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heat transfer mystery

heat transfer mystery

heat transfer mystery

(OP)
Can anybody explain why my numbers are not equally out?

At work, I am trying to determine the heat transfer taking place in a water cabinet. Both loops have water in them. There is a counter current flow heat exchanger where the transfer takes place.

I take the temperature measurements with a Microscanner D1001 IR thermometer. With this device I contact the surface of a pipe and take the measurement. I take about 3-5 measurements and average them out. I do this at the same spot on the pipes for the other inlet/outlet to the heat exchanger. Each pipe has the same diameter and the same material. (I know this is not a 100% correct method, but I am hoping it gets me close enough, because the thermocouple connected to the pipes are not calibrated. I have done this step about 50 times to get enough values to minimize any outliers.

Next I take the flow on the two loops.

From there I use the equation:
     Heat tranferred = specific heat of water * flow * delta temperature. (I have no issues with the units).

When I compare the two loops, the numbers are off on average of 5kW (the average of the loops is (30kW and 25kW). The same loop is almost always higher. And this is the same case when I take the measurements off of duplicate machines. I got my data from over 5 water cabinets that were the same.

Why are my two loops not equally? Any guesses?

Thanks!

RE: heat transfer mystery

Two ideas come to mind. Perhaps you are experiencing ambient heat losses. That would affect the hotter stream more than the cooler one. The other "guess" is just inaccuracy in the measurement. Taking measurements from the exterior of the pipe is bound to be less accurate than if dedicated thermocouples were used. Try checking your readings with another instrument if possible. Also look for any ways in which one measurement could have more error than another.

RE: heat transfer mystery

No idear what a water cabinet is - but:

Does both loop have the same pumping rate/pump?

Energy is added for pumping the liquid round in the loop maybe theres a difference here?

Best regards

Morten

RE: heat transfer mystery

djack77494 (Chemical)has made a good point"Taking measurements from the exterior of the pipe is bound to be less accurate"
Regards

RE: heat transfer mystery

There are a number of sources of error:
>  The thickness of the pipe itself can lead to a temperature delta.

>  The emissivity of the pipe may be questionable.  This can easily lead to errors of 5% to 20% in the measured heat, which is what your IR sensor is measuring.

>  If the temperatures are high, then there will be substantial differences in the radiated losses.

TTFN

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RE: heat transfer mystery

(OP)
For emissivity concerns, the manual says that placing the IR thermometer on metal results in innaccurate data. But, if something like a black magic marker spot was drawn on the metal, then the recording should be accurate.

the temperatures are low, 70-95 degrees F.

I understand that taking temperatures from the exterior of the pipe are less accurate, but when i take the temp from the exterior of 4 pipes that are the same, then my calculations should still equal.

The pump is not adding heat at the place I am measuring because the pump is located about 2 meters away. I am taking the measurements just outside of the heat exchanger.

From hearing you all, it sounds like the best way to do this is to get a calibrated thermocouple and flowmeter and go from there, because my IR thermometer/flowmeter method will never work?

RE: heat transfer mystery

Have you considered that the hotter loop might have a higher potential of transferring energy to the environment that the colder one due to the higher delta T between the loop and whichever environment is in?

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying " Damn that was fun!" - Unknown>>

RE: heat transfer mystery

How are you measuring flow rates on the loops?  Are the flow instruments calibrated?  Are they calibrated at the temperatures you are operating at?

RE: heat transfer mystery

(OP)
i think i have a plan. anybody see an issue with it?

i am going to bring the machine down, and close the valve allowing water to it.

there are already thermocouples hooked up to the system, but for whatever reason the temperatures displayed on the gauges are not accurate. if i hook up a flute meter to the thermocouple, this should provide an accurate reading, correct?

for the flow, i am going to hook up a circuit setter to get a more accurate flow.

any issues?

RE: heat transfer mystery

(OP)
this is to edit the previous post. it is "fluke meter," not "flute meter."

and i will be collecting the trend for both temperature flow.
 

RE: heat transfer mystery

I am confused with the original post.  Are the two loops within a heat exchanger casing and the casing baffled to circuit its cooling/heating water? or is one loop counterflowing the other loop with both them in intimate contact?

RE: heat transfer mystery

(OP)
the two loops are within a heat exchanger.

RE: heat transfer mystery

I missed the start of this thread so I need to enter in the middle.  For what you are trying to do, the external temperature indicator should be good enough since the bias will tend to be in the same direction and you are using dT so errors should cancel.  

The really hard measurement is the flow rate of the two streams.  I'm not sure what a "circuit setter" is, but anything that improves flow measurement will improve the calculation.

Once you nail all of your measurements, the two loops will still be different heat loads because of:  (1) unmeasured losses to ambient; and (2) heat transfer efficiency of the cabinet.  You should do better than 17%, but I'd be really surprised if you do much better than 10%.

David

RE: heat transfer mystery

(OP)
a circuit setter is just an improved flow meter. installing it will be somewhat of a hassle.

on average it was 17%. sometimes it is near 0%, but others it is as high as 100%.

the thing i do not like is the chilled loop is the one reporting a greater heat transfer. with heat transfer efficiency, would it seem more logical that the warmer loop should provide data that suggests greater heat transfer taking place?

thanks.
 

RE: heat transfer mystery

If the chilled loop is accepting more energy than the warm loop is giving up then you have MAGIC.  Or maybe the chilled loop is farther from ambient and it is chilling the room more than the warm loop is?

David

RE: heat transfer mystery

(OP)
well i heard once an engineer waved his wand over the heat exchanger so who knows...

RE: heat transfer mystery

After reading several times, I'm still unclear as to what the problem is.  I think you're saying that the cool side gains 30 kW, while the hot side loses 25 kW?

What are the cold side and hot side temperatures, exactly?  What is the temperature of the room

The choices for error are limited, and have been repeated a few times already:
> errors in temperature measurement
> errors in flow measurement
> errors in modeling, i.e., parasitic loads, or differences in the piping itself

 

TTFN

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RE: heat transfer mystery

(OP)
the average values are as follows.

Loop 1. inlet: 77.0 F   outlet: 90.8 F   flow: 14.9 gpm
Loop 2. inlet: 92.0 F   outlet: 86.3 F   flow: 29.1 gpm
Room temperature: 78.8 F

a weird note, the Loop 1(ECW loop that provides water to the entire facility) inlet temp varies between 72-82 F dependent on machine. measurements taken off the same machine under the same conditions on different days have a precision of +/- 1 F. however, when compared against other machines (same model) the precision between machines is poor. the facility engineers have no logical explanation for this.

i believe this is an temp accuracy issue, and i will be getting some more reliable numbers before this week ends.

 

RE: heat transfer mystery

I concur with the opinions of others that this is a mostly a measurement issue, although losses to the air from the heat exchanger or the inlet/outlet pipes may also contribute to the appearance of a 1st law violation.

I ran your numbers myself and given the delta T's you report and GPM of the flow, I get about 26 and -21 BTU/sec from the low and high sides respectively.  Since 1 BTU = 1.06 KJ, I don't see how you get 25 KW and 30 KW for the two sides (I used 7.5 lbm/gal for the water and 1 BTU/lbm-F for its specific heat).

But, that issue aside, You need to take into account the systematic uncertianty of the thermocouple (T/C) as well as its random uncertainty.  A basic text on T/C will put the accuracy of a T/C temperature measurmement system as +/- 1.5 F or so.  Since you are taking the difference between two temperatures, the total uncertainty is the "Root sum square" of the two, or SQRT(2)*1.5 F.  This total uncertainty in temperature difference alone is in excess of 2 F (since your delta T for the hot side is only 5.7 F, this means a relative uncertainty in temperature alone of 37%!).  You can continue this using the uncertainty in the mass flow rate (I assumed +/- 0.5 GPM) and the total uncertainty is a whopping +/- 8.2 BTU/sec (8.7 KW)!  Hence, your "hot side" heat tranfer could easily be as high as 29 BTU/sec (using my numbers) or as high as 32 KW (using your numbers).

Note that all my analysis so far does not even address random uncertainty, which you state is on the order of +/- 5 F for the inlet loop.  Honestly, I am surprised your heat transfer rates are as CLOSE as they are!

If you do go ahead and use a Fluke meter, be sure to use an ice bath for a reference junction if your meter does not have on-baord compensation, sicne T/C's actually read a temperature difference.

I suggest visiting www.Omega.com and getting their "Temperature Handbook" (free of charge) if you don't have it already (I have no stake in the company but they have always been very helpful).  I can suggest some good books on total uncertainty analysis if you like.

Good luck!

RE: heat transfer mystery

What exactly is in the water cabinet?  How much pipe, etc.?  Anything else in there?

As for the calculations, I get results similar to the OP:

(86.3 F - 92 F)(29.1 gal/min)(993.3 kg/m^3) = -24.2 kW
(90.8 F - 77 F)(14.9 gal/min)(993.3 kg/m^3) = +30.0 kW

gal US = 3.7854E-3 m^3
 

TTFN

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RE: heat transfer mystery

IRStuff:

Dimensionally, I can't figure out what you are doing.

The units of your answer are in F-Kg/min, which is not a kW (and computationally the magnitude is wrong as well).

I suggest (for the first one):

(86.3 F - 92 F)(29.1 gal/min)(7.5 lbm/gal)(1 min/60 sec) = -20.7 BTU/sec

Observing that 1 BTU= 1.06 kJ, this yields about 22 kJ/sec = 22 kW.

Hope that helps.
Dave

RE: heat transfer mystery

(OP)
dave great post on the uncertainty. your calculations are off because you used 7.5lbm/gal as the weight of water. however, that is the weight of ice (water). the weight of water is 8.33 lbm/gal. in addition you needed to convert from BTUs to Watt to get my values. one BTU = 1054.35 joules.

inside of the cabinet is the following:
heat exchanger
7 thermocouples (ECW Supply, ECW Return, one for each of 5 branches that the other loop breaks into).
3 flow meters (ECW Supply, ECW Return, Other Loop Supply)
1 pump (for the hotter loop)
1 water reservoir (for the hotter loop) (located above teh pump and heat exchanger).

after leaving the heat exchanger, the hot loop branches into 5 (going to various parts of the machine, coil, generator, etc).

overall, there is about 7 meters of pipes. all piping is part of one of the loops.

the pump is located about 3 ft from the heat exchanger.




 

RE: heat transfer mystery

IR stuff:

I stand corrected.  My mass density was wrong (7.5 lbm/gal) should actually be 8.3 lbm/gal.

Sorry for the confusion and the tone of my post.

Dave

RE: heat transfer mystery

No matter, I left off the BTU/lb-F SI-equivalent, myself.  There's no convenient way to copy/paste stuff from Mathcad.

So, what's the power consumption of the pump?  I know that smaller pumps that we've used can crank about 200 W for 1 gal/min, so 29 gal/min would get you 5800 W, which, if all of that gets sucked into the cold side, would account for the heat load discrepancy.

TTFN

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RE: heat transfer mystery

fiberguy02:

Just out of curiosity, how big is this "water cabinet?"

If the counter flow heat exchanger is itself jacketed by another volume of water, then the thermal mass of the external jacket of water should be taken into account as well.

So far I have been assuming that Q_in=Q_out, but really it is Q_in-Q_out=d/dt(E_internal)= m*c*dT/dt=UADeltaT

Is there any cyclical fluctuation of the flows or the temperatures--like if the system runs at a lower load at night?  If so, there could be significant lag between the power in and the power out due to a particularly long time constant mc/UA (I won't do the math because I obviously have difficulty doing that).

Just a long-shot guess based on what this system sounds like.  I could easily be mistaken since I can't see this thing.  Good luck.

Dave

RE: heat transfer mystery

I think IRstuff got it, the pump is adding energy to the hot side loop; how much depends on the efficiency of the pump at its operating point.

RE: heat transfer mystery

I think i mentioned that possibility a week ago smile

Best regards

Morten

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