Your heart pumps blood into your aorta (diameter 2.5 cm) with a maximum flow rate of about 500 cm^3/s. Assume that blood flow in the aorta is laminar (which is not a very accurate assumption) and that blood is a Newtonian fluid with a viscosity similar to that of water.

a. Find the pressure drop per unit length along the aorta. Compare the pressure drop along a 10 cm length of aorta to atmospheric pressure (105 Pa).
b. Estimate the power required for the heart to push blood along a 10 cm length of aorta, and compare to the basal metabolic rate of 100 W.
c. Determine and sketch the velocity profile across the aorta (assuming laminar flow). What is the velocity at the center

Answer:

a. i) The pressure drop per unit length is 52,151.89 Pa

ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta

b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts

ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta

c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel

ii) The velocity at the center is approximately 2.04 m/s

Explanation:

The given diameter of the aorta, D = 2.5 cm = 0.025 m

The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s

Assumptions;

The blood flow is laminar

The blood is a Newtonian fluid

The viscosity of water ≈ 0.01 poise = 1 cp

a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;

[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]

Where;

Q = The flow rate = A·v

A = The cross sectional area

R = The radius = D/2

Δp/L = The pressure drop per unit length of the pipe

Therefore, we have;

[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]

The pressure drop per unit length ΔP/L = 52,151.89 Pa

ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;

∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa

Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;

[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175

Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta

b. i) The power, P = Q × ΔP

Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts

ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;

P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35

The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length

c. i) The velocity profile across the aorta is given as follows;

[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]

Where;

[tex]v_m[/tex] = The velocity at the center

We get;

[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]

The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s

ii) The velocity profile, v(r), is given by the following formula;

[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]

Therefore, we have;

[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]

The velocity profile of the pipe is created with Microsoft Excel

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