Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery.

If the original diameter of the artery is D , what should be the new diameter (in terms of D ) to accomplish this for the same pressure gradient?

To double the flow rate of blood through the partially clogged artery, we need to reduce the resistance to flow caused by the cholesterol buildup. According to Poiseuille's Law, the resistance to flow in a cylindrical pipe is directly proportional to the viscosity of the fluid, the length of the pipe, and inversely proportional to the fourth power of the radius of the pipe (assuming laminar flow).

Let's assume the original diameter of the artery is D.

To achieve double the flow rate, we need to reduce the resistance by half. Since the resistance is inversely proportional to the fourth power of the radius, reducing it by half means increasing the radius to the power of 1/4.

So, we need to find the new radius in terms of the original radius. Let's call the new radius R.

According to the relationship:

Resistance ∝ 1 / (R^4)

Setting the resistance after the change equal to half of the resistance before the change, we have:

1 / (R^4) = 1 / (2 * D^4)

Now, let's solve for R:

1 / (R^4) = 1 / (2 * D^4)

Cross-multiplying:

(R^4) = 2 * D^4

Taking the fourth root of both sides:

R = (2 * D^4)^(1/4)

Simplifying:

R = 2^(1/4) * D

Therefore, the new diameter of the artery should be approximately 1.19 times the original diameter (D) to double the flow rate of blood through the artery for the same pressure gradient.

To find the new diameter of the artery, we need to consider the relationship between the flow rate, the diameter, and the viscosity of the blood.

We can use a principle called Poiseuille's law, which relates the flow rate of a fluid through a pipe (in this case, the artery) to the pressure gradient, viscosity, and diameter. Poiseuille's law states that the flow rate (Q) is directly proportional to the fourth power of the radius (r) of the pipe, and inversely proportional to the viscosity (η) of the fluid.

Mathematically, Poiseuille's law can be expressed as:

Q ∝ r^4 / η

If we want to double the flow rate (Q), we need to find the new radius (r') that will achieve this. Let's call the original diameter D and the new diameter D' (in terms of D):

D' = x * D

where x represents the factor by which the diameter changes. Since diameter is twice the radius, we can rewrite the equation as:

r' = (x * D) / 2

Substituting this value into Poiseuille's law, we get:

Q' ∝ [(x * D) / 2]^4 / η

Since we want to double the flow rate, Q' = 2Q. Therefore, we can rewrite the equation as:

2Q ∝ [(x * D) / 2]^4 / η

Simplifying the equation, we find:

2 ∝ (x^4 * D^4) / η

Now, let's find the relationship between the new diameter factor x and the original diameter factor D. We can rearrange the equation as follows:

2 ∝ x^4 / D^4

To find x, we need to solve for it. Taking the fourth root of both sides of the equation, we have:

∛2 ∝ x / D

x ∝ ∛2 * D

Therefore, the new diameter (D') in terms of the original diameter D to achieve a doubled flow rate is:

D' = ∛2 * D

Hence, the new diameter should be the cube root of 2 times the original diameter D.

dV/dt = L/(R^4) for a constant pressure gradient

dV/dt = L/(1/2*D)^4 = 16L/(D^4)

2dV/dt = 16L/Dnew = 2*16L/(D^4)

Dnew = 4th root{(2*D^4)}