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Cardiac Physiology for Lab

Cardiac Physiology for Lab. Cardiac Output (CO) Blood pressure Vessel resistance. 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 .

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Cardiac Physiology for Lab

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  1. Cardiac Physiology for Lab

  2. Cardiac Output (CO) • Blood pressure • Vessel resistance

  3. Blood Flow (L/min) • Blood flowis 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 • CO= HR X SV • 70 b/min x 70 ml/beat =4900ml/min

  4. Ventricular Ejection Volume = Stroke Volume • Stroke Volume (SV) • amount ejected, ~ 70 ml • End Diastolic Volume (EDV) • ~120 ml • SV/EDV= ejection fraction, • 70/120 = 58% • at rest ~ 60% • during vigorous exercise as high as 90% • diseased heart < 50% • End-systolic volume: amount left in heart (50ml)‏

  5. Cardiac Output (CO)‏ • Amount ejected by a ventricle in 1 minute • CO = HR x SV • Resting values, 4- 6 L/min • Vigorous exercise, 21 L/min • Cardiac 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

  6. 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.

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

  8. 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 120-50 = 70 ml • Ejection Fraction EF = SV/End diastolic volume. Normal is 70/120 = 58%

  9. 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

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

  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

  12. 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 120 + 80 = 200 200/2 = 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.

  13. 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.

  14. 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

  15. 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.

  16. 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 (140 + 100)/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.

  17. 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 (60 + 40)/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.

  18. 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.

  19. 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.

  20. 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.

  21. 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.

  22. 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.

  23. 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.

  24. 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.

  25. 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.

  26. 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.

  27. 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.

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