Recent content by Caccioppoli

The problem is actually of diffusion. It starts with Fick's First Law of Diffusion, an ODE (which is steady), after it uses a PSEUDO-Steady State Approximation (small t) to get an approximated expression for fluxes. To understand more about this kind of approximation I should read the main...

Managed to reach the solution :D eq.#1 is aq^3+b=q eq.#1 would be simpler if b=0, the zero-order approximation is q_0=aq_0^3 so q_0=a^{-0.5} The next order (1st order) approximation is q=q_0 + \delta It is assumed that \delta is small in comparison to q_0 so that all the terms in \delta^2...

There is a clearer way to reach the result :smile:eq.#1 is aq^3+b=q eq.#1 would be simpler if b=0, the zero-order approximation is q_0=aq_0^3 so q_0=a^{-0.5} The next order (1st order) approximation is q=q_0 + \delta It is assumed that \delta is small in comparison to q_0 so that all the...

Simon Bridge, I have managed to reach the solution, THANK YOU SO MUCH! Premise: Having a look at eq 27 and using your notation we have that δ is -b/2 not b/3 (sorry for the mistake :smile:) So we have two equations, basically: aq^3-q+b=0 and q=a^{-0.5}+\delta with a small δ Writing those...

You are right in stating there is a relation between x' and time. The only thing I can tell reading the article is that delta x' are small if I consider small times (still not having an explicit relationship between x' and t)... but how could this relate with (x')^-3 or q^3 ? Anyway, q^3 is...

I can update the problem since I've done some progress. The following equation q=aq^3+b [eq#1] can be approximated with q=a^{-0.5} + \delta [eq#2] with \delta=-b/2 Where does this approximation come from and why is \delta=-b/2? Thank you very much.

I haven't still figured out how to arrive at the result... I have q=aq^3+b [eq#1] that can somehow (?) be approximated with q=a^{-0.5} + \delta [eq#2] with \delta=b/3 Why is \delta=b/3?

It was my mistake in the substitution of q+delta inside eq24, I've found that using eq26 and eq27 do actually bring to eq28, so no problem between eq26 and 28. Last thing I would like to know is how that fitted term came out... I knew that every approximation should be based on Taylor series...

Ahahaha for the "squint" , and thank you for giving others a readable input (sorry if I had not uploaded a bigger image). By the way: 1) I can't figure why delta is beta/3 (using your notation for the terms) and I have no clue on how to fit delta 2) Also if I use that expressions in eq26 and...

Sorry, I've uploaded a bigger version of the image, split in two figures.

Hello, may someone be so kind to explain how to arrive, step by step, from equation 23 to 28? Most of all I would like to understand the approximation with delta: if I substitute eq26 in 25 I get a different result (e.g. delta^3 terms). See the attached image. Thank you very much. PS eq24...

I would like to understand only the approximation. Maybe I could have got the logic behind: "approximating a cubic polynomial with a quadratic"; but if I substitute eq26 in 25 I get a different result (e.g. delta^3 terms). eq24 may be taken as it is, take phi as just

Hello, may someone be so kind to explain how to arrive, step by step, from equation 23 to 28? I simply cannot arrive at the result, no matter what, and I cannot explain how they got δ. See the attached image. Thank you very much. Mario

Thank you very much for the answer. An european guide line (etag) says that I have to prepare a solution with pH 13.2 having Ca(OH)2 powder and distillated water. Since theoretically and experimentally I have verified that it is impossible to reach that pH, I asked for information. They said...

Hi to all, I'm in this problem: I've got to prepare a solution of Ca(OH)2 with pH=13.2 . On Perry I read that the solubility value of Ca(OH)2 at T=25°C is 1.65 g/L, so a solution with water and Calcium Hydroxide cannot reach pH=13.2. So I was suggested to prepare the mixture in the following...