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Chapter 9: Problem 87

If the cardiac output of a small dog is $4.1 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s}\( the radius of its aorta is \)0.50 \mathrm{cm},$ and the aorta length is \(40.0 \mathrm{cm},\) determine the pressure drop across the aorta of the dog. Assume the viscosity of blood is $4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}$

Short Answer

Expert verified
Answer: The pressure drop across the aorta of the small dog is 4999.28 Pa.

Step by step solution

01

Collect the needed information

We are given: - Cardiac output (flow rate) Q = \(4.1 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s}\) - Aorta radius r = \(0.50 \mathrm{cm}\) (convert to meters) - Aorta length L = \(40.0 \mathrm{cm}\) (convert to meters) - Blood viscosity \(\eta\) = \(4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\)
02

Convert radius and length to meters

To work with SI units, convert the radius and length to meters: r = \(0.50 \mathrm{cm}\) = \(0.005 \mathrm{m}\) L = \(40.0 \mathrm{cm}\) = \(0.40 \mathrm{m}\)
03

Apply Poiseuille's Law

Poiseuille's law relates pressure drop, flow rate, radius, length, and fluid viscosity as follows: \(\Delta P = \frac{8 \eta Q L}{\pi r^4}\) Plug in the given values: \(\Delta P = \frac{8(4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s})(4.1 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s})(0.40 \mathrm{m})}{\pi (0.005 \mathrm{m})^4}\)
04

Calculate the pressure drop

Evaluate the expression to find the pressure drop: \(\Delta P = \frac{8(4.0 \times 10^{-3})(4.1 \times 10^{-3})(0.40)}{\pi (0.005)^4} = 4999.28 \,\mathrm{Pa}\)
05

Final Answer

The pressure drop across the aorta of the small dog is 4999.28 Pa.

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