Fluid Mechanics: pressure gradient of a tube bifurcation

[laminarflowon]

The question is as follows: A circular tube of radius 'a' bifurcates into two tubes with equal radii 'ka', where k is a dimensionless coefficient. Derive an expression for the ratio of the pressure gradient in each bifurcated tube to that in the initial tube in terms of 'k'.

I'm not sure how to approach the problem. I believe I might need to use the Navier-Stokes equation, but we haven't covered Reynold's numbers in class. The only other way I can think of approaching the derivation problem is by first describing the jump in pressure across the three-dimensional interface (the splitting point) as being equal to the product of the surface tension and twice the mean curvature of the surface (cylindrical).

Thanks in advance!
-Josh

 

[KataKoniK]

Dear Josh,

Thank you for your question. The problem you have described can be solved using the continuity and Bernoulli's equations. Let's assume that the initial tube has a constant flow rate Q and the pressure gradient is dP/dx. After the bifurcation, the flow splits into two equal tubes with a flow rate of Q/2 in each tube.

Using the continuity equation, we can write:

Q = Q/2 + Q/2

This means that the total flow rate before and after the bifurcation remains the same. Now, let's apply Bernoulli's equation to the initial tube and the two bifurcated tubes separately:

Initial tube:
P + 1/2ρv^2 + ρgh = constant
where P is the pressure, ρ is the density, v is the velocity, and h is the height above some reference point.

Bifurcated tubes:
P + 1/2ρ(v/2)^2 + ρgh = constant
where v/2 is the velocity in each bifurcated tube.

Since the height above the reference point is the same for all three tubes, we can equate the two constants in the equations above. This gives us:

P + 1/2ρv^2 = P + 1/4ρv^2

Simplifying and rearranging, we get:

P = 1/2ρv^2

Now, we can calculate the pressure gradient in each tube using the equation for pressure:

dP/dx = 1/2ρ(dv/dx)^2

Since the flow rate remains the same before and after the bifurcation, we can write:

dP/dx = 1/2ρ(dv/dx)^2 = 1/2ρ(d(v/2)/dx)^2 = 1/4ρ(dv/dx)^2

This means that the pressure gradient in each bifurcated tube is half of the pressure gradient in the initial tube. In terms of k, we can write:

dP/dx (bifurcated tube) = 1/2ρ(dv/dx)^2 = (1/2ρ(dv/dx)^2)/k^2 = (1/4ρ(dv/dx)^2)/k = (dP/dx (initial tube))/k

Therefore, the ratio of the