Cardiac Physiology for Lab 1

Cardiac Output (CO) Blood pressure Vessel resistance 2

3 Blood Flow (L/min) Blood flow is the quantity of blood that passes a given point in the circulation in a given period of time. Overall flow in the circulation of an adult is 5 liters/min which is the cardiac output. HR = heart rate SV = stroke volume (how much blood is ejected from the left ventricle) CO= HR X SV 70 b/min x 70 ml/beat =4900ml/min

Ventricular Ejection Volume = Stroke Volume Stroke Volume (SV) –amount ejected, ~ 70 ml End Diastolic Volume (EDV) ~120 ml (max amount the left ventricle can hold) SV/EDV= ejection fraction (what percentage of blood is ejected from the left ventricle) EF = SV/EDV 70/120 = 58% –at rest ~ 60% –during vigorous exercise as high as 90% –diseased heart < 50% End-systolic volume: amount left in heart (50ml)‏ = 50ml 4

Cardiac Output (CO)‏ Amount ejected by a ventricle inAmount ejected by a ventricle in 1 minute 1 minute CO = HR x SVCO = HR x SV Resting values, 4- 6 L/minResting values, 4- 6 L/min Vigorous exercise, 21 L/minVigorous exercise, 21 L/min Cardiac reserve: difference between maximum and resting COCardiac reserve: difference between maximum and resting CO If resting CO = 6 L/min and after exercise increases to 21 L/min, what is the cardiac reserve? CR = 21 – 6 L/min CR = 15 L/min 5

Volumes and Fraction End diastolic volume = 120 ml End systolic volume = 50 ml Ejection volume (stroke volume) = 70 ml Ejection fraction = 70ml/120ml = 58% (normally 60%)‏ If heart rate (HR) is 70 beats/minute, what is cardiac output? Cardiac output = HR * stroke volume = 70/min. * 70 ml = 4900ml/min. 6

Questions If EDV = 120 ml and ESV = 50 ml: What is the SV? = 70 ml What is the EF? 70/120 = 58% What is the CO if HR is 70 bpm? 70/bpm * 70 ml = 4900ml/min. 7

Formulas to Know Cardiac Output CO= HR X SV Cardiac Reserve –If resting CO = 6 L/min and after exercise increases to 21 L/min, what is the cardiac reserve? CR = 21 – 6 L/min Stroke Volume SV = (End diastolic volume) – (End systolic volume). Normal is = 70 ml Ejection Fraction EF = SV/End diastolic volume. Normal is 70/120 = 58% 8

Ohm’s Law Formulas Q= P/R P = QR R = P/Q Q is cardiac output P is average blood pressure of the aorta R is resistance in the blood vessels 9

Factors Affecting CO More blood viscosity (causes decreases CO) Total vessel length (longer decreases CO) Vessel diameter (larger increases CO) 10

11 Ohm’s Law Q=  P/R Flow (Q) through a blood vessel which is the same thing as saying Cardiac Output (CO) through the heart, is determined by: 1) The pressure difference (  P) between the two ends of the circulatory tube (arteries and veins) –Directly related to flow 2) Resistance (R) of the vessel –Inversely related to flow

Clinical Significance Normal blood pressure is 120/80 mm Hg. 120 represents systolic pressure, and 80 represents diastolic pressure. The average of these two pressures is 100 mm Hg = /2 = 100 (the average) Therefore, the average pressure in the first vessel leaving the heart (the aorta) is 100 mm Hg. “100 mm Hg” means the amount of pressure required to lift a column of mercury 100 mm in the air. This is how the original blood pressure cuffs work. 12

Clinical Significance In a normal person, the arterial pressure is 100 and the pressure in the veins is 0 (if there were any pressure in the capillaries, they would blow out, so blood pressure drops to zero by the time it gets there, and stays at zero in the veins. The pressure difference (  P) is normally 100 – 0 = 100 Remember, Cardiac Output (CO) is normally about 5 liters per minute. 13

Clinical Significance Therefore, applying Ohm’s Law (Q=  P/R) to a normal person, we get this: 5 = 100/R Solving for R: R = 100/5 R = 20 PRU That means that the normal amount of resistance in the blood vessels is 20 PRU (peripheral resistance units). Overall, the values for a normal person are: 5 = 100/20 14

Now let’s solve for  P (change in pressure) instead of Q (cardiac output)  P means subtracting the pressure in the veins (P2) from the average pressure in the arteries (P1). Therefore,  P = P1 – P2 Since P2 (blood pressure in the veins) is always 0, for our purposes, you could just write P instead of  P. P symbolizes blood pressure. Since BP is written systolic/diastolic, you add up both pressures and take the average. The average person’s blood pressure is 120/80, so the overall average pressure in the aorta is about 100 mm Hg. 15

Clinical Significance: Solve for Q (cardiac output) A person might have blood pressure higher than normal. –They ate too much salt, so they are retaining water A patient has blood pressure of 140/100, yet their last BP reading a few weeks ago was 120/80. When questioned, the patient said they ate a lot of salt and drank a lot of water yesterday. Problem: What is their cardiac output right now? We can assume the resistance in their blood vessels is normal since their BP was normal recently. Solution: First find the average arterial pressure ( )/2 = 120 Then apply Ohm’s Law (Q=  P/R) Q = 120/20 Q = 6 (Cardiac output increases) The heart is pumping with more force than normal. Since it takes more time to pump a larger bolus, the heart rate is slower. 16

Clinical Significance A person might have blood pressure lower than normal. –They are dehydrated A patient has blood pressure of 60/40, yet their last BP reading a few weeks ago was 120/80. When questioned, the patient said they just got back from a hike and they are thirsty. Problem: What is their cardiac output right now? We can assume the resistance in their blood vessels is normal since their BP was normal recently. Solution: First find the average arterial pressure ( )/2 = 50 Then apply Ohm’s Law (Q=  P/R) Q = 50/20 Q = 2.5 (Cardiac output decreases) The heart is pumping with less force than normal. Since it takes less time to pump a smaller bolus, the heart rate is faster. 17

Clinical Significance What is CO if you change the resistance? A patient might have higher vessel resistance if they have clogged arteries (atherosclerosis) or calcium deposits in the arteries (arteriosclerosis). Problem: What is the cardiac output in a patient with BP of 120/80 and a higher than normal resistance? Let’s say R = 30. Apply Ohm’s Law (Q=  P/R) Q = 100/30 Q = 3.3 (Cardiac output decreases) The heart is pumping with less force than normal. Since it takes less time to pump a smaller bolus, the heart rate is faster. 18

Clinical Significance Problem: What is the cardiac output in a patient with BP of 120/80 and a lower than normal resistance, perhaps they are athletes who have developed large arteries? Let’s say R = 10. Apply Ohm’s Law (Q=  P/R) Q = 100/10 Q = 10 (Cardiac output increases) The heart is pumping with more force than normal. Since it takes more time to pump a larger bolus, the heart rate is slower. Therefore, someone with a slow heart rate and large cardiac output might have a condition relating to low peripheral resistance, such as an aerobic athlete. 19

Solve for P What would you expect the blood pressure to be in a person who has increased peripheral resistance (clogged arteries)? Let’s say R = 50 P = QR P = (5)(50) P = 250 (normal would be 100) The person would have high blood pressure. 20

Solve for P What would you expect the blood pressure to be in a person who has decreased peripheral resistance (athlete)? Let’s say R = 10 P = QR P = (5)(10) P = 50 (normal would be 100) The person would have low blood pressure. 21

Solve for P What would you expect the blood pressure to be in a person who has increased cardiac output (over-hydration)? Let’s say Q = 6 P = QR P = (6)(20) P = 120 (normal would be 100) The person would have higher blood pressure. 22

Solve for P What would you expect the blood pressure to be in a person who has decreased cardiac output (dehydration)? Let’s say Q = 4 P = QR P = (4)(20) P = 80 (normal would be 100) The person would have lower blood pressure. 23

Now let’s solve for R What would you expect the peripheral resistance to be in a person who has decreased cardiac output (dehydration)? Let’s say Q = 4 R = P/Q R = 100/4 R = 25 (normal would be 20) The person would have higher than normal resistance. 24

Solve for R What would you expect the peripheral resistance to be in a person who has increased cardiac output (over-hydration)? Let’s say Q = 6 R = P/Q R = 100/6 R = 16 (normal would be 20) The person would have lower than normal resistance. 25

Solve for R What would you expect the peripheral resistance to be in a person who has decreased blood pressure (athlete)? Let’s say P = 80 R = P/Q R = 80/5 R = 16 (normal would be 20) The person would have low peripheral resistance. 26

Solve for R What would you expect the peripheral resistance to be in a person who has increased blood pressure (clogged arteries)? Let’s say P = 120 R = P/Q R = 120/5 R = 24 (normal would be 20) The person would have higher peripheral resistance. 27