I'm planning on building a plant growth chamber inside a sealed/insulated 55 g rain barrel, and I'm trying to figure out how much of the heat from my lights will enter the chamber, and how much will leave through the heat sinks.

I plan on attracting 3x3000k 80cri Vero 29 gen 7 COB LEDs to the lid and running around 120-150 W/h total through them. According to their datasheet, their efficiency runs around 130-190 lumens/W depending on how hard you drive them which I have read is around 40-55% efficient (although Wiki seems to disagree with that). I believe this means that they will produce 54-82.5 W/h of heat.

I will be attaching both COBs to their own Arctic apline 11 plus active heatsinks rated for 100 W (see datasheet). They are aluminium and the fan moves 37.8cfm over its fins, but surface area and mass values for the aluminum portions are not provided by the manufacturer. That being said it’s total mass is 428 g.

I really would like to know what fraction of heat will be diverted into the growth chamber vs what fraction of heat will exit through the heat sink, so I can figure out if using DIY thermoelectric cooling is a viable option to combat this heat and keep the inside of the chamber no warmer than 85°F/30°C.

If so this will allow me to build a sweet 55 gallon growth chamber that can utilize CO2 injection while being temperature controlled.

Other variables:

  • Max ambient temperature 35°C
  • Inside walls of container have an Emissivity value of 0.044
  • $\begingroup$ There are better heatsinks to be found, also I'd be (conservatively) assuming that 100% of the energy will get dissipated as heat through the heat sink. $\endgroup$ – ratchet freak Aug 9 '18 at 9:19

I believe that you have two possible approaches to an energy balance.

picture of system in two cases

System A

The walls allow heat flow. The defining equations are

$$ P_i = q_x + q_a + P_o \\ q_x = \epsilon_{HS} (1 - \epsilon_L) P_i \\ q_a = (T_c - T_a) / R_T \\ R_T = R_c + R_w + R_a \\ R_c = 1/h_c A_w\ \ R_w = w_w / A_w k_c\ \ R_a = 1/h_a A_w $$

The balance here is with power input, heat flow out of the fins, heat flow across the container, and cooling power output. The heat flow from the fins is the heat sink efficiency times the heat produced by the lamps as determined from the lamp efficiency (for light production). The heat flow across the wall is a thermal resistance calculation. In this case, you assume / calculate the overall thermal resistance $R_T$. You define the fin efficiency $\epsilon_{HS}$ and lamp efficiency $\epsilon_L$. You set the air temperature $T_a$ and power input $P_i$. What is left is the cooling power that you need $P_o$. The maximum is when $\epsilon_{HS} = 0$ (all lamp heat is dumped to the container) and $\epsilon_L = 0$ (the lamp produces only heat and no light). This approach allows you to size the cooling unit for the maximum required cooling rate $P_{o,max}$ (W) at the output.

System B

The system is adiabatic. The defining equations are

$$ q_L = q_c \\ q_L = (1 - \epsilon_{HS}) (1 - \epsilon_L) P_i \\ q_c = (T_c - T_x) / R_c \\ R_c = 1/h_c A_c $$

The balance here is heat flow from the lamps and heat flow to the cooling unit. Assume that all heat produced by the lamps is dumped perfectly to the system. The heat flow to the cooling unit is defined by the thermal resistance. This approach allows you to design the temperature, efficiency (convection coefficient), and area of the cooling unit. This will again give a maximum design when $\epsilon_{HS} = 0$ and $\epsilon_L = 0$.


Here are the questions that I find are left to address:

  • Which system model is more appropriate, conducting walls or insulated walls (System A or System B)?
  • How much heat will the lamps dump to the container (what are $\epsilon_{HS}$ and $\epsilon_L$)?
  • How effective is the cooling unit (what are its nominal values of $h_c$ and $A_c$)?


  • System A assumes the chamber is hotter than the surrounding air. System B does not care.
  • Neglect radiation, especially in System A. Unless something in the system is above a few hundred oF, radiation is of third order consequence.
  • This assumes that all heat produced by the lamps is dumped perfectly to the system. When the lamps are hot above the system, convection is nearly non-existent. The chamber will have a hot zone at the top and a cool zone at the bottom. You must adjust the placement of hot and cold elements accordingly.
  • This analysis only gives a picture of how to find the maximum design parameters. The next step may be to decide how much control you want / need on the cooling unit.

Edit (After the Fact)

The original question was ...

I really would like to know what fraction of heat will be diverted into the growth chamber vs what fraction of heat will exit through the heat sink,

The variable you want to determine is $\epsilon_{HS}$. Based on the system picture, I can suggest a simple experiment. Set up the chamber with everything as desired but exclude the cooling system. Put a thermocouple in the chamber. Run the lamps until the internal temperature stabilizes. Use the equations to back-calculate a value or estimate for $\epsilon_{HS}$.

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  • $\begingroup$ Would it not or be a combination of both? I plan to insulate the walls of the container, and have the heat sink sticking out of the top of the container. Thus the walls should be Adiabetic, yet heat from the COB should be able to dissipate out through the heat sink. $\endgroup$ – Jefferson Gately Aug 11 '18 at 16:02
  • $\begingroup$ Also I plan to instal internal fans to circulate the air inside the chamber to prevent a large gradient between the hot and cold regions of the container. $\endgroup$ – Jefferson Gately Aug 11 '18 at 16:06
  • $\begingroup$ With insulated walls, the process is adiabatic. This is System B. In this case, $q_x = \epsilon_{HS} P_i$. When $\epsilon_{HS} = 1$, all heat generated by the lamps will exit through the heat sink fins. An additional option to using fans can be to put the cooling unit around the lamps. In a simple picture, I think of a copper pipe with chilled water that surrounds the lamps. $\endgroup$ – Jeffrey J Weimer Aug 12 '18 at 16:46
  • $\begingroup$ @JeffersonGately Does my answer now address your question? Please do the courtesy to up vote and/or check it appropriately. Otherwise, comment how this can be improved. $\endgroup$ – Jeffrey J Weimer Sep 4 '18 at 15:02

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