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Carlo Hansen
17-10-2005, 03:30 PM
Hi to all

I am working with a problem concerning RH in cold stores
and with freezing. -20 to -30C.
What is the limit for a TEV and a EEV in thise cases, without any aditional components.
It sems to me that the limit will be about 65% RH with a
TEV.
Do someone have calculations about this or some practical facts.

Best regards
Carlo Hansen

TXiceman
17-10-2005, 06:45 PM
I do not understand your question. The RH in the space is determined by the coil characteristics and the rom conditions. You have to plot the room and coil conditions on a Psy chart to determine where you are. If you are trying to dry a rom out, you may need to add some reheat to get to your final room conditions.

Ken

chemi-cool
17-10-2005, 07:51 PM
I thought that using electric heaters to raise temp to decrease RH is a waist of energy.


Today, dehumidifiers are more efficient and much more common.

Chemi:)

frank
17-10-2005, 08:39 PM
Hi Chemi

I have a friend who works for York and is currently installing -45C blast freezers on a food production line. They are pre-heating the incoming air to lower the %RH (nearly dry air) prior to it hitting the evaporator. The benefits are reduced ice buildup and shorter defrosts.
When you look at the benefits and increased product manufacture against increased use of energy, I think that this project is coming out on the plus side.

Peter_1
17-10-2005, 10:14 PM
Preheating or re-heating Frank?
Preheating to dry is very unefficient.

Plot this once out in a psychometric chart and you will see that preheating doens't work as well as re-heating.

Preheating is only used when you need a total enthalpy control like you find in a cleanroom.

To remove moisture, you need energy to achieve this and if you do it with a heater or a dehumidifier, the total energy input remains almost the same as far as I can remember (given that the temperature has to remain at the same level)

Carlo Hansen
18-10-2005, 07:06 PM
Thanks for the input,

I maybe ask this question a litlle dificult to understand, my point is that a TEV will see 5-8 K between inlet and outlet
of the evaporator, beside this you have a Delta T in the
cold or frezing room between the air and the evaporator temperature.
My question is then, what is the max possible RH% you can reach vith a TEV or a EEV, ( with a realy big evaporator.)

Best regards
Carlo Hansen

frank
20-10-2005, 07:47 PM
Hi Peter

From what I recall of the conversation it was pre-heating, i.e. using gas fired heaters to raise the temp of the incoming air. Obviously, this had an effect on the total cooling load but from what i understand they were experimenting with ways of increasing production - not saving money :confused:

US Iceman
20-10-2005, 08:06 PM
Frank,

If the desire was to increase production, i.e., shorter or no defrost, how does pre-heating the air help. I agree the RH would decrease, but wouldn't the amount of moisture (grains) be the same at either condition?

If the moisture content of the air does not change, this would have no effect on the quantity of frost formed on the coils.

Were they using a desiccant dehumidifier to remove the moisture from the airstream? And could the "pre-heating" actually have been the need to cool the air as it exits the desiccant system?

I'm curious as to what they were actually doing and the goals they hoped to achieve.

Regards,

botrous
20-10-2005, 08:07 PM
what i understand they were experimenting with ways of increasing production - not saving money


Was this a governomental project :)

frank
20-10-2005, 08:44 PM
Iceman

It wasn't one of my projects but one that a friend of mine was involved in.
The project was a food manufacturer (Pies, pasties etc) and it was a blast freezer. The theory behind the project was that if the air could be heated to a dry state then the moisture content entering the blast freezer would be greatly reduced and therefore frost buildup on the coil would be minimal and defrost periods reduced.
As I say, that was the basis of the discussion (over a pint or two, sad I know) but I cannot comment on the outcome as I have not seen him for a while so I do not have any results to pass on.
I must say that the theory seemed logical at the time but, in practice, I don't know.

Peter_1
20-10-2005, 09:17 PM
If you plot this in a Mollier chart, when heating, you're going straight up.

So the moisture content of the air - the absolute humidity - isn't changing when you heat it.

When you have to cool th preheated air afterwards, then you have a huge increase in heating load for the cooling system.

So you first have to pass anyway through the dewpointline before you can cool it further down to the desired lower temperature.

Especially for a freezer I don't understand at all what they were experimenting and what they thought they could achieve with this.

US Iceman
20-10-2005, 11:40 PM
Frank,

My belief is that the concept your friend was trying to implement was for the right reasons, but perhaps maybe the wrong approach.

To reduce the amount of frost on the freezer coils, they would have to dehumidify the air entering the coils. Or, dehumidify the entire blast freezer volume.

As Peter said,


So the moisture content of the air - the absolute humidity - isn't changing when you heat it

Since the absolute humidity is not changing, the same amount of frost will form on the coils. The RH will change, but not the absolute humidity.

The only way to get the lower moisture content (absolute humidity) is to use a pre-cooling coil to reduce the moisture, or to use a desiccant dehumidifier. The pre-cooling coil would still frost, but the frost volume on the main cooling coil might operate for longer periods of time without defrosting.

One method we use here is to use variable fin-spacing. The first couple of rows in the coil may have a wider fin-spacing, which allows frost to buildup, but not block the airflow. The remainder of the coil rows would have a slightly tighter fin spacing.

Some of the best discussions happen in a pub. ;) These are great places to brainstorm ideas and what-ifs.

It would be interesting to hear back from you on your friends progress.

Regards,
US Iceman

US Iceman
20-10-2005, 11:54 PM
I think this got off topic a little bit.


My question is then, what is the max possible RH% you can reach with a TEV or a EEV, ( with a really big evaporator.)

I think TXiceman had the right idea.

You will need the sensible heat ratio of the coil and the bypass factor and then plot the conditions on the psychrometric chart to determine the RH of the air.

If I understand what Carlo is looking for is the actual coil temperature to use. I have not done this before, but my guess is to use the saturated evaporating temperature of the refrigerant and disregard the evaporator superheat.

Once all of the data points are know you could then determine the room RH and decide if re-heat is required.

Any ideas from you other guys??

Dan
21-10-2005, 03:17 AM
I maybe ask this question a litlle dificult to understand, my point is that a TEV will see 5-8 K between inlet and outlet
of the evaporator, beside this you have a Delta T in the
cold or frezing room between the air and the evaporator temperature.
My question is then, what is the max possible RH% you can reach vith a TEV or a EEV, ( with a realy big evaporator.)


Carlo, I think I understand your question now. You have started some interesting side conversations, too. :)

ASHRAE and others say you can maintain 90% RH with certain products by sizing coils at 10 deg F or so TD. I think it is more complex than that. Not that I disagree.

Decreasing the temperature difference between the coil surface and the air passing over it will decrease the rate of moisture removal from the air. Thus you would want to choose a larger coil surface compared to a smaller coil surface if you want to maintain a high humidity.

I don't think it is so simple, however. This ignores the product loads and respiration rates, refrigeration run time, air velocities, and the limitations of direct expansion, forced air evaporator units.

Victorman
25-10-2005, 03:27 AM
Frank :

I Think that the preheating of the air prior entering the Blast Freezer coils is a part of a process when drying air with desiccant wheels.
Gas or Electric heaters are used to evaporate the moisture ( Chrystals ) from the part of the wheel that is saturated into the air ;achieving temperatures over 100 F.
Since the wheel is hot and turning very slow, the dry air becomes hot. ( around 90 F plus ).
Air Dryers can achieve very low Dew points , therefore avoiding any frost on the coils.
I will add that there are two streams of air leaving the Wheel , one full of moisture and hot , and the other very dry and also hot.Please look at followind site.

http://cargocaire.com/

Carlo Hansen
25-10-2005, 05:23 PM
Thanks for all the Reply`s.

I think i got some new ideas, thanks to you.
Dan you could understand my question, but as you say,
it is not so simple.
My calculation shows that with a DT about 6K in a cold
room, The RH can not exed 68% RH and for a freezing room with same conditions 58%. Because if you are using a big evaporator, only a small part will work and the rest will almost take room temp using a TEV.
I am trying to make a cold room with 90-95% of RH for
bakeryes and meat storring.
I have some test now, but they are to big to set in to this forum.

I can mail the test if someone like to see it.

Best regards
Carlo Hansen

chemi-cool
25-10-2005, 08:30 PM
Hi Carlo.

If the problem is too low RH, you can always employ a humidifier that release the wet air in front of the fans.

I did something like this in a cold room for flowers.
also the evaporator fans, run only with the compressors.

Chemi:)

Peter_1
25-10-2005, 08:37 PM
My calculation shows that with a DT about 6K in a cold
room, The RH can not exed 68% RH and for a freezing room with same conditions 58%. Because if you are using a big evaporator, only a small part will work and the rest will almost take room temp using a TEV.
I am trying to make a cold room with 90-95% of RH for
bakeryes and meat storring.
I can mail the test if someone like to see it.

I would like to see your test.
As said, we had a room at 95% RH, select a DT of 4 to 5 K and a unit which runs +/- 6 hours/day.
You will notice a high RH, a lot higher then 68%.

It's almost impossible to plot this or to calculate this: you have first the bypass-factor of the evaporator (air passing trough the fins which wasn't cooled by the fins and which didn't release it's moisture) and after shutting down the compressor, the remaining water on the fins will re- evaporate again, you have the moisture content of the products which humidifies the air, opening of the doors,...

thing
26-10-2005, 05:01 AM
Frank, I think you have mix up RH and moisture content. RH is relative term. Heating up air can change RH but not moisture content. In another word, you can reduce RH by heating up the air but the moisture content is remain constant.

frank
26-10-2005, 08:51 PM
Frank, I think you have mix up RH and moisture content. RH is relative term. Heating up air can change RH but not moisture content. In another word, you can reduce RH by heating up the air but the moisture content is remain constant.


It wasn't one of my projects
:) :) :) :)

lana
01-11-2005, 06:22 AM
Hi to all

I am working with a problem concerning RH in cold stores
and with freezing. -20 to -30C.
What is the limit for a TEV and a EEV in thise cases, without any aditional components.


What I understand from your question is this :

When you cool the air with dehumidification (like you have in the cold room) then the RH% will increase and the value depends on the evaporator DT chosen. i.e. the size of the evaporator.

TD 4-5C ----- RH 90%
TD 6-7C------RH 80%
TD 7-9C ----- RH 75%

The above values are approx.

About the reheat or preheat I must say that heating the air only reduces its RH% value but the actual moisture content in the air stays exactly the same.
Cheers.:)

SHERYCOOL
01-11-2005, 11:29 AM
I Think Reheat Coil Reduces The Room Rh.

wambat
01-11-2005, 07:58 PM
I thought that using electric heaters to raise temp to decrease RH is a waist of energy.


Today, dehumidifiers are more efficient and much more common.

Chemi:)
That's right Chemi for example: if you raise say 55*F at 50% Rh to 95*F the wet bulb will go up 25%, the RH will go down 73%, the VP will stay the same, the DP will stay the same, the GR/lb will stay the same, the volume will increase 7%, and the big one-Enthalpy will increase 14%. So it definately is a (waist) size 40 of time. :D

Gary
02-11-2005, 11:26 AM
I Think Reheat Coil Reduces The Room Rh.

Reheat does not remove moisture. It lengthens the run time for the compressor, giving the system more time to get the moisture removal job done.

Given enough run time, how low the humidity goes depends upon the temperature of the coil. The lower the coil temperature, the more moisture is removed, and the lower the humidity goes.

If the coil is not cold enough to achieve the humidity level you want, then reheat accomplishes nothing.

coolman
02-11-2005, 08:00 PM
At those temperatures trying to dry with heaters and longer evaporator running times is very expensive on electrics.

Best way to remove moisture is using a desicant dehumidifier.
These dehumidifiers do have a very high efficiency at low temperatures.

Victor

Lc_shi
03-11-2005, 02:40 AM
Effiecient dehumidification method is using dessicant to de-couple sensible and latent heat. No need to reheat.
Now the liquid dessicant dehumidifier can process air directly to supplied point.

rgds
LC

Lc_shi
04-11-2005, 02:58 AM
Hi Mark,
Yes. Same enthalpy for same start and end point. Convention cooling dehumidification need to cool air below dewpoint and reheat(can use condensing heat);but liquid dehumifier doesn't need to be below dewpoint. Attached picture for ref.

rgds
LC

Manoj Menon
04-11-2005, 07:38 AM
Hi to all

I am working with a problem concerning RH in cold stores
and with freezing. -20 to -30C.
What is the limit for a TEV and a EEV in thise cases, without any aditional components.
It sems to me that the limit will be about 65% RH with a
TEV.
Do someone have calculations about this or some practical facts.

Best regards
Carlo Hansen


Dear All,

I don't think anything to do with TEV / EEV. The humidity can be controlled by using Higher TD cooling units. Or Higher FPI units. OR use de humidifiers which will :) icrease your heat load.

alfred_yagne
03-03-2010, 05:50 AM
Hi, I must thanks you guys for the explainations i have got from your forum. But can anyone help me show how to plot in psychrometric chart, for example my current project requires 15KW heatload, 60% relative humidity, 15 C room temp. I used 5 C as evaporating temperature. Now I want to find the reheat load. Please help me. I still have no idea what airflowrate to use if it is needed can you help me compute and plot it? i am really having a hard time doing it. I regret that i did not study refrigeration well in college. Anyway, I would like to use commercial refrigeration, dx here.

Gary
03-03-2010, 06:18 AM
Hi, I must thanks you guys for the explainations i have got from your forum. But can anyone help me show how to plot in psychrometric chart, for example my current project requires 15KW heatload, 60% relative humidity, 15 C room temp. I used 5 C as evaporating temperature. Now I want to find the reheat load. Please help me. I still have no idea what airflowrate to use if it is needed can you help me compute and plot it? i am really having a hard time doing it. I regret that i did not study refrigeration well in college. Anyway, I would like to use commercial refrigeration, dx here.

Here is a psych chart:

http://www.uigi.com/UIGI_SI.PDF

Find your room temp (15C) along the bottom. Trace a vertical line upwards to 60%RH. Now a horizontal line to the left until you reach the 100%RH curve. Now straight down to 7C.

In order to achieve your target humidity at your target temperature you will need an evap air off temp of 7C.

Peter_1
03-03-2010, 06:33 AM
Hi, I must thanks you guys for the explainations i have got from your forum. But can anyone help me show how to plot in psychrometric chart, for example my current project requires 15KW heatload, 60% relative humidity, 15 C room temp. I used 5 C as evaporating temperature. Now I want to find the reheat load. Please help me. I still have no idea what airflowrate to use if it is needed can you help me compute and plot it? i am really having a hard time doing it. I regret that i did not study refrigeration well in college. Anyway, I would like to use commercial refrigeration, dx here.

Depends mainly on your internal load and moisture added in the room and also your fresh air rate.

alfred_yagne
03-03-2010, 06:34 AM
Thanks Gary, I appreciate it. Does that mean that I can have approx 60% of rh out of 7 C evaporating temperature and 15C room temperature? If that's the case TD should be 8C. Is my understanding correct? Thanks a lot. I asked help from a aircooler manufacturere, and he gave me aircooler with hotgas reheat of 19 KW from the parameters listed on my formeer post. What do you think. He gave me instructions how he got the reheat load, but unfortunately i did not get it. Do you have rules of thumb on reheat loads for decreasing humidities?

Gary
03-03-2010, 06:52 AM
Thanks Gary, I appreciate it. Does that mean that I can have approx 60% of rh out of 7 C evaporating temperature and 15C room temperature? If that's the case TD should be 8C. Is my understanding correct? Thanks a lot. I asked help from a aircooler manufacturere, and he gave me aircooler with hotgas reheat of 19 KW from the parameters listed on my formeer post. What do you think. He gave me instructions how he got the reheat load, but unfortunately i did not get it. Do you have rules of thumb on reheat loads for decreasing humidities?

That 7C is air off temp, not evaporating temp. The evaporating temp will need to be somewhat lower.

The point here is that 7C air @ 100%RH, when warmed to 15C will be @ 60%RH.

alfred_yagne
03-03-2010, 07:45 AM
[quote=Gary;179868]That 7C is air off temp, not evaporating temp. The evaporating temp will need to be somewhat lower.

What is air off temp? Sorry, I am not very familiar with the term. Thnaks

Gary
03-03-2010, 02:42 PM
[quote=Gary;179868]That 7C is air off temp, not evaporating temp. The evaporating temp will need to be somewhat lower.

What is air off temp? Sorry, I am not very familiar with the term. Thnaks

Air off temp is the temperature of the air as it leaves the evaporator coil.

rk tigere
04-03-2010, 09:56 AM
:)Alfred if you can be able to get your hands on the Ashrae Fundamentals handbook,it will take you 20 pages & a few hours to understand how the Psychometric chart works,u'll see a chapter entitled Psychometrics,it helped me a lot pal:)...thanks Gary for the link ,Nice chart....:D

SHERYCOOL
05-03-2010, 02:22 AM
Reheat does not remove moisture. It lengthens the run time for the compressor, giving the system more time to get the moisture removal job done.

Given enough run time, how low the humidity goes depends upon the temperature of the coil. The lower the coil temperature, the more moisture is removed, and the lower the humidity goes.

If the coil is not cold enough to achieve the humidity level you want, then reheat accomplishes nothing.


That 7C is air off temp, not evaporating temp. The evaporating temp will need to be somewhat lower.

The point here is that 7C air @ 100%RH, when warmed to 15C will be @ 60%RH.



Gary : y say : Reheat does not remove moisture.

but here y say : The point here is that 7C air @ 100%RH, when warmed to 15C will be @ 60%RH

i say Reduce not remove.

Gary
05-03-2010, 03:01 AM
Gary : y say : Reheat does not remove moisture.

but here y say : The point here is that 7C air @ 100%RH, when warmed to 15C will be @ 60%RH

i say Reduce not remove.

As the air moves through the evaporator coil its temperature is reduced/RH increased, until it is holding more moisture than it is capable of holding (over 100%RH).

Excess moisture then condenses out of the air until the RH is held at 100%.

If the leaving air temp is 7C and its RH is 100%, then when the temperature of that air is raised to 15C its RH will be 60%.

If the system runs long enough for all of the room air to flow through the coil, then all of the room air will be 15C@60%RH.

If needed, reheat can be used to extend the run time, so that the system runs long enough to achieve the humidity target.

Hoping this explanation is clearer.

Victorman
07-03-2010, 01:18 AM
Hi Carlo
It seems to me that what you need is a copy of
" The Properties of Mixtures of Air and Saturated Water
Vapor " at the proper Barometric Pressure.
You are dealing with Specific Humidity , If the
air is leaving the coil at 7C ( 52.5 grains of Moisture
per pound of Dry Air ) and you reheat it to 15C you will
have 52.5 Grains of Moisture per pound of Dry Air . It
is the same amount of moisture in the air with different
Relative Humidity.

I would like to see your tests , if I may.
Another thing is ,if you are working with temps. of
-28.5C ( for example ) you will have approx. 2.914 grains of moisture Lb/Dry Air . The moisture in the air
entering the coil has to be below the above number otherwise the excess moisture will end up on the coil.

Greetings
Victorman

anna.savvy
07-03-2010, 09:22 AM
Hello, everyone i was following the discussion meanwhile i cleared several doubts of mine too...

thanks everyone...

alfred_yagne
08-03-2010, 10:04 AM
Hi Gary,

I am really interested on your discussion about reheat. Ammmm, may I ask then any advise of what evaporating temperature to use or any technique how to size it? if for example air temp off is 7C from RT is 15 C with 60% rh? I currently use 1C as my evaporating temperature. Any reaction on what I selected? I somehow got your point that it will matter with the aircooler manufacturer. But how can I be more precise on selecting evaporating temperature and how will i know if i need reheat. I would also like to ask what is more effecient. Electric reheat or hotgas reheat. I really appreciate your posts gary :):)