Due to the internal resistance of battery cells, they produce heat when a current is applied to them. We must dissipate this heat somewhere to stop the cells from heating up.
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We measured the airflow of our Noctua IPPC fan when pulling air through 1 or our prototype modules (see this page). We will be contacting Noctua to obtain a more accurate P-Q curve for their fan.
| Condition | Airflow (m3/h) | Airflow (CFM) | Static Pressure (from Cooling Technique Testing) | Static Pressure (from Noctua P-Q Curve) |
|---|---|---|---|---|
| Just fan duct | 26.97 | 15.87 | 5.55 | |
| 1 Module and fan duct | 15.52 | 9.13 | 5.6 |
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| Variable | Description | Value (Units) |
|---|---|---|
| H | Least amount of heat removed | (W) |
| Cp | Specific heat capacity of the air | 1005 (J/Kg℃) |
| M | Mass of the air | (Kg) |
| ∆T | Temperature difference | Tc - Tamb (℃) |
M = Q x ρ
| Variable | Description | Value |
|---|---|---|
| M | Mass of the air | |
| Q | Flow rate of the air | |
| ρ | Density of the air | ρ = 1.18 Kg/m3 |
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Now we need to ensure that the fans we spec will be able to produce the required amount of airflow with the amount of static pressure they must overcome.
Comparing against
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measured results
Discharge Current | Condition | Rate of Temperature Rise (degC / s) | Expected module heat production | Heat Production based on Rate of T Rise |
|---|---|---|---|---|
| 100A | No Fan | 0.015011184 | 570W / 18 = 31.67W | 532W / 18 = 26.56W |
| 100A | Noctua Fan | 0.00384364035 | 570W / 18 = 31.67W | 136W / 18 = 7.56W |
| 0A | No Fan (Cooldown) | -0.0047593750 | 0W | -168W / 18 = -9.373W |
| 0A | Noctua Fan (Cooldown) | -0.0159833 | 0W | -566.6W / 18 = -31.48W |
With the fan, we were able to remove 26.56 - 7.56 = 19W of heat with a Noctua iPPC 3000 fan. The 50W of heat removal that was previously calculated is assuming that the system has reached a steady state and that the difference in temperature from the input to the output is in fact 10 degrees C. With the 5 modules in a row (giving the air a longer time to heat up) and more heat to dissipate in the first place, I believe that the calculation of 50W is accurate to within 20% - we should be able to remove 40W of heat.
The other way that I would still like to compare is looking at the cool down rate of temperature decrease with and without the fan. This should give us a similar value. As can be seen on the cooldown slope of the noctua fan graph above, the rate of cooldown decreases as the temperature difference decreases. For the rate of decrease with the fan, I tried to capture the steepest part of the slope at the beginning of the cooling phase. With the addition of the fan on the cooldown portion, we were able to remove 31.48 - 9.373 = 22.1W of heat with the Noctua fan.
Please note, that these values for the rate of temperature rise were calculated using the average of the rate of temperature rise of the individual cells. Because there is such a large temperature variation between the cells (which affects the amount of cooling potential due to the delta between the cell temp and the air temp), these values are not to be taken as 100% true - they are experimental and could be improved with better airflow paths, better measuring equipment, etc.
As a conclusion to this testing data, I will restate what I mentioned earlier - that I believe we will be able to remove more heat once we stack all 5 modules. However, I will add that due to the uncertainty in the measurements, temperature distributions, and difficulty in simulating everything, I strongly advise to allow a safety factor of 2 when specifying the fan airflow requirements.
Might be useful, sometime and thought I would throw a link here - an interesting way to measure airflow in a closed pipe or duct:
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