Cardiac Physiology for Lab
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Transcript of Cardiac Physiology for Lab
Cardiac Physiology for Lab
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Cardiac Output (CO)Blood pressureVessel resistance
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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)
120-70 = 50ml4
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
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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?• 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.
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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
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
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Factors Affecting CO
• More blood viscosity (causes decreases CO)• Total vessel length (longer decreases CO)• Vessel diameter (larger increases CO)
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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 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.
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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.
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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
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(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
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
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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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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