How Does Constriction in the Bronchus Affect Air Pressure?

[JamesL]

We are studying fluid dynamics in class right now...

Here is a problem that came up on my homework that I am not sure how to solve:

When a person inhales, air moves down the bronchus(windpipe) at 10.4 cm/s. The average flow speed of the air doubles through a constriction in the bronchus. Use (density of air) = 1.29 kg/m^3. Assuming an incompressible flow, determine the pressure drop in the constriction. Answer in units of Pa.
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When i first looked at the problem i took "The average flow speed of the air doubles through a constriction" to mean that it goes from 10.4 cm/s to 20.8 cm/s.

Then i just plugged the values into Bernoulli's equation and then plugged the following into my calculator:

x + (1/2)(1.29)(.104) = (x-y) + (1/2)(1.29)(.208)... only solving for y, which would be the pressure drop. i got .020929 Pa as my answer, and this is not correct.

So, unless i made some mistake in the above equations, i assume that "The average flow speed of the air doubles through a constriction" means something else... but I am not sure where to start.

Any help is appreciated.

 

[enigma]

Aside from too much precision in your answer, it looks right to me (you missed the squares typing it in, but you didn't in your calculations).

How do you know it's not correct? If it's entered into a computer, it may not like the 5 sig figs, or it may be looking for the change in pressure, and not the pressure drop (switch negative sign). It's possible I'm missing something too, but I don't think I am...

 

[M98Ranger]

First of all, great job on attempting to solve the problem using Bernoulli's equation! However, there are a few things to consider in order to arrive at the correct answer.

Firstly, the statement "The average flow speed of the air doubles through a constriction" means that the velocity of the air increases from 10.4 cm/s to 20.8 cm/s at the constriction. This does not necessarily mean that the average flow speed doubles, as the velocity profile of the air may not be uniform throughout the cross-section of the bronchus.

Secondly, it is important to note that Bernoulli's equation is only applicable for incompressible flows, which means that the density of the fluid should remain constant throughout the flow. However, in this case, the air is being compressed as it passes through the constriction. Therefore, we cannot use Bernoulli's equation directly.

To solve this problem, we can use the continuity equation, which states that the mass flow rate of a fluid is constant throughout a pipe or tube. This can be written as:

ρ1A1V1 = ρ2A2V2

Where ρ is the density of the fluid, A is the cross-sectional area, and V is the velocity at two different points in the flow (subscript 1 for initial conditions and subscript 2 for final conditions).

In this case, we can use the continuity equation to relate the velocities at the two different points in the bronchus. We can also assume that the density of air remains constant at 1.29 kg/m^3.

Therefore, we have:

(1.29 kg/m^3)(A1)(10.4 cm/s) = (1.29 kg/m^3)(A2)(20.8 cm/s)

We can also relate the cross-sectional areas using the fact that the cross-sectional area is inversely proportional to the square of the velocity (A1V1 = A2V2).

Substituting this relationship into the continuity equation, we get:

(1.29 kg/m^3)(10.4 cm/s)(V1^2) = (1.29 kg/m^3)(20.8 cm/s)(V2^2)

Simplifying, we get:

V1^2 = 2V2^2

Therefore, the velocity at the constriction (V2) is equal to the square root of half