A specimen of LaNi5 containing hydrogen is placed in a vacuum furnace. After 1 hour, at what depth from the surface of the specimen has the concentration of hydrogen reached 1/3 the initial concentration?
The diffusion coefficient of hydrogen in the alloy has a value of 3.091*10-6 cm2/s. Assume that the initial concentration of hydrogen is uniform throughout the specimen and that the concentration of hydrogen is maintained at zero in the vacuum furnace.
DATA: Error Function Values (given without regard as to whether you need these data to solve the problem) for values of z less than 0.6, use the aproximation erf(z) = z; erf(1.0) = 0.843; erf(2.0) = 0.998
Express your answer in units of cm:
0.0703
To solve this problem, we can use Fick's second law of diffusion, which describes how the concentration of a diffusing species changes with time and distance.
The equation is given by:
∂C/∂t = D * (∂^2C/∂x^2)
Where:
- C is the concentration of hydrogen
- t is time
- x is the distance from the surface
- D is the diffusion coefficient
We can rearrange this equation to solve for x:
(∂^2C/∂x^2) = (∂C/∂t) / D
Assuming that the concentration of hydrogen decreases linearly with depth, we can integrate this equation with appropriate boundary conditions.
Let's denote the initial concentration of hydrogen as C0 and the concentration of hydrogen at depth x as C(x). We're looking for the depth x at which C(x) = C0/3.
We can integrate the equation above twice to get:
∫∫ (∂^2C/∂x^2) dx dx = ∫∫ (∂C/∂t) / D dx dx
Taking into account the initial concentration C(x=0) = C0 and the constant concentration at the surface C(x=infinity) = 0, we have:
∫∫ (∂^2C/∂x^2) dx dx = ∫∫ (∂C/∂t) / D dx dx
0 to x ∫(0 to x) (∂^2C/∂x^2) dx dx = 0 to t ∫(0 to x) (∂C/∂t) / D dx dx
Integrating the left side:
0 to x ∫(0 to x) (∂^2C/∂x^2) dx dx = 0 to x (∂C/∂x) - [0 to x (∂C/∂x)] dx
0 to x ∫(0 to x) (∂^2C/∂x^2) dx dx = x * (∂C/∂x)
Integrating the right side:
0 to t ∫(0 to x) (∂C/∂t) / D dx dx = [0 to t x * (∂C/∂t) / D] dx
0 to t ∫(0 to x) (∂C/∂t) / D dx dx = (1/D) * [0 to t x * (∂C/∂t)] dx
0 to t ∫(0 to x) (∂C/∂t) / D dx dx = (1/D) * [0 to t C] dx
Since the initial concentration C(x=0) = C0 and the constant concentration at the surface C(x=infinity) = 0, we have:
0 to t ∫(0 to x) (∂C/∂t) / D dx dx = (1/D) * [0 to t C0] dx
Plugging these results back into the integral equation:
x * (∂C/∂x) = (1/D) * [0 to t C0] dx
Integrating both sides:
∫∫ x * (∂C/∂x) dx dx = (1/D) * [0 to t C0] dx
∫∫ x dC = (1/D) * [0 to t C0] dx
Integrating the right side:
∫∫ x dC = (1/D) * [0 to t C0] dx
∫∫ x dC = (C0/D) * [0 to t] dx
∫∫ x dC = (C0/D) * t * x
∫∫ x dC = (C0/D) * t^2 * x/2
Applying the boundary conditions:
0 to x ∫(C0 to C0/3) x dC = (C0/D) * t^2 * x/2
Simplifying the integration:
0 to x [C0 * x - (C0/3) * x] dx = (C0/D) * t^2 * x/2
0 to x (2/3) * (C0/3) * x dx = (C0/D) * t^2 * x/2
(2/9) * x^3 = (C0/D) * t^2 * x/2
Simplifying further:
x^3 = (9C0/4D) * t^2 * x
Simplifying again:
x^2 = (9C0/4D) * t^2
x = sqrt((9C0/4D) * t^2)
Now we can plug in the values given in the problem:
C0 = initial concentration of hydrogen =Given
D = diffusion coefficient = 3.091 * 10^-6 cm^2/s
t = 1 hour = 3600 seconds
x = sqrt((9C0/4D) * t^2)
x = sqrt((9 * C0) / (4 * D) * t^2)
x = sqrt((9 * C0) / (4 * (3.091 * 10^-6)) * (3600^2))
Evaluate this expression to find the depth from the surface of the specimen at which the concentration of hydrogen has reached 1/3 its initial concentration.