27
Cardiac Physiology for Lab 1

Cardiac Physiology for Lab

  • Upload
    barny

  • View
    49

  • Download
    0

Embed Size (px)

DESCRIPTION

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 . - PowerPoint PPT Presentation

Citation preview

Page 1: Cardiac Physiology for Lab

Cardiac Physiology for Lab

1

Page 2: Cardiac Physiology for Lab

Cardiac Output (CO)Blood pressureVessel resistance

2

Page 3: Cardiac Physiology for Lab

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

Page 4: Cardiac Physiology for Lab

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)

120-70 = 50ml4

Page 5: Cardiac Physiology for Lab

Cardiac Output (CO)

• Amount ejected by a ventricle in Amount ejected by a ventricle in 1 minute1 minute• CO = HR x SVCO = HR x SV• Resting values, 4- 6 L/minResting values, 4- 6 L/min• Vigorous exercise, 21 L/min Vigorous exercise, 21 L/min • Cardiac reserve: difference Cardiac reserve: difference

between maximum and resting CObetween 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/minCR = 15 L/min

5

Page 6: Cardiac Physiology for Lab

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

Page 7: Cardiac Physiology for Lab

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.

7

Page 8: Cardiac Physiology for Lab

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%8

Page 9: Cardiac Physiology for Lab

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

Page 10: Cardiac Physiology for Lab

Factors Affecting CO

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

10

Page 11: Cardiac Physiology for Lab

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

Page 12: Cardiac Physiology for Lab

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 Hg120 + 80 = 200200/2 = 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

Page 13: Cardiac Physiology for Lab

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 normally100 – 0 = 100

Remember, Cardiac Output (CO) is normally about 5 liters per minute.

13

Page 14: Cardiac Physiology for Lab

Clinical Significance

• Therefore, applying Ohm’s Law (Q=P/R) to a normal person, we get this:

5 = 100/RSolving for R:R = 100/5R = 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

Page 15: Cardiac Physiology for Lab

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

Page 16: Cardiac Physiology for Lab

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 = 120/20Q = 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

Page 17: Cardiac Physiology for Lab

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 = 50/20Q = 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

Page 18: Cardiac Physiology for Lab

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

Page 19: Cardiac Physiology for Lab

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

Page 20: Cardiac Physiology for Lab

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

Page 21: Cardiac Physiology for Lab

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

Page 22: Cardiac Physiology for Lab

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

Page 23: Cardiac Physiology for Lab

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

Page 24: Cardiac Physiology for Lab

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

Page 25: Cardiac Physiology for Lab

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

Page 26: Cardiac Physiology for Lab

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

Page 27: Cardiac Physiology for Lab

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