Answer:
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)
Explanation:
Given data:
Diffusion constant for nitrogen is ![= 1.85\times 10^{-10} m^2/s](https://tex.z-dn.net/?f=%3D%201.85%5Ctimes%2010%5E%7B-10%7D%20m%5E2%2Fs)
Diffusion flux ![= 1.0\times 10^{-7} kg/m^2-s](https://tex.z-dn.net/?f=%3D%201.0%5Ctimes%2010%5E%7B-7%7D%20kg%2Fm%5E2-s)
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law
![J = D \frac{C_A C_B}{X_A X_B}](https://tex.z-dn.net/?f=J%20%3D%20D%20%5Cfrac%7BC_A%20C_B%7D%7BX_A%20X_B%7D)
Take C_A as point on which nitrogen concentration is 2 kg/m^3
![x_B = X_A + D\frac{C_A -C_B}{J}](https://tex.z-dn.net/?f=x_B%20%3D%20X_A%20%2B%20D%5Cfrac%7BC_A%20-C_B%7D%7BJ%7D)
Assume X_A is zero at the surface
![X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}](https://tex.z-dn.net/?f=X_B%20%3D%200%20%2B%20%28%2012%5Ctimes%2010%5E%7B-11%7D%20%29%20%5Cfrac%7B2-0.5%7D%7B1%5Ctimes%2010%5E%7B-7%7D%7D)
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)
The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25
<h3>How to find the Instantaneous rate of change?</h3>
The annual inventory cost C for a manufacturer is given as;
C = (1012000/Q) + 7.5Q
where Q is the order size when the inventory is replenished.
Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341
Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23
Instantaneous rate of change in C is first order derivative C':
C'(Q) = -1,012,000/(Q²) + 7.5
C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25
Read more about Instantaneous rate of change at; brainly.com/question/14666106
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