Can over-pumping a heating or cooling coil actually give you less heat transfer? 2

As a matter of fact,it may increase the heat transfer since higher flow rate will raise the value of the convective heat transfer coefficient within the water side.

But, you are limited by the air side. Assuming EAT of, say, 70F, your effective heat transfer on the air side is essentially (180+160)/2 - 70, which means that you can only get about 10% more convective transfer on air side if the entire exchanger is at 180, but you'd need infinite flow rate to get that. So, it is better, but the increased flow rate, pump sizing, additional power, etc., are all downsides for a relatively minor improvement.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

I understand that there are practical limits, but it is generally correct to say that more water flow will give you more heat transfer, correct?

I was being told the exact opposite that higher water flow would result in less heat transfer, which didn't make sense to me.

Correct. The misconception may be caused by the exit T going down as flow goes up, but the total heat transferred goes up. Q = m*.Cp.dT

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

The only unknown would be if power draw required to increase the flow would add more heat than is gained by the capacity increase.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

Thanks all. I knew the answer. Just needed a reality check.

Hmmm...

I'll readily admit doing the actual math on this is not my forte, but if you'll bear with me anyway I could use the education [I havent't been in school for years, but I still try to learn something new every day, plus I've never been afraid of seeming to be an idiot...after all, it has happened so many times by now, and I'm still here]... ;o)

So here goes:

Where in the above thread is the allowance for the time it takes for the actual heat transfer to take place?

Based on some of the things I've seen, the 'seasoned tradesman' may not have articulated his understanding all that clearly, but he may have had a point...

Would not the quantity of heat transferred be expressed as an inverse parabola? It would seem to me that if the flow of the fluids is low, admittedly there will be less actual thermal energy xfer, but the 'efficiency' of the HXR will be optimum since there will be ample time for the exchange to occur. As the flows are increased, the quantity of heat transferred will increase, but the 'efficiency' will drop slightly due to decreasing dwell time.

To my understanding, pushing more and more fluids through a given HXR would eventually level off the parabola, then cause it to logarithmically / exponentially decline since despite the presence and maintenance of the required delta T there would simply not be enough time available for the energy flow between fluids to actually take place.

Just wondering...

CR

For math not being your forte, your argument sure uses a lot of math terms.

Your basis, "Would not the quantity of heat transferred be expressed as an inverse parabola", is wrong. As a heat exchanger is pushed harder and harder it asymptotes to a maximum heat duty that is bound by the 2nd Law of thermodynamics. It's easier to see if you mentally fix the flows and imagine larger and larger heat transfer areas, trying to squeeze out every BTU of heat possible. Since, the 2nd Law will not let the exit temperature of the hot stream be less than the entrance temperature of the cold stream (countercurrent flow), there is a limit to the amount of heat that can be transferred no matter how big the heat exchanger is. I think searching and reading about heat exchanger effectiveness and looking at some heat exchanger effectiveness curves would help.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

Hi Latexman, thanks for the response.

After sending my OR I began to recognize the holes in the way I had articulated myself; hindsight is frequently described as 20/20. You quite correctly pointed up my errors, which I appreciate.

So I'll try again...

I was attempting to use the inverse parabola concept to express the value of, um, let's see if I can get it right this time, what you would get if, for a heat exchanger of fixed surface area, you plotted the fluid flows [keeping them proportional throughout] on the X axis of a graph and the actual heat energy transferred on the Y axis.

Using your terminology, and hoping I'm doing so correctly, as the fluid flows are increased, the curve would reach a point of asymptote [what I was trying to convey by 'levelling off'] at precisely the point at which the design engineer wanted it to, namely where the HXR does what it was made to do at no more cost than necessary [other than for any built-in margins to accommodate for surface fouling, etc].

If fluid flows are then increased beyond this point, would not the line fall away from asymptote more and more rapidly as the loiter time of the fluids in the HXR is decreased?

CR

No, because the exchange will be then limited by a constant exchanger shell temperature and the convection of the air. The efficiency may go down the tubes, but not the heat transferred. Here, the efficiency could be defined as ratio of power transferred to power supplied.

So, it's possible that the tradesman is referring to the system efficiency, which will go down.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

Your logic is sound, but it must be carried further to reach the right conclusion.

Yes, if fluid flows are increased beyond the design point, the “loiter time” of the fluids in the HXR decrease. The inlet temperatures and heat transfer area are, of course, fixed, but
everything else changes, so follow along:

The hot outlet temperature increases, due to the “loiter time” decrease. It’s easier for me to think it had less time to get colder.
Also along those lines, the cold outlet temperature decreases.
Since the hot got hotter and the cold got colder, there is more driving force, more [Δ]T, between the hot stream and the cold stream.
Also, since flows increased, so did turbulence.
Therefore, the heat transfer coefficient will increase too. The change may be small, but it will be there.

Here’s the basic formulas:
Q = U.A.[Δ]T_LM = (w.Cp.(T_IN-T_OUT)_HOT = (w.Cp.(T_OUT-T_IN)_COLD

If U and [Δ]T_LM increase and A remains the same, Q must increase!

This means the increase in fluid flow must be larger than the subsequent decrease in T_IN-T_OUT (or T_OUT-T_IN), since Q increased.


Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

Some of you are way over thinking this and getting confused. More flow will always result in more heat transfered per minute,approaching a limit asymtopically. The heat transfered per gallon will go down, approaching a limit of zero. So what is important to you? In most cases it is the btu's per minute.

As an aside, there are, however, cases in which the heat flux may drop with increasing [Δ]T between the "hot" and "cold" fluids, and these refer to boiling conditions, not the case in hand.