Fluid and Fluid Mechanic Fluid in motion Dynamic Equation of Continuity After having worked on fluid at ret we turn to a moving fluid To decribe a moving fluid we develop two equation that govern the motion of the fluid through ome medium, like a pipe Thee two equation are the equation of continuity (which ay that the fluid flow rate i continuou) and the econd i Bernoulli equation (which i a tatement of conervation of energy) Conider the pipe in Figure #1 with varying cro-ectional area At point 1 let the cro-ectional area be A 1 and at point let the cro ectional area of the pipe be A Fluid enter point 1 from the left moving at velocity v 1 What we d like to determine, i with what velocity will it leave point To do thi we need to determine the ma flow rate i a contant throughout the pipe That i, the ma of fluid per unit time entering at point 1 and the ma of fluid per unit time leaving point Providing that there are no leak in the ytem (the pipe) the ma of fluid entering the pipe at point 1 (per unit time) ha to equal the ma of fluid leaving the pipe at point (per unit time) To quantify thi we write Δm 1 Δm Thu we can write uing the denity of the fluid, Δm 1 ρ 1ΔV 1 for the rate of change of the ma of fluid flowing pat point 1 and imilarly Δm ρ ΔV for the ma of fluid flowing pat point ΔV i the volume of fluid paing any point in pace and equating the two expreion give Δm 1 Δm ρ 1ΔV 1 ρ ΔV ρ 1A 1 ρ A Δl Figure #1: Diagram howing the relevant quantitie ued to determine the equation of continuity D Giancoli, Phyic: Principle with Application, 7 th Ed, Prentice Hall
The denity of the fluid at point 1 and point could be different allowing for the fact that the fluid may be compreible If the fluid i incompreible, a we will aume for everything that we will do, then ρ 1 ρ and we identify Δl v a the peed of the fluid at any point Therefore, we have ρ 1A 1 ρ A Δl A 1 v 1 A v, which we call the equation of continuity The equation of continuity give u a way of determining the velocity of the fluid moving at point By the equitation of continuity, the amount of fluid moving pat point 1 per unit time i the ame a that moving pat point per unit time Thi implie that ρ 1 V 1 ρ V and ince the fluid i incompreible the volume of fluid in (per unit time) at point 1 i equal to the volume of fluid out (per unit time) at point Thi i an alternate verion of the equation of continuity and ay the ame thing Bernoulli Equation: Conervation of Energy in a Moving Fluid Having the equation of continuity, we are now in a poition to decribe the moving fluid Let aume that we have a pipe a hown in Figure # The left end of the pipe at point 1 i at a height y 1 above the ground and the fluid enter point 1 with a peed v 1 through the pipe with cro-ectional area A 1 The fluid at point leave with a peed v through a ection of pipe with cro-ectional area A and point i at a vertical height y above the ground We d like, again, to determine the peed of the fluid when it leave point To do thi we need to get the fluid moving into point 1 Thi take work and the work done to move the fluid into point 1 i given by ome applied force and how far into the pipe you need to move the ma of fluid Thu, W 1 F 1 P 1 A 1, where P 1 i the preure of the fluid at point 1 and i the ditance it take to move a ma of fluid m 1 pat point 1 To keep the fluid moving at point, we have to puh on the fluid ahead of the incoming fluid that i moving toward point from the left The fluid immediately to the right of point offer reitance againt the incoming fluid Thi take energy out of the fluid and cot work The work done againt the incoming fluid i W F Δl P A Δl, where P i the preure of the fluid at point and Δl i the ditance it take to move a ma of fluid m pat point The net work done i equal to the change in energy of the ytem Thu,
ΔE ytem W 1 + W P 1 A 1 P A Δl ΔK T + ΔK R + ΔU g + ΔU We aume that there i no change in pring potential energy and that the fluid i not rotating Figure #: Diagram howing the relevant quantitie ued to determine the Bernoulli equation D Giancoli, Phyic: Principle with Application, 7 th Ed, Prentice Hall Thu, we have ΔE ytem P 1 A 1 P A Δl ΔK T + ΔU g 1 m v 1 ( m 1 v 1 ) + ( m gy m 1 gy 1 ) Noting that i AΔl the volume of the fluid at any point, we have P 1 V P V 1 m v 1 ( m 1 v 1 ) + ( m gy m 1 gy 1 ), where by the equation of continuity the volume of fluid in at point 1 equal the volume of fluid out at point, namely A 1 A Δl V 1 V V Now we divide both ide by the volume of fluid and note that the denity of the fluid i imply the ratio of the ma of the fluid divided by it volume Further, we aume again that the fluid i incompreible o that ρ 1 ρ ρ We have P 1 P 1 ρv 1 ( ρv 1 ) + ( ρgy ρgy 1 ), or ΔE 0 ( P P 1 ) + 1 ρv 1 ( ρv 1 ) + ( ρgy ρgy 1 ), in the abence of any external force Thi i called Bernoulli equation and it a tatement of conervation of energy in a moving fluid The firt term on the right i the difference in preure between point 1 and The econd term i the difference in the kinetic energy per unit volume of fluid and the third i the difference in gravitational potential energy per unit volume of fluid So, we
identify the volume kinetic energy a K Volume 1 ρv and the volume gravitational potential energy a U g Volume ρgy In fact, ince Bernoulli equation i a tatement of conervation of energy, it hould apply to any ituation How about a fluid at ret? Conider Figure #3 below in which the egment of fluid i at ret with repect to the urrounding fluid Thu v v 1 0 The bottom of the cylinder of fluid i at a depth y below the urface while the top of the cylinder i at depth y 1 below the urface, o that y y 1 h Therefore from Bernoulli equation the preure at the cylinder bottom P i: ΔE 0 ( P P 1 ) + ( ρgy ρgy 1 ) P P 1 + ρg( y 1 y ) P 1 + ρgh Thi i exactly the reult we obtained when we looked the force on a tatic fluid, namely that the preure increae linearly with increaing depth Figure #3: Variation of preure with depth D Giancoli, Phyic: Principle with Application, 7 th Ed, Prentice Hall Now let apply thi to a few example problem
Example #1: Fluid exiting a yringe The body of a yringe i held horizontally and ha a cro ectional area 5 10 5 m while the needle ha a cro-ectional area of A needle 10 10 8 m Suppoe that a force of F applied N i applied to the plunger on the body ide of the yringe, what i the peed of the fluid a it leave the needle tip? Aume that the fluid inide ha a denity of 1050 kg m 3 and that in the abence of the applied force, the preure on all ide of the yringe i 1Atm 101 10 5 N m Solution: We apply Bernoulli equation to the ituation Defining poition to be the needle and poition 1 to be the body of the needle we have: ΔE 0 P needle P body energy (per unit volume) of the fluid in the needle and we have 1 ( ) + 1 1 ( ρ body v body ) We olve thi for the kinetic P body P needle + 1 ρ body v body The preure applied at the body of the needle i due to atmopheric preure and the force applied over the croectional area of the body of the needle Solving for the peed of the fluid in the needle we have P body P needle + v body F applied Abody + P atm P atm + v body F applied + v body To olve for the peed of the fluid exiting the needle, we need to know how fat the fluid i moving in the body of the needle Here we will ue the equation of continuity, which ay that A 1 v 1 A v v body A needle Therefore olving for the velocity of the fluid in the body of the needle we have v body A needle v Inerting needle
thi into our equation we can olve for the velocity of the exiting fluid Subtituting we have F applied + v body F applied + 1 A needle F applied A needle F applied 1 A needle ρ fluid 1 N ( 10 10 8 m ) 1 5 10 5 m 1050 kg m ( 5 10 5 m ) 3 1 13 m Comment: Notice A needle << that o that when we formed the ratio, the ratio wa o mall compared to one that we could have jut ignored the term 1 A needle 1 Example #: Lift on an airplane wing Conider the airplane hown below What i the aerodynamic lift produced on the wing if the wing ha a 60m urface area and the velocity of the airflow acro the top and bottom of the wing are 340 m and 90 m repectively? Aume that the wing i thin enough o that the airflow over and under the wing doe not have an appreciable change in height And in fact, one could argue that the front and back edge of the wing are the ame height, o what ever increae in gravitational potential the air get going over the front edge it loe when it come back down to the back edge, effectively experiencing no net change in height P top P bottom Figure #4: Photograph of one of the US Air Force Thunderbird aircraft The red arrow how the preure on the upper and lower urface of the wing The difference in preure give rie to lift Photo: S LaBrake
Solution: We tart with Bernoulli equation and ue the tated aumption ΔE 0 ( P P 1 ) + 1 ρv 1 ( ρv 1 ) + ( ρgy ρgy 1 ) ( P P 1 ) + 1 ρv 1 ( ρv 1 ) We notice that ince energy i conerved, where the velocity i higher, by Bernoulli equation, the preure mut be lower at that point in order to keep the energy a fixed quantity Thu ince the velocity of the air i lower on the bottom of the wing, the preure mut be higher there than on top of the wing and thee preure are hown in figure #4 above Next we define poition a the bottom of the wing and poition 1 the top of the wing We have (uing the denity of air a ρ air 13 kg m 3 ) ( ) + 1 ρv bottom 1 ( ρv top ) 1 ρ air v bottom ( ) 340 m 0 P bottom P top P bottom P top 1 ρ air v top 1 13 kg m 3 (( ) ( 90 m ) ) 0475 N m Now, the preure i related to the force that applied divided by an area We define the preure on the bottom and top of the wing to be F bottom A wing and F top A wing repectively The difference in force between the top and bottom of the wing define the lift F lift F bottom F top, which i perpendicular to the wing urface Here ince the preure on the wing bottom i greater than on the wing top, the net force i directed vertically upward and the force of lift act vertically upward to raie the plane up in the air Thu we have P bottom P top 0475 N m F bottom A wing F top A wing F lift A wing We can olve for the force of lift and we get F lift A wing ( P bottom P top ) 60m 0475 N 13 10 6 N m Comment: If the lift force i larger than the weight of the airplane then the airplane accelerate upward at ome rate, call it a plane, and the airplane rie If the weight of the airplane i greater than the lift (ay by the airplane lowing down it forward velocity) then the plane fall at a plane If the weight of the airplane and the lift force are equal the plane neither rie nor fall but travel forward at a contant velocity Thi i provided of coure that the thrut produced by the engine puhing the plane forward equal the drag on the airplane produced by air friction
Example #3: An old Wet water tower Conider a water tower, like one might find in the old Wet, to be circular with a top that ha a cro-ectional area A top and the ide a depth d Further let the water tower be open to air on the top A Wild Wet outlaw hoot a hole in the ide of the water tower at a depth h < d below the urface and thi hole ha a cro-ectional area A ide The water flow out of the ide of the water tower through the mall hole What i the peed of the water flowing out of the mall hole on the ide of the water tower if h 5m? Solution: We apply Bernoulli equation and define poition to be the hole on the ide of the tank and poition 1 to be the top of the tank a hown in figure #6 below v v 1 We have ΔE 0 P ide P top Figure #6: A tank of water, open to the air on top, with water exiting through a ide hole a depth h below the top Thi i a model of the water tower D Giancoli, Phyic: Principle with Application, 7 th Ed, Prentice Hall ( ) + 1 ρv ide 1 ( ρv top ) + ( ρgy ide ρgy top ) Next we note that the top i open to air and that the water flowing out of the hole on the ide i flowing into air Thu the preure on the top and ide of the tank i due to air and P ide P top P air Returning to Bernoulli equation we have olving for v ide : v ide v top + ( gy ide + gy top ) Defining the zero of the gravitational potential energy to be at y top, we can olve for v ide Thu v ide v top g( h) v top + gh Again, we need to relate the peed of the water moving acro the top urface of the water tower to the peed of the water coming out of the ide of the tower We return to the equation of continuity and we get A 1 v 1 A v A ide v ide A top v top v top A ide Atop v ide Auming that
A top >> A ide we have A ide << 1 and v top A ide A Atop top v ide ~ 0 So, the peed of the water exiting the hole on the ide of the water tower i approximately v ide v top + gh ~ gh 98 m 5m 99 m Comment: Notice that the denity of the fluid diappeared here Thi i exactly analogou to dropping a ma m from ret and letting it fall through a height h The velocity of the falling ma doe not depend on the ma, but only the height through which it fell Example #4: Fluid flow in arterie and vein Suppoe that the aorta ha a radiu of about r aorta 15cm and that the typical blood velocity i around 30 cm with an average denity of ρ 1050 kg A implified model m 3 of the human circulatory ytem i hown in Figure #7 The human circulatory ytem i a cloed ytem, o the flow rate of blood out of the heart ha to be the ame a the flow rate of blood coming back to the heart Quetion and Solution: a What i the total flow rate of blood through the aorta? The total flow rate i determined from the equation of continuity We have Q A aorta v aorta ( πr aorta )v aorta π ( 0015m) 03 m 15 10 4 m 3 b What i the average blood velocity in the major arterie if the total croectional area of the major arterie i 0cm? Firt we convert 0cm into quare meter We have ( ) 10 3 m Then we ue the equation of continuity to Figure #7: Simplified model of the human circulatory ytem D Giancoli, Phyic: Principle with Application, 7 th Ed, Prentice Hall 0cm 100cm 1m determine the blood velocity in the major arterie From the equation of Q continuity we have Q A aorta v aorta A arterie v arterie v arterie 15 3 10 4 m 0075 m A arterie 10 3 m 75 cm c On the aumption that all the blood in the circulatory ytem goe through the capillarie, what i the total cro ectional area of the capillarie if the average velocity of the blood in the capillarie i 003 cm? Again we ue the equation of continuity We have for the total cro-ectional area of all of the capillarie Q Q A arterie v arterie A capillarie v capilaie A capillarie 15 3 10 4 m 05m v capilarie 3 10 4 m
d If a typical capillary ha a cro ectional area of A capillary 3 10 11 m, about how many capillarie are there in the human body? The approximate number of capillarie in the human body i # A capillarie A capillary roughly 17 billion 05m 3 10 11 m 17 1010 or e If a capillary ha an average length of l 075mm what i the average time that a red blood cell pend in a capillary? Auming that the blood velocity i contant in the capillarie then the average time a red blood cell pend in a capillary i v capilary l t t l 075 10 3 m 5 v capilary 3 10 4 m f What are the kinetic energy per unit volume for blood in the aorta, the major arterie, and the capillarie? The kinetic energy per unit volume of blood in the major ytem i given by Evaluating thi for each ytem we find: K Volume 1 mv V 1 ρv Aorta: K Volume 1 ρv 1 1050 kg m ( 03 m ) 473 J m 3 Arterie: K Volume 1 ρv 1 1050 kg m ( 0075 m ) 95 J m 3 Capillarie: K Volume 1 ρv 1 1050 kg m 3 10 4 m ( ) 473 10 5 J m 3 Example #5: The Heart a a mechanical pump The human heart can be modeled a a mechanical pump The aorta i a large artery that carrie oxygenated blood away from the heart to variou organ in the body For an individual at ret, the blood ( ρ blood 1050 kg m 3 ) in the aorta of radiu r aorta 15cm flow at a rate of 5 10 4 m 3 In what follow, we will be talking about power Power i min the rate at which work i done or the rate at which energy i tranferred into or out of a ytem by an external force Thu we can define the work done on an object a P ΔE ΔW FΔx Fv in unit of Joule per econd or equivalently Watt, where 1W 1 J
Quetion and Solution: a With every beat, the heart doe work moving the blood into the aorta The heart doe work at a rate of 05 J Derive an expreion for the energy per unit volume of blood aociated with the blood flow into the aorta? We tart with the expreion for power and we ee that the quetion ak u to determine an expreion for the energy per unit volume Thu we will take our equation for power and multiply it by a factor of one Thi will generate the energy per unit volume we eek In addition we generate an extra term that, the volume of blood flow per unit time Thi i imply the flow rate, which we will call Q Thu we have P ΔE Volume Volume ΔE Volume Volume ΔE ΔE Q and the energy per unit volume i Volume Uing the number given in the problem we can calculate a value for thi quantity, namely Volume P Q ΔE Volume P Q 05 J 5 10 4 m 3 60 60000 J m 3 60000 N m b The energy per unit volume (or kinetic energy per unit volume) of a moving fluid correpond to a difference in preure between two different point in pace Suppoe you have the condition called atherocleroi Atherocleroi i a dieae in which plaque build up inide wall of your arterie Arterie are blood veel that carry oxygen-rich blood to your heart and other part of your body Plaque i made up of fat, choleterol, calcium, and other ubtance found in the blood Over time, plaque harden and narrow your arterie and contrict blood flow Thi limit the flow of oxygen-rich blood to your organ and other part of your body Atherocleroi can lead to eriou problem, including heart attack, troke, or even death) Suppoe that you had the condition of atherocleroi, what would be the radiu of the opening that remain and what percent of the aorta would be blocked? Aume that the buildup on the wall of the aorta i uniform o that the opening that remain i circular and that the patient i lying horizontal The ituation i hown in Figure #8 below
From Heart to Body Blockage Blockage Figure #8 Cartoon model of the human aorta with a blockage for a peron lying horizontally In order to anwer the quetion of how much of the aorta i blocked, we firt calculate the peed of the blood in the unblocked portion of the aorta We can calculate thi peed from the flow rate and the croectional area of the aorta From the information given in the problem, we have the flow rate where Q A aorta v aorta and thu v aorta Q A aorta 54 10 4 m 3 60 π ( 0015m) 0017 m Next we turn to Bernoulli equation and for a peron who i lying horizontally, the difference in preure between the unblocked and blocked portion of the aorta i P 1 + 1 ρv 1 + ρgy 1 P + 1 ρv + ρgy Defining poition 1 to be in the unblocked aorta and poition in the blockage, we have P aorta + 1 ρv aorta the blood flow in the blockage we get P blockage + 1 ρv blockage Solving for the peed of v blockage ρ P P ( aorta blockage) + v aorta ( 60000 ) N m + ( 0017 m 1050 kg ) 101 m Having now calculated the m 3 peed of the blood in the blockage and ince the flow rate i continuou, we can determine the area of the blockage From the area we can determine the radiu of the blockage and what percent i blocked The area of the blockage can be determined from the flow rate The flow rate and area of the blockage are Q Q A blockage v blockage A blockage 54 10 4 m 3 60 841 10 7 m Therefore the radiu of the blockage, v blockage 107 m auming that the blockage i circular i given by A blockage πr blockage r blockage A blockage π 841 10 7 m π 00005m 005cm
Latly the percent of the aorta that i blocked i % r r i f 15cm 005cm 100% 15cm 100% 96% r i