Air cooler for high humidity room.

[longminh]

Please help me! I made a room for keep Nobashi soak 11x20x5m, operated in +5 degree. I choice Guentner cooler 2 set. See below for specifications.
-Model: S-GHK 045.2E/110-WLD/16P.E
-Cooling area:27.4 m2
-Cooling fan: 0.28kWx380Vx50Hzx1set
-Defrost: water spray
-Air flow: 4071 m3/h
-Feed NH3: circulation pump
Now allway make ice in cooler. temp can not down below +10 degree. Humidity >85%.
Please show me what happend? And how to solve this problem.
Thanks and best regards.

[mbc]

what is the suction line pressure ,(separator )

[josei]

Longminh,

Welcome to R-E.

Your question does not say if product is fresh , on ice? (the temp of the room, not frozen, right?).With regard to the size of the room, temperature and r.h. typical of Vietnam, there is a high load of infiltration, then hardly the cooling load is met with your 2 evaporators Guntner each with 5.5 kW of capacity , for sure requiring more NH3 evap. to achieve the conditions of 0.5 ° C / 90% rh for the product Nobayashi shrimp as usualy required by customer .

[Peter_1]

What's Nobashi soak? Actual load (means temp in/out, kg's, time ...), temperature at low side, ...
2 evaporators of each 4.000 m³/h in a room of 11 x 20 x 5 m?
How did you selected these 2 evaporators? What parameters did you use to calculate those?
We need a lot more data to help you

[josei]

Longminh,

Your question does not say if product is fresh , on ice ? (looking the temp of the room,is not frozen, right?).With regard to the size of the room, temperature and r.h. typical of Vietnam, there is a high load of infiltration, then hardly the cooling load is met with your 2 evaporators Guntner each with 5.5 kW of capacity , for sure requiring more NH3 evap. to achieve the conditions of 0.5 ° C / 90% rh for the product Nobayashi shrimp as usualy required by customer .

This was updated today

[longminh]

Thank for all. More some data for this room:
+ Product Shrimp fresh
+ Daily Shrimp come in an out is: min 24 ton
+ Humity in: 94%
+ Load temp: 15oC (Shrimp)
+ Final Temp: 5oC (Shrimp)
+ Door open: min 120 times/ day; one time 5 to 10 min (estimate)
+People inside: 8
+ Insulation: 100mm K=0.018kcal/mhC

[Peter_1]

I then repeat my questions....what's the actual load you've used to calculate your 2 Güntner evaporators?
What's the condition outside when opening the doors?

You need max +/-70 kW, average of 50 kW. You now have +/- 10 kW !!!
We should appreciate it if you keep posting with information and upgrades

[longminh]

I calculate with out side temp+30oC,total heat load: 36.3kW.
And one Güntner evaporators capacity: 18.3kW
I used 2 evaporators, i think enough capacity.

[mad fridgie]

Heat load is OK for cooling over 24 hr period (with allowance for defrost)
How ever the evap coils are just incorrect,
You need approx 4-5 times more airflow and the coils to match

[Peter_1]

Are you sure about your capacity? Look once here http://www.guentner.ru/fileadmin/lit...heet_DE_EN.pdf
Think you possible made a fault in your selection

[josei]

I calculate with out side temp+30oC,total heat load: 36.3kW.
And one Güntner evaporators capacity: 18.3kW
I used 2 evaporators, i think enough capacity.

Hello friends of Lonminh:

I'm in the same load range as you show, my figures (kWp) calculated according to the ASHRAE handbook 2010, as follow:

Load	kWp	%	kWs
Transmission	3.7	9%	3.7
Product	9.4	22%	9.4
Infiltration	21.9	51%	2.2
Ilumin	0.7	2%	0.7
People	1.1	3%	0.3
Evap.fan mot	3.2	7%	1.6
Safety.Factor	3.2	7%	1.8
Total Load	43.2	100%	19.7

Some comments to your project:

1. Your packaging should organize separately processing and storage operations. The storage area must be unique for processed product.
2. As a result of the above, the load should (look at kWs above) be reduced considerably, because infiltration (down from 51% to about 10% when the door opening is near nule)
3. NH3 evaporation temperature should be 0.5 to 1 ° C, preventing freezing of the coil and achieve the highest heat transfer for 24 hours a day, without needing to defrost.Note that the dif.temp no more than 5°C to get 90-95% rh that shrimp requires
4. If you process only during the day, can improve the cooling of the product during night when the load would be only 49% (enough operating one evaporator alone).
5. The actual load of product is taken by the ice provided in the process, so this load is oversized (I would take only 3 kW instead of 9.4 kW).

[Peter_1]

Josei, if you take a DT of only 5K, then your RH will indeed high but the capacity of the evaporators will be seriously reduced and they're even now a lot too small.
Longminh, did you checked my Guntenr link? I see a capacity of +/- 5.4 kW for 8DT at an average RH of 85%.
You're working at 95%, so a huge load of latent heat and perhaps even a smaller DT.
18 kW and 4.700 m³/h airflow isn't possible.
I take - averages- the surface in m² x 1.75 to 2.0 is your kW and airflow x1 (bigger coils) to 2 (smaller coils) is also kW.
So at 18 kW you need at least 100 m² surface and 9000 m³/h to 12.000 airflow. You have 4.750 m³/h and +/- 27m². So something must be wrong.

[longminh]

Thanks Josei and Peter_1.
I inspected on the side. And I think main reason temp not down because the factory used as a processing room, not storage. They made chemical and shrimp mixture inside room, more heat load.
And more information for this cooler system: NH3 system with circulation pump, low receiver -31oC (also used for cold storage -25oC).
And I checked deforst, with 20min water spray can not remove all ice inside cooler coil.

[mad fridgie]

You can not expect high humidity if you are using the NH3 unregulated, no wonder the evap is a ball of ice.

[longminh]

Mad fridgie, I understood you. I used one regulation valve to control liquid supply (Danfoss valve, open 1/4 circle). I think this control mode is OK.

[mad fridgie]

Are you feeding -31C NH3 liquid to the evaporators, from you pump system?

[Peter_1]

Longminh, have you already checked the capacity in the link I gave you?
I would add to MF's post..injecting NH3 liquid of -32°C in a coil in a process room which is at 2 or 5°C?
Is this correct?

[longminh]

Peter_1, I already check your link, but I have a specification document from Guentner. Exactly capacity for this cooler model.
And I upload my design for this cooler in the link below. please check and help me. Thanks.
http://www.mediafire.com/?ya4azcdpl2dui32

[Peter_1]

Can you scan that document and post it please?

[josei]

Top capacity given by Evaporator GHK 045.2E/110 at Evap. Temp. of -15.5 C (see PDF attached). Any lower temp. (i.e. -16 C and less) is rejected by Guntner´s software with this advise:

TopCapacByHighVelocNH3.png

With Ammonia T.E. = -31°K is mandatory to use a EPR valve to control it. I suggest u consult to Danfoss Rep.

[Peter_1]

Josei, you even gave in 80% and it's in fact 93%. How much sensible heat will finally cool the room due to the huge latent heat of the air? Very little.
You see the capacity of 14,5 kW, far away from the 20 to 25 kW he needs but that's total capacity, in this case mostly 'latent capacity'
And a DT of 20K, this is something not done. Frost problems, very low COP,

[mad fridgie]

Nice info Josei, but I have agree with Peter, the load is latent, not sensible and this is a high humidity room, not a drying room.
You can make most heat exchangers read what you want ,by changing the numbers.
We still do not know if the OP is feeding the evap with -31C liquid from the LT pump system.
he simply needs more airflow, more surface area, high pressure/temp liquid feed, EPR control, he wants to tie into the LT system. If he is lucky the system is 2 stage, he could look at teeing into the second stage.

Longminh, draw a picture of the whole system. Then we can help. Sometimes we have to bite the bullet and just admit that we have made a mistake, I would think that there is no one on this site who has not made a mistake, it seems to be just part of the job!

[josei]

... my figures (kWp) calculated according to the ASHRAE handbook 2010, are as follow:

Load	kWp	%	kWs
Transmission	3.7	9%	3.7
Product	9.4	22%	9.4
Infiltration	21.9	51%	2.2
Ilumin	0.7	2%	0.7
People	1.1	3%	0.3
Evap.fan mot	3.2	7%	1.6
Safety.Factor	3.2	7%	1.8
Total Load	43.2	100%	19.7

...
1. Your packaging should organize separately processing and storage operations. The storage area must be unique for processed product.
2. As a result of the above, the load should (look at kWs above) be reduced considerably, because infiltration (down from 51% to about 10% when the door opening is near nule)
3. NH3 evaporation temperature should be 0.5 to 1 ° C, preventing freezing of the coil and achieve the highest heat transfer for 24 hours a day, without needing to defrost.Note that the dif.temp no more than 5°C to get 90-95% rh that shrimp requires
4. If you process only during the day, can improve the cooling of the product during night when the load would be only 49% (enough operating one evaporator alone).
5. The actual load of product is taken by the ice provided in the process, so this load is oversized (I would take only 3 kW instead of 9.4 kW).

According to what I saw in the design of Longminh (LM) (which is his room´s layout attached) indicates data and operating conditions of "now".

ShrimpJuiceHotRoomStory.pngSo then I have tried to show (using Guntner) what could be achieved with the his evaporators installed, the result showed what Peter and Mad commented: evaporators required (according to my calculation of loads) has a very high component of infiltration = 21.9 kW outdoor air latent load [see psychrometric data and additional attached chart].
My recommendations in my previous post (above) only point to "amend" what is done: to avoid a lot of heat load from the room by changing its operation.

What LM could do?: increase the number of evaporators and operate it at 0.5 C TE; (The weight loss of shrimp is very expensive, but that's another discussion for later).

[Peter_1]

Josei, very informative posts you made.

[longminh]

Many thank for your help,josei and every people.
I already view your calculate. Now, I had some new inform. Factory will install more some air condition cooler around this room. It's meaning air income room will be only 20oC.
I also calculate again. And i think 2 cooler enough for this room, no need more numbers.
Please see attach file: http://www.mediafire.com/?s58tgllnrbejnzj

[mad fridgie]

-31 liquid NH3 entering, high humidity??????????????????????????

[josei]

...
I already view your calculate. Now, I had some new inform. Factory will install more some air condition cooler around this room. It's meaning air income room will be only 20oC.

Precool air proposed is a step towards a better condition from one side (room) of the system. Another is the refrig. side (ammonia) where the condition should improve as well.

Who is your refrigeration contractor?; you must raise the temperature of the refrigerant. I'm with Mad who said "evaporator Will Be a ball of ice"...always...but too low E.T..???

I also found some data that are not correct, according to what you have designed,(see below). Figures are: Air inlet temp. and pump feed rate. Do you agree?

[longminh]

Dear all.

Last week I try to modify regulation valve for 2 cooler. And inspection, room temp lowest only 8oC.

And I decide must add more 1 cooler for this room if it want to operate in 0->5oC.

Many thanks for all help.