Question 6.5: The human heart is a four-chambered pump where valves allow ......

The general form of Bernoulli’s equation (6-11) is given below, where the subscript 1 refers to a position upstream in the atrium or ventricle and subscript 2 refers to the position at the valve opening.

g y_{1}+{\frac{v_{1}^{2}}{2}}+{\frac{P_{1}}{\rho_{1}}}=g y_{2}+{\frac{v_{2}^{2}}{2}}+{\frac{P_{2}}{\rho_{2}}}   (6-11)

Presuming the patient is lying down on a table, z_{1} = z_{2}:

P_{1}+\frac{\rho_{1}v_{1}^{2}}{2}=P_{2}+\frac{\rho_{2}v_{2}^{2}}{2}  (6-17)

Rearranging,

P_{1}-P_{2}={\frac{\rho}{2}}(v_{2}^{2}-v_{1}^{2})    (6-18)

Blood has a density of 1070\,{\mathrm{kg/m}}^{3}, which is equal to {\mathrm{8~mmHg}}(s^{2}/{\mathrm{m^{2}}}). Inserting this value into equation (6-18) results in the final equation used to understand pressure drop and velocity in heart valves, an equation that is used daily in hospitals around the world:

P_{1}-P_{2}=4(v_{2}^{2}-v_{1}^{2})     (6-19)

As always, it is instructive to keep in mind the units of each parameter in the equation. The units of pressure are mmHg (common pressure units used in medical applications), and the number 4 and velocity, have units of (s^{2}/{ m}^{2})\mathrm{mmHg}\mathrm{~and~m/s}, respectively.

A patient is being examined for possible mitral valve stenosis by echocardiography. The mitral valve is located between the left atrium and left ventricle of the heart and maintains flow in one direction. In mitral valve stenosis, the valve does not open fully, resulting in decreased blood flow. A patient was examined by echocardiography and was found to have a blood flow velocity in the left atrium of 0.45 m/s, where the area is measured to be 0.0013 m and the maximum velocity through the valve was 1.82\;\mathrm{m/s_{\mathrm{}}}^{2}.

We wish to determine the pressure drop between the left ventricle and the valve as well as the valve diameter. The pressure is determined by equation (6-19) as

P_{1}-P_{2}=4(v_{2}^{2}-v_{1}^{2}) = 4\frac{s^{2}}{\mathrm{m^{2}}}\mathrm{mmH}{\mathrm{g}}\Bigl[(1.82\,\mathrm{m/s})^{2}-(0.45\,\mathrm{m/s})^{2}\Bigr]=12.44\,\mathrm{mmH}{\mathrm{g}}

The cross-sectional area of the open mitral valve is

\rho_{1}v_{1}A_{1}=\rho_{2}v_{2}A_{2}

Since blood is at a constant density in the problem, the equation reduces to

v_{\mathrm{atrium}}A_{\mathrm{atrium}}=v_{\mathrm{valve}}A_{\mathrm{valve}}

Solving for the valve area,

A_{\mathrm{valve}}={\frac{v_{\mathrm{atrium}}A_{\mathrm{atrium}}}{v_{\mathrm{valve}}}}={\frac{(0.45~\mathrm{m/s})(0.0013~\mathrm{m})}{(1.82\,\mathrm{m/s})}}=3.2\times10^{-4}\,\mathrm{m^{2}}

As a result, this patient would be further evaluated for possible mild mitral valve stenosis.