For this device answer the following questions. a. Approximately how long will i

Question

For this device answer the following questions.

a. Approximately how long will it take for the flux through the barrier to reach steady-state (d)?

b. What will be the flux of pheromone through the barrier and into the air immediately after the barrier is put in place (mg/cm2s)?

c. At steady-state what is the flux of pheromone out of the barrier (mg/cm2s)? How fast is the pheromone released by the device (mg/s)?

d. If the chamber were full of water instead of air, would the flux out of the barrier change? Explain. (Hint: think about the D.O. meter demonstration).

e. How fast is the pheromone released by the device (mg/s)?

f. If the chamber were full of water instead of air would the flux out of the barrier change? In what direction? Explain.

g. Would you expect wind speed to impact the rate of release by the device? Explain.

h. If you wanted to increase the rate of release by the device what design change(s) could you make?

Impermeable holder Solid pheromone coPA 0.2 mg/L C Pw 5 mg/L PAT 0.06 cm2/s DPW 3.4E-6 cm2/s PB 1.9E-8 cm2/s O O 5 cm3 of air Polymeric diffusion barrier (B), thickness 0.06 cm, area 2 cm PAB 1E-5

Explanation / Answer

Solution:

The computation are expressed below,

(a) t = x2/2D =[ (-0.06)2 / 2*0.06 + (-0.06)2 /2*3.4E-6 + (-0.06)2 / 2*1.9E-8] = 95266.28 seconds= 26.5 hrs (approx)

(b) Flux through barrier is,

J = - D dC/dx = [- 0.06*(0.2 / -0.06)] + [-3.4E-6*(5 / -0.06)] = 0.2 gm / cm2.sec

(d) by changing water inside the chamber the flux will change as diffusivity is fast in water medium than air, but in outside it remains the same representing chambr flux > outside flux.

Note: Kindly post question in to two parts for quick answering.

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