Question: From 80°C to 5°C for a jelly factory

[ysh1317]

Hi all, it's me again, from Indonesia :)
Here is the case:
This is a request from a local jelly factory, the owner wants to build a cooling system for his jelly production. The jelly material will run on a 3 meter long conveyor belt, and will go through a water-cooled chiller in the middle. The jellies temperature will be 80°C when starting at the beginning of the conveyor belt, and needs to be 5°C when coming out of the chiller.
So my question is, how to calculate how much compressor power do I need to cool it? Also, (this what confuse me more) how to calculate how fast the conveyor belt should run to properly cool those jellies to 5°C?

Sorry if I didn't put the question into words clear enough.
Thank you in advance for any reply or input :)

[frank]

More importantly, we would need to know the mass of jelly to be cooled and the time span for cooling, i.e. how much jelly and how long to cool it?

[Magoo]

First do a sample trial of actual product with a probe in it and one monitoring the space temperature around the sample, put it in a chiller and monitor temp change with different space temps etc., with the gelletene content of jelly, then you can start to establish the cooling enviroment required and retention time through a cooling tunnel.
As Frank suggested, product volumes/weight/mass is missing from your initial posting

[ysh1317]

Thank you all for the replies so far, I will get back to the owner to gather more data.

[ysh1317]

The tray that sit on the conveyor belt would be 50x50cm in dimension and the weight of the product itself is 50 gram.
I wonder if there is a calculation formula for a starting point for the trial (to save time)?

Thanks :)

[Tesla]

Hi ysh1317
Watt you need is Q=M SHC TD and 1 Watt = 1 Joule per second.
0.05kG * 4.2kJ/kGK * 75K = 15.75kJ
To cool in 1 minute 15.75kJ/(60seconds) = 0.2625kW plus losses from tray and heat transfer losses.
You can play with the values here - SHC of jelly will be less than water possibly around 3kJ/kGK and it would be prudent to do a seperate calculation for the tray as that will need cooling too and will likely be a lot more than the jelly. You can change the time as well. This will only give you the product load and it will depend on what method of heat transfer you are intending to use like are the trays coming into direct contact with the chilled water and if so the chilled water would need to be brine of glycol as the water will need to be below 0°C to achieve 5°C of jelly product. 50 Grams seems to be very small too. Personally I would use two stages of cooling - first stage would use a dry cooler to reduce the product to around 40 or 50°C then use a secondary chilled water loop to reduce product to 20°C, then primary chilled water loop for the final product temp of 5°C. Once you have calculated the product load then you could size the refrigeration system with further calculation. I know this is not exactly right but should give you an idea and start in the right direction. You could search this forum for more calculation examples then work out what you think you need and post it on this thread

[ysh1317]

@Tesla thank you for the great info, surely give me a headstart... much appreciate it!

[mad fridgie]

start at the start, how much jelly does your client make an hour (mass) then per batch. if your conveyer is 3meter, then production rate will determine conveyer speed.
Next cooling rate, this not so much determined by mass but more the physical shape of the jelly product, does it cover the whole tray like a pancake (very thin) or are there lumps of products, the first will cool very quickly the second not so quickly (with same system)