heat transfer mystery

[fiberguy02]

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!

 

[Dave41A]

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

 

[fiberguy02]

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.

 

[Dave41A]

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

 

[IRstuff]

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

FAQ731-376

 

[Dave41A]

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

 

[btrueblood]

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.

 

[MortenA]

I think i mentioned that possibility a week ago

Best regards

Morten