problems in echocardiography 2

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HEMODYNAMIC PROBLEMS HEMODYNAMIC PROBLEMS IN ECHOCARDIOGRAPHY IN ECHOCARDIOGRAPHY Daniel W Mason MD FACC FASE Daniel W Mason MD FACC FASE

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Page 1: Problems in Echocardiography 2

HEMODYNAMIC HEMODYNAMIC PROBLEMS IN PROBLEMS IN

ECHOCARDIOGRAPHYECHOCARDIOGRAPHYDaniel W Mason MD FACC FASEDaniel W Mason MD FACC FASE

Page 2: Problems in Echocardiography 2

CW, PW, and Color FlowCW, PW, and Color Flow

1. This is all doppler, you are all familiar 1. This is all doppler, you are all familiar with this, with a frequency shift of reflected with this, with a frequency shift of reflected ultrasound waves proportional to the ultrasound waves proportional to the speed of the reflecting medium.speed of the reflecting medium.

ΛΛF = 2fF = 2f00v/c x cosv/c x cosΘΘ

2. Increasing or decreasing frequency is 2. Increasing or decreasing frequency is related to direction of flow. related to direction of flow.

Page 3: Problems in Echocardiography 2

CW vs PWCW vs PW

1. The magnitude of frequency shift we can 1. The magnitude of frequency shift we can measure depends on the frequency with measure depends on the frequency with which we can sample.which we can sample.

2. CW requires 2 transducers, one is 2. CW requires 2 transducers, one is continually sending, the other is continually sending, the other is continually receiving. Therefore, sampling continually receiving. Therefore, sampling frequency is rapid so high shifts can be frequency is rapid so high shifts can be measured. However, there is no ranging.measured. However, there is no ranging.

Page 4: Problems in Echocardiography 2

CW vs PWCW vs PW

2. PW sends out very brief pulses, and 2. PW sends out very brief pulses, and then just listens with the one transducer. then just listens with the one transducer. Since we know the speed of sound in Since we know the speed of sound in tissue, we can range. This allows us to tissue, we can range. This allows us to localize in space where the shift occurs, localize in space where the shift occurs, but dramatically reduces the frequency but dramatically reduces the frequency shifts we can measure before aliasing shifts we can measure before aliasing occurs. occurs.

Page 5: Problems in Echocardiography 2

So the doppler shift will look like a So the doppler shift will look like a sine wave, but we only get a brief look at sine wave, but we only get a brief look at each wave with each pulse (sampling)each wave with each pulse (sampling)

Page 6: Problems in Echocardiography 2

So as long as we sample So as long as we sample ≥ than 2 x ≥ than 2 x the doppler shift frequency, we can the doppler shift frequency, we can determine it’s frequency and direction. determine it’s frequency and direction. Otherwise, if sampling is less, then we get Otherwise, if sampling is less, then we get velocity and direction ambiguity. The velocity and direction ambiguity. The Nyquist Limit is the place where aliasing Nyquist Limit is the place where aliasing begins, and that is why it is = ½ PRF.begins, and that is why it is = ½ PRF.

Page 7: Problems in Echocardiography 2

Color FlowColor Flow

Color flow is an extension of PW into a 2 Color flow is an extension of PW into a 2 dimensional image just as 2D is to M dimensional image just as 2D is to M mode echo. Ranging allows us to see the mode echo. Ranging allows us to see the location, and frequency shift is location, and frequency shift is represented by arbitrary colors. I will not represented by arbitrary colors. I will not get into resolution issues today. get into resolution issues today.

Page 8: Problems in Echocardiography 2

CalculationsCalculations

Getting comfortable with the Getting comfortable with the formulas formulas

Doing our best to understand what a Doing our best to understand what a TVI or VTI really is (Time Velocity TVI or VTI really is (Time Velocity Integral)Integral)

Solving problemsSolving problems

Page 9: Problems in Echocardiography 2

FORMULASFORMULAS

ππR R 22 = area of circle = area of circle

ππR R 2 =2 = ππ(1/2D)(1/2D)22 = 1/4 = 1/4ππDD22 = 0.785D= 0.785D22

Area of circle = 0.785DArea of circle = 0.785D22

Page 10: Problems in Echocardiography 2

FORMULASFORMULAS

Area x Length = VolumeArea x Length = Volume

LL

Page 11: Problems in Echocardiography 2

FORMULASFORMULAS

If flow through a cylinder were If flow through a cylinder were constant, we can easily measure constant, we can easily measure volume if we know velocity, time, volume if we know velocity, time, and diameter.and diameter.

Page 12: Problems in Echocardiography 2

FORMULASFORMULAS

L = V x TL = V x TVV

TT

Page 13: Problems in Echocardiography 2

FORMULASFORMULAS

Then, if we know diameter, we can calculate Then, if we know diameter, we can calculate area of the cylinder. Multiply this by the length area of the cylinder. Multiply this by the length and we get a volumeand we get a volume

V = area x length.V = area x length.

In this case, L would be the length of the cylinder In this case, L would be the length of the cylinder described by flow over that time (T), and could described by flow over that time (T), and could be referred to as stroke distance, the distance be referred to as stroke distance, the distance the fluid traveled over that period of time.the fluid traveled over that period of time.

Page 14: Problems in Echocardiography 2

FORMULASFORMULAS

But flow is not constant, it’s velocity But flow is not constant, it’s velocity is constantly changing, so how can is constantly changing, so how can we measure volume. What would Sir we measure volume. What would Sir Isaac Newton have done? Calculus? Isaac Newton have done? Calculus? Integrate?Integrate?

Page 15: Problems in Echocardiography 2

VTIVTI

VV

aa

TT

Page 16: Problems in Echocardiography 2

VTIVTI

The intergral of the curve in essence is taking The intergral of the curve in essence is taking the average velocity during that time, and the average velocity during that time, and multiplying it by the time of the systolic period, multiplying it by the time of the systolic period, giving us stroke distance. giving us stroke distance.

If this were the LVOT, and the aorta had the If this were the LVOT, and the aorta had the same diameter, it would tell us how far the blood same diameter, it would tell us how far the blood traveled, stroke distance. Since the aorta is traveled, stroke distance. Since the aorta is larger, the distance will be smaller. However, larger, the distance will be smaller. However, multiplying the VTI x area will still give stroke multiplying the VTI x area will still give stroke volume. volume.

Page 17: Problems in Echocardiography 2

TVI or time velocity integralTVI or time velocity integralArea x length = volumeArea x length = volume

Area x TVI = volumeArea x TVI = volume

TVI = lengthTVI = length

There is nothing in the physical There is nothing in the physical world that is measured by the world that is measured by the distance given by VTI, but the distance given by VTI, but the mathematical model gives a mathematical model gives a number that is proportional to the number that is proportional to the stroke volume.stroke volume.

Page 18: Problems in Echocardiography 2

MORE FORMULASMORE FORMULAS

Flow = Velocity x Area Flow = Velocity x Area (cm(cm33/sec) = (cm/sec x cm/sec) = (cm/sec x cm22) )

Therefore Area = Flow/VelocityTherefore Area = Flow/Velocity

Don’t confuse flow used in PISA continuity Don’t confuse flow used in PISA continuity with stroke volumes used with AVA by with stroke volumes used with AVA by continuity.continuity.

P = 4VP = 4V2 2 Simplified Bernoulli Equation Simplified Bernoulli Equation

Page 19: Problems in Echocardiography 2
Page 20: Problems in Echocardiography 2

MORE FORMULASMORE FORMULAS

AVA x TVIAVA x TVIavav = Area = Arealvotlvot x TVI x TVIlvotlvot

AVA = (AreaAVA = (Arealvotlvot x TVI x TVIlvotlvot)/ TVI)/ TVIavav

Using Maximal velocities of the aortic Using Maximal velocities of the aortic valve and lvot gives a reasonable valve and lvot gives a reasonable approximation of the area.approximation of the area.

Ratio’s of TVI’s or maximal velocities are Ratio’s of TVI’s or maximal velocities are most helpful when it is difficult to measure most helpful when it is difficult to measure a diameter (calcium or prosthesis)a diameter (calcium or prosthesis)

Page 21: Problems in Echocardiography 2

PRESSURE HALF TIMESPRESSURE HALF TIMES

By definition, this is the time it takes By definition, this is the time it takes for the for the pressure to decrease by ½. to decrease by ½.

But pressure and But pressure and velocity are not are not linear, so it is not the time it takes linear, so it is not the time it takes the velocity to decrease by ½.the velocity to decrease by ½.

P = 4VP = 4V22

Page 22: Problems in Echocardiography 2

PRESSURE HALF TIMESPRESSURE HALF TIMES

PP

VV

Page 23: Problems in Echocardiography 2

PRESSURE HALF TIMESPRESSURE HALF TIMES

PP1 1 = 4V= 4V1122

1/2P1/2P11 = 4V = 4VXX22

½(4V½(4V1122) = 4V) = 4Vxx

22

1/2V1/2V112 = 2 = VVxx

22

VVx x = V = V11/sqrt 2 =V/sqrt 2 =V11/1.4/1.4

The time it takes to go from VThe time it takes to go from V11 to V to Vxx is the pressure half timeis the pressure half time

Page 24: Problems in Echocardiography 2

PRESSURE HALF TIMESPRESSURE HALF TIMES

VVmaxmax

V Vt1/2t1/2 = V = Vmaxmax/1.4/1.4

PHT PHT

Page 25: Problems in Echocardiography 2

PRESSURE HALF TIMESPRESSURE HALF TIMES

DT is deceleration time, from peak to DT is deceleration time, from peak to zero.zero.PHT = .29 x DTPHT = .29 x DTMVA = 220/PHTMVA = 220/PHTThis was empirically derivedThis was empirically derivedAnything above 220 suggest severe Anything above 220 suggest severe MS.MS.Less than 200 for AI suggest it may Less than 200 for AI suggest it may be severebe severe

Page 26: Problems in Echocardiography 2

Pitfalls of Pressure ½ TimePitfalls of Pressure ½ Time

Diastolic dysfunction (severe) with Diastolic dysfunction (severe) with high filling pressures or acute severe high filling pressures or acute severe aortic insufficiency will shorten the ½ time, aortic insufficiency will shorten the ½ time, making the MVA to high.making the MVA to high.

An ASD drops LA pressures quickly, An ASD drops LA pressures quickly, giving the same result. giving the same result.

Page 27: Problems in Echocardiography 2
Page 28: Problems in Echocardiography 2
Page 29: Problems in Echocardiography 2

PISAPISA

Proximal Isovelocity Surface AreaProximal Isovelocity Surface Area

rr

Page 30: Problems in Echocardiography 2

PISAPISA

Area of a sphere= 4 Area of a sphere= 4 ππrr22

Area of a hemisphere= 2 Area of a hemisphere= 2 ππrr22

Flow = VelocityFlow = Velocityaa x Area x Area

Flow = VelocityFlow = Velocityaa x 6.28r x 6.28r2 2

ERO (area) = Flow/VelocityERO (area) = Flow/Velocitypp

Regurg. Volume = ERO x TVIRegurg. Volume = ERO x TVImrmr

Page 31: Problems in Echocardiography 2

PISAPISA

MVA = (6.28rMVA = (6.28r22 x V x Vaa)/MS)/MSVp Vp x x αα°/180°°/180°

Page 32: Problems in Echocardiography 2
Page 33: Problems in Echocardiography 2

QQpp/Q/Qss

.785D.785D22rvotrvot x TVI x TVIrvot rvot = SV= SVrv rv = Q= Qpp

.785D.785D22lvotlvot x TVI x TVIlvot lvot = SV= SVlv lv = Q= Qss

QQpp/Q/Qss = Pul. Flow/Sys. Flow = Pul. Flow/Sys. Flow

Page 34: Problems in Echocardiography 2

Pulmonary Vascular ResistancePulmonary Vascular Resistance

R = Pressure / COR = Pressure / CO

Peak TR Velocity ~ Peak TR Velocity ~ RVSP RVSP ~ Pressure~ Pressure

TVITVIrvot rvot ~ CO therefore~ CO therefore

R ~ Peak TR Velocity / TVIR ~ Peak TR Velocity / TVIrvot rvot

Page 35: Problems in Echocardiography 2

If If TVITVIrvot rvot is in cmis in cm

and TRand TRpeak V peak V is in M/secis in M/sec

RR(Woods)(Woods) = TR = TRpeak V peak V / TVI/ TVIrvot rvot x 10 +.16x 10 +.16

Down and dirty a TRDown and dirty a TRpeak V peak V / TVI/ TVIrvot rvot ratio ofratio of

< 0.2 suggest low R< 0.2 suggest low R

Page 36: Problems in Echocardiography 2

Problem 1Problem 1

58yom with previous MI and mitral 58yom with previous MI and mitral regurgitationregurgitationDDlvotlvot = 2.0cm TVI = 2.0cm TVIlvotlvot = 0.19m = 0.19mDDmama = 3.4cm TVI = 3.4cm TVImama = 0.176m = 0.176m

What is the regurgitant volume?What is the regurgitant volume?

What is the regurgitant fraction?What is the regurgitant fraction?

Page 37: Problems in Echocardiography 2

Answer 1Answer 1

SVSVlvot lvot = .785 x 2= .785 x 222 x19 = 59.96 x19 = 59.96

SVSVmama = .785 x 3.4 = .785 x 3.422 x 17.6 = 159.7 x 17.6 = 159.7

Regurg. Volume = 160 – 60 = 100Regurg. Volume = 160 – 60 = 100

Regurg. Fraction = 100/160 = .625 = Regurg. Fraction = 100/160 = .625 = 63%63%

Page 38: Problems in Echocardiography 2

Problem 2Problem 2

On TEE, the following was foundOn TEE, the following was foundpisa r = 0.9cm pisa r = 0.9cm aliasing velocity = 40cm/secaliasing velocity = 40cm/secMV Regurg. Velocity = 4.5m/secMV Regurg. Velocity = 4.5m/sec

TVITVImvr mvr = 227cm= 227cm What is the ERO and Regurg. What is the ERO and Regurg.

Fraction? Fraction?

Page 39: Problems in Echocardiography 2

Answer 2Answer 2

ERO = (6.28 x (.9)ERO = (6.28 x (.9)22 x 40)/ 450 x 40)/ 450 = .45cm= .45cm2 2 = 45mm = 45mm22

Regurg. Volume = .45 x 227 = 102ccRegurg. Volume = .45 x 227 = 102cc

Page 40: Problems in Echocardiography 2

Problem 3Problem 3

72 yof with a murmur and syncope72 yof with a murmur and syncopeHeart Rate 60Heart Rate 60TVITVIlvotlvot = .21M/sec LVOT = .21M/sec LVOTdd =2.2cm =2.2cmTVITVIavav = 1.15M/sec = 1.15M/sec

What is the AV area?What is the AV area?

What is the cardiac output?What is the cardiac output?

Page 41: Problems in Echocardiography 2

Answer 3Answer 3

AVA = AreaAVA = Arealvot lvot x TVIx TVIlvotlvot/TVI/TVIavav

= .785 x (2.2) = .785 x (2.2)22 x 21/115 = .69 x 21/115 = .69

CO = .785 x (2.2)CO = .785 x (2.2)22 x 21 x 60 = x 21 x 60 = 4.79L/min4.79L/min

Page 42: Problems in Echocardiography 2

Problem 4Problem 4

65yom with AI65yom with AILVOTLVOTdd = 2.6cm TVI = 2.6cm TVIlvotlvot =16cm =16cmMVMVannulusannulus = 3.5cm TVI = 3.5cm TVImvmv = 6.0cm = 6.0cm AR peak velocity = 5.14M/secAR peak velocity = 5.14M/secTVITVIavravr = 2.77M = 2.77MWhat is the regurgitant volume?What is the regurgitant volume?What is the regurgitant fraction?What is the regurgitant fraction?What is the effective regurg. orifice What is the effective regurg. orifice area?area?

Page 43: Problems in Echocardiography 2

Answer 4Answer 4

LVOTLVOTsvsv = .785 x (2.6) = .785 x (2.6)22 x 16 = 85cc x 16 = 85cc

MVMVsvsv = .785 x (3.5) = .785 x (3.5)22 x 6 = 58cc x 6 = 58cc

Regurgitant Volume = 85-58 = 27ccRegurgitant Volume = 85-58 = 27cc

Regurgitant Fraction = 27/85 = 32%Regurgitant Fraction = 27/85 = 32%

ERO = 27/277 = 0.1 cmERO = 27/277 = 0.1 cm22

Page 44: Problems in Echocardiography 2

PROBLEM 5PROBLEM 5

What is the PA end diastolic What is the PA end diastolic pressure in a patient whose PR pressure in a patient whose PR endiastolic velocity is 1.5m/sec?endiastolic velocity is 1.5m/sec?

What is the LA pressure when the What is the LA pressure when the velocity over a PFO is 1.5m/sec?velocity over a PFO is 1.5m/sec?

What view would we get that velocity What view would we get that velocity with a TTE?with a TTE?

Page 45: Problems in Echocardiography 2

Answer 5Answer 5

(1.5 x 1.5 x 4) + 10 = 19mmHg(1.5 x 1.5 x 4) + 10 = 19mmHg (1.5 x 1.5 x 4) + 10 = 19mmHg(1.5 x 1.5 x 4) + 10 = 19mmHg subcostalsubcostal

Page 46: Problems in Echocardiography 2

PROLBLEM 6PROLBLEM 6

If the CW tracing of an aortic insufficiency If the CW tracing of an aortic insufficiency jet shows a maximum velocity of jet shows a maximum velocity of 3.0m/sec, what is the velocity at the 3.0m/sec, what is the velocity at the pressure half time?pressure half time?

If that velocity occurs 240 msec latter, If that velocity occurs 240 msec latter, what is the DT?what is the DT?

If the end-diastolic velocity of this jet is If the end-diastolic velocity of this jet is 2.7m/sec and the BP is 136/60, what is the 2.7m/sec and the BP is 136/60, what is the LVEDP? LVEDP?

Page 47: Problems in Echocardiography 2

ANSWER 6ANSWER 6

3/1.4 = 2.14m/sec3/1.4 = 2.14m/sec

240/.29 = 828msec240/.29 = 828msec

60 – (4 x (2.7)60 – (4 x (2.7)22) = 31mmHg) = 31mmHg

Page 48: Problems in Echocardiography 2

PROBLEM 7PROBLEM 7

An ASD measured 1 cm in diameter on TTE. An ASD measured 1 cm in diameter on TTE.

HR 70bpm. TVI of flow / beat was 0.6M.HR 70bpm. TVI of flow / beat was 0.6M.

LVOTLVOTTVI TVI was 0.25 M, LVOT was 0.25 M, LVOTDD was 2 cm. was 2 cm.

The shunt was L to R. The shunt was L to R.

What is the flow / minute through the ASD?What is the flow / minute through the ASD?

What is the Qp/Qs.What is the Qp/Qs.

Page 49: Problems in Echocardiography 2

Answer 7Answer 7

QQshunt shunt = .785 x 1= .785 x 122 x 60 x70 / 1000 = 3.3L/min x 60 x70 / 1000 = 3.3L/min

QQS S = .785 x 2= .785 x 222 x 25 x 70 / 1000 = 5.5L/min x 25 x 70 / 1000 = 5.5L/min

QQPP = Q = QS S + Q + Qshunt shunt = 5.5 + 3.3 = 8.8L/min= 5.5 + 3.3 = 8.8L/min

QQPP/ Q/ QSS = 8.8 / 5.5 = 1.6 = 8.8 / 5.5 = 1.6

Page 50: Problems in Echocardiography 2

The EndThe End