# Thread: axisymmetric jet formula derivation

1. ## axisymmetric jet formula derivation

hi all
plz take alook to this pdf(attachment)...
section 4.3.1(axisymmetric jet)
i am lookng for the full derivation(with all details) of equation 4.3.16b
anybody can help me or give me a hint...
it is an emergency....
thanks

2. ## Getting 4.3.16b from 4.3.16a

I'm not sure how far back you need to go. However, if you look at 4.3.16a and go from there, the authors claim that $\displaystyle \lim_{\eta\to\infty}\frac{f'(\eta)}{\eta}=0.$
This is most definitely true if it is known that $\displaystyle f'(\eta)$ is bounded. $\displaystyle f(\eta)$ is called the stream function. If it represents a physical parameter, then most likely you can assume $\displaystyle f'(\eta)$ is bounded on physical grounds. The same goes for $\displaystyle f(\eta)$ itself. But now, if $\displaystyle f(\eta)$ and $\displaystyle f'(\eta)$ are both bounded, then it is the case that $\displaystyle \lim_{\eta\to\infty}\frac{f(\eta)f'(\eta)}{\eta}=0$.
The authors also claim that $\displaystyle f''(\eta)$ goes to $\displaystyle 0$ as $\displaystyle \eta$ goes to infinity. If that is the case, then if you take limits in the entire Equation 4.3.16a, you will get $\displaystyle 0$ on the LHS, which implies you must get $\displaystyle 0$ on the RHS. Hence, $\displaystyle \frac{f f'}{\eta}+f''-\frac{f'}{\eta}=0$. Multiplying through by $\displaystyle \eta$ gives you Equation 4.3.16b.

So, in getting 4.3.16b from 4.3.16a, the problem reduces down to knowing that $\displaystyle f$ and $\displaystyle f'$ are both bounded, and that $\displaystyle \lim_{\eta\to\infty}f''(\eta)=0$.

With regard to the second derivative, you might be able to get that from the "note" in-between Equations 4.3.10 and 4.3.11: "... noting that $\displaystyle F=f'/\eta$..." This tells you that $\displaystyle F\eta=f'$, and hence $\displaystyle f''=F'\eta+F$. Is it the case that $\displaystyle \lim_{\eta\to\infty}\left(\eta F'(\eta)+F(\eta)\right)=0$?

These are just the beginnings of some ideas.

3. thanks dear dr. ackbeet
i try solving the problem , in the case of any problem i'll leave another message here...
good luck

4. Originally Posted by Ackbeet
I'm not sure how far back you need to go. However, if you look at 4.3.16a and go from there, the authors claim that $\displaystyle \lim_{\eta\to\infty}\frac{f'(\eta)}{\eta}=0.$
This is most definitely true if it is known that $\displaystyle f'(\eta)$ is bounded. $\displaystyle f(\eta)$ is called the stream function. If it represents a physical parameter, then most likely you can assume $\displaystyle f'(\eta)$ is bounded on physical grounds. The same goes for $\displaystyle f(\eta)$ itself. But now, if $\displaystyle f(\eta)$ and $\displaystyle f'(\eta)$ are both bounded, then it is the case that $\displaystyle \lim_{\eta\to\infty}\frac{f(\eta)f'(\eta)}{\eta}=0$.
The authors also claim that $\displaystyle f''(\eta)$ goes to $\displaystyle 0$ as $\displaystyle \eta$ goes to infinity. If that is the case, then if you take limits in the entire Equation 4.3.16a, you will get $\displaystyle 0$ on the LHS, which implies you must get $\displaystyle 0$ on the RHS. Hence, $\displaystyle \frac{f f'}{\eta}+f''-\frac{f'}{\eta}=0$. Multiplying through by $\displaystyle \eta$ gives you Equation 4.3.16b.

So, in getting 4.3.16b from 4.3.16a, the problem reduces down to knowing that $\displaystyle f$ and $\displaystyle f'$ are both bounded, and that $\displaystyle \lim_{\eta\to\infty}f''(\eta)=0$.

With regard to the second derivative, you might be able to get that from the "note" in-between Equations 4.3.10 and 4.3.11: "... noting that $\displaystyle F=f'/\eta$..." This tells you that $\displaystyle F\eta=f'$, and hence $\displaystyle f''=F'\eta+F$. Is it the case that $\displaystyle \lim_{\eta\to\infty}\left(\eta F'(\eta)+F(\eta)\right)=0$?

These are just the beginnings of some ideas.

hello
http://www.mathhelpforum.com/math-he...erivation.html
I want to ask you to write the full derivation for obtaining eq. 4.3.16b from the first with all details...
i have not enough time to spend on it..
i am ready to pay you for solving this problem...
i just want you to solve

morteza08@gmail.com

5. I'm sorry, Morteza, but that's not the way we operate in this forum. We don't just solve your problems for you. You have to do the main work; we'll help you get unstuck if you need a little push here or there. Otherwise, you see, you won't own the answer for yourself. So ask away on specific questions all you like, where you've shown you've done some work, and we'll help you out.

6. ## thanks

Originally Posted by Ackbeet
I'm sorry, Morteza, but that's not the way we operate in this forum. We don't just solve your problems for you. You have to do the main work; we'll help you get unstuck if you need a little push here or there. Otherwise, you see, you won't own the answer for yourself. So ask away on specific questions all you like, where you've shown you've done some work, and we'll help you out.
dera dr.
I know this is not the rule here to solve a peoblem , that's why i did not ask you to solve the prblem at my first thread,,, but please make me hopeful, could you please help me if i send you my work and giv me hints if i need,,,,
thanks...it is a vital project and you are the only dear person that helped me to start solving the problem, i want to be quarranteed in the rest of the work..

1. Tell me exactly which equation you want to start from (along with any other assumptions, conditions, etc., that go along with that equation). You've already said you want to derive Equation 4.3.16b. So, together, the starting point and ending point give us a trajectory of thought and define the problem.

2. Show me the first place in the logical train of thought, from the starting equation(s) that you determined in Step 1 to Equation 4.3.16b, where you are stuck.

8. Originally Posted by Ackbeet

1. Tell me exactly which equation you want to start from (along with any other assumptions, conditions, etc., that go along with that equation). You've already said you want to derive Equation 4.3.16b. So, together, the starting point and ending point give us a trajectory of thought and define the problem.

2. Show me the first place in the logical train of thought, from the starting equation(s) that you determined in Step 1 to Equation 4.3.16b, where you are stuck.
ok dr.
i will do it soon. i try to organise all of my works and will send you them.... i consider all you want me to do
thanks

9. Have you made any progress while the site was down?

10. Hi Dr. Keister
Yes I wrote down the derivation process and i circled where i have problem with. that is not with me right now, i will show you it as soon as i go home..
thx for your compassion and follow-up

11. Have you got any updates?

12. Originally Posted by Ackbeet