### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Independent Study 101 325

UI

GPA 3.72

### View Full Document

## 19

## 0

## Popular in Course

## Popular in Physical Science

This 56 page Class Notes was uploaded by Sigrid Strosin on Friday October 23, 2015. The Class Notes belongs to 101 325 at University of Iowa taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/228105/101-325-university-of-iowa in Physical Science at University of Iowa.

## Reviews for Independent Study

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/23/15

57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 1 Chapter 2 Pressure and Fluid Statics Pressure For a static uid the only stress is the normal stress since by de nition a uid subjected to a shear stress must deform and undergo motion Normal stresses are referred to as pressure p For the general case the stress on a uid element or at a point is a tensor t stress tensor Txx Txy sz E E E Yx W Y1 sz sz Tzz i force j direction For a static uid Tij 0 i j shear stresses 0 Iii p Exx tyy In i j normal stresses p Also shows that p is isotropic one value at a point which is independent of direction a scalar 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 2 De nition of Pressure 8F dF 2 P 11m5 A d A Nm Pa Pascal 8Aao F normal force acting over A As already noted p is a scalar which can be easily demonstrated by considering the equilibrium of forces on a wedgeshaped uid element MM Geometry 39 AA 2 Ay lt prAsina AX A E cosoc 11N3i8ht AzAlsina AZ A E sinoc I AxAlcosa pzAA cosa W mg ZFX 0 psvzg pnAA sin or pXAA sin or 0 y pn Z px 12 AXAZAy 2F 0 pnA Ay cosoc pZ A Ay cos 0c pnAA COS 06 pZAA COS OL W O A 2 cosocsinocAy 0 W 1M cosocA sinocAy MAycosoc 2 H H Y AX AZ pnpZ A sinoc0 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 3 pn pZ A sinoc 0 pnzpZ for A690 ie pn px py pz p is single valued at a point and independent of direction A bodysurface in contact with a static uid experiences a force due to p Ep SIpndA e 1 53 fANA B g J t W i To var i 4quot 53 Note if p constant FF 0 for a closed body Scalar form of Green s Theorem I fnds j Vfdv s V f constant 2Vf 0 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 4 Pressure Transmission Pascal39s law in a closed system a pressure change produced at one point in the system is transmitted throughout the entire system Absolute Pressure Gage Pressure and Vacuum pg gt 0 pa atmospheric gt a pressure pA p pglt0 101325 kPa pA lt pa pA O absolute zero For pAgtpaa pg 2 pA pa gage pressure For PAltpa pvac 2 pg pa pA vacuum pressure 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2005 5 Pressure Variation with Elevation Basic Differential Equation For a static uid pressure varies only with elevation within the uid This can be shown by consideration of equilibrium of forces on a uid element A 4 A AXJ 1 3 t 15 order Taylor serles estimate for pressure X variation over dz W BAXA BAi Newton s law momentum principle applied to a static uid XE ma 0 for a static uid ie ZFX ZFy ZFZ 0 ZFZ 0 dedy p 2 pdzdxdy pngdydz 0 2 6p 62 Pg V Basic equation for pressure variation with elevation 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 6 ZFyzO ZFXZO dedZ p gdydxdz O pdydz p gd dydz O 20 L0 6y 6X For a static uid the pressure only varies with elevation z and is constant in horizontal xy planes The basic equation for pressure variation with elevation can be integrated depending on whether p constant or p pz ie whether the uid is incompressible liquid or lowspeed gas or compressible highspeed gas since g constant Pressure Variation for a UniformDensity Fluid 2 pg 2 y p constant for liquid r dz g1 AP YAZ T p2p1VZ2Z1 Alternate forms p1 yZ1 p2 yzz conStant p yz constant p1ezometr1c pressure pz O O gage p yz increase linearly with depth decrease linearly with height p vz B z constant piezometric head 57020 Fluid Mechanics Professor Fred Stem Fall 2006 Chapter 2 7 Oil with a specific gravity of 080 forms a layer 090 deep in an open tank that is otherwise filled with water The total depth of water and oil is 3 In What is the gage pressure at the bottom of the tank pyzconstant P1YZ1P2 YZ2 pZ p1yzlizz pl pm 0 p3 pZ Wm Z2 Zr 70609810x 21 277 277kPa Solution First determine the pressure at the oilrwater interface staying within the oil and then calculate the pressure at the bottom h h 7 7 where p is the pressure at free surface of oil 2 is the elevation of free surface of oil p is the pressure at interface between oil and water and 21 is the elevation at interface between oil andwater For this example p 0 7 080 X 9810 Nmz z 3 m and z 2J0 mi Therefore pg 3 090 m X 080 X 9810 Nmx 706 kPa gage Now obtain p from 71 22 P3 z 7 7 where p1 has already been calculated and 7 9810 Nm i 7060 p 9809810 210 277 kPa gage pZ ymlAz 8x 9810xi9 706kPa 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 8 Pressure Variation for Compressible Fluids Basic equation for pressure variation with elevation v vpz pg dz Pressure variation equation can be integrated for ypz known For example here we solve for the pressure in the atmosphere assuming ppT given from ideal gas law Tz known and g 7 gz p pRT R gas constant 287 Jkg OK dry air pT in absolute scale dP m dz RT E g dz which can be integrated for Tz known p R Tz 57020 Fluid Mechanics Professor Fred Stern Fall 2006 Pressure Variation in the Troposphere T To OLZ zo linear decrease To Tz0 where p p0z0 known OL lapse rate 65 oKkm Chapter 2 9 dp dz Z39ZT0O ZZ0 p R To aZZ0 dZ39 XdZ lnp iln o ocz z0 constant ocR use reference condition zo i Earth surface lnpo iInTO constant p0 1013 kPa ocR T 15 C solve for constant a 2 65 oKkm mg g lnToOKZZo pooc R T 0 T0O ZZo gQR p0 T 0 ie p decreases for increasing z 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 1 0 Pressure Variation in the Stratosphere TTs 55 C dingy p RTS lnp iz constant RT S use reference condition to nd constant 1 e z zogRTS 130 p p0 eXp Z lagR1 ie p decreases exponentially for increasing z Pressure Measurements Pressure is an important variable in uid mechanics and many instruments have been devised for its measurement Many devices are based on hydrostatics such as barometers and manometers ie determine pressure through measurement of a column or columns of a liquid using the 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 1 1 pressure variation with elevation equation for an incompressible uid Differential manometer More modern devices include Bourdon Tube Gage mechanical device based on de ection of a spring and pressure transducers based on de ection of a exible diaphragmmembrane The de ection can be monitored by a strain gage such that voltage output is oc Ap across diaphragm which enables electronic data acquisition with computers Bourdon Tube Gage In this course we will use both manometers and pressure transducers in EFD labs 2 and 3 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 12 Manome 1 Barometer pv ngh pm pm th pV 0 ie vapor pressure Hg nearly zero at normal T h 76 cm pm 101 kPa or 146 psia Note patm is relative to absolute zero ie absolute pressure patm pm ocation weather Consider Why water barometer is impractical YthHg YHZOhHZO YHg hHg SthHg 136gtlt76 10336cm34ft 11H20 H20 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 13 2 Piezometer pm Patm Yh ppipe p absolute p vh gage Simple but impractical for large p and vacuum pressures ie pabs lt patm Also for small p and small d due to large surface tension effects could be corrected using Ah 4GYd but accuracy may be problem if p y Aha 3 U tube or differential manometer rmmanometev liquid p1vah vlp4 p1patm p4 vah vl gage YWlSmAh S l for gases S ltlt SIn and can be neglected ie can neglect Ap in gas compared to Ap in liquid in determining p4 ppipe Example 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 14 Air at 20 C is in pipe with a water manometer For given conditions compute gage pressure in pipe 1 140cm Ah70cm P4 7 gage 162 P1 0 Pressure same at 2amp3 since same elevation amp Pascal s 21 Ah 2 P3 Step39bY39SteP methOd law in closed system p3 Yairl p4 pressure change produce at yAh one part transmitted throughout entire system p1 Ah yairl p4 complete circuit method 7Ah valr1 p4 gage ywater20 C 9790 Nm3 2 p3 Ah 6853 Pa Nmz 7m pg pabs or p p3patm 68531013001I286kgm3 could p 2 E R C 273 28720 273 use OK 7 1286 X 981ms2 1262 Nm3 316 note mI ltlt water p4 p valr1 6853 1262 gtlt 14 6835 Pa 17668 if neglect effect of air column p4 6853 Pa 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 15 A differential manometer determines the difference in pressures at two points and C2 when the actual pressure at anV point in the sv stem cannot be determined 620 29 12 0 9 i M K it l glllp2gzjlvm1llh Wf Wf KYf a difference in piezometric head iffluidisagas yfltlt ym pl p2ymAh if uid is liquid amp pipe horizontal E1 E2 P1 P2 Vm Yr Ah 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 16 Hydrostatic Forces 0n Plane Surfaces For a static uid the shear stress is zero and the only stress is the normal stress ie pressure p Recall that p is a scalar which when in contact with a solid surface exerts a normal force towards the surface t LFrquot BJA39 Fp IpndA A For a plane surface n constant such that we can separately consider the magnitude and line of action of FF EpFIpdA A Line of action is towards and normal to A through the center of pressure XCP yep 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 17 Unless otherwise stated throughout the chapter assume patm acts at liquid surface Also we will use gage pressure so that p O at the liquid surface Horizontal Surfaces horizontal surface with area A p constant szpdAsz Line of action is through centroid of A ie ch yep i y 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 1 8 Inclined Surfaces dp dz Y Ap yAz FIGURE 1 10 Distribution uf hydrostatic pressure on a MW 06 plane Eurace y I39i39cc quince 1 l Rexullnnl hurt I K b Plan mu m u l lml plum mit dFpdAW sinoch H 1 y and sin 0L are constants P F jpdA ysinocjydA A A yA 1 2 dA y Aly 1St moment of area 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 1 9 F ysin 05A HJ 5 pressure at centroid of A Center of Pressure Center of pressure is in general below centroid since pressure increases with depth Center of pressure is determined by equating the moments of the resultant and distributed forces about any arbitrary axis Determine yEB by taking moments about horizontal axis 00 ych lde A IypdA A jyyysinocdA A ysinocijdA L 10 211d moment of area about 00 moment of inertia 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 20 transfer equation I 2A l 0 ych ysinoc 2Al 2 ycppA YSIII0 Y AH ycpysina A ysinoc 2Al ycp A 2Ai ycp is below centroid by l A ycp gt for large For po 7 O y must be measured from an equivalent free surface located poy above y 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 21 Determine xQB by taking moment about y axis chF deF A jxpdA A ch yy sin OLA jxyy sin OLdA A ch yA IxydA L IXy product of inertia lxy yA transfer equation chyA Ly A For plane surfaces with symmetry about an axis normal to 00 IXy 0 and ch X 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 22 1 bu I getquot L IIIr I I 39239 392 Iw 0 III Rectangle II Circle A I g A s I I512 I 7 0109558 1 In 2d 1 2 33927 9 a 3 I o I 7 i V k I 1 T I t39 Semicilclc Ill Triangle nk A T M I 1 Iquot I answer 3 39 Iquot eolaun H Huang clrcle y A IL A 2m le R4 39 I 12 113 L 3 III 0 I 0 I I i I F 3 u 7 FBI M 2 2L quot 7 A n5 3 I I z LU In 0109751J 36 I I L NI3A L3 M 3 I I 77 I I F if I g 4 E I R I39 R 37 I 2 I 57 020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 Hvdrostatic Forces on Curved Surfaces Free surface 3 pyh x E lp dA h distance below A free surface Horizontal Components X and y components FX Fi jppidA dAX projection of pdA onto 2 plane J to Xdirection dAy zg dA projection pdA onto plane J to ydirection Therefore the horizontal components can be determined by some methods developed for submerged plane surfaces The horizontal component of force acting on a curved surface is equal to the force acting on a vertical projection of that surface including both magnitude and line of action 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 24 Vertical Components 2 FZ k ipgde I pdAz p vh AN A2 V hdistance x P A below free surface 3 y jhdA Z yV AZ weight of uid above surface A The vertical component of force acting on a curved surface is equal to the net weight of the column of uid above the curved surface with line of action through the centroid of that uid volume 57 020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 25 39 Example Drum Gate 39 i Pressure Diagram p yh yR1cosG n sin6i cosGk dA E Rde E ijl cos G sin 9i cosG 12ERdG 0 W p n dA F 2 FX 2 y szl cosGsinGd9 o 2 2ysz MR2 cosGlc0529 4 0 yR2R gt same force as that on projection of p A area onto vertical lane FZ y szl cosecosede 0 TE y R2 sinG 9 11126 2 4 0 7t TERZ sz z z 2 X Y 2 YL 2 j Y 3 net weight of water above surface 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 26 Buoyancy Archimedes Principle U FB 2 Ba Fvl uid weight above Surface 2 ABC uid weight above Surface 1 ADC uid weight equivalent to body volume FB pg submerged volume Line of action is through centroid of center of buoyancy Net Horizontal forces are zero since FBAD FBCD 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 Hydromet A hydrometer uses the buoyancy principle to determine speci c weights of liquids 3 lm ampquotamp w k39o Stem 2 guy A lofew Ah QMquot 39 LJ l BulbrI 15 w l 9 a M w w ggxl T quotCXw S 4X w Wmgyf Syw W yw 0 SYWGJ0 A SJM 310 aAh Yf a cross section area stem 3108 0 aAh aAh O OS Ah l jAhS calibrate scale using uids of known S 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 28 Example apparent weight King Hero ordered a new crown to be made from pure gold When he received the crown he suspected that other metals had been used in its construction Archimedes discovered that the crown required a force of 47 to suspend it when immersed in water and that it displaced 189 in3 of water He concluded that the crown was not pure gold Do you agree WK39 TM 1L 3 3g w ZFVE 0waFb w02waw Fbyc yw Wvc Fbvw or 2 21 Wavw Y0 3 Yw 3 Yo 47624gtlt1891728 4921 peg 1891728 3 p0 153 slugsft3 psteel and since gold is heavier than steel the crown can not be pure gold 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 29 Stabilit 0f Immersed and Floatin Bodies Here we ll consider transverse stability In actual applications both transverse and longitudinal stability are important Immersed Bodies FIGURE 3 15 Conditlom Ofslabzlity Weight for immersed bodies 41 Stable 2 Neutral c Unstable a w W Static equilibrium requires ZFV 0 and 2M 2 0 EM 0 requires that the centers of gravity and buoyancy coincide ie C G and body is neutrally stable If C is above G then the body is stable righting moment when heeled If G is above C then the body is unstable heeling moment when heeled 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 Floating Bodies For a oating body the situation is slightly more complicated since the center of buoyancy will generally shift when the body is rotated depending upon the shape of the body and the position in which it is oating Positive G M Negative G M The center of buoyancy centroid of the displaced volume shifts laterally to the right for the case shown because part of the original buoyant volume ACE is transferred to a new buoyant volume EOD The point of intersection of the lines of action of the buoyant force before and after heel is called the metacenter M and the distance GM is called the metacentric height If GM is positive that is if M is above G then the ship is stable however if GM is negative the ship is unstable 57020 Fluid Mechanics Professor Fred Stern Fall 2006 31 Chapter 2 Floating Bodies OL small heel angle X 2 CC lateral displacement of C C center of buoyancy W ie centroid of displaced volume Solve for GM find g using 1 basic definition for centroid of 3 and 2 trigonometry m Fig 317 1 Basic definition of centroid of volume IXd Z XiASVZi moment about centerplane XIV moment V before heel moment of AOB moment of EOD 0 due to symmetry of original V about y axis ie ship centerplane j XCPVL j XCPVL tan OL yX AOB EOD d ydAxtanoch j x2 tanoch j x2 tanoch AOB EOD 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 3 2 Q71 tanocIXZdA ship waterplane area moment of inertia of ship waterplane about z axis OO ie 100 100 moment of inertia of waterplane area about centerplane axis 2 Trigonometry X tanocIOO I CC39 X M CMtanoc CM 2 100 GM CM CG I00 GM CG 1 GM gt 0 Stable GM lt O Unstable 57020 Fluid Mechanics Professor Fred Stern Fall 2006 Fluids in RigidBody Motion For uids in motion the pressure variation is no longer hydrostatic and is determined from application of Newton s 2nd Law to a uid element 9 tx 13 4X PaWVquot P K 951 AxX AaAt E lt4 392 In Viscous stresses p pressure Ma inertia force X 2 610 weight body force Gt 6 6 ay T Newton s 2nd Law pressure net surface force in X direction 62 a jar M ZEEBES per unit volume p21 fb fs The acceleration of uid particle See Chapter 4 gz w Dt 6t body force pgk f5 surface force Q fv a 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 3 4 fp surface force due to p Vp fV surface force due to viscous stresses ti Neglected in this chapter and pg fb f included later in Chapter 6 p when deriving complete Navier Stokes equations pa pgk VP inertia force body force due surface force due to to gravity pressure gradients Where for general uid motion ie relative motion between uid particles Z a V VV Dr 3 sub stantl a1 der1vat1ve bacccaelleration 51ng832 X p m Dt 8X au au au au 8p p u v w 6t 8x 8y dz 8x Dv Y D Dt 6y 6v smznnmmcnms 71 me meessaandStzm mm 35 DW gt13 a Note forV0 z 7 7 7 2 th pg 02 azqa Y WWW 210 9x 9y gypgyv oz pa 7Vp 72 Euler s equation for inviscid ow Vy 0 Continuity equation for incompressible ow See Chapter 6 4 equations in four unknowns y and p Euler s equation can be integrated to get Bernoulli equation See Chapter 3 Streamline coordinates Str ines are the lines that are tangent to the velocity vectors throughout the ow eld m r lm nzvlltclg n my in Flow in the xrz lane b Flow in terms of streamline and normal coordinates 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 3 6 Along a streamline p pV2yZC Across the streamline V2 dn Z C 19 pl m 7 But in this chapter rigid body motion ie no relative motion between uid particles For rigid body translation 61 ax 2quot For rigid body rotating 2 492g 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 37 Examples of Pressure Variation From Acceleration Uniform Linear Acceleration pa pgk Vp Vppagkpga ggk sz plaxigazlAltJ azaxiaZlA 3p 3p a a ax p x az pg z 1 ax lt O p increase inX 2 ax gt O p decrease in X 6 6 pgaz 1 0 gt O p decrease in z 2 0 lt0 and 5 ltg p decreaseinz but slower thang 3 0 lt0 and 5 gtg pincreaseinz 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 38 unit vector in direction of Vp Vp l Vp l 2 laxigaz1 J 2 2 2 lax g 212 l n unit vector in direction of p constant A a J gtlt 39 i39ki39k J J J i to Vp ax g az by definition lines of constant p are aX gaz normal to Vp 6 tan391 aX g a2 angle between f1 and X dp A 2 2 2 V s a a gt ds p p X gr Jl Pg p st constant gt pgage pGS 57020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2006 Rigid Body Rotation Consider a cylindrical tank of liquid rotating at a constant rate Q 21 it v LX gt v csavsh SHM QM c mquot a 9x 9x r0 centripetal acceleration rQZer 7 V2 M Me sfngs 2 er r 6 A l 6 A 6 A V a V er e eZ p 5 6r r669 6z pgk prQZ r grad in cylindrical coordinates 6p 2 6p 5p 16 r9 6r p 6z pg 69 C 0 pressure distribution is hydrostatic in z direction and p ErZQZ fz 0 2 Pz 2 39Pg p pgz Cr c 2 p Er292 pgz constant B z constant 2 v 2g V r9 57020 Fluid Mechanics Chapter 2 Professor Fred Stem Fall 2006 40 The constant is determined by specifying the pressure at one point say p p0 at r z 0 0 1 2 2 ppopgz rQ Note pressure is linear in z and parabolic in r Curves of constant pressure are given by abr2 2 2 Zpl porQ 2g Pg which are paraboloids of revolution concave upward with their minimum point on the axis of rotation Free surface is found by requiring volume of liquid to be constant before and after rotation The unit vector in the direction of Vp is a g pglAlt pr 22 r 2 12 aw ng pr92 e A Y E tanezj Zz slope ofs r I W sz 1e r Clexp equatlon of Vp surfaces g 57020 Fluid Mechani c s Chapter 2 Professor Fred Stem Fall 2006 F ri y rotation top uid at rest streamers hang vertically upward bottom rigid body rotation 5 ig Experimental demonstration with buoyant streamer of the uid force eld in gid od streamers are aligned with the direction of maxIInum prcssurc gradient From Ra outresy of R Ian Fletcher 94 PRESSURE DISTRIBUTION IN A FLUID 57 020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 1 Chapter 3 Pressure and Fluid Statics 31 Pressure For a static uid the only stress is the normal stress since by de nition a uid subjected to a shear stress must deform and undergo motion Normal stresses are referred to as pressure For the general case the stress on a uid element or at a point is a tensor Tij stress tensor TXX Txy TXZ TyX Tyy Tyz TZX sz TZZ i force j direction For a static uid W 0 i j shear stresses 0 Th 2 p TXX Tyy TZZ i j normal stresses p Also shows that p is isotropic one value at a point which is independent of direction a scalar 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 2 De nition of Pressure 8F dF 2 P 11m5 A d A Nm Pa Pascal 8Aao F normal force acting over A As already noted p is a scalar which can be easily demonstrated by considering the equilibrium of forces on a wedgeshaped uid element MM Geometry 39 AA 2 Ay lt prAsina AX A E cosoc 11N3i8ht AzAlsina AZ A E sinoc I AxAlcosa pzAA cosa W mg ZFX 0 psvzg pnAA sin or pXAA sin or 0 y pn Z px 12 AXAZAy 2F 0 pnA Ay cosoc pZ A Ay cos 0c pnAA COS 06 pZAA COS OL W O A 2 cosocsinocAy 0 W 1M cosocA sinocAy MAycosoc 2 H H Y AX AZ pnpZ A sinoc0 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 3 pn pZ A sinoc 0 pnpZ for A690 ie pn px py pz p is single valued at a point and independent of direction since 06 arbitrary and independent pH of 06 A bodysurface in contact with a static uid experiences a force due to p p ngdA ghee SB v quotquotquot u hrs 7quot l gz g39a 3 r Note if p constant E 0 for a closed body Scalar form of Green s Theorem if ds Vi Vfdv f constant 2Vf 0 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 4 Pressure Transmission Pascal39s law in a closed system a pressure change produced at one point in the system is transmitted throughout the entire system Absolute Pressure Gage Pressure and Vacuum pg gt 0 pa atmospheric gt a pressure pA p pglt0 101325 kPa pA lt pa pA O absolute zero For pAgtpaa pg 2 pA pa gage pressure For PAltpa pvac 2 pg pa pA vacuum pressure 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 5 32 Pressure Variation with Elevation Basic Differential Equation For a static uid pressure varies only with elevation within the uid This can be shown by consideration of equilibrium of forces on a uid element A 4 A AXJ 1 3 1st order Taylor series estimate for pressure X variation over dz W BAXA BAi Newton s law momentum principle applied to a static uid XE ma 0 for a static uid ie ZFX ZFy ZFZ 0 ZFZ 0 dedy p 2 pdzdxdy pngdydz 0 2 6p 62 Pg V Basic equation for pressure variation with elevation 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 6 ZFy O ZFX 0 dedZ p dydxdz O pdydz p dxdydz 0 9y 8X 20 20 6y 8X For a static uid the pressure only varies with elevation z and is constant in horizontal xy planes The basic equation for pressure variation with elevation can be integrated depending on whether p constant or p pz ie whether the uid is incompressible liquid or lowspeed gas or compressible highspeed gas since g constant Pressure Variation for a UniformDensity Fluid 2 pg 2 y p constant for liquid r dz g1 AP YAZ T p2p1VZ2Z1 Alternate forms p1 yZ1 p2 Yzz constant p yz constant piezometric pressure pz O O gage p yz increase linearly with depthl decrease linearly with height T p vz B z constant piezometric head 57020 Fluid Mechanics Professor Fred Stem Fall 2005 Chapter 3 7 Oil with a specific gravity of 080 forms a layer 090 deep in an open tank that is otherwise filled with water The total depth of water and oil is 3 In What is the gage pressure at the bottom of the tank pyzconstant P1YZ1P2 YZ2 pZ p1yzlizz pl pm 0 p3 pZ Wm Z2 Zr 70609810x 21 277 277kPa Solution First determine the pressure at the oilrwater interface staying within the oil and then calculate the pressure at the bottom h h 7 7 where p is the pressure at free surface of oil 2 is the elevation of free surface of oil p is the pressure at interface between oil and water and 21 is the elevation at interface between oil andwater For this example p 0 7 080 X 9810 Nmz z 3 m and z 2J0 mi Therefore pg 3 090 m X 080 X 9810 Nmx 706 kPa gage Now obtain p from 71 22 P3 z 7 7 where p1 has already been calculated and 7 9810 Nm i 7060 p 9809810 210 277 kPa gage pZ ymlAz 8x 9810xi9 706kPa 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 8 Pressure Variation for Compressible Fluids Basic equation for pressure variation with elevation v vpz pg dz Pressure variation equation can be integrated for ypz known For example here we solve for the pressure in the atmosphere assuming ppT given from ideal gas law Tz known and g 7 gz p pRT R gas constant 287 Jkg OK dry air pT in absolute scale dP m dz RT E g dz which can be integrated for Tz known p R Tz 57020 Fluid Mechanics Professor Fred Stern Fall 2005 Pressure Variation in the Troposphere T To OLZ zo linear decrease To Tz0 where p p0z0 known OL lapse rate 65 oKkm Chapter 3 9 dp dz Z39ZT0O ZZ0 p R To aZZ0 dZ39 XdZ lnp iln o ocz z0 constant ocR use reference condition zo i Earth surface lnpo iInTO constant p0 1013 kPa ocR T 15 C solve for constant a 2 65 oKkm mg g lnToOKZZo pooc R T 0 T0O ZZo gQR p0 T 0 ie p decreases for increasing z 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 1 0 Pressure Variation in the Stratosphere TTs 55 C dingy p RTS lnp iz constant RT S use reference condition to nd constant 1 e z zogRTS 1 p p0 eXp Z lagR1 ie p decreases exponentially for increasing z 3 3 Pressure Measurements Pressure is an important variable in uid mechanics and many instruments have been devised for its measurement Many devices are based on hydrostatics such as barometers and manometers ie determine pressure through measurement of a column or columns of a liquid using the 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 1 1 pressure variation with elevation equation for an incompressible uid Differential manometer More modern devices include Bourdon Tube Gage mechanical device based on de ection of a spring and pressure transducers based on de ection of a exible diaphragmmembrane The de ection can be monitored by a strain gage such that voltage output is oc Ap across diaphragm which enables electronic data acquisition with computers Bourdon Tube Gage In this course we will use both manometers and pressure transducers in EFD labs 2 and 3 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 12 Manome 1 Barometer pv ngh pm pm th pV 0 ie vapor pressure Hg nearly zero at normal T h 76 cm pm 101 kPa or 146 psia Note patm is relative to absolute zero ie absolute pressure patm pm ocation weather Consider Why water barometer is impractical YthHg YHZOhHZO YHg hHg SthHg 136gtlt76 10336cm34ft 11H20 H20 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 13 2 Piezometer pm Patm Yh ppipe p absolute p vh gage Simple but impractical for large p and vacuum pressures ie pabs lt patm Also for small p and small d due to large surface tension effects could be corrected using Ah 4GYd but accuracy may be problem if p y Aha 3 U tube or differential manometer rmmanometev liquid p1vah vlp4 p1patm p4 vah vl gage YWlSmAh S l for gases S ltlt SIn and can be neglected ie can neglect Ap in gas compared to Ap in liquid in determining p4 ppipe Example 57020 Fluid Mechanics Chapter 3 Professor Fred Stem Fall 2005 14 Air at 20 C is in pipe with a water manometer For given conditions compute gage pressure in pipe 1 140cm Ah70cm P4 7 gage 162 P1 0 Pressure same at 2amp3 since same elevation amp Pascal s 21 Ah 2 P3 Step39bY39SteP methOd law in closed system p3 Yairl p4 pressure change produce at yAh one part transmitted throughout entire system p1 Ah yairl p4 complete circuit method 7Ah valr1 p4 gage ywater20 C 9790 Nm3 2 p3 Ah 6853 Pa Nmz 7m pg pabs or p p3patm 68531013001I286kgm3 could p 2 E R C 273 28720 273 use OK 7 1286 X 981ms2 1262 Nm3 316 note mI ltlt water p4 p valr1 6853 1262 gtlt 14 6835 Pa 17668 if neglect effect of air column p4 6853 Pa 57020 Fluid Mechanics Chapter 3 Professor Fred Stern Fall 2005 15 A differential manometer determines the difference in pressures at two points and C2 when the actual pressure at anV point in the sv stem cannot be determined 620 29 12 0 9 i M K it l glllp2gzjlvm1llh Wf Wf KYf a difference in piezometric head iffluidisagas yfltlt ym pl p2ymAh if uid is liquid amp pipe horizontal E1 E2 P1 P2 Vm Yr Ah

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.