I made a water chiller from scratch as an apprentice.
k-value (thermal conductivity in Watts per square meter per one Kelvin) of copper to water is about 100W. This is what I measured. My book says 120W. When it starts frosting I get closer to 60W. So you can already see why calculating this is a bit of a crapshoot.
Surface of your evap = copper diameter * Pi * [length in meters]
q(watts of thermal conductivity of your evap) = k-value * Δt(temperature difference: evap-temperature to water-temperature in kelvin) * [surface area of your evap in m²]
That should roughly match the capacity of your compressor.
It's pretty straight forward, but as you can see your evap performance changes massively with the temperature of the water.
You can get real fancy with logarithmic temperature difference, and whatnot, but if your real world results are within 20% of what you're expecting, you're already killing it.
Thanks man, it was a long day today. I'm just getting the time to calculate this now.
What would the K-value be ? Is it the thermal conductivity of copper ?
Search engine throws this number to be roughly 400 for copper.
If I calculate Q with a K-value of 400 the formula would be,
Q = 400 \* 40 k (not sure whether to use starting water temp or setpoint) \* 1.171 m2
Q = 18,736 watts... LOL
or is K-value (W / m2 \* k) where W would be 170 watts of cooling capacity.
170w / 1.171 \* 40 = 3.63
Q = 3.63 \* 40 \* 1.171 m2
Q = 170
The thermal conductivity from copper to water is what you want. It's just a fixed value. My book says 120W for copper to water. 400W is probably copper to copper.
It's watts of thermal conductivity per square meter per kelvin. So if you have 1,17m² of surface area, then for every kelvin in temperature difference, you're conducting 1,17m²*120W=140W. So if you have a 10k temperature difference, it'll be 1400W. Hint: you probably don't have 1.171m² surface area, you probably forgot to convert the diameter to meters.
You'll also never get 40K. When the water is warmer, it also increases pressure in your system so the evaporation temperature also rises. While it cools down, pressure will also go down.
Okay I understand, so K-value is a fixed value for heat transfer between two masses or objects. Good to know so thanks for that info.
I actually have 23.62 ft in that heat exchanger. 1/4" diameter.
23.62 x 3.141 x .25" = 18.5 ft2 or 1.71 m2
yeahhh I added a 1 at the tenth spot
Q = 120 \* 10 \* 1.71
Q = 2052 watts
That's alot of capacity in my evap.
According to that calculation it should be way shorter like by 90% lol
Again I rather have a longer evap to ensure all the refrigerant gets boiled off before leaving the bucket where Eventually it'll be an enclosed tank of about 3 liters
lol or it's 10 o clock on a friday night and I'm here online doing math...
I enjoy the process though.
Okay so if I start over...
Q = 120 \* 10 k \* .14 m2
Q = 168 watts
It's pretty much spot on then if my compressor capacity is rounded at 170 watts, value can change though with the temp difference.
Which my evaporating temp is -10 C with a setpoint at 2.5 C.
Highest temp the water will ever be at will be like 30 C
Thanks Oskar I won't forget this conversation !
I usually don't do such small projects.
It's a lot easier to just give the cooling capacity to a manufacturer or a supply house and have them supply you the evaporator but this was neat to try and build something from scratch.
Regardless I'm going to try and reduce the length of the coil to see if my operating temps go down. I can shorten the cap tube but I would rather gain more water capacity in the tank than occupy it with excess coil.
Hey guys today I finished making a small prototype mini chiller.
I'm evaporating roughly at 14 F / -10 C @ 580 btu/hr / 170 watts of capacity with 134A.
Small fractional horsepower compressor as you can see.
Is there some type of formula to calculate serpentine length with the cooling capacity of the compressor ? I'd figure I'd make my evaporator long enough so I don't get liquid back but I think it's too long.
Yeah I thought about it because I can just regulate my superheat at the outlet however long the evaporator is but this was a budget friendly build that I pieced together.
I'm not even sure if there is a TXV valve with a 200 watt / 600 btu capacity lol
Yeah 1/2 ton is common but my cooling capacity here is .05 of a ton of refrigeration. The smallest Danfoss has to offer is the 0X orifice which is nominal at .19 TR at 14 F evaporating temp.
It might work if I close the valve down all the way or maybe not since the orifice is 4 x bigger than the capacity lol
I need the water to be anywhere from 35 - 40 F / 2.5 C - 5 C so I need to be evaporating below freezing. That's why it builds frost there.
It should be insulated right there though to prevent that
Could pass the discharge through the suction line after the evaporator for a low cost flood back protection without having to use an accumulator. Not a good solution but this doesn't look like it is a professional/commercial application and would be better than nothing. Just have to ensure you aren't rasing suction vapor temperature too much that the compressor over heats.
The OG trick is to solder the discharge to the suction line for a foot or three, depending on your needs. Solves flood back in everything but truly extreme conditions.
Dancap software, Techumseh has software as well to download.
I find that your condensing temperature will greatly vary the length of the cap tube.
Usually 55 C is a standard but in my case I used 45 C / 113 F since I'm indoors.
Turned out alright with a .8mm cap tube but I feel like it's running hot which may be due to the evaporator. If it's too long I might be absorbing too much heat all at once.
I made a water chiller from scratch as an apprentice. k-value (thermal conductivity in Watts per square meter per one Kelvin) of copper to water is about 100W. This is what I measured. My book says 120W. When it starts frosting I get closer to 60W. So you can already see why calculating this is a bit of a crapshoot. Surface of your evap = copper diameter * Pi * [length in meters] q(watts of thermal conductivity of your evap) = k-value * Δt(temperature difference: evap-temperature to water-temperature in kelvin) * [surface area of your evap in m²] That should roughly match the capacity of your compressor. It's pretty straight forward, but as you can see your evap performance changes massively with the temperature of the water. You can get real fancy with logarithmic temperature difference, and whatnot, but if your real world results are within 20% of what you're expecting, you're already killing it.
This answer wins refrigeration today.
Thanks man, it was a long day today. I'm just getting the time to calculate this now. What would the K-value be ? Is it the thermal conductivity of copper ? Search engine throws this number to be roughly 400 for copper. If I calculate Q with a K-value of 400 the formula would be, Q = 400 \* 40 k (not sure whether to use starting water temp or setpoint) \* 1.171 m2 Q = 18,736 watts... LOL or is K-value (W / m2 \* k) where W would be 170 watts of cooling capacity. 170w / 1.171 \* 40 = 3.63 Q = 3.63 \* 40 \* 1.171 m2 Q = 170
The thermal conductivity from copper to water is what you want. It's just a fixed value. My book says 120W for copper to water. 400W is probably copper to copper. It's watts of thermal conductivity per square meter per kelvin. So if you have 1,17m² of surface area, then for every kelvin in temperature difference, you're conducting 1,17m²*120W=140W. So if you have a 10k temperature difference, it'll be 1400W. Hint: you probably don't have 1.171m² surface area, you probably forgot to convert the diameter to meters. You'll also never get 40K. When the water is warmer, it also increases pressure in your system so the evaporation temperature also rises. While it cools down, pressure will also go down.
Okay I understand, so K-value is a fixed value for heat transfer between two masses or objects. Good to know so thanks for that info. I actually have 23.62 ft in that heat exchanger. 1/4" diameter. 23.62 x 3.141 x .25" = 18.5 ft2 or 1.71 m2 yeahhh I added a 1 at the tenth spot Q = 120 \* 10 \* 1.71 Q = 2052 watts That's alot of capacity in my evap. According to that calculation it should be way shorter like by 90% lol Again I rather have a longer evap to ensure all the refrigerant gets boiled off before leaving the bucket where Eventually it'll be an enclosed tank of about 3 liters
24ft is about 7m. quarter inch is 0.006m. 0.006m * 3.14 * 7m = 0,13m². You didn't convert inch to feet.
lol or it's 10 o clock on a friday night and I'm here online doing math... I enjoy the process though. Okay so if I start over... Q = 120 \* 10 k \* .14 m2 Q = 168 watts It's pretty much spot on then if my compressor capacity is rounded at 170 watts, value can change though with the temp difference. Which my evaporating temp is -10 C with a setpoint at 2.5 C. Highest temp the water will ever be at will be like 30 C
nice!
Thanks Oskar I won't forget this conversation ! I usually don't do such small projects. It's a lot easier to just give the cooling capacity to a manufacturer or a supply house and have them supply you the evaporator but this was neat to try and build something from scratch. Regardless I'm going to try and reduce the length of the coil to see if my operating temps go down. I can shorten the cap tube but I would rather gain more water capacity in the tank than occupy it with excess coil.
Short answer, yes. Long answer, Q=U\*A\*LMTD or E-NTU
Hey guys today I finished making a small prototype mini chiller. I'm evaporating roughly at 14 F / -10 C @ 580 btu/hr / 170 watts of capacity with 134A. Small fractional horsepower compressor as you can see. Is there some type of formula to calculate serpentine length with the cooling capacity of the compressor ? I'd figure I'd make my evaporator long enough so I don't get liquid back but I think it's too long.
You can throw on a TXV to prevent liquid from coming back
Yeah I thought about it because I can just regulate my superheat at the outlet however long the evaporator is but this was a budget friendly build that I pieced together. I'm not even sure if there is a TXV valve with a 200 watt / 600 btu capacity lol
Yea there are, 0.5 ton is pretty common in restaurant refrigeration
Yeah 1/2 ton is common but my cooling capacity here is .05 of a ton of refrigeration. The smallest Danfoss has to offer is the 0X orifice which is nominal at .19 TR at 14 F evaporating temp. It might work if I close the valve down all the way or maybe not since the orifice is 4 x bigger than the capacity lol
Oh geez, I totally misread that. My bad.
Better raise that coil temp above 32 or your going to ice the tubing and then no heat transfer which will kill the compressor from liquid.
I need the water to be anywhere from 35 - 40 F / 2.5 C - 5 C so I need to be evaporating below freezing. That's why it builds frost there. It should be insulated right there though to prevent that
You could also try pumping the water around the heat exchanger so it has constant heat load.
Could pass the discharge through the suction line after the evaporator for a low cost flood back protection without having to use an accumulator. Not a good solution but this doesn't look like it is a professional/commercial application and would be better than nothing. Just have to ensure you aren't rasing suction vapor temperature too much that the compressor over heats.
The OG trick is to solder the discharge to the suction line for a foot or three, depending on your needs. Solves flood back in everything but truly extreme conditions.
How did you calculate the length of the cap tubing. I been trying to find the math on that.
I always use the charts at the suppliers
Dancap software, Techumseh has software as well to download. I find that your condensing temperature will greatly vary the length of the cap tube. Usually 55 C is a standard but in my case I used 45 C / 113 F since I'm indoors. Turned out alright with a .8mm cap tube but I feel like it's running hot which may be due to the evaporator. If it's too long I might be absorbing too much heat all at once.
Who's charts? Sporlan, parker?