To reduce the loss of helium, a pressure vessel is lined with a membrane made of tantalum. The concentration of helium at the inner surface of the membrane is held constant at 4.44*10-5 kg/m3, while the concentration at the outer surface of the membrane is held constant at zero. The area of the membrane is 2.22 m2, and the diffusivity of helium in tantalum is DHe = 3.091 * 10-11 m2/s.
What must be the minimum membrane thickness that will ensure that at steady state the helium leak rate not exceed 1.11*10-6 kg/h?
Express your answer in m:
9.88*10^-6 m
Thank you!
It´s wrong :(
its ri8 man..
To find the minimum membrane thickness that will ensure the helium leak rate does not exceed a certain value, we need to use Fick's Law of Diffusion.
Fick's Law states that the diffusion flux (J) is equal to the product of the diffusion coefficient (D), the concentration gradient (∆C/∆x), and the area (A) through which diffusion occurs.
J = -D * (∆C/∆x) * A
In this case, we are given the diffusion coefficient of helium in tantalum (DHe) as 3.091 * 10^(-11) m^2/s and the concentration at the inner surface (Cin) as 4.44 * 10^(-5) kg/m^3.
Let's assume the membrane thickness is represented by ∆x. At steady state, the diffusion flux into the membrane should be equal to the helium leak rate.
J = Helium leak rate = 1.11 * 10^(-6) kg/h
∆C/∆x = (Cin - Cout) / ∆x = Cin / ∆x (since the concentration at the outer surface is zero)
Substituting the given values into Fick's Law equation, we can rearrange it to solve for ∆x.
Helium leak rate = -DHe * (Cin / ∆x) * A
1.11 * 10^(-6) kg/h = - (3.091 * 10^(-11) m^2/s) * (4.44 * 10^(-5) kg/m^3) * (2.22 m^2) / ∆x
To find the minimum membrane thickness (∆x), we can isolate it on one side of the equation by rearranging:
∆x = - (3.091 * 10^(-11) m^2/s) * (4.44 * 10^(-5) kg/m^3) * (2.22 m^2) / (1.11 * 10^(-6) kg/h)
Evaluating this expression will give us the minimum membrane thickness required to ensure the helium leak rate does not exceed the given value.