Calculating Heat Exchange Surface for Fluid Flow

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Discussion Overview

The discussion revolves around calculating the heat exchange surface area for a fluid flow in a heat exchanger, specifically when the fluid enters and exits at different temperatures while exchanging heat with steam at a constant temperature. Participants explore the implications of using the log mean temperature difference (LMTD) method in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions which temperature value of the fluid should be used in the heat exchange formula, given that the fluid temperature changes.
  • Another suggests using the log mean temperature difference (LMTD) as a standard approach for heat exchangers.
  • A participant raises a concern about using LMTD when the steam remains at a constant temperature, questioning if it affects the calculation.
  • There is a discussion about the implications of the fluid's final temperature being close to the steam temperature, which could lead to undefined values in the LMTD formula.
  • One participant notes that if the temperatures are equal, no heat transfer occurs, but acknowledges that in practice, there is usually a non-zero temperature difference.
  • Another participant emphasizes the need for a mean temperature value for accurate calculations, given the fluid's varying temperature during the heat exchange process.
  • Participants discuss the sensitivity of the LMTD method in cases where the steam temperature is constant and the fluid temperature approaches it closely.
  • There is mention of deriving the dependence of heat transfer from a variable along the heat exchanger, although uncertainty remains about its usefulness.
  • A participant expresses anxiety about an upcoming test, indicating that their thoughts may not be entirely clear, and later reflects on their understanding of the topic after some consideration.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using LMTD in this scenario, particularly regarding the implications of constant steam temperature and the fluid's temperature approaching it. The discussion remains unresolved, with multiple competing perspectives on the best approach to calculate the heat exchange surface area.

Contextual Notes

Participants highlight limitations in applying the LMTD method when the steam temperature is constant and the fluid temperature approaches it, leading to potential undefined values. There are also concerns about the assumptions made in the calculations and the practical implications of the theoretical models discussed.

dRic2
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Fluid comes in at ##T_1## and comes out at ##T_2## and I know that heat is exchanged with steam at constant ##T_{steam}##.

If I want to know the Surface of the exchanger then

##US(T_{steam}-T_{fluid}) = \dot m_{fluid} cp_{fluid} (T_2-T_1)##

My question is, since ##T_{fluid}## changes, which value of ##T_{fluid} ##should I use in the above formula?

PS: Since I don't know the surface, I don't know how the exchanges is designed so I have really know clue on what to do
 
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But how do I use LMTD if the steam remains at constant T? Does it make any difference? Or should I just do ##LMTD = \frac {(T_2 - T_{steam})-(T_1 - T_{steam})} {ln \frac {T_2-T_{steam}} {T_1 - T_{steam}}}##?
 
Yep
 
Thank you! :)
 
Excuse me, what if ##T_2 ≈ T_{steam}##? The above formula breaks because ## ln \frac {{T_2} - {T_{steam}}} {T_1 - T_{steam}} = -\infty##

Btw, I check how LMTD is derived for countercurrent heat exchanger... I don't know if I can apply it if the steam remains at the same T during the exchange.
 
If temperatures are equal, no heat transfer takes place. Just like the expression says. Not very realistic: in reality there is always a non-zero approach.
 
The final temperatures are equal, but the fluid comes in at a lower temperature. The heat transfer is maximum at the beginning and zero at the end. I need a mean value, so I need a mean value of temperature. But I can't use the above formula.

Say the fluid comes in at 100 °C and comes out at nearly 200°C (let's say 199°C). To heat it up I use steam (at equilibrium) at 200°C so it condenses and it remains at 200°C. How am I suppose to find the surface of the exchanger ?
 
You can ask someone to dig out the equipment drawings :smile:

From the temperatures it appears that the surface area is at least $$Q\over U\; {\rm LMTD}$$ with lmtd = 21.5 for 1 degree and 8.7 for .001 degree approach.
 
  • #10
BvU said:
with lmtd = 21.5 for 1 degree and 8.7 for .001 degree approach.

yeah, LMTD is very sensitive in this very case, so I was thinking that maybe it is not the best approach here.

I have a test in two days, so I can't think to much about it now, later I will try to derive the dependence of Q from x:
##dQ = h(T_{steam} - T(x))\pi Ddx## and see if it leads to something... but I'm not to sure it will be of any help
 
  • #11
dRic2 said:
yeah, LMTD is very sensitive in this very case, so I was thinking that maybe it is not the best approach here.

I have a test in two days, so I can't think to much about it now, later I will try to derive the dependence of Q from x:
##dQ = h(T_{steam} - T(x))\pi Ddx## and see if it leads to something... but I'm not to sure it will be of any help
It will lead to the LMTD. Why are you saying that there is sensitivity of the use of LMTD when the steam temperature is constant? In the example you gave, you are not going to let the fluid temperature approach the steam temperature to 1C since that would require too large a heat transfer area.
 
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  • #12
dRic2 said:
I have a test in two days,
I hope it went well -- and I guess there's no connection between the test and this thread...

Is this a theoretical exercise ? If not, is your expedition UTCWAP (up the creek without a paddle) because it is based on plant measurements ? Could you post a sketch of the actual setup ?
 
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  • #13
BvU said:
I hope it went well
Thanks, I'm waiting for the results

BvU said:
and I guess there's no connection between the test and this thread...

Is this a theoretical exercise ?

It's just something I was thinking about.

Chestermiller said:
It will lead to the LMTD.

Yeah, but I had to check myself :)

Chestermiller said:
Why are you saying that there is sensitivity of the use of LMTD when the steam temperature is constant?

No reason. Sorry for the trouble. I tend to be anxious before a test and I don't actually think straight. I've checked with my mind clear today and I got it immidiately.

Sorry, and thanks for the replies!
 
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