1. Physics of the Cardiovascular System

2. Work done by the heart The work W done by a pump working at a constant pressure P is equal to the product of the pressure and the volume pumped W = P Δ V The physical work done by the heart = average pressure x volume of blood pumped

3. Example 1 Suppose that the average pressure =100 mm Hg The gauge pressure = ρ g h = 13.6 x 980 x10= 1.4 x 10 ⁵ dynes / cm ² If 80 ml of blood is pumped each second W = P ΔV The work per second = 80 x 1.4 x 10 ⁵ 7 =1.1 x 10 ergs /sec = 1.1 J/sec = 1.1 watts

4. Example 2 A heart discharges 80 cm3 of blood at each beat against a mean pressure of 10 cm of mercury the pulse frequency being 72 per minute. Calculate the rate of working in watts. pressure = 10 x 13.6 x 980 = 1.335 x 10 ⁵ dynes / cm² ⁵ Work done per beat = 80 x 1.33 x10 work done per second =72 x 80 x 1.335 x 10 ⁵ (ergs/sec ) 60 5

5. Work done per second in Joules =72 x 80 x 1.335 x 10 ⁵ = 1.28 Joule /sec 60 x 10 Work done per second = 1.28 watts power = 1.28 watts 7

6. Pressure across blood vessel wall The greatest pressure drop in the cardiovascular system occurs in the region of the arteries and capillaries. The capillaries have very thin walls (1 µm ) that permit easy diffusion of O 2 and CO 2. In order to understand why they do not burst we discuss the law of Laplace. According to the Lapplace ’s law, the tension in the wall is related to the radius of the tube and the pressure inside the tube

7. T = RP where T is the wall tension P is the pressure R is the radius of the tube

8. For along tube of radius R with blood pressure P we can calculate the tension in the walls. The tension is very small for very small vessels, and thus their thin wall of radius R, do not break. Consider along tube of radius R carrying a blood at pressure P. The tension T in the wall can be calculated. The pressure is uniform on the wall. The force per unit length pushing upward =2RP

9. Figure 1 : For a long tube of radius R with blood pressure P (a ) we can calculate the tension in the walls (b ) the tension is very small for very small vessels, and thus their thin walls do not break. (a ) ( b)

10. The tension T in the wall can be calculated. The pressure is uniform on the wall. The force per unit length pushing upward =2RP There is a tension force T per unit length at each edge that holds the top half of the tube to the bottom half. Since the wall is in equilibrium, the forces holding them together. 2T = 2 RP T = RP

11. Blood Flow -Laminar and Turbulent Laminar : is a streamline flow which is present in most blood vessels. Turbulent : is the flowing of blood rapidly fast. the heart valves,for example. In laminar flow the blood that is contact with the walls of the blood vessel is essentially stationary, the layer of blood next to the outside layer is moving slowly, and successive layers move rapidly at the center of the vessel,figure2.

12. Figure 2 : Blood flow in vessels (a ) In the laminar flow in most of the vessels there is a greater velocity at the center as indicated by the longer arrow. (b ) The distribution of red blood cells is not uniform ; they are more dense at the center so the blood that flows into small arteries has smaller percentage of red blood cells than the blood in the main artery.

13. Critical Velocity If you gradually increase the velocity of a fluid flowing in a tube by reducing the radius of the tube, it will reach a critical velocity V c when laminar flow changes into turbulent flow as shown in figure 2. The critical velocity will be lower if there are restriction or obstructions in the tube.

14. Figure 3: If the fluid is flowing in a long tapering tube, the velocity will gradually increase to the point where it exceeds the critical velocity Vc, producing turbulent flow.

15. Reynold’ s found that the critical velocity is proportional to the viscosity of the fluid and is inversely proportional to the density of the fluid and radius R of the tube. V c = K η critical velocity ρ R Where K is constant and is called Reynold’ s number. K = 1000 for many fluids including blood

16. η is the viscosity of fluid R is the radius ρ is the density of the fluid In aorta, which has a radius 1 cm in adults, the critical velocity V c = K η ρ R V c =( 1000 )( 4 x 10¯³ Pas) ( 1000 Kg/m ³) (10¯² m) V c = 0.4 m /sec

17. Viscosity Viscosity :is the ratio of the stress to velocity gradient. Viscosity = stress/velocity gradient ( Pas ) Viscosity = pressure (Pa ) velocity gradient [(m/sec )/m ] Velocity gradient = velocity /Δ L

18. The velocity in the aorta ranges from 0 to 0.5 m /sec, and thus the flow is turbulent during part of the systole. During heavy exercise the amount of blood pumped by the heart may increase four or five times and the critical velocity will be exceeded for longer period of time. The heart sounds of a person doing heavy exercise are different from those of a person at rest.

19. Laminar flow is more efficient than turbulent flow.Figure (3 ) shows that the slope of the curve in laminar flow region is greater than that in the turbulent flow region. That is, a given increase in pressure causes a greater increase in the laminar flow rate than in turbulent flow rate. The reduction in efficiency is apparent in the blood flow through an artery with an obstruction ( figure 3 b ).

20. Figure 4: (a ) When the flow in a tube becomes turbulent ( at pressure Pc ) The slope of the flow rate versus pressure decreases so that compared to laminar flow a greater increases of pressure is necessary to obtain given increase in flow rate. (b) In a obstructed artery the pressure needed to produce a given flow rate is greater than in a normal artery of the same size.

21. For the flow rate Va a pressure of P1 is needed for the normal artery and a some what higher pressure P2 is needed for the obstructed artery. If both arteries are required to deliver a new flow rate Vb, the increase in pressure ΔP2 ( and thus the work ) will be much greater for the obstructed artery since the flow will be turbulent.

22. Blood Pressure and its Measurement A. Direct method of measurin g blood pressure. Figure 5: Direct blood measurement. A hollow needle is inserted in the blood vessels, and a catheter ( hollow plastic tube ) is thread through the needle.The catheter transmits the blood pressure transducer.The blood pressure deflects the diaphragm,causing a change of resistance in the four strain gauge wires.The T wires undergo tension and the C wires undergo compression.

23. By placing the tension and compression wires in opposite arms of a bridge, a voltage output is obtained that operates a meter or displays pulsatile waveforms on a scope or recorder

25. In this method a hollow needle is inserted in the blood vessel and a catheter ( hollow plastic tube ) is threaded through the needle. The catheter placed in the arm ; during catheterization of the heart, the catheter is advanced into the chambers of the heart Every few minutes, the stopcock is rotated so that a few milliliters of flushing solution pass through the catheter, thus preventing a clot from forming at the tip.

26. The liquid enters the dome of the pressure transducer, where it pushes down on the metal diaphragm. The resulting bending of the diaphragm moves an armature, around which are wound fine strain gauge wires. The wires are all under tension to begin with, but downward movement of the armature increases the tension T of the two wires.

27. The increased tension stretches these wires, makes them narrower, and increases their resistance. The other two wires undergo slight compression C. Compression slackens these wires, makes them fatter, and decreases their resistance. By placing the tension and compression wires in opposite arms of a bridge, a voltage output is obtained that operates a meter or displays pulsatile waveforms on a scope or recorder (figure 5).

28. Direct measurement of blood pressure on a standing person and on a person under gravity The pressure in the circulating system varies throughout the body.Even in major arteries the pressure varies from one point to another because of gravitational forces (figure 6a ) shows schematically direct measurements of blood pressure made on a standing person ; open glass tube manometers are shown connected to arteries in the foot, upper arm, and head.

29. Figure 6: Pressure in standing person (a ) If glass capillaries were connected to the arteries at different location the blood would rise to about the same level. (b) If the body were accelerated upward at 3 g the blood would not reach the brain and blackout would result.

30. In this situation the blood rises to essentially the same level in all three manometers. The greater pressure P in the foot is due the gravitational force ( ρhg ) produced by the column of blood ( of height h ) between the heart and foot added to the pressure at the head is due to the elevation of the head over the heart. Since mercury is about 13 times as dense as blood ( ρ =13.6 gm/cm ³, ρ = 1.04 gm / cm ³ ) a column of mercury would be only one – thirteen as height as a given column of blood.

31. That is, if your blood pressure is 120 /80 mm Hg (120 mm Hg systolic and 80 mm Hg diastolic ) it would be 1560/ 1040 measured in millimeters of blood. If the average pressure at your heart is 100 mm Hg rise to an average height of 1300 mm or 1.3 m above your heart.

32. If gravity on earth suddenly became three times greater, blood rise only about 43 cm above the heart and it would not reach the brain of a standing person. This situation can be produced artificially by accelerating the body at 3 g in a vertical direction (figure 6b ).

33. It can also occur in an airplane pulling out of a dive, causing the pilot to blackout. These conditions also producing pooling of blood in the legs. Special tight fitting suits that compress the legs have been designed to reduce this pooling.

34. Indirect method of measuring Blood pressure The most common measurement of blood pressure is the indirect measurement of arterial pressure by means of the sphygmomanometer. This method actually measures the effect of an externally applied pressure upon the circulation.

35. Figure 7: Blood pressure measurement using a sphygmomanometer.

36. Sphygmomanometer Sphygmomanometer consists of a pressure cuff and gauge wrapped around the patients left arm at the level of the heart, and a stethoscope placed over the brachial artery at the elbow. The pressure cuff is inflated rapidly to a pressure sufficient to stop the flow of blood and the air is gradually released. As the pressure in the cuff drops below the systolic blood pressure the turbulent flow of blood squirting through the artery causes sound vibrations that can be heard in the stethoscope.They are called Korotkoff or K sounds. This on set of K sounds indicates the systolic pressure level.As the pressure falls further, K sounds become louder and then begin to fade.

37. The point at which the k sounds die out or change indicates the diastolic pressure. The gauge pressure is read on a mercury manometer calibrated in mm Hg.

38. Bernoulli’s Principle applied to the cardiovascular system Bernoulli’s principle is based on the law of conversion of energy. Pressure in fluid is a form of potential energy. Since it has the ability to perform work. In a moving fluid there is kinetic energy (K.E ) due to the motion. The normal volume flow rate of the blood = 5 liter /minute

39. The volume rate of flow = volume Time Q = Length x area Time Q = velocity x area

40. average velocity of flow ( v ) = volume rate of flow cross sectional area Average velocity of flow v = Q A

41. In a closed system, the average velocity of flow is inversely proportional to the area of the pipes through which it flows. The total fluid flowing through the system remains constant.

42. The volume of the fluid passing through the tube of cross sectional area A 1 is equal to the volume of fluid passing through A 2 in figure ( 1 ). Q 1 = Q 2 v 1 A 1 =v 2 A 2

43. P1 Fluid v1 A1 P2 v2 A2 P1 v1 A1 Figure 1 : Fluid passing through tube with a constricted region. Q1 Q2 Q1Q1 Q1Q1

44. The velocity of the fluid increases in narrow section of the tube, and the pressure decreases. in a fluid is a form of potential energy = ρ g h In a moving fluid there is kinetic energy=½ m v² Kinetic energy per unit volume = ½ mv² V k. E per volume = ½ ρ v² Kinetic energy density = ½ ρ v²

45. Example : Calculate the average kinetic energy per unit volume of 1 gm (1 cm³ ) of blood as it leave the heart,with average velocity Is about 30 cm /sec. Kinetic energy = ½ m v² = ½ x 1 x 30² = 450 ergs Kinetic energy density = 450 ergs / cm³

46. Bernoulli’s effect Bernoulli’s investigated that the pressure in a flowing fluid is lowest where it velocity is greatest. The fluid velocities greatest at the constriction because the liquid must flow faster through the restricted area to transport the same volume of fluid in a given time as shown in figure 2.

47. fluid Bernoulli’ s effect Figure 2

48. Venturi tube Venturi tube is a device used to measure the velocity of flowing liquid. It consists of a conical section in a pipe. The pressure difference between the main tube and the constriction tube is measured by a manometer (figure 3 ).

49. Fluid P1 vvv1vvvv1v v1 A1 P1 v1 A1 P2 v2 A2 Figure 3 : Venturi tube h manometer Mercury ( Hg ) Density ρ´

50. According to Bernulli’s equation P 1 + ½ ρ v² 1 = P 2 + ½ ρ v² 2 where ρ is the density of the fluid P 1 – P 2 = ½ ρ ( v² 2 –v² 1 ) Since v 1 A 1 = v 2 A 2 P 1 – P 2 = ½ ρ [ v² 1 ( A 1 /A 2 )² – v² 1 ] V 2 = v 1 (A 1 /A 2 ) P 1 – P 2 = ½ ρ v² 1 [ ( A 1 /A 2 )² - 1 ]

51. V² 1 = 2 ( P 1 – P 2 ) ρ [ ( A 1 / A 2 )² – 1 ] P 1 –P 2 = ρ´ g h Velocity of flow is V² 1 = 2 ρ´ g h ρ [(A 1 /A 2 )² – 1 ]

52. Poiseuille’s Law for the Flow of Fluids Poiseuille’s law states that the flow through a given tube depends on the pressure difference from one end to the other (Pa - Pb ), the length L of the tube, R is the radius of the tube, and η is the viscosity of the fluid. If the pressure difference is doubled, the flow rate also doubles, figure (4).

53. Poiseuille’s findings Figure 4 :. The flow rate through a tube depends on the pressure difference from one end of the tube to the other, the length of the tube, the viscosity of the fluid, and the radius.

54. Poiseuille’s equation Flow rate Q = (P a – P b ) π R ⁴ 8 η L The unit of flow rate is ( m³ / sec ) where P a –P b is the pressure difference between two points a and b (Newton / m² ) η is the viscosity of the fluid ( Pas ) R is the radius of the tube ( meter ) L is the length of the tube ( meter ) Flow rate is Q = (P a – P b ) π R ⁴ 8 η L

55. If ( P a –P b ) = ρ g h Then the volume rate of flow is : Q = π (ρ g h ) R ⁴ 8 η L

56. Newtonian and Non –Newtonian Liquid A liquid that obeys Poiseuille’s law called a Newtonian liquid. Fluid for which the viscosity is independent of pressure are called Newtonian fluids, and the flow the flow of Newtonian fluids such as water. Blood is a non –Newtonian liquid. Figure (5) shows a plot of the rates of flow of water and blood against the pressure gradient. The characteristic straight line plot for water becomes a distinct curve in the case of blood..

57. .The explanation for this anomalous behavior lies in the shape of the molecules in suspension shows departures from poiseuilles law, that is in small vessels the large red blood cells tend to accumulate in the faster axil part of the flow, so that there are fewer cells close to the walls to contribute to wall friction

58. Figure (5) : The rates of flow of water and blood as functions of the pressure gradient Rate of flow Water ( Newtonian liquid ) Pressure gradient Blood ( Non –Newtonian liquid )

59. Starling s Law of capillarity Starling s Law of capillarity describes the flow of fluids into out of the capillaries. Fluid movement through the capillary wall is the result of two pressures. 1.The hydrostatics pressure P across the capillary wall forcing fluids out of the capillary. 2.The osmotic pressure bringing fluids in.

60. The capillary pressure varies from about 30 mm Hg where the blood flows in at the arterial end to about 15 mm Hg where the blood leaves the capillary at the venous end. The osmotic pressure is estimated to be about 20 mm Hg into the capillary.

61. Q.12 Q. 12 An artery with a 3 mm radius is partially blocked with plaque ; in the constricted region the effective radius is 2mm and the average blood velocity is 50 cm /sec. (a) What is the average velocity of the blood in the un constricted region. (b ) Would there be turbulent flow in either region ? (c) For the blood in the constricted region, find the equivalent pressure due to the kinetic energy of the blood.

62. Answer of q.12 (a) R 1 = 3 mm R 2 =2mm A = π R² A 1 /A 2 =[ (3mm )² / (2 mm )² ] A 1 V 1 = A 2 V 2 V 1 = (4/9) 50 cm/sec =22 cm /sec (b) No

63. Answer of q.12 ( c ) ½ ρ v² = ½ (1.04) (50) ² = 1300 ergs / cm³ = 1 mm Hg

64. Q.15 Q.15 If the radius of an arteriole changed from 50 to 40 μm, how much would the flow rate through it decrease ?

65. Answer of Q.15 Flow rate Q = (P a – P b ) π R ⁴ 8 η L New Flow rate Q= (Pa – Pb )π(40μm) ⁴ 8 η L Old flow rate Q = (Pa – Pb ) π(50) ⁴ 8 η L