# Chemical Forums

## Specialty Chemistry Forums => Chemical Engineering Forum => Topic started by: gera19 on April 10, 2006, 07:47:38 AM

Title: Heat Transfer for Forced Convection
Post by: gera19 on April 10, 2006, 07:47:38 AM
See Q  5.Can something tell me how to continue. 1st i find average velocity. then i find Re. And sub into dittus boelter eqn to find Nu. then i find h.
Then i finally find the inside wall temp abt 465.5deg. However i didnt even use the Gnielinski eqn at all. Can someone teach me what is the Q asking when they say "use e dittus boelter n gnielinski correlation and compare the results."

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.chemicalforums.com%2Fusers%2Fgeodome%2Flilly_forced_convection_q.JPG&hash=7b73479b83f0ea13fb3a324bdc5a432a)
Title: Re: hard heat transfer Q
Post by: Donaldson Tan on April 11, 2006, 03:36:35 PM
Using Dittus-Boelter Correlation,
1. Calculate the average velocity using u = Q/A
where Q is volumetric flowrate and A is cross-section area.
2. Calculate Re and Pr
3. Find Nu using the Correlation
4. Express the heat transfer coeffecient U, in terms of Nu, Characteristic Length and conductivity.
5. substitute the expression from (4) into q = U.(Tw - 280) where q is the heat flux.
6. Find Tw (Tw is wall temperature)
Title: Re: Heat Transfer for Forced Convection
Post by: Donaldson Tan on April 11, 2006, 04:00:15 PM
the same procedure above will be used when you use the gnielinski correlation.

the gnielinski correlation should give you a more accurate answer because it takes in account of entry effect, which will be significant for a short pipe. The Dittus correlation works fine for an infinitely long pipe.

Pr = 0.6929 so the first Gnielinski Correlation will be used.

Step(4) will be more complicated in the case of using the Genilinski Correlation, because the heat transfer coefficient U will not only be expressed in terms of characteristic length and conductivity, but also the length of the pipe and the inner wall temperature (T = 280C), ie. U = f(D,L,k,T,Tw,Re,Pr).

Since q = U.(Tw - 280), then q = f(D,L,k,T,Tw,Re,Pr).(Tw - 280). This equation requires iteration to solve, to find the value of Tw.
Title: Re: Heat Transfer for Forced Convection
Post by: gera19 on April 12, 2006, 11:19:00 PM
However when i used the iteration to sub Tw=495.5deg into the gnielinski correlation i got a new Tw of 501 deg. Then i repeat e iteration and got 506deg..its getting further and further from 495.5deg. However i thought that the 2 correlation shouldnt differ tt much.
I emailed my lec and this is what he replied:

However i didnt even use the Gnielinski eqn at all. Can someone teach me what is the Q asking when they say "use e dittus boelter n gnielinski correlation and compare the results."
If you have more than one correlation for a similar case, you can use them to see if the result is similar (NOTE: these correlations do not give exact results, but estimates to perhaps 5 to 10% accuracy. Also, each correlation is for a specific case, and the results will be more accurate if your situation is most closely to the case used to derive the correlation).

The situation speaks about a compressed gas, so you use the first form of G. eq (small Pr). However, in the equation appears the term (T/Tw). You have already calculated Tw with DB, so you can put it into the correlation. You will find Nu, and h, and {delta T}, and a new value of Tw. If the new value is the same as the one you used before, you are happy, and can stop. If the new value is very different, you can use the new Tw in the equation, and repeat the calculation. This gives you yet a newer Tw, and you can continue this "iteration" until you are happy with the agreement. You were to comment on how much h (and delta T or Tw) are different with the two correlations. If the agreement is within 5 degree, you will be quite content that the results can be trusted.

Keep in mind that you calculated a wall temperature of 465 C, and the bulk fluid temperature is 280 degree. It does not make much difference if the wall temperature turns out to be 460 or 470 degree, but you see that it is much hotter than the inside 280 degree. Also, keep in mind that the inside temperature is not everywhere 280 degree - you heat your gas stream, and you add 11 kJ/m^2 per s to the gas. Therefore, as the gas flows through the pipe, it goes hotter. It is therefore not useful to analyze a situation to better than 1 degree - a few cm away up or down your pipe, the temperature will be different anyways.

Pls help.thanks.if u dun understand what i talking,u wan me to type all my working out?
Title: Re: Heat Transfer for Forced Convection
Post by: Donaldson Tan on April 13, 2006, 09:20:28 AM
Pls help.thanks.if u dun understand what i talking,u wan me to type all my working out?
I know you are totally capable of typing in proper English, so please use proper English in future.

However i didnt even use the Gnielinski eqn at all. Can someone teach me what is the Q asking when they say "use e http://www.chemicalforums.com/index.php?action=post2;start=0;board=33dittus boelter n gnielinski correlation and compare the results."
If you have more than one correlation for a similar case, you can use them to see if the result is similar (NOTE: these correlations do not give exact results, but estimates to perhaps 5 to 10% accuracy. Also, each correlation is for a specific case, and the results will be more accurate if your situation is most closely to the case used to derive the correlation).
You are supposed to compare the Tw derived from each correlation.

However when i used the iteration to sub Tw=495.5deg into the gnielinski correlation i got a new Tw of 501 deg. Then i repeat e iteration and got 506deg..its getting further and further from 495.5deg. However i thought that the 2 correlation shouldnt differ tt much.hanks.if u dun understand what i talking,u wan me to type all my working out?
What is your final answer for the iteration procedure? The problem with iteration is that the process may converge when the trial values are less than a particular critical value, but process may diverge if your trial values exceed the same critical value. Try iteratating with a trial value less than 495.5K, eg. 490K.

The situation speaks about a compressed gas, so you use the first form of G. eq (small Pr). However, in the equation appears the term (T/Tw). You have already calculated Tw with DB, so you can put it into the correlation. You will find Nu, and h, and {delta T}, and a new value of Tw. If the new value is the same as the one you used before, you are happy, and can stop. If the new value is very different, you can use the new Tw in the equation, and repeat the calculation. This gives you yet a newer Tw, and you can continue this "iteration" until you are happy with the agreement. You were to comment on how much h (and delta T or Tw) are different with the two correlations. If the agreement is within 5 degree, you will be quite content that the results can be trusted.
The 2 correlations are meant for different situation. The correlation whose situation is more similar to the one in question will be the more accurate one. Is the flow turbulent or laminar? If the flow is laminar, entry effect is significant, especially since the pipe length (0.6m) is quite short. If that is so, the Gnielinski Correlation would be more accurate. Your lecturer did mention that correlation are accurate within 10%. This error margin is not always valid for all engineering correlations. For instance, Zuber's correlation for forced convection boiling is accurate within 100%. The point is no correlation is accurate, but they should be able to produce a reliable estimate.