(Noob Need Help) Simple Refrigeration Design

[eSwYeL]

Hello, I want to make a simple chillier but got no knowledge on the engineering side. (Please refer to my humble drawing):-

- The room temp is 38c
- The evaporator unit is just a simple cooper tube that i coil together
- The Condenser also made from simple coiled cooper tube.
- The Condenser is submerged in high flow rate water to chill.
- I would like to turn 4x6ft off water into ice block.

My Questions is:-

1) What is the right length and size of the cooper tube for the condenser and evaporator side?
2) What type of TXV do i use for 1 hp Compressor?

Please help me with your comment and improvement that i can make to make the simple chillier more efficient.

Thank you in advance.

[NoNickName]

I'm under the impression that you reversed the cycle.
Despite that, in order to answer your question, we need to know the heat load to be removed (or the room size and appliances), the expected time for cooling and the final desired temperature in the room.

[eSwYeL]

I'm under the impression that you reversed the cycle.
Despite that, in order to answer your question, we need to know the heat load to be removed (or the room size and appliances), the expected time for cooling and the final desired temperature in the room.

Thank you for the fast replies, sorry because i got the cycle wrong :P . The purpose of the unit is to make an ice cube inside a 4ft x 4ft or 4ft x 6ft container which will be made of fiberglass and polystyrene insulation. The maximum room temperature can reach up to 38 degree Celsius. The expected time won’t matter much but faster is better :p. and the desire temperature are minus 10c i guess. Thanks again

[NoNickName]

A cube 4x4x4 ft contains (in metric units) 1.8 m3 of water (1800 liters).
In order to free 1800 liters @ 38°C down to -10°C there are three cycles:
1) sensible cooling from 38°C to 0°C
2) latent cooling (freezing)
3) sensible cooling from 0°C to -10°C

1) 38 x 1 kcal/Kg x 1800 Kg = 68400 kcal = 79,5 kWh
2) 79 kcal/Kg x 1800 = 142200 kcal = 165 kWh
3) 10 x 0.47 kcal/Kg x 1800Kg = 8460 kcal = 10 kWh

Your total energy to be removed is 254.5 kWh.
A 1HP compressor is capable of delivering (at best) 2kW of cooling capacity at te = -5°C. Check with manufacturer data.
The time it will take for the 64ft3 to freeze is 254.5kWh / 2kW = 127 hours (5 days).

Your homework is to find the length of copper required to transfer 2kW with a minimum TD of 10K between refrigerant and water, starting from the heat transfer capability of refrigeration copper piping.

[eSwYeL]

A cube 4x4x4 ft contains (in metric units) 1.8 m3 of water (1800 liters).
In order to free 1800 liters @ 38°C down to -10°C there are three cycles:
1) sensible cooling from 38°C to 0°C
2) latent cooling (freezing)
3) sensible cooling from 0°C to -10°C

1) 38 x 1 kcal/Kg x 1800 Kg = 68400 kcal = 79,5 kWh
2) 79 kcal/Kg x 1800 = 142200 kcal = 165 kWh
3) 10 x 0.47 kcal/Kg x 1800Kg = 8460 kcal = 10 kWh

Your total energy to be removed is 254.5 kWh.
A 1HP compressor is capable of delivering (at best) 2kW of cooling capacity at te = -5°C. Check with manufacturer data.
The time it will take for the 64ft3 to freeze is 254.5kWh / 2kW = 127 hours (5 days).

Your homework is to find the length of copper required to transfer 2kW with a minimum TD of 10K between refrigerant and water, starting from the heat transfer capability of refrigeration copper piping.


Thanks for the info, but how do i calculate the cooper length and size. Please help me or point me to some clue on how to do it. FYI i don’t have any knowledge in this field. thanks again :P

[NoNickName]

Ok. Heat transfer coefficient of refrigeration grade copper pipe is around 0.081 kW/m2K.
Your TD between refrigerant and water is decreasing along the cooling process, but let's say it's around 10K
Your cooling capacity is 2kW.
10 / 2 * 0.081 = 2.46 m2 of exchange surface required.
I'm sure you know how to calculate the surface of a cilinder.
eg: a 16mm wide pipe has a circumference of 50.2mm = 0.050 m.
In order to reach 2.46 m2 you need 2.46 / 0.05 = 49 meters of submersed coiled pipe.
Increase diameter to reduce length of coiled pipe and associated pressure drop.

EDIT: once the block is frozen, the pipe is locked within the cube. Hope this is not an issue for you.

[eSwYeL]

Ok. Heat transfer coefficient of refrigeration grade copper pipe is around 0.081 kW/m2K.
Your TD between refrigerant and water is decreasing along the cooling process, but let's say it's around 10K
Your cooling capacity is 2kW.
10 / 2 * 0.081 = 2.46 m2 of exchange surface required.
I'm sure you know how to calculate the surface of a cilinder.
eg: a 16mm wide pipe has a circumference of 50.2mm = 0.050 m.
In order to reach 2.46 m2 you need 2.46 / 0.05 = 49 meters of submersed coiled pipe.
Increase diameter to reduce length of coiled pipe and associated pressure drop.

EDIT: once the block is frozen, the pipe is locked within the cube. Hope this is not an issue for you.


;)
Thanks Friend... So for 16mm wide pipe i need to have 49mtr of cooper pipe submerge underwater. What about the hot side/condenser side? How much pipe do i need? Is it the same as the one on the cool side? What type of Thermal expansion valve do i need and can you suggest any improvement that i need to make it more efficient? Do i need a dryer or anything else? Thank you....
FYI... I’m thinking of using the ice for ice sculpturing, since here in Sabah we don’t have winter :p. As for the pipe getting stuck inside the ice cube, I’m going to coil the tube on the outer rim of the box then when the ice formed i just cut it with a saw.

[NoNickName]

No, it's not the same: you have to add the power of the motor.
Now, I'm not going to help you any further unless I'm payed. I gave you plenty of hints.
And no, if you don't submerge completely the pipe, then you have to increase the length accordingly to the actual exposed surface of the pipe.

[eSwYeL]

No, it's not the same: you have to add the power of the motor.
Now, I'm not going to help you any further unless I'm payed. I gave you plenty of hints.
And no, if you don't submerge completely the pipe, then you have to increase the length accordingly to the actual exposed surface of the pipe.

Thanks friend, you help me a lot already. Sorry because i can’t pay you since I’m in Kota kinabalu. I’m going to do some more research to figure out how it works.

[Rayin]

Dear NoNickName
Thanks for your inputs. Is the evaporator design so simple? Is it just theory or can we do it in real life situation? How about designing a fin and tube evaporator with forced air cooling. Could you please give me some guide lines.!! Also may I know how to calculate the heat load of open type display coolers?

Thanks in advance
With regards
Rayin

[NoNickName]

Dear NoNickName
Thanks for your inputs. Is the evaporator design so simple?

Is it? Good for you.



Is it just theory or can we do it in real life situation?

It depends on how far away your real life situation is from the theoretical one.



How about designing a fin and tube evaporator with forced air cooling.

How about it?



Also may I know how to calculate the heat load of open type display coolers?

I think you should give yourself an education on thermodynamics first.

[Rayin]

Dear NoNickName,
Thank you for your prompt reply.
I feel it easy your calculation because, it seems that you did not consider the Phase cahnge process of refrigerant in the evaporator coil. As we know that the phase of refrigerant at the exit of evaporator will not be the same as that at the entrance. So is it correct to consider the heat transfer coefficient as a constant for the entire coil?
I know how to do heat load calculations for a closed unit having 4 walls, floor and top like walk-in cold room, rerigerator etc. But how to do it for a space that does not have a wall on one side or do not have a top, like super market freezers or Grab and Go type coolers etc.
Thanking you once again
With regards
Rayin

[NoNickName]

My calculation is simply a heat load calculation. It's correct, as long as you know what you are doing with it.

The heat transfer coefficient of the coil depends on a lot of different things, including grade of cleanliness and amount and distribution of frost. For an ideal coil, yes, the heat transfer coefficient is constant along a 2-phase transformation.

Calculating the heat load for a cabinet, it's exactly the same than for a 4-wall room, replacing the heat transfer coefficient of a wall, with that of an air curtain in front of the cabinet, and the boundary temperature with the temperature of the aisle.

[Rayin]

Dear NoNickName
Thank you very much indeed for your reply.
Regards
Rayin

[eSwYeL]

Still havent figure out how long is the hot and cold side cooper or brass tubing i should use. help plz... tq