40% glycol solution and ton of refigeration 1

[rk27]

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?

 

[quark]

rk27,

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.

 

[25362]

On a different subject: one must be careful when using the third equation by rk27.

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 ?

 

[quark]

Yes, you are right as usual. The constant 500 comes from 8.34 lb/gal x 60min/hr.

 

[25362]

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.

 

[rk27]

Thank you to everyone for their speedy replys. The main reason for my post was I have a primary-secondary loop. The calculation for the secondary loop , using an ultrasonic meter permanently attached to the piping, is approximately 96 tons. If I apply the same formula to the primary loop, I get 126 tons. I have to guess that the flow is at design conditions in the primary because it was balanced recently, or maybe I should measure the pressure drop myself. My problem is that during the summer the plant brings a second chiller on-line even though the secondary loop is only using 356 gallons and a single 200 ton chiller is capable of producing 461 gallons. Also, I am wondering if the use of tons is becoming antiquated because it really is a reference point under certain conditions, and that the kBtuH can vary significantly when operating off of design conditions.

 

[quark]

RK27,

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)

http://www.automatedbuildings.com/news/apr02/art/htmn/htmn.htm

We can be helpful if you have any specific questions.

25362,

Can you please explain the method?

Regards,

 

[25362]

Quark:

The method of what ? Units' cancellation ?

 

[quark]

Yes, unit cancellation and the tongue twister.

 

[25362]

Quark: there is nothing to it really. I mentioned the old twister to remind that plenty of arithmetical problems sound like it. The puzzle of deciding what to do with the numbers given in the above query is easily solved by using the technique of unit cancellation.

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:

CH₄ + 2O₂ => CO₂ + 2H₂O​

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.