kg/m 3

continued… SG SG Pb= 11.3 Pb= 11.3 Alcohol =.79 Alcohol =.79 Al= 2.7 Al= 2.7 SG will tell you if substance floats or not >1 sink 1 sink <1 Float

Atmospheric Pressure Pressure due to the Atmosphere, changes with depth Pressure due to the Atmosphere, changes with depth Earth’s Atmosphere is complicated Earth’s Atmosphere is complicated P for air changes P for air changes No distinct top surface to measure h No distinct top surface to measure h Atmospheric Pressure is x 10 5 Pa 14.7 psi 10 -3

This is another unit: Atm 1 atm = x 10 5 N/m 2 1 atm = x 10 5 N/m 2 (Pa) = kPa (Pa) = kPa another unit is the bar 1 bar= 1 x 10 5 N/m 2 1 bar= 1 x 10 5 N/m 2 1 bar= 100 kPa 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. Gauges measure pressure. They measure over and above atmospheric pressure. To get absolute pressure, one must add atmospheric pressure to gauge pressure. P= P atm + P G

Example car tire gauge reads 220 kPa, Absolute pressure within the tire is 220 kPa kPa car tire gauge reads 220 kPa, Absolute pressure within the tire is 220 kPa kPa =321 kPa =321 kPa 33 psi psi = 47.7 psi 33 psi psi = 47.7 psi

Pressure P= f/A Force/ Area Force applied to area P= f/A Force/ Area Force applied to area SI unit is N/m 2 Pascal, Pa SI unit is N/m 2 Pascal, Pa 1 Pa = 1 N/m 2 PSI? 1 Pa = 1 N/m 2 PSI? Feet Feet psi psi

Fluids exert a pressure equal in all directions Fluids exert a pressure equal in all directions *overhead picture* car tire, swimming pool In fluids, force always acts to the surface In fluids, force always acts to the surface as depth increases within a fluid, so does pressure 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 Pressure is directly proportional to density of liquid and to depth within liquid This is just for the liquid- NOT any external force on the liquid This is just for the liquid- NOT any external force on the liquid Example Prob.

States that pressure applied to a confined fluid increases the pressure throughout by the same amount 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) Pascal's Principle carries with it hydraulics (pg. 280) Changing the Area changes the force Changing the Area changes the force For this to be true, the fluid must not compress (effectively they don’t) For this to be true, the fluid must not compress (effectively they don’t) Blaise Pascal, French

P in = P out Input Output F out F in A out A in F in = F out (A in ) F in = F out (A in ) A out A out

A small force can be used to exert a larger force by making the area of one piston larger than the area of another 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 Small input area, Large output area greatly multiplies the input force F= 200 A= 100 A= 1000 F = 2000 in out

F out F in Mechanical advantage If area is 20x greater then output force will be 20x greater MA Homework!

3 types of buoyancy: 3 types of buoyancy: + rise + rise - sink - sink neutral equilibrium neutral equilibrium Buoyancy

All objects appear to weigh less when submerged in a fluid All objects appear to weigh less when submerged in a fluid Why a buoyant force? pg. 283 F2F2F2F2 F1F1F1F1 F 2 > F 1 F 2 is greater b/c there is more pressure at the lower depth This is how we derive the formula…

F B = F 2 – F 1 F B = F 2 – F 1 F B = Ap f gh 2 -Ap f gh 1 F B = Ap f gh 2 -Ap f gh 1 = Ap f g(h 2 - h 1 ) = Ap f g(h 2 - h 1 ) Ap f gh (Ah = Vol) Ap f gh (Ah = Vol) p f gV p f gV Recall F= AP F= AP p= m/v p= m/v w= mg w= mg Vol= Ah Vol= Ah Also… pV = mass so…. m f g=f B

F buoyant = p f gV= m f g This is Archimedes Formula (Principle) This is Archimedes Formula (Principle) In English= In English= The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the object 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… 1. Find m of object ____g weight ___N 2. Find V of Object ____cm 3 _____m 3 3. Find m of submerged ____ g weight ____N 4. Find buoyant force on object f B = weight (air) – weight (submerged) f B = weight (air) – weight (submerged) = _______ N - _______ N = _______ N - _______ N f B = _______N f B = _______N Now check w/ formula Now check w/ formula f B = p f gV f B = p f gV = 1 x 10 3 kg/m 3 (9.8m/s 2 ) __________m 3 = 1 x 10 3 kg/m 3 (9.8m/s 2 ) __________m 3 = ______ = ______ % error = 5.7% x What about f B = M f g

Why do steel ships float? Why do steel ships float? How can fish suspend themselves in H 2 0 How can fish suspend themselves in H 2 0 How do submarines work? How do submarines work?

Restatement : Restatement : f B = W f = p f Vg f B = W f = p f Vg The volume of an object can also be found by: The volume of an object can also be found by: V f = m = w V f = m = w p f gp p f gp

Archimedes Principle applies to both submerged and floating objects f B = weight of object --for floating objects f B = weight of object --for floating objects p f V disp g = p o V o g --g cancels p f V disp g = p o V o g --g cancels V disp = p o V disp = p o V o p f V o p f

Shortcut– sometimes applied (if p is known) Shortcut– sometimes applied (if p is known) Density of floating object = fraction of object that density of fluid floating in is submerged. Density of floating object = fraction of object that density of fluid floating in is submerged. wood Log water 900 kg/m kg/m 3 =.9 9/10 of log is underwater 90 % of log is submerged

fluid dynamics (hydrodynamics if water) fluid dynamics (hydrodynamics if water) If flow is smooth- streamline or Laminar If flow is smooth- streamline or Laminar if flow is erratic, currents present – turbulent if flow is erratic, currents present – turbulent Viscosity- internal resistance within fluid Viscosity- internal resistance within fluid Syrup (high viscosity) Syrup (high viscosity) Water (low viscosity) Water (low viscosity)

Flux- term used to describe Volume of fluid passing through a given area each second Flux- term used to describe Volume of fluid passing through a given area each second Rate of Flow - assume ideal fluid, frictionless, laminar Rate of Flow - assume ideal fluid, frictionless, laminar Velocity Area D= Vt

flow rate R flow rate R units = m 3 /s (volume/ time) units = m 3 /s (volume/ time) m/s x m 2 (m 3 /s) m/s x m 2 (m 3 /s) R = VA Velocity x Area R = VA Velocity x Area

Equation of continuity states that rate of flow remains constant. Velocity and area will change (inversely), but rate (m 3 /s) stays the same. Equation of continuity states that rate of flow remains constant. Velocity and area will change (inversely), but rate (m 3 /s) stays the same. Area 1 Area 2 Velocity 1 Velocity 2

because R(m 3 /t) stays the same… because R(m 3 /t) stays the same… A 1 V 1 = A 2 V 2 A 1 V 1 = A 2 V 2 V 1 = A 2 V 2 V 1 = A 2 V 2 A 1 A 1 Large Velocity – Small Area Small Velocity- Large Area River Ex River Ex. Fast Slow Narrow Wide

Blood Flow Example pg. 288 Blood Flow Example pg. 288 Rate of flow of blood in human body stays same –Big Aorta to Small Capillaries 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. Follow along w/ the book to see how the sample problem is solved.

We can also find Power of a moving fluid We can also find Power of a moving fluid Power = PR Power = PR Power = pressure x flow rate Power = pressure x flow rate N/m 2 x m 3 /s Nm/s J/s N/m 2 x m 3 /s Nm/s J/s Power = work / time Power = work / time

Where the velocity of a fluid is high, the pressure is low, where the velocity is low, the pressure is high. 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 P high V 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 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 Car’s Convertible Top -- jeep Tarp in back of truck Tarp in back of truck carburetor carburetor airplane wing airplane wing chimneys/ outhouses chimneys/ outhouses perfume atomizer perfume atomizer Ventilation in burrows Ventilation in burrows hanging ping pong balls hanging ping pong balls ping pong ball = funnel ping pong ball = funnel

Here’s the equation P 1 + ½ pV pgy 1 = P 2 + 1/2pV pgy 2 P 1 + ½ pV pgy 1 = P 2 + 1/2pV pgy 2 When solving for P 2 the formula looks like this: When solving for P 2 the formula looks like this: P 2 = P 1 + pg(y 1 -y 2 )+ ½ p(v 2 1 -v 2 2 ) P 2 = P 1 + pg(y 1 -y 2 )+ ½ p(v 2 1 -v 2 2 ) P= pressure P= pressure p= density of flowing fluid p= density of flowing fluid g= gravity g= gravity y= height y= height V= velocity V= velocity

**Overhead Picture** Bernoulli’s Equation is really an expression of the law of energy conservation. Bernoulli’s Equation is really an expression of the law of energy conservation. The formula is derived from work energy principle - - pg. 290 The formula is derived from work energy principle - - pg. 290

Go to overhead for equation… P Vel A h