Lecture 8 luids or a long tie now we have een talking aout classical echanics part of physics which studies acroscopic otion of particle-like ojects or rigid odies Using different ethods we have considered translational otion rotation and equiliriu However all this tie we have een talking aout solid sustances only You know of course that solids are not the only for of atter existing in nature Physics of fluids studies liquids and gases The area of physics which studies otion of fluids is known as hydrodynaics The area of physics which studies fluids in equiliriu is known as hydrostatics We shall discuss siple applications of the oth luids include oth liquids and gases They oth are of enorous significance for our everyday life They are used everywhere starting fro drinking water and air which everyody needs for living and ending up with different types of technical fluids involved in alost any technological process So what are the ain differences etween fluids and solids? In contrast with solids fluid can flow It does not have fixed shape Usually fluid takes the shape of the oundary (container) which it has een placed in This happens ecause as any other structure fluid consists of olecules However the interolecular onds in fluids are uch weaker than in solids Solids have a crystalline structure where every olecule has its own place in the crystalline lattice Interolecular onds do not allow olecules to leave their places They can only oscillate near fixed positions luids on the other hand do not have these strong interolecular interactions Molecules can ove relatively free under the influence of external forces Liquid still has soe close-range order in its structure and it can aintain its continuity Gases do not even have that Since fluids are so uch different fro solids we will need a different set of variales to descrie the We already know that to find the correct coordinate syste soeties eans to solve the prole It is also true aout finding the correct set of variales In the case of a rigid ody which has fixed shape and with the sae properties for each of its parts we used ass and force as fundaental variales to solve echanical proles However when we are dealing with fluids we have to reeer that they can change their properties throughout the volue They do not even have a fixed volue Different parts of fluid can ove with different speeds relative to each
other This eans that instead of considering fluid as a whole we have to study sall parts of this fluid These parts can e treated as relatively unifor and oving with certain speeds This is why we have to introduce a different set of variales: instead of ass we shall use density; instead of force we shall use pressure We shall consider a sall eleent of fluid which has volue V and ass In this case the density of the eleent is going to e V (8) At this point we have ade an assuption of classical hydrodynaics that we can treat fluid as if it has continuous density which changes soothly in space rather than considering fluid as coplicated olecular structure This eans that volue V is large enough copared to the size of one olecule so it includes any olecules and density introduced y eans of the equation 8 is soewhat averaged density over local volue V Even though this picture is not perfect especially when talking aout very sall ojects like sall liquid droplets consisting of only a few olecules ut it works fine for liquid in larger volues In fact it is all right to use this approach even for acroscopic droplets which we can e oserved in the everyday life such as rain droplets Since we are going to study only fundaentals of hydrodynaics we shall go even further and liit our attention y unifor fluids only These fluids have the sae density everywhere throughout their volue so V (8) This is quite an accurate approxiation as long as we are only considering liquids and do not study liquid-gas oundaries Liquids are alost incopressile However it is not accurate to treat gases this way Equation 8 can only give us the average density of gas or volue needs to e very sall so one can ignore effects of gravity SI unit for density is kilogra per eter cued ( kg 3 ) Tale 4- in the ook shows densities for soe coon sustances If we consider a sall eleent of fluid and insert a force sensor there we can easure the force acting on the surface area A of the sensor In this case we can introduce pressure of fluid inside of this eleent as
p A (83) Everything which we have said aout local density is also true aout pressure which is considered to e changing soothly throughout the fluid In ideal case we can even treat it as constant everywhere in fluid if it has flat surface and unifor external force exerted perpendicular to this surface so that p (84) A This assuption is ore accurate for gases in sall volues where we can ignore the influence of gravitational field (ecause of the sall density of gas) while it is not very accurate for liquids where pressure changes with depth due to gravitational effects It is an experiental result that in unifor fluid pressure is a scalar so equation 84 only involves asolute value of force SI unit for pressure is Pascal which is Newton per eter squared (Pa N ) Other coon units are related as 5 at 0 0 Pa 760torr 4 7l in Hydrostatics As it was already entioned hydrostatics studies fluids at rest Let us consider a liquid which is placed into a tank and stays there at rest (see the picture) Everyody knows that pressure in water gets larger with increase of depth It ay ecoe enorous at the on the otto of the ocean On the other hand pressure in air gets saller with increase of height It is very low at the top of ountains Let us see what the reason for that is We need to find pressure in fluid as function of the altitude We usually refer to this pressure as hydrostatic pressure The y-axis in the picture is directed upward so the value of y-coordinate y at level is larger than y at level We
shall consider a liquid eleent in the for of the cylinder (it is shown y a different color in the picture even though it is still the sae liquid) It has the sae area A of the top surface as of the otto surface This eleent is in static equiliriu which eans that the vector su of all the forces acting on it is zero There are three forces acting on this cylinder: the gravitational force g in the downward direction the force in the downward direction due to the liquid's pressure aove the cylinder and the force in the upward direction due to the liquid's pressure elow the cylinder The condition of equiliriu gives us g The ass of this cylinder can e calculated as V Ah A y yg where h y y is the cylinder's height and is the density of liquid With account of equation 84 we have p A p A A y y g p p g y y g g (85) The last equation relates pressure at two different levels If we choose the ground level with pressure p 0 equal to the atospheric pressure at the surface of this tank then the pressure p at any depth h is going to e p p0 gh (86) We can also apply equation 85 to calculate the pressure of air at certain height which is going to e less than its value at the ground level Since the gas density is uch less than the density of liquid the pressure inside of gas can e considered as alost constant in the case when the size of the container h is sall So the pressure at point in a fluid in static equiliriu depends on the depth of that point ut not on any horizontal diension of the fluid or its container Pressure p in the equation 86 is said to e asolute pressure It consists of two ters: the atospheric pressure p 0 and the gauge pressure gh due to the liquid itself Exaple 8 The U-tue contains two liquids: water of density w 3 998kg is in the right ar and oil of unknown density x is in the left ar Measureent gives l 35 d=3 (see the picture) ind the density of this oil
Since pressure at the sae level is the sae we can calculate pressure at the level of water-oil interface y two different ethods irst using the oil ar which gives p p g l d 0 x g Then using the water ar which gives p p0 wgl After that we have p g l d p gl 0 x 0 g l d gl x l d l x g w g w g w g g kg kg x wl l d 998 35 35 3 95 3 3 The aove exaple gives us an idea of how one can uild a device to easure pressure in particular a device which allows easuring the atospheric pressure Since even sall changes in this pressure can cause a lot of change in weather this device is very iportant It is known as the ercury aroeter which was first introduced y Italian scientist Evangelista Torricelli in the 7 th century This siplest aroeter consists of the long glass tue filled with ercury It is closed at one end and open at the other end which is placed in the dish with ercury The pressure at the dish level is equal to the atospheric pressure On the other hand this pressure is the sae as the gauge pressure of the colun of ercury so p 0 gh Oserving the height of the colun we can find what the atospheric pressure is This is why the unit of pressure known as illieter of ercury colun was introduced Noral atospheric pressure gives the height of this colun to e 760
Exercise: Why do you think Torricelli used expensive and (as we now know) dangerous sustance of ercury instead of coon water to uild his aroeter? Another iportant principle which is related to any applications is the Pascal's principle This principle states that a change in the pressure applied to an enclosed incopressile fluid is transitted undiinished to every portion of that fluid and to the walls of its container The ost useful application of the Pascal's principle is hydraulic lever shown on the picture elow If one applies an external force of agnitude to the left saller piston with the area A it will produce change in the liquid s pressure p This change will e the sae A everywhere inside of liquid So the pressure acting on the second larger piston fro the liquid will also change y the sae aount This eans that it will e additional force acting on the second piston fro the liquid which is pa A A Since the area of the second piston is larger than the area of the first piston the force So one can apply relatively sall force at one side of the syste and otain a very large force at the other side to lift soe heavy oject You ight think that this would e ipossile ecause of the law of conservation of energy In fact there is nothing wrong with conservation of energy here Even though the force gets larger ut the second piston will ove for a distance d which is saller than distance d for which the first piston has oved Indeed this liquid is incopressile so it can not change its total volue which eans that changes of the volue are the sae on oth sides d A d A
d d A A In this case work perfored y the second force is the sae as the work perfored y the first force W d A A d A d W A and energy is conserved So a given force applied over a given distance can e transfored to a grater force applied over saller distance Exaple 8 In the hydraulic syste shown on the picture efore the piston on the left has a diaeter of 45 c and a ass of 7 kg The piston on the right has a diaeter of c and a ass of 3 kg If the density of the fluid is 750 kg/ 3 what is the height difference h etween the two pistons? Pressure at the sae level in liquid is the sae This pressure can e found either as the pressure of the piston on the left p or as the pressure of the piston on the right p plus the pressure of the liquid with height h This eans that p p gh Taking into account that pressure can e found as the force acting per unit of area one has g g gh A A A h h A HG I KJ HG I KJ 4 4 A A d d 750kg HG 7 kg kg 0 045 3 g 0 g 3 I KJ 05 Let consider another well-known effect which occurs in fluid the uoyant force Everyody knows that if you put soething heavy into water you can lift it with uch less effort copared to that in the air This is ecause water as well as any other fluid exerts a uoyant force on any oject placed in it If we consider soe ody placed in liquid we can notice that pressures are different at the top and at the otto of this ody since we have already learned that pressure in liquid depends on depth only This pressure difference is the reason for the uoyant force The force acting on the top surface
of the ody downwards is p A and on the otto surface of the ody upwards p A so the net uoyant force acting on this ody fro the fluid is where g (87) A p p A gh g f f f Ah is the ass of the fluid displaced y the ody So we have arrived to f the Archiedes' principle: When a ody fully or partially suerged in a fluid a uoyant force fro the surrounding fluid acts on the ody This force is directed upward and has a agnitude equal to weight of the fluid that has een displaced y the ody The existence of the uoyant force is the reason why floating ecoes possile Indeed let us consider soe oject partly suerged into fluid and floating in it If it floats it is in equiliriu so the net force acting on this oject in the vertical direction is zero On the other hand the only two forces acting on it are gravitational force and uoyant force so g g g 0 f 0 This would only e possile if ass of the displaced fluid is equal to the ass of the ody However if the ass of the ody is ore than the ass of the displaced fluid then the ody will sink If the ody is copletely suerged then its volue equal V the volue of the displaced fluid and we have f V f f V so if copletely suerged ody floats in fluid it has the sae density as this fluid If the density of the ody is less than the one of the fluid it will float ut it will not e copletely suerged If the density of the ody is larger than the density of fluid it sinks V f Exaple 83 A geologist finds that a oon rock whose ass is 80 kg has an apparent ass of 68 kg when it is suerged in water What is the density of the rock? irst of all we have to understand what apparent ass is This ass is easured y soe scale when the rock is suerged in water The scale usually shows the tension force not the real weight When in the air this tension is equal to the weight of the oject
In the water this tension T will e equal to the apparent weight T g a where a stands for the apparent ass Let us use Newton's second law for the rock suerged into water Since this rock is in alance one has g T B 0 According to Archiedes' principle the uoyant force g a g watervg 0 V 0 a water B water Vg so Here V is the volue of the rock so a water 0 rock water a rock water rock rock water a a rock kg 8 0kg 3 kg 998 4 050 3 3 8 0kg 68 kg Hydrodynaics So far we have een talking aout stationary fluids Now let us consider fluids in otion We shall only consider ideal fluids There are several assuptions we will ake aout these fluids Those are the following a) luid undergoes a steady (lainar) flow The velocity of this fluid at any point in space is not changing with tie neither in agnitude nor in direction ) This fluid is incopressile (it is liquid rather than gas) It has the constant value of density everywhere c) This flow is nonviscous There is no friction etween different layers of this fluid If you ever used a garden hose you know that speed of water eerging fro this hose depends on the area of the hose's cross section The saller the area (if you partially close the hose) the higher the speed of the eerging water Let us see how this speed is related to the area of the cross section Suppose we have a tue of length L and area of its cross section varies along the tue It has area A at the eginning of the tue and liquid flows with speed v through this cross section and it has area A at the end of the tue and liquid flows with speed v through that cross section The liquid is incopressile
and the tue has a fixed volue This eans that the aount (volue) of liquid entering the tue at one end should e equal to the aount (volue) of liquid leaving the tue at the other end Let us find what volue of liquid is passing the tue's cross section for a sall tie t This volue can e calculated as V Ax where x vt the distance for which the fluid oves for that tie Since this volue is the sae at oth ends of the tue one has So V A v t A v t A v V A x A x A v Av const (88) This last relation is known as the equation of continuity which states that the quantity R v Av the volue flow rate reains constant for any tue of flow in ideal fluid Exaple 84 A roo easures 30 y 45 y 60 If the heating and air conditioning ducts to and fro the roo are circular and have a diaeter of 30 c what is the speed of the air flow in the ducts necessary for all of the air in the roo to e exchanged every 0 in? (Assue that the air s density is constant) Since the density of the air is constant one can write continuity equation in the for v A v A where the left side of the equation refers to the duct and v is the unknown speed of air in the duct while A 4 d is the area of the cross section for a the duct The right side of the equation refers to the roo where v is the speed of t the air in the roo which has distance a etween walls and A cross section for this roo So c is the area of the v d 4 a t c v 4ac 4 30 4 5 6 0 5 9 d t 0 30 0in in g s The content of the continuity equation is siple It eans conservation of atter in the tue Let us see what we can find fro the other iportant conservation law the law of conservation of energy We shall consider a tue which is not necessarily horizontal
so one end of the tue is elevated for height h and another end is elevated for height h Let us use the energy conservation in the for of work-kinetic energy theore for the fluid oving along the tue In this case the work perfored on this fluid is equal to the change of the fluid s kinetic energy which is W K W K K W v v where V is the ass of fluid which enters the tue at one end and leaves it at the other end for tie interval t The work done on this fluid eleent coes fro the two sources: fro the gravitational force and fro the external forces at oth ends of the tue The work of gravitational force on the ass eleent when it oves fro one end of the tue to the other end is taken with inus-sign change of its gravitational potential energy which is g Wg g h h Work is also done on the ass eleent y the force at the first end of the tue to push liquid into the tue this work is W x p A x p V Another work is done y the liquid in the tue itself to push the fluid outside of the tue at the other end this work is W x p A x p V Coining all this together we arrive to Wg W W v v gh hg pv pv v v pv gh v pv gh v (89) p V gh V v p V gh V v p gh v p gh v This equation is known as Bernoulli's equation and it states that quantity
p gh v const reains constant along the tue of flow Exaple 85 What is the lift (in Newtons) due to the Bernoulli's principle on a wing of area 80 if the air passes over the top and otto surfaces at speeds 340/s and 90 /s respectively? Let us apply Bernoulli's equation One can ignore the thickness of the wing so the top part of the wing and the otto part are at the sae height which eans P v P v where suscript refers to the top and suscript refers to the otto So the pressure difference etween the otto part and the top part is P P P v v To find the lift force acting on the wing one has to ultiply this pressure difference y the area of the wing so AP A v d vi kg Taking into account that it is the air with density 9 3 HG I KJ kg we have H G I N s K J 3 H G I 80 9 340 90 s K J 63 6 0 flowing around the wing