Overall heat transfer coefficient for water cooling

Click For Summary
SUMMARY

The discussion focuses on calculating the overall heat transfer coefficient (U) for a water cooling system in a fermenter, specifically analyzing the cooling process from 121°C to 35°C. The user employs the equation Q = A*U*dtm but observes that U varies significantly between the initial extreme cooling phase and the subsequent fermentation stage. Suggestions include considering the heat produced by bacterial fermentation and using the logarithmic mean temperature difference (LMTD) for more accurate calculations. The variability of U is confirmed, indicating it is not a constant value in this context.

PREREQUISITES
  • Understanding of heat transfer principles, specifically the equation Q = A*U*dtm
  • Knowledge of the logarithmic mean temperature difference (LMTD) method
  • Familiarity with the effects of fermentation on temperature control
  • Basic thermodynamics, particularly concepts related to cooling systems
NEXT STEPS
  • Research the application of logarithmic mean temperature difference (LMTD) in heat transfer calculations
  • Explore the impact of biological processes on thermal dynamics in fermentation systems
  • Study the principles of heat exchangers and their design considerations
  • Investigate advanced heat transfer coefficient calculations for varying conditions
USEFUL FOR

Engineers, process designers, and researchers involved in fermentation technology, thermal management, and heat transfer analysis will benefit from this discussion.

andrejen88
Messages
1
Reaction score
0
Hi all!

I'm analyzing a fermenter and the water cooling system that cools it down.

In analyzing this I've found kW and kWh required to cool down the fermenter using Q = dt*cp*m

However when I try to find the overall heat transfer coefficient U using the data, I find that it varies with either Q or M.

I use the equation: Q = A*U*dtm

At first Q is high due to an extreme cooling from 121 C to 35 C. After that during bacterial fermentation the temperature should stay the same. However this second stage gives me a lower U-value (twice) than during the extreme cooling in the beginning. Does any of you have a suggestion to why this is happening?

Can you at all use this formula to calculate a heat transfer coefficient for water cooling systems, or can it only be applied for heat exchangers?
 
Engineering news on Phys.org
Hello andrejen, :welcome:

Could it be the bacterial fermentation itself produces heat ?

You want to be a bit clearer in your typing: m means the same as M (I think) and dtm is not dt times m but the mean delta t.

Could also be that mean delta t is not good for the expression and you need the log mean delta t (aka LMTD).

Furthermore, U doesn't have to be an all-over constant and in your case depends on variables that vary.

There is a very thorough book by the VDI : http://link.springer.com/referencework/10.1007%2F978-3-540-32218-4 that might help you if you can get hold of it. I think the text is parallel in german and english.
 
Last edited by a moderator:
andrejen88 said:
At first Q is high due to an extreme cooling from 121 C to 35 C. After that during bacterial fermentation the temperature should stay the same. However this second stage gives me a lower U-value (twice) than during the extreme cooling in the beginning. Does any of you have a suggestion to why this is happening?

+1 to what BvU said about the fermentation producing heat. If you want the temperature to remain constant it shouldn't need any cooling unless there is a heat source.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
18
Views
2K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K