An artery branches into two smaller veins, identical arteries as shown in the diagram. The largest artery has a diameter of 10 mm. The two branches have the same diameter = 5 mm. The average speed of the blood in the main artery is given in the diagram. The blood has a density of 1004 kg/m^3.
Main artery Vo = 0.50 m/s before it branches out into two veins.
QUESTION: Calculate v, the average blood flow speed in the two branches of smaller radius.
What you need to apply here is called the contnuity equation. It says that the product of velocity and TOTAL flow area is constant, for incompressible flow. Liquids exhibit incompressible flow.
Since the two smaller arteries have, TOGETHER, half the area of the main artery, the velocity in each smaller artery must be twice that in the main artery, or 1.0 m/s.
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To calculate the average blood flow speed in the two branches of smaller radius, we can make use of the principle of conservation of mass.
According to the principle of conservation of mass, the rate of flow of blood at any point in the circulatory system is constant. This means that the total flow rate of blood in the main artery before branching must be equal to the total flow rate of blood in the two smaller branches after branching.
Mathematically, we can express this as:
A1 * V1 = A2 * V2 + A3 * V3
Where:
- A1 is the cross-sectional area of the main artery
- V1 is the average blood flow speed in the main artery before branching
- A2 and A3 are the cross-sectional areas of the two smaller branches
- V2 and V3 are the average blood flow speeds in the two smaller branches
We are given:
- A1 = π * (d1/2)^2, where d1 is the diameter of the main artery
- V1 = 0.50 m/s
- d2 = d3 = 5 mm
Now, let's calculate the cross-sectional area for the main artery and the two smaller branches:
A1 = π * (10 mm / 2)^2
A2 = A3 = π * (5 mm / 2)^2
Now substitute the known values into the equation:
(π * (10 mm / 2)^2) * 0.50 m/s = (π * (5 mm / 2)^2) * V2 + (π * (5 mm / 2)^2) * V3
Simplifying the equation, we get:
(π * 5 mm^2) * V2 + (π * 5 mm^2) * V3 = (π * 10 mm^2) * (0.50 m/s)
Cancel out the common term on both sides:
V2 + V3 = 0.50 m/s * (10 mm^2 / 5 mm^2)
V2 + V3 = 0.50 m/s * 2
V2 + V3 = 1.00 m/s
Now, since the two smaller branches are identical, we can assume their average flow speeds are equal. So, we can rewrite the equation as:
2 * V = 1.00 m/s
Therefore, the average blood flow speed in the two branches of smaller radius (v) is:
v = (1.00 m/s) / 2
v = 0.50 m/s
Hence, the average blood flow speed in the two branches of smaller radius is 0.50 m/s.