Hemodynamic Calculations
Martin Ingi Sigurdsson
Nathan H. Waldron
1. An echocardiographic assessment of left ventricular (LV) output revealed HR 100/min, BP 130/76 mm Hg, and LV outflow tract velocity time integral (LVOT VTI) 10 cm. Following this a bolus of 500 cc of crystalloid was given. Reassessment shows HR 95/min, BP 128/78 mm Hg, and LVOT VTI 12.5 cm. What statement best describes the volume responsiveness of the patient?
A. The patient is not volume responsive given minimum change in blood pressure.
B. The patient is considered to be volume responsive based on the change in VTI.
C. The patient is not volume responsive based on the change in VTI.
D. Measurement of the LVOT radius is not needed to calculate cardiac output (CO).
View Answer
1. Correct Answer: B. The patient is considered to be volume responsive based on the change in VTI.
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate. Since the diameter of the LVOT presumably does not change with a fluid bolus, changes in CO can be assessed by change in HR and the VTI alone. Here, the total change was 
=1.19, indicating that the stroke volume increased by 19% with the fluid bolus. This patient would therefore be considered fluid responsive (more than 15% increase in CO).
=1.19, indicating that the stroke volume increased by 19% with the fluid bolus. This patient would therefore be considered fluid responsive (more than 15% increase in CO).Selected References
1. Lamia B, Ochagavia A, Monnet X, Chemla D, Richard C, Teboul JL. Echocardiographic prediction of volume responsiveness in critically ill patients with spontaneously breathing activity. Intensive Care Med. 2007 Jul;33(7):1125-1132. doi: 10.1007/s00134-007-0646-7. Epub 2007 May 17. PMID: 17508199.
2. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
2. A patient in mixed cardiogenic and septic shock is assessed by echocardiography. HR is 120/min, mean arterial pressure (MAP) is 70 mm Hg, central venous pressure (CVP) is 10 mm Hg, LVOT diameter is 2.0 cm, and LVOT VTI is 20 cm. What is the patient’s systemic vascular resistance (SVR)?
A. 637 dynes × sec × cm-5
B. 742 dynes × sec × cm-5
C. 159 dynes × sec × cm-5
D. 1152 dynes × sec × cm-5
View Answer
2. Correct Answer: A. 637 dynes × sec × cm-5
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the time velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate. Using the given values, CO=20 cm ×(1 cm)2 × π × 120 beats/min= 7536 cm3/min= 7.54 L/min
Using Ohm’s equation MAP-CVP = CO × SVR and a conversion factor of 80 to convert from peripheral resistance units to the more traditional centimeter-gram-second units (dynes × sec × cm-5), we can therefore calculate 
dynes × sec × cm-5. This patient’s SVR is therefore slightly reduced, suggesting that a vasopressor rather than an inotrope would be helpful for management of his shock.
dynes × sec × cm-5. This patient’s SVR is therefore slightly reduced, suggesting that a vasopressor rather than an inotrope would be helpful for management of his shock.Selected Reference
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
3. An adult male patient in mixed cardiogenic and septic shock is assessed by echocardiography. HR is 80 mm Hg, MAP is 50 mm Hg, CVP is 15 mm Hg, LVOT diameter is 2.0 cm, and LVOT VTI is 12 cm. There is no increase in LVOT VTI with a passive leg raise, no valvulopathies are identified, and there are no signs of pericardial tamponade. What is the appropriate next step in his treatment?
A. Give a 1 L fluid bolus
B. Initiate esmolol infusion
C. Initiate norepinephrine infusion
D. Initiate dobutamine infusion
View Answer
3. Correct Answer: D. Initiate dobutamine infusion
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate. Using the given values, CO = 12 cm × (1cm)2 × π × 80 beats/min = 3014 cm3/min = 3.0 L/min. The average body surface area of an adult male is 1.9 m2, giving a cardiac index of 1.57 L/min/m2. Using Ohm’s equation MAP – CVP = CO × SVR and a conversion factor of 80 to convert from peripheral resistance units to the more traditional centimeter-gram-second units (dynes × sec × cm-5), we can calculate 
dynes × sec × cm-5. This patient’s SVR is therefore normal but the CO is severely reduced, suggesting that an inotrope such as dobutamine would be an appropriate next step in treatment.
dynes × sec × cm-5. This patient’s SVR is therefore normal but the CO is severely reduced, suggesting that an inotrope such as dobutamine would be an appropriate next step in treatment.Selected Reference
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
4. Following the original treatment chosen for the patient in Question 3, the patient remains hypotensive. HR is 100, MAP is 55 mm Hg, CVP is 15 mm Hg, LVOT diameter is 2.0 cm, LVOT VTI is 20 cm, and other echocardiographic findings are unchanged. What is the logical next step?
A. Give 1 L fluid bolus
B. Initiate esmolol infusion
C. Initiate norepinephrine infusion
D. Increase dobutamine infusion
View Answer
4. Correct Answer: C. Initiate norepinephrine infusion
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate. Using the given values, CO = 20 cm × (1 cm)2 × π × 100 beats/min = 6280 cm3/min = 6.3 L/min. The average body surface area of an adult male is 1.9 m2, giving a cardiac index of 3.31 L/m2. Using Ohm’s equation MAP – CVP = CO × SVR and a conversion factor of 80 to convert from peripheral resistance units to the more traditional centimeter-gram-second units (dynes × sec × cm-5), we can calculate 
dynes × sec × cm-5. This patient’s SVR is therefore low but the patient’s CO is in the normal range, suggesting that the patient is vasodilated. A vasopressor such as norepinephrine would be an appropriate next step in treatment.
dynes × sec × cm-5. This patient’s SVR is therefore low but the patient’s CO is in the normal range, suggesting that the patient is vasodilated. A vasopressor such as norepinephrine would be an appropriate next step in treatment.Selected References
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
2. Porter TR, Shillcutt SK, Adams MS, et al. Guidelines for the use of echocardiography as a monitor for therapeutic intervention in adults: a report from the American Society of Echocardiography. J Am Soc Echocardiogr. 2015;28:40-56.
5. A hypotensive patient is evaluated for the etiology of shock. HR is 100/min, MAP is 55 mm Hg, LVOT diameter is 1.8 cm, LVOT VTI is 26 cm, and inferior vena cava (IVC) diameter is 2.4 cm without respiratory variation. What is the most likely etiology of this patient’s shock?
A. Hypovolemic shock
B. Obstructive shock
C. Distributive shock
D. Cardiogenic shock
View Answer
5. Correct Answer: C. Distributive shock
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate. Using the given values, CO=26 cm × (0.9 cm)2 × π × 100 beats/min = 6612 cm3/min = 6.6 L/min. Given an IVC size of 2.4 cm without a respiratory variation, CVP can be estimated as 15 (10-20) mm Hg. Using Ohm’s equation MAP – CVP = CO × SVR and a conversion factor of 80 to convert from peripheral resistance units to the more traditional centimeter-gram-second units (dynes × sec × cm-5), we can calculate 
dynes × sec × cm-5. This patient’s right-sided cardiac filling pressure is therefore normal/high, CO is normal/high but SVR is low, suggestive of distributive shock.
dynes × sec × cm-5. This patient’s right-sided cardiac filling pressure is therefore normal/high, CO is normal/high but SVR is low, suggestive of distributive shock.Selected References
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
2. Porter TR, Shillcutt SK, Adams MS, et al. Guidelines for the use of echocardiography as a monitor for therapeutic intervention in adults: a report from the American Society of Echocardiography. J Am Soc Echocardiogr. 2015;28:40-56.
6. A patient has the following quantification of their intracardiac shunt: LVOT diameter 2.0 cm, LVOT VTI 22 cm, MAP 70 mm Hg, right ventricular outflow tract (RVOT) diameter 2.4 cm, RVOT VTI 24 cm. Which shunt physiology, among the following, could explain this quantification?
A. Atrial septal defect with left → right shunt
B. It cannot be determined without HR
C. Patent foramen ovale with right → left shunt
D. Ventricular septal defect with right → left shunt
View Answer
6. Correct Answer: A. Atrial septal defect with left → right shunt
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
This can be done on both the right and left sides of the heart, and if done with the same HR measurements, VTI and radius of LVOT and RVOT can be used to assess intracardiac shunts. Here, both the diameter and VTI of the RVOT are larger than the LVOT, indicating that blood flow through the RVOT is more than the blood flow through the LVOT. This indicates that blood is flowing from the left side of the heart to the right side of the heart prior to the LVOT, making atrial septal defect with left-to-right shunt the most likely answer.
Selected References
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
2. Porter TR, Shillcutt SK, Adams MS, et al. Guidelines for the use of echocardiography as a monitor for therapeutic intervention in adults: a report from the American Society of Echocardiography. J Am Soc Echocardiogr. 2015;28:40-56.
7. Figure 19.1 shows a continuous VTI assessment over the LVOT in a mechanically ventilated patient in sinus rhythm.
 Figure 19.1 |
2D assessment ruled out tamponade and RV failure. What hemodynamic assessment can be made based on the tracing?
A. The patient is likely fluid responsive.
B. Fluid responsiveness cannot be predicted based on the image.
C. The patient has low LV ejection fraction.
D. The patient has a high afterload.
View Answer
7. Correct Answer: A. The patient is likely fluid responsive.
Rationale: Figure 19.1 shows a continuous VTI assessment of the LVOT during a full respiratory cycle. This reveals that Vmax, the maximum velocity obtained during each contraction, changes by at least 20% during the respiratory cycle (where Vmax increases during inspiration). This indicates that the patient has a substantial stroke volume variation and is likely fluid responsive.
Selected Reference
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
8. Figure 19.2 shows a continuous Doppler pulse wave assessment of the VTI in the LVOT during a respiratory cycle in a mechanically ventilated patient with a prior history of chronic heart failure admitted currently in septic shock.
 Figure 19.2 |
The patient is in a sinus rhythm. 2D assessment ruled out tamponade and RV failure. Following a fluid bolus of 1000 cc, the VTILVOT is reassessed and changes by 5% during the respiratory cycle. The patient is still hypotensive and the average LVOTVTI is 21 cm, the LVOT diameter is 2 cm, and the HR is 66/min. What is the next appropriate step?
A. Administer another fluid bolus since the patient is still fluid responsive
B. Administer a diuretic, the patient is now fluid overloaded
C. Initiate an inotrope
D. Initiate a vasopressor
View Answer
8. Correct Answer: D. Initiate a vasopressor
Rationale: Figure 19.2 shows a continuous VTI assessment of the LVOT during a full respiratory cycle. This reveals that Vmax, the maximum velocity obtained during each contraction, changes by at least 20% during the respiratory cycle (where Vmax increases during inspiration). This indicates that the patient has substantial stroke volume variation and is likely fluid responsive. After a fluid bolus the VTILVOT variability goes down to 5% (VTI and Vmax can both be used). Therefore the patient is likely no longer fluid responsive. The calculated CO is, CO = VTILVOT × r2LVOT × π × HR = 16 × 1 × π × 100 = 4.3 L/min. This is likely an adequate CO, hence a vasopressor is a logical next choice.
Selected Reference
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
9. A patient with a pulmonary artery (PA) catheter and continuous CO measurement has a measured SVR of 2200 dynes ×sec × cm-5, MAP of 80 mm Hg, CVP of 10 mm Hg, LVOT diameter of 2 cm, and LVOT VTI of 10 cm. What is their HR?
A. 75/min
B. 51/min
C. 101/min
D. 81/min
View Answer
9. Correct Answer: D. 81/min
Rationale: Using Ohms equation MAP – CVP = CO × SVR and a conversion factor of 80 to convert from peripheral resistance units to the more traditional centimeter-gram-second units (dynes × sec × cm-5), we can calculate 
cc/min.
cc/min.CO (in cc/min) can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
where VTI is the velocity time integral of the LVOT, r is the radius of the LVOT, and HR is the heart rate.
Using the given numbers 
.
.Selected Reference
1. Miller A, Mandeville J. Predicting and measuring fluid responsiveness with echocardiography. Echo Res Pract. 2016;3:G1-G12.
10. Figure 19.3 is a still from a continuous wave Doppler assessment of the blood flow through a stenotic aortic valve in a critically ill patient.
 Figure 19.3 |
The LVOT diameter is 2.0 cm and the HR is 80/min, and the patient’s body surface area is 1.9 m2. What is the cardiac index?
A. 3.00 L/min/m2
B. 1.59 L/min/m2
C. 6.36 L/min/m2
D. 2.54 L/min/m2
View Answer
10. Correct Answer: B. 1.59 L/min/m2
Rationale: CO can be quantified using echocardiography by using the equation
CO = VTILVOT × r2LVOT × π × HR
Figure 19.3 shows a “double envelope” flow through a stenotic aortic valve, where one pattern (smaller envelope) of dense velocities describes the flow in the LVOT and the second flow pattern (larger envelope) describes the increased velocity flow through the stenotic valve. Therefore the VTI through the LVOT is the smaller envelope and CO = 12 × 12 × π × 80 = 3014 cc/min = 3.0 L/min. Given the surface area of 1.9 m2, the cardiac index is 
L/min/m2.
L/min/m2.Stay updated, free articles. Join our Telegram channel
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