Cooling down a storage tank with commodity 1

[daviddor]

Hii everyone,
since the last post was gone, don't know why actually, i have some more issues so i will ask here.
So we have a storage tank with a ton of fruits, i know all the data inside and outside including materials . Besides the respiration heat, that i know already, if i want to keep the tank in let's say 10c , the outside temperature and the fruits temperature are 30c , obviously the fruits temperature will change with the time. How much power do i need to keep it 10c for 2 hours. Note that i don't want to change the temperature of the fruits completely but just keep the environment at a certain temperature .
Thank you very much

 

[LittleInch]

How are the fruits stored?

If open in trays and you can get air flow then blow our circulate air at 10C.

Power will be dependant on your A/C system.

If it's all loose packed then you will need to add a jacket, maybe liquid which circulates under some insulation.

All a bit vague.

Don't know why the last one went.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Latexman]

I don't see why the last one was deleted either. daviddor, did you delete the first post in that thread? That would take down the whole thread. If not, I think we should ask Admin for it to be resurrected.

Good Luck,
Latexman

 

[daviddor]

Didn't delete anything , but never mind.

The cooling system is going to be like blowing cool air to the inside , with power that we choose .
The fruits i would say , lets take the worst case scenario, going to be stored as a big pile in the tank. Its a small tank something like 1 [m3] so one ton inside gonna fill most of the volume , and some left for air . The idea is to keep the air/atmosphere inside at 10c temperature, like in this article:

i added the file also.

So my question is how to calculate an approximate energy/power that i need for that. Thank you very much

 

[LittleInch]

daviddor,

Sorry I misread your original post and now it doesn't make sense.

You say "I want to KEEP the tank at 10C" [my emphasis], for 2 hours, but then say the ambient and fruit temp are at 30C.

But then you don't say how much temp drop the fruits should be be over time.

What is 10C? Clearly not the "fruits".

I'm getting very confused here and maybe that's me, but I really don't think you've really thought about this and what it is exactly you're trying to find out.

In the attachment what is meant by "pre coooling"? Doesn't make sense.

I think we've tried to draw out form you what it is you're trying to do and I for one just don't understand.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[daviddor]

Yeah, Right, I didn't talk about The temperature drop in the fruits, but just to keep the air inside at 10c. The ambient 30c is from outside the tank and the initial temperature of the fruits that come from outside. The conductivity of the tank Walls I have already. So it's not The typical removement of field heat ( changing The complete temperature of the fruits). But more an initial treatment of putting them in a cool atmosphere,
Thank you for your time

 

[LittleInch]

You have indeterminate things here so your calculation will also be indeterminate.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[georgeverghese]

You have 2 cases /steps for heat removal here:
Case 1 : Reducing the temp of the insulated tank from initial 30degC to 10degC
Case 2: Maintaining the temp in the tank at 10degC

For case 2, heat removal load / hr is merely the sum of heat leak from ambient per hour + heat of respiration of the fruit per hour
For case 1, you need to state the time permitted to chill the tank material, insulation and fruit from 30degC to 10degC - for this step, coolant air will have to be less than 10degC, else this step will take forever. Say let coolant air be 5degC.
Coolant air may be raised at 10degC for step 2.

Work out heat load for these 2 steps.

 

[daviddor]

Hii

LittleInch, if you need any more details about the system please tell me which ones.


georgeverghese. thank you for that. lets talk about case 2 for a moment, the heat leak and the respiration i know how to determine , but then i was thinking, one moment , most of the volume in the tank contains fruits with an initial temperature of 30c , they must "absorb" some of the coolness in the air. since heat moves from high temperature to low temperature then , it comes from outside the tank , but also moves from lets say the center of the tank(fruits at 30c) to the air inside (10c). So there is more heat added to the air(10c) from the inside . Now another way to look at that is , What would be the change of the internal energy of the fruits after 2 hours of staying inside the tank at 10c . This change is the heat added to the air from inside. Now i don't know yet how to determine this.

 

[Snickster]

I was reading your design problem and what I get from it is that you are starting with a tank in a room with fruit inside of the tank and everything is at equiullibrium temperature of 30C. Now you want to cool down the room to 10C for 2 hours. So basically it is a HVAC problem where you have internal heat gain to the room from the tank with fruit and you have heat gain through the walls and windows of the room duringg summer like a HVAC design system, and maybe also other internal heat gain from motors, electrical equipment, people, etc.

You need to size your HVAC system for all the loads present, then select a specific HVAC unit which will be rated for a specific Tons of refrigeration and power consumption.

However things are not totally straight forward:

Since you are starting off at 30C (86F) inside the room and tank you will need to pull down the temperature in the room over a reasonable time to 10C while heat is being transfered from tank, from outside and from equipment,etc. inside the room. The larger the unit the faster the trmperature will drop from 30C to 10C desired.

In this case in the beginning there will be a greater heat load to pull the temperature down. I think ASHRAE has some guides as to sizing HVAC equipment for pull down loads.

Also you will have to assume a heat flow mechanism from tank to room. For the worse case would be room instantly drops to 10C while tank is still at 30C so this would give you the maximum possible heat transfer from the tank. To be conservative you may just want to add this entire maximum load to you cooling load requirements for sizing your HVAC unit.

 

[FMJalink]

Hi daviddor,

I am afraid it will be difficult to obtain here the requested detail information, As such things as mangoes treatment and storage with larger operations, is quite specialized.

You are talking about putting mango fruits in a 1 m3 tank. That would something else than expected with fruits packed each with its own protection packed in boxes for transport and storage. As you will know mangoes are quite delicate an do not like too low temperatures.

Generally speaking the mango fruit can be seen as a sphere cooled from the outside. After the initial cooling on the outside the heat release rate drops, as the temperature gradient reduces. Problem is the heat transfer coefficient on the outside.

To have an average temperature of 10 ○C in the room, There has to be a difference in outlet temperature and inlet temperature in order to absolvdpe the heat given by the fruits during cooling. One should keep sufficient distance from the minimally allowed temperature, as the mangoes become worthless after having been too long in a too cool place.

The design basically comes down to knowing the heat release rate in time from the fruits, approximate mean temperature, the acceptable minimum temperature and the air flow rate.

I hope this hd!ps somewhat,
FMJalink

 

[daviddor]

Hii,

Snickster: thank you for your time. one thing you got wrong i think. The room is in 30c,the tank is inside the room , the air in the tank 10c, the fruits in the tank are at 30c(initially)

FMJalink: thank you for your time. Its not going to be only for mangoes, it is just an example, anyway the fruits part is less my business . i just want to determine the energy that is needed for that, calculating the mean tempeture of the mangoes after two hours in 10c can be good, do you know how to do it?

Maybe an approximate calculation would be for one mango with this formula:

 

[pierreick]

Hi,
Consider the resource attached:

https://nzifst.org.nz/resources/unitoperations/httrtheory4.htm

Hope this is helping you.
Pierre

 

[FMJalink]

Hi pierreick,

Yes, it will be a formula as you mention. Formulas will give an indication only, as they idealize the situation. The fruits will not be all of the same size and will have different heat conduction coefficient, etc.

Personally, I would try following approach to obtain data that I could use to evaluate the cooling process. Pending on the size the fruits, they will get cooled throughout or not during the pre-cooling. The fruits will probably be placed in cooled storage after pre-cooling. There fruits could be further cooled in nearly stagnant conditions.

I would try to couple theoretical approach with test results. For the approach I would use Excel with VBA software.
Theoretically you have three parts in the cooling process.

First the cold air flow entering the tank, most likely on the bottom, leaving heated up on the top. This air flow along the fruits heats up. The heating rate is to be coupled with the heat release from the fruits on elevation y from the bottom.
The fruits are stapled on each other in the tank. This leaves a free cross section Afree. The air flows through the net free cross section. This gives an average air velocity v. Per elevation y, a heat balance between air and fruit can be set up. First approach this could be left out and entrance temperature could be used.

Secondly would be the heat transfer between cooling air and outside of the fruit. Here standard approaches as Nu = f(Nu, Re, mu) for spheres could be used.

Third would be the heat conduction inside the fruit, that could be approached with e.g. 1/alfa *dT/dt = 1/r^2*d/dr(r^2*dT/dr). The outside temperature and heat transfer rate should be coupled with that on the outside.
Using here a spreadsheet with VBA would allow all kinds of changes in the process as higher temperature and lower air velocity after pre-cooling, etc.

Also the heat release can be easily investigated, also on different elevations and for different tank dimensions.
After the set up with formulas the approach should be tested with an experiment or data from a published experiment. There will be other people that have done this already. With addition of parameters and adjusting of coefficients, this should give an instrument to get more insight in the design.

Kind regards,
FMJalink

 

[daviddor]

Hi pierreick,
Wow amazing, this is exactly what i needed, even the example inside is 2 hours of cooling. i guess now i will make several calculations . The best case scenario would be for a temperature of a normal size mango, then assuming this temperature for all of them . The worst case scenario would be treat 1 ton as one big sphere since they are all in one big pile without much space between them. And then some cases in between those extremes .
After coming up with the temperature change, easy to calculate the change in the internal energy( which is the heat) adding the heat respiration and the leak from the outside will sum up the approximate power needed.

Thank you very much again , i will start processing

 

[daviddor]

Hi pierreick,
I tried to work with the link you added , assuming a sphere with diameter of 0.08m for one mango , something is wrong when i am trying to use the charts i will add may calculation maybe you see something wrong thank you

 

[pierreick]

Hi,
Review the example 6-10 from the link above
Pierre

 

[daviddor]

Thank you very much , i guess this is exactly my case , it just my h of convection is somehow 300[w/m^2*s] and then my 1/biot is going to zero which is ok, but it doesn't fit the x axis (

there is not much gap there when the biot is very big

I calculated the h(convection) with the long formula two posts above ...

 

[LittleInch]

Well a 2m/s air velocity is pretty high.

I didn't see an answer in the post so not sure what you were trying to calcualte.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[daviddor]

Really? I guess a normal fan coming out from a cooling system. Trying to calculate the temperature of the fruits after 2 hours in 10c