Right Ventricular Inefficiency in Repaired Tetralogy of Fallot: Proof of Concept for Energy Calculations From Cardiac MRI Data

Abstract

Repaired tetralogy of Fallot (rTOF) patients develop right ventricular (RV) dilatation and dysfunction. To prevent their demise, pulmonary valve replacement is necessary, though appropriate timing for it is challenged by a paucity of reliable diagnostic parameters. In this pilot study, we hypothesized that stroke work (SW) and energy calculations would delineate the inefficiency of RV performance in rTOF. RV SW was calculated for both an rTOF and a normal subject by utilizing RV pressure and volume measurements obtained during cardiac catheterization and MRI studies. Energy transfer rate and ratio were computed at the main pulmonary artery (PA). Compared to the normal RV, the rTOF RV had higher operating pressure, lower computed SW (0.078 J vs. 0.115 J for normal), and higher negative energy transfer at the PA (0.044 J vs. 0.002 J for normal). Furthermore, the energy transfer ratio was nearly twice as high for the normal RV (1.06) as for the rTOF RV (0.56). RV SW and energy transfer ratio delineate important operational efficiency differences in blood flow from the RV to the PA between rTOF and normal subjects. Our pilot data suggest that the rTOF RV is significantly less efficient than normal.

Appendix

RV SW Computation

The Fourier series of the co-registered and synchronized RV pressure and volume curves (refer to the “Methods” section on co-registration of the catheterization and the CMR data) were sampled at a series of closely spaced consecutive time points, t ₁, t ₂,… ,t _n , over one cardiac period, T, progressing from t = 0 s to t = T s. Thus, the P–V loop points evaluated at the sampled time points, t ₁, t ₂,… ,t _n are (p ₁, V ₁), (p ₂, V ₂),…,(p _n , V _n ).

The SW given by Eq. 1, which is the area enclosed by the P–V loop, is computed by straightforward application of Gauss theorem, reducing Eq. 1 to

$$ {\text{SW}} = \, \oint\limits_{C} {pdV} = \frac{1}{2}\oint\limits_{C} {\left( {pdV - Vdp} \right)} , $$

(6)

a cyclic integral over the closed path, C, and then reducing the right-hand side of Eq. 6 to a summation over the closed path formed by the sampled P–V loop points, (p ₁, V ₁), (p ₂, V ₂),… ,(p _n , V _n ), (p ₁, V ₁) to

$$ {\text{SW}} = \left[ {\sum\limits_{i = 1}^{n - 1} { \frac{1}{2}\left( {p_{i} V_{i + 1} - p_{i + 1} V_{i} } \right)} } \right] + \frac{1}{2}\left( {p_{n} V_{1} - p_{1} V_{n} } \right) $$

(7)

Derivation of the Formulae for Energy Transfer Rate at MPA

Equation 2 for the rate of energy transfer at MPA can be derived in a straightforward manner from basic fluid mechanics. In the analysis of CMR images, identifying the arterial wall location on the image plane at all the phases of MRI gives the distribution of blood velocity, $\vec{V}_{\text{m}} (\vec{x},t),$ at any point x inside the arterial section and at any instant of time t. The arterial blood velocity distribution, $\vec{V}_{\text{m}} (\vec{x},t),$ varies both spatially and temporally. The rate of the total energy transferred to the blood at MPA $(\dot{E}_{\text{m}} ),$ neglecting potential energy changes because of gravity, is given by

$$ \dot{E}_{\text{m}} = \iint \limits_{{A_{\text{m}} }} {\left( {p_{\text{m}} + \frac{1}{2}\rho \vec{V}_{\text{m}} \cdot \vec{V}_{\text{m}} } \right)}\vec{V}_{\text{m}} \cdot ndA_{\text{m}} $$

(8)

where the subscript m stands for the MPA location; A _m the MPA cross-sectional area; n the unit normal to the cross section; p _m the pulsatile MPA pressure; and ρ the density of the blood.

Equation 8 can be simplified to Eq. 2 by approximating the velocity distribution $\vec{V}_{\text{m}} (\vec{x},t)$ over the MPA cross section by an average MPA velocity vector with magnitude v _m given by v _m = Q _m/A _m, and by noting that

$$ Q_{\text{m}} = \iint \limits_{{A_{\text{m}} }} {\vec{V}_{\text{m}} \cdot ndA_{\text{m}} } , $$

(9)

is simply the volumetric flow rate at the MPA section.

Result for the Intermediate Case

The results for our intermediate subject with an atrial septal defect and partial anomalous pulmonary venous return abnormal and overloaded RV but a normal PV are presented in this section.

Figure 7 shows the RV pressure and volume variation with time. The peak pressure and peak volume are 26.6 mmHg and 111.8 mL, respectively, and fall somewhere between the rTOF and the normal subject (Table 1). The volume overloading is evident from the pressure–volume curve (Fig. 7) compared with that of rTOF and the normal subject (Figs. 4a and 4b). The extent of volume overloading is lower than our rTOF subject. As shown in Fig. 8, the P–V loop for the intermediate subject also falls in between the rTOF and the normal subject and hence the computed SW (0.8365 J) follows the same trend. Figure 8 shows that the intermediate subject has a volume overloaded RV with patterns of filling that mirror rTOF and yet operates at a relatively normal operating pressure (approx. 27 mmHg). As such, the increased pressure in rTOF RV is the result of regurgitant backflow from the PA. Interestingly, the quadrant of this PV loop between end systole and end diastole does not remain isovolumic despite the fact that the higher confirmed competency of PV and TV. The reasons for these are unclear, though possibly secondary to delayed TV given the excess flow though it compared with flow through the mitral valve. In addition, other investigators have found that the RV is never truly isovolumic.29,30 More importantly, Fig. 8 shows despite operating with an abnormally higher volume than the normal subject RV, the intermediate subject RV does not operate at higher pressure than the normal subject RV. The variation of the energy transfer rate at the MPA for the intermediate subject (Fig. 9) also is in between the normal and the rTOF subject, with minimum and the maximum values at 0.033 and 0.482 J/s, respectively. The presence of the functioning PV prevents the backflow of the blood into the RV and thus the graph for energy rate has no negative contribution. The time-averaged energy transfer rate over one cardiac cycle $(\tilde{\dot{E}}_{\text{net}} )$ was calculated to be 0.198 J/s. When compared with rTOF subject, the normal subject RV operates at a lower pressure, and as such, we conclude that the increased pressure in the rTOF RV is necessary to overcome the energy losses resulting from the backflow and regurgitation because of the incompetent PV.

FIGURE 7

Variation of RV pressure and volume with time for the intermediate subject. Our intermediate subject has overloaded RV but functioning PV

FIGURE 8

RV PV diagram for our intermediate subject contrasted with those of our normal and rTOF subject. The SW value also falls in between the normal and the rTOF subject

FIGURE 9

MPA power output for the intermediate subject. Comparison of total rate of power output at MPA for the normal, rTOF, and the intermediate subject