There is a steady laminar blood flow in a tapered cylindrical vessel of circular cross-section, as shown in the figure below. Two cross-sections, labeled A and B in the figure, have radii $0.5$ centimeter (cm) and $0.45$ cm, respectively. If the average flow speed of the blood at cross-section A is $1$ cm/second, the average flow speed at cross-section B is cm/second. (Round off to one decimal place).

Assume blood to be an incompressible fluid.

Question

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Answer 1

Rather than computing both cross-sectional areas explicitly, use the fact that for a circular pipe the area scales with the square of the radius, so the continuity equation reduces directly to a ratio of radii.

For an incompressible fluid in steady flow, $A_A v_A = A_B v_B$. Since $A = \pi r^2$ for a circle, this becomes $\pi r_A^2 v_A = \pi r_B^2 v_B$, and the $\pi$ cancels on both sides, leaving:

\[ v_B = v_A \left( \frac{r_A}{r_B} \right)^2 \]

This form is convenient because it needs only the ratio of the two radii, not their individual areas. Here $r_A = 0.5$ cm and $r_B = 0.45$ cm, so the ratio is:

\[ \frac{r_A}{r_B} = \frac{0.5}{0.45} = 1.1111 \]

Squaring this ratio gives the factor by which the speed increases:

\[ \left( \frac{r_A}{r_B} \right)^2 = (1.1111)^2 = 1.2346 \]

Multiplying by the known speed at A:

\[ v_B = 1 \times 1.2346 = 1.2346\ \text{cm/s} \approx 1.2\ \text{cm/s} \]

Let's summarize:

Continuity for an incompressible fluid in a circular pipe reduces to $v_B = v_A (r_A/r_B)^2$.
A vessel that narrows always speeds the flow up, since the radius ratio $r_A/r_B$ is greater than $1$ for a taper.
Squaring the radius ratio directly gives the speed multiplication factor, without ever computing the actual areas in $\text{cm}^2$.

The average flow speed at cross-section B is $1.2$ cm/second.

Answer 2