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.
Calculate v, the average blood flow speed in the two branches of smaller radius.
Main artery Vo = 0.50 m/s before it branches out into two veins.
To calculate the average blood flow speed in the two branches of smaller radius, we can use the principle of continuity. According to this principle, the flow rate of a fluid remains constant as it passes through different sections of a pipe or vessel, assuming the fluid is incompressible.
The flow rate (Q) is given by the equation: Q = A * v
Where:
Q is the flow rate (constant)
A is the cross-sectional area of the vessel
v is the average blood flow speed
In this case, we know the cross-sectional area of the main artery (A1) and the cross-sectional area of the two smaller branches (A2). We also know the average blood flow speed in the main artery (v1) before it branches out.
To begin, we can calculate the flow rate in the main artery (Q1) using the equation Q1 = A1 * v1. Since Q1 is constant, it will be equal to the flow rate in the two smaller branches, which we will call Q2. Hence, Q2 = Q1.
We can rearrange the equation to solve for the average blood flow speed in the two smaller branches (v2):
Q2 = A2 * v2
Since Q2 = Q1:
A1 * v1 = A2 * v2
Now, let's apply this information to the given values in the question.
Given:
Diameter of the main artery = 10 mm
Diameter of the two smaller branches = 5 mm
Average blood flow speed in the main artery (v1) = 0.50 m/s
First, let's convert the diameters to radii by dividing them by 2:
Radius of the main artery (r1) = 10 mm / 2 = 5 mm = 0.005 m
Radius of the two smaller branches (r2) = 5 mm / 2 = 2.5 mm = 0.0025 m
The cross-sectional area of a vessel is given by the equation: A = π * r^2
Calculating the cross-sectional areas:
A1 = π * (0.005 m)^2
A2 = π * (0.0025 m)^2
Now substitute these values into the equation A1 * v1 = A2 * v2:
π * (0.005 m)^2 * 0.50 m/s = π * (0.0025 m)^2 * v2
Simplify and solve for v2:
(0.005 m)^2 * 0.50 m/s = (0.0025 m)^2 * v2
0.00125 m^2/s = 0.00000625 m^2 * v2
v2 = 0.00125 m^2/s / 0.00000625 m^2
v2 = 0.00125 m/s
Therefore, the average blood flow speed in the two branches of smaller radius (v) is 0.00125 m/s.