40% glycol solution and ton of refigeration
40% glycol solution and ton of refigeration
(OP)
I am trying to work out if you can use the rule of thumb formula for tonnage when using 40% glycol.
ton = gpm X delta T / 24. This is translated to
ton = gpm X delta T X 500 / 12000 or
ton = gpm X delta T X Sp Gravity of fluid X Sp Heat of fluid X 60 / 12000
Well for water every thing is great hence the 24 constant or the 500 constant, the problem with the 40% ethylene glycol is the tonnage is reduced by ~10% due to its physical properties. My question is: is there a problem using the 12000 btu per ton as related to the amount of energy required to melt a ton of ice in 24 hours: 144 btu X 2000 lbs / 24 hours since the characterisrtics of the fluid are different from water?
ton = gpm X delta T / 24. This is translated to
ton = gpm X delta T X 500 / 12000 or
ton = gpm X delta T X Sp Gravity of fluid X Sp Heat of fluid X 60 / 12000
Well for water every thing is great hence the 24 constant or the 500 constant, the problem with the 40% ethylene glycol is the tonnage is reduced by ~10% due to its physical properties. My question is: is there a problem using the 12000 btu per ton as related to the amount of energy required to melt a ton of ice in 24 hours: 144 btu X 2000 lbs / 24 hours since the characterisrtics of the fluid are different from water?





RE: 40% glycol solution and ton of refigeration
TR is TR irrespective of the secondary refrigerant and same definition can be used with any medium. Secondly, there is a little misunderstanding. The apparent capacity reduction is due to the fact that specific heat is low for glycol solutions when compared to water. You will get the same tonnage if you increase mass flowrate of glycol solution. Suppose if water flowrate is x kg/hr then you require to put [x/cp] kg/hr of glycol solution to acheive same tonnage.
cp is specific heat of glycol solution.
RE: 40% glycol solution and ton of refigeration
When the heat capacity is in Btu/(lb*oF), and when using a non dimensional specific gravity of the fluid, one has to multiply by 8.34 lb/gal (water's density) or, alternatively, express the fluid's density in lb/gal.
Quark: am I right ?
RE: 40% glycol solution and ton of refigeration
RE: 40% glycol solution and ton of refigeration
To quark, thanks.
As an answer to the old tongue twister:
"If Peter Piper picked a peck of pickled peppers, how many pecks of pickled peppers did Peter Piper pick ?"
This is the subject of units cancellation which I frequently use to check myself.
RE: 40% glycol solution and ton of refigeration
RE: 40% glycol solution and ton of refigeration
It is very difficult to assess what is happening without knowing the things in detail. It is better if you do energy audit for this installation.
Here is a good article(also check two other articles by the same author)(Courtesy - imok2)
ht
We can be helpful if you have any specific questions.
25362,
Can you please explain the method?
Regards,
RE: 40% glycol solution and ton of refigeration
Quark:
The method of what ? Units' cancellation ?
RE: 40% glycol solution and ton of refigeration
RE: 40% glycol solution and ton of refigeration
To return to Peter Piper for an illustration, suppose he steadily picks 3/8 of a peck of pickled peppers in an hour, how much will he have picked in two-thirds of an hour ?
The answer is obtained by multiplying:
3/8 peck/hour x 2/3 hour . Hour cancels out, giving the answer: 1/4 peck.
Another common, almost trivial, example. What mass of water forms when 8 g of methane burn in air ?
Stoichiometry gives us:
Knowing the molecular masses for methane (m) and water (w), are 16 and 18, respectively.
We write the multiplication so as to cancel the units:
18 g w/mol w x 2 mol w/mol m x mol m/16 g m x 8 g m = 18 g w
Rk27 wrote:
(gal/min)(oF)[Btu/(lb.oF)](min/h)(TR.h/Btu) = (TR) (gal/lb), meaning that lb/gal is missing to complete the cancellation.
It is clear you expected a more sophisticated presentation, but this is what I could show. Sorry.