# Physics 153 Introductory Physics II. Week One: FLUIDS. Dr. Joseph J. Trout

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1 Physics 153 Introductory Physics II Week One: FLUIDS Dr. Joseph J. Trout

2 States (Phases) of Matter: Solid: Fixed shape. Fixed size. Even a large force will not readily change the shape or volume (incompressible). Fluids: Liquid: Takes on shape of container. Incompressible. Volume can be changed by a significantly large force. Gas: Fills to fit container. compressible.

3 Density: mass per unit volume. = m V Table 10-1 Substance Density kg/m 3 Solids Liquids Gas Aluminum 2700 Iron 7800 Copper 8900 Lead Concrete 2300 Bone 1800 Ice (water) 917 Water ( 4 O C) 1000 Sea Water 1025 Blood (Plasma) 1030 Mercury Air 1.29 Helium 0.18 Steam (Water) 0.6

4 Specific Gravity: Ratio of density of substance to density of water. Table 10-1 Substance Density kg/m 3 SG SG= substance water Solids Liquids Gas Aluminum Iron Bone Ice (water) Water ( 4 O C) Blood (Plasma) Mercury Steam (Water) water =1000 kg m 3

5 Pressure: Force per unit area. P= F A 1 Pa=1 N /m 2 1atm=1.013 X 10 5 N /m 2 =101.3 kpa 1 bar=1.00 X 10 5 Pa 1atm=760 mm Hg 30 in Hg

6 A fluid can exert pressure in all directions. In a static fluid (fluid at rest ), the pressure must be the same in all directions. If not, there would be a non-zero net force, and the fluid would be in motion.

7 Pressure at a Depth h

8 A P A weight of column P=P A area of column P=P mg A A h A P

9 A P A P h weight of column P=P A area of column P=P mg A A But... m= V P=P V g A A P=P A Ah g A P=P A g h

10 Another look... Pressure at a Depth

11 Water Level h=40 m P=1000 kg m 3 P= g h 9.8 m s 40 m=3.92 X Pa

12 50 m 30 m 1 m 20 km Dam 20 m P=P A g h P X 10 5 Pa 1000 kg m 9.8 m 3 s 30.5m=4.0 X Pa F =PA= 4.0 X m 1m=8 X 10 6 N

13 50 m 30 m 1 m 20 km Dam 20 m If the lake was 50 km long, instead of 20 km long, would the dam need to be redesigned?

14 P A Measuring Pressure Pressure to be Measured. P Manometer P=P A Mercury

15 P A Measuring Pressure Pressure to be Measured. P Manometer h P P A P=P A Hg g h Mercury

16 P A Measuring Pressure Pressure to be Measured. P Manometer h P P A P=P A Hg g h Mercury

17 Barometer Measuring Pressure Open End P=0 Closed End P A P A h Closed End Column filled with mercury is inverted in a dish of mercury (Hg).

18 Barometer Measuring Pressure P=0 Closed End P A = g h P A =13600 kg m 9.8 m 3 s 2 76 cm P A =1.013 X 10 5 N =1.00 atm 2 m P A =1atm Mercury h=76 cm

19 Barometer Measuring Pressure P=0 Closed End P A = g h P A h When atmospheric pressure is falling, there will be a change in weather. We say the mercury is falling. Mercury

20 Barometer Measuring Pressure Why use mercury? P=0 Closed End P A = g h h= P A g X 10 5 N m 2 h= kg m 3 h=0.76m=76cm 9.8 m/ s2 Mercury P A h

21 Measuring Pressure Closed End Barometer Why use mercury? P=0 P A = g h h= P A g X 10 5 N m h= 2 kg m/ m 3 s2 h=10.33 m=33.91 ft Water P A h

22 Measuring Pressure Closed End Barometer BTW, every scuba diver knows that every meters (approximately 33 ft) is equivalent to to adding 1.0 atm. P=0 P A h P A = g h h=10.33m Water

23 Bourdon gauge - Bourdon gauge - uses a coiled tube, which, as it expands due to pressure increase causes a rotation of an arm connected to the tube. Measuring Pressure

24 Measuring Pressure Secondary transducer * resistive (strain gauge) * inductive * capacitive - The deflection of the piston is often one half of a capacitor, so that when the piston moves, the capacitance of the device changes. This is a common way (with proper calibrations) to get a very precise, electronic reading from a manometer, and this configuration is called a capacitive manometer vacuum gauge. This is also called a capacitance manometer, in which the diaphragm makes up a part of a capacitor. A change in pressure leads to the flexure of the diaphragm, which results in a change in capacitance. These gauges are effective from 10?3 Torr to 10?4 Torr. * piezoelectric/piezoresistive

25 Gauge Pressure Many pressure gauges measure the pressure above atmospheric pressure. P=P A P G

26 Absolute Pressure Pressure Gauge reads: 200 kpa P=P A P G P=101kPa 200 kpa=301kpa Pressure of Tire.

27 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. A F 1 =P 1 A h 1 h 2 F B =F 2 F 1 F B =P 2 A P 1 A F B = P 2 P 1 A h=h 2 h 1 F 2 =P 2 A F A V = A h h

28 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. A F 1 =P 1 A h 1 h 2 F B =F 2 F 1 F B =P 2 A P 1 A F B = P 2 P 1 A h=h 2 h 1 F B = F g h 2 F g h 1 A F 2 =P 2 A F A V = A h h

29 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. A F 1 =P 1 A h 1 h 2 F B =F 2 F 1 F B =P 2 A P 1 A F B = P 2 P 1 A h=h 2 h 1 F B = F g h 2 F g h 1 A F 2 =P 2 A F A F B = F g h 2 h 1 A F B = F V F g F B =m F g V = A h h

30 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. A F 1 =P 1 A h 1 h 2 F net =F B w obj F net = F V F g mg F net = F V F g obj V obj g h=h 2 h 1 F 2 =P 2 A F

31 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. A F 1 =P 1 A h 1 h 2 h=h 2 h 1 F net =F B w obj F net = F V F g mg F net = F V F g obj V obj g F net = F V F g obj V obj g F 2 =P 2 A F net = F V F obj V obj g F

32 Archimedes' Principle the magnitude of the buoyant force on an object immersed in a liquid is equal to the magnitude of the weight of the fluid displaced. F net = F V F obj V obj g A F 1 =P 1 A h=h 2 h 1 F 2 =P 2 A F h 1 h 2 If completely submerged: V F =V obj F net = F obj V obj g

33 A plastic cylinder of volume 0.2 m 3 and a density of 800 kg/m 3 and is completely submerged in water and released. What is the nest force on the object? If completely submerged: V F =V obj F net = F obj V obj g = kg kg F net m m 3 F net = 392 N 0.2m m/ s 2 BTW... w obj =m obj g= obj V obj g=1568 N

34 A copper cylinder of volume 0.2 m 3 and a density of 8900 kg/m 3 and is completely submerged in water and released. What is the nest force on the object? If completely submerged: V F =V obj F net = F obj V obj g = kg kg F net m m m3 9.8m/s 2 F net = N BTW... w obj =m obj g= obj V obj g=17444 N

35 Body Fat F net =W app W F B =0.0 N W =W app F B m P g=w app F V F g F

36 Body Fat F net =W app W F B =0.0 N W =W app F B m P g=w app F V F g V P P g=w app F V F g But... V F =V P F

37 Body Fat F net =W app W F B =0.0 N W =W app F B m P g=w app F V F g V P P g=w app F V F g But... V F =V P V F P g=w app F V F g P = W app F V F g V F g F

38 Body Fat F net =W app W F B =0.0 N W =W app F B m P g=w app F V F g V P P g=w app F V F g But... V F =V P V F P g=w app F V F g P = W app F V F g V F g Up to 1% error. F Brozek Formula: BF= Siri Formula: BF= 4570 kg/ m3 P 4950 kg/ m3 P

39 Body Fat Once the overall density of the person is known: fat =900 kg /m 3 lean =1100 kg /m 3 Given: x f is the fraction of body fat and M is the total mass of the person Then: x l =1 x f is the fraction of lean muscle and bone. m f =x f M m l = 1 x f M

40 Body Fat Once the overall density of the person is known: fat =900 kg /m 3 lean =1100 kg /m 3 Given: x f is the fraction of body fat and M is the total mass of the person Then: x l =1 x f is the fraction of lean muscle and bone. m f =x f M m l = 1 x f M V p =V f V l = m f f m l l = x f M f 1 x f M l

41 Body Fat Once the overall density of the person is known: fat =900 kg /m 3 lean =1100 kg /m 3 Given: x f is the fraction of body fat and M is the total mass of the person Then: x l =1 x f is the fraction of lean muscle and bone. m f =x f M m l = 1 x f M V p =V f V l = m f f m l l = x f M P = M V p = x f M f M 1 x f M l f = 1 x f M 1 l x f 1 x f f l x f = 1 P l f l f f l f = 4950kg /m3 P 4.50

42 Body Fat There are other, less costly methods ( which have little to do with this chapter ). : The Bioelectrical impedance analysis (BIA) - Two conductors are attached to a person's body and a small electrical current is sent through the body. The resistance between the conductors will provide a measure of body fat, since the resistance to electricity varies between adipose, muscular and skeletal tissue. Fat-free mass (muscles) is a good conductor as it contains a large amount of water (approximately 73%) and electrolytes, unlike fat which is anhydrous and a poor conductor of electrical current. Anthropometric Methods - There exist various anthropometric methods for estimating body fat. The term anthropometric refers to measurements made of various parameters of the human body, such as circumferences of various body parts or thicknesses of skinfolds. Height and Circumference Methods - There also exist formulas for estimating body fat percentage from an individual's weight and girth measurements. For example, the U.S. Navy Circumference method compares abdomen or waist and hips measurements to neck measurement and height.

43 Pascal's Principle if an external pressure is applied to a confined fluid, the pressure at every point within the fluid is increased by that amount. F in F out P in =P out A F in in = F out A in A out P in P out Aout m car =20,000 kg A in =0.02m 2 A out =2.00 m 2 F in = A in A out w car = kg 9.8 m/s 2 =1.96 X 10 5 N =F out F out

44 Pascal's Principle if an external pressure is applied to a confined fluid, the pressure at every point within the fluid is increased by that amount. F in F out P in =P out A in P in P out m car =20,000kg A in =0.02 m 2 A out =2.00 m 2 Aout = 0.02 m2 F in 2.00 m 2 F in = F in = 0.02 m m 2 F in = F out A in A out A in A out F out kg 9.8 m/ s2 =1960 N N =1960 N

45 Pascal's Principle if an external pressure is applied to a confined fluid, the pressure at every point within the fluid is increased by that amount. A in V in =V out A in h in = A out h out hin V in V out A out h out

46 Pascal's Principle if an external pressure is applied to a confined fluid, the pressure at every point within the fluid is increased by that amount. A in V in =V out A in h in = A out h out hin V in V out A out h out W in =W out F in h in =F out h out A in F A out out A out h A in out =F out h out A in A out F out A out A in h out =F out h out So... W in =W out F out h out =F out h out

47 Pascal's Principle if an external pressure is applied to a confined fluid, the pressure at every point within the fluid is increased by that amount.

48 Fluids in Motion

49 Ideal Fluid no internal friction no viscosity. Real Fluid with viscosity. Laminar Flow Viscosity measure in the internal friction (between layers) of a fluid.

50 Turbulent Flow Viscosity measure in the internal friction (between layers) of a fluid.

51

52 Continuity Equation = m V v 1 v 2 Mass flow rate = m ṁ= dm dt t

53 Continuity Equation = m V A 1 A 2 v 1 v 2 m 1 = m 2 t t

54 Continuity Equation = m V m= V A 1 A 2 v 1 v 2 x 1 x 2 m 1 = m 2 t t

55 Continuity Equation V 1 V 2 = m V m= V A 1 A 2 v 1 v 2 x 1 x 2 m 1 t = m 2 t 1 V 1 t = 2 V 2 t

56 Continuity Equation V 1 V 2 = m V m= V A 1 A 2 v 1 v 2 x 1 x 2 A V = A h h m 1 t = m 2 t 1 V 1 t = 2 V 2 t 1 A 1 x 1 = 2 A 2 x 2 t t 1 A 1 v 1 = 2 A 2 v 2

57 Continuity Equation V 1 V 2 = m V m= V A 1 A 2 v 1 v 2 x 1 x 2 1 A 1 v 1 = 2 A 2 v 2

58 Continuity Equation V 1 V 2 = m V m= V A 1 A 2 v 1 v 2 x 1 x 2 1 A 1 v 1 = 2 A 2 v 2 If 1 = 2 then: A 1 v 1 = A 2 v 2

59 5 Liters 1.00 cm diameter garden hose takes seconds to fill a 5.0 liter container. What is the mass flow rate? m t = V t = 1000 kg /m3 5l 80.0 s m = t = kg s 1000 kg /m 3 5l 1 X 10 3 m s 1.0 l

60 m 1 t = kg s A 0.24 cm diamter nozzle is placed on the 1.00 cm diameter garden hose. How far will the water travel if held horizontally, 0.8 meters off the ground? m 1 t m 1 t = m 2 t = 2 A 2 v 2 y=0.8 m v 2 = kg s 1000 kg /m m 2 v 2 =13.82 m/ s x=?

61 m 1 t = kg s A 0.24 cm diamter nozzle is placed on the 1.00 cm diameter garden hose. How far will the water travel if held horizontally, 0.8 meters off the ground? y= y o v o y t 1 2 g t 2 v ox =v 2 =13.82m/s t= 2 y g y= 1 2 g t 2 = m 9.8 m/ s 2=0.4 s y=0.8 m x=?

62 m 1 t = kg s A 0.24 cm diamter nozzle is placed on the 1.00 cm diameter garden hose. How far will the water travel if held horizontally, 0.8 meters off the ground? y= y o v o y t 1 2 g t 2 v o x =v 2 =13.82m/s y=0.8 m t= 2 y g y= 1 2 g t 2 = m 9.8 m/s 2=0.4 s v x = x t x=v x t=13.82m/ s 0.4 s =5.32 m x=?

63 P 2 A 2 P 1 A 1 h 1 h 2

64 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 x 1 W 1 =F 1 x 1 =P 1 A 1 x 1 W 2 = F 2 x 2 = P 2 A 2 x 2

65 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 x 1 W 1 =F 1 x 1 =P 1 A 1 x 1 W 2 = F 2 x 2 = P 2 A 2 x 2 W 3 = U = m 2 gy 2 m 1 gy 1 W net =W 1 W 2 W 3 =P 1 A 1 x 1 P 2 A 2 x 2 m 2 gy 2 m 1 gy 1

66 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 x 1 W net =P 1 A 1 x 1 P 2 A 2 x 2 m 2 gy 2 m 1 gy 1 W net = K = 1 2 m 2 v m 1 v 1 2

67 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 x 1 W net =P 1 A 1 x 1 P 2 A 2 x 2 m 2 gy 2 m 1 gy 1 W net = K = 1 2 m 2 v m 1 v 1 2 P 1 A 1 x 1 P 2 A 2 x 2 m 2 gy 2 m 1 gy 1 = 1 2 m 2 v m 1 v 1 2

68 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 P 1 A 1 x 1 V 1 x 1 P 1 A 1 x m 1v 2 2 m 1 gy 1 =P 2 A 2 x 2 m 2 gy m 2 v m 1 V 1 v 2 1 m 1 V 1 V 1 =V 2 gy 1=P 2 A 2 x 2 V m 2 V 2 v 2 1 m 2 V 2 gy 1

69 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 P 1 A 1 x 1 V 1 x 1 P 1 A 1 x m 1 v 2 2 m 1 gy 1 =P 2 A 2 x 2 m 2 gy m 2 v m 1 V 1 v 2 1 m 1 V 1 V 1 =V 2 gy 1=P 2 A 2 x 2 V m 2 V 2 P v 2 1 gy 1 =P v 2 2 gy 2 v 2 2 m 2 V 2 gy 2

70 v 2 A 2 P 2 v 1 P 1 A 1 x 2 h 2 h 1 x 1 P v 2 gy =P v 2 gy 2 2 P 1 2 v2 gy=constant

71 v 2 =0 P 2 P 1 =P 2 y= y 2 y 1 v 1 =? P 1 P v 2 gy =P v 2 gy v 2 gy =gy v 1 = 2g y 2 y 1

72 Fluid Flow Rate in a Round Tube depends on: Viscosity of Fluid. Pressure difference. Dimensions of tube. Q= R4 P 1 P 2 8 L

73 Surface Tension = F L

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