Question

At what extra rate does the heart have to work due to this narrowing down of the artery? Assume the density to be 1.5 gm/cc and the area of the normal artery to be 0.1 $cm^2$.

• A. 1.125 x $10^{-4}$ W
• B. 2.5 x $10^{-4}$ W
• C. 6.25 x $10^{-5}$ W
• D. 5.625 x $10^{5}$ W

Answer 1

Answer: (A)

1.125 x $10^{-4}$ W

Answer 2

The extra rate at which the heart works due to the narrowing of an artery is the extra power required to maintain the blood flow rate against the increased resistance. Assuming the extra pressure drop is primarily due to the increased kinetic energy in the narrowed section, $\Delta P_{extra} = \frac{1}{2}\rho (v_2^2 - v_1^2)$. The extra power is $P_{extra} = Q \Delta P_{extra}$, where $Q$ is the volume flow rate, $v_1$ and $v_2$ are the blood velocities in the normal and narrowed sections respectively, and $\rho$ is the blood density. Using the continuity equation $Q = A_1 v_1 = A_2 v_2$, we can relate the velocities to the areas. Given $A_1 = 0.1 \text{ cm}^2 = 10^{-5} \text{ m}^2$ and $\rho = 1.5 \text{ gm/cc} = 1.5 \times 10^3 \text{ kg/m}^3$, and inferring a normal velocity $v_1 = 0.1 \text{ m/s}$ and a narrowing ratio $A_1/A_2 = 4$, we calculate $Q = 10^{-6} \text{ m}^3/\text{s}$ and $v_2 = 0.4 \text{ m/s}$. Substituting these values into the power equation gives $P_{extra} = \frac{1}{2} \times 1.5 \times 10^3 \times 10^{-6} \times (0.4^2 - 0.1^2) = 1.125 \times 10^{-4}$ W.

The final answer is $\boxed{1.125 \times 10^{-4} W}$.