Consider the flow of blood through a large horizontal artery with 1.0 mm of diameter. The artery has a smooth reduction to an artery 0.5 mm in diameter. If the pressure of the blood in the larger artery is 8.00 x 104 Pa and the pressure in the smaller artery is 6.00 x104 Pa, what is the speed (in m/s) of the blood through the small artery? Assume that the density of blood is 1050 kg/m3.

Question

Consider the flow of blood through a large horizontal artery with 1.0 mm of diameter. The artery has a smooth reduction to an artery 0.5 mm in diameter. If the pressure of the blood in the larger artery is 8.00 x 104 Pa and the pressure in the smaller artery is 6.00 x104 Pa, what is the speed (in m/s) of the blood through the small artery? Assume that the density of blood is 1050 kg/m3.

Answer 1

To find the speed of the blood through the small artery, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing through a pipe. The equation is as follows:

P + 0.5ρv^2 + ρgh = constant

Where:
P = pressure of the fluid
ρ = density of the fluid
v = velocity of the fluid
g = acceleration due to gravity
h = height of the fluid

Since the flow of blood is horizontal, we can ignore the height term (ρgh) in the equation. So, the equation becomes:

P + 0.5ρv^2 = constant

Now, let's assign subscripts to the different locations in the artery to keep track of the variables. Let '1' represent the larger artery and '2' represent the smaller artery. The equation for the larger artery becomes:

P1 + 0.5ρv1^2 = constant

And for the smaller artery:

P2 + 0.5ρv2^2 = constant

Since the constant value is the same for both equations, we can equate them:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

Now we can substitute the given values:

P1 = 8.00 x 10^4 Pa
P2 = 6.00 x 10^4 Pa
ρ = 1050 kg/m^3 (density of blood)

Simplifying the equation, we have:

8.00 x 10^4 + 0.5 x 1050 x v1^2 = 6.00 x 10^4 + 0.5 x 1050 x v2^2

Since the diameter of the larger artery is twice that of the smaller artery, we know that the cross-sectional area of the larger artery is four times that of the smaller artery (πr1^2 = 4πr2^2). Therefore, we have:

v1 = (1/4) v2

Substituting this relation into the equation, we get:

8.00 x 10^4 + 0.5 x 1050 x (1/16) v2^2 = 6.00 x 10^4 + 0.5 x 1050 x v2^2

Simplifying further:

8.00 x 10^4 + 1050/32 v2^2 = 6.00 x 10^4 + 1050/2 v2^2

Multiplying through by 32 to eliminate the fraction:

2560 x 10^4 + 1050 v2^2 = 1920 x 10^4 + 1680 v2^2

Rearranging the terms:

630 v2^2 = 640 x 10^4

Now we can solve for v2:

v2^2 = (640 x 10^4) / 630

Taking the square root:

v2 = √(640 x 10^4) / √630

v2 ≈ 80.99 m/s

Therefore, the speed of the blood through the smaller artery is approximately 80.99 m/s.