Archived Oxygen Concentration: Insects & Tracheae

[ElikuAberts]

Homework Statement

Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface of the insect's body and penetrate into the interior. Suppose that a trachea is 1.9 mm long with a cross-sectional area of 2.1 10-9 m2. The concentration of oxygen in the air outside the insect is 0.23 kg/m3, and the diffusion constant is 1.1 10-5 m2/s. If the mass per second of oxygen diffusing through a trachea is 1.7 10-12 kg/s, find the oxygen concentration at the interior end of the tube.

Homework Equations

m/t = (DA(deltaC)) / L

The Attempt at a Solution

m/t = (DA(deltaC)) / L
1.7 E-12 = ((1.1 E-5 m^2/s)(2.1 E-9 m^2)(change in C)) / .0019 m
change in c = .3698268
(inside concentration - outside concentration) = .3698268
(inside concentration - .23 kg/m^3) = .3698268
inside concentration = .5998 kg/m^3

Thats not the right answer... but that is what I keep getting. Someone at school told me that the equation m/t = (DA(deltaC)) / L was correct, so that means I must be messing up after that, but I don't know where and I can't figure it out! If anyone would help me, I would reeeally appreciate it! :-)

 

[MexChemE]

For this system, the mass flux of oxygen is given by
j = D \frac{(c_o - c_i)}{L}
Where j = \frac{f}{A}, and f is the mass flow rate of oxygen. I can derive the above equation if needed to. What we need is to find the concentration of oxygen inside the insect, so
c_i = c_o - \frac{fL}{DA}
c_i = 0.23 \frac{kg}{m^3} - \frac{\left(1.7 \cdot 10^{-12} \frac{kg}{s} \right) (1.9 \cdot 10^{-3} m)}{ \left( 1.1 \cdot 10^{-5} \frac{m^2}{s} \right) (2.1 \cdot 10^{-9} m^2)} = 0.09 \frac{kg}{m^3}
This result is logical, if oxygen diffuses from the exterior into the body of the insect, then the concentration of oxygen in the outside must always be greater than the concentration in the inside.