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A small artery has a length of 1.1 x 10����� m and a radius of 2.5 x 10������ m. If the pressure drop across the artery is 1.55 kPa, what is the flow rate through the artery? Assume that the temperature is 37 degrees Celsius and the viscosity of whole blood is 2.084 x 10����� Pa��s.

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Final answer:

The flow rate through the artery is approximately 2.61 x 10^{-14} {m}^3/s.

Step-by-step explanation:

The flow rate (Q) through a cylindrical tube (like an artery) can be calculated using Poiseuille's Law, which is given by the formula:

Q = {π . r^4 , ΔP} / {8 . η .L}

where:

- Q is the flow rate,

- π is the mathematical constant pi (approximately 3.14159),

- r is the radius of the tube,

- ΔP is the pressure drop across the tube,

- η is the viscosity of the fluid,

- L is the length of the tube.

Given values:

- r = 2.5 x 10^{-6} m (radius),

- Δ P = 1.55 x 10^3 Pa (pressure drop),

- η = 2.084 η 10^{-3} Pa·s (viscosity),

- L = 1.1 x 10^{-3} m (length).

Now, substitute these values into the formula:

Q = {3.14159 x (2.5 x 10^{-6})^4 x (1.55 x 10^3)} / {8 x 2.084 x 10^{-3} x 1.1 x 10^{-3}}

Let's calculate the flow rate (Q) using the provided values:

Q = {3.14159 x (2.5 x 10^{-6})^4 x (1.55 x 10^3)} / {8 x 2.084 x 10^{-3} x 1.1 x 10^{-3}}

Q ≈ {3.14159 x (2.5 x 10^{-6})^4 x (1.55 x 10^3)} / {8 x 2.084 x 10^{-3} x 1.1 x 10^{-3}}

Q ≈ {3.14159 x 9.765625 x 10^{-26} x 1.55 x 10^3} / {1.81648 x 10^{-8}}

Q ≈ {3.065041012 x 10^{-25} x 1.55 x 10^3} / {1.81648 x 10^{-8}} \]

Q ≈ {4.74658652 x 10^{-22}} / {1.81648 x 10^{-8}}

Q ≈ 2.61384 x 10^{-14} m^3/s

So, the flow rate through the artery is approximately \(2.61384 \times 10^{-14} \, \text{m}^3/\text{s}\).

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