The Nutra has a fan (I’m guessing something like a 12V PC cooler), a ¼HP agitator motor, a small pump motor, a water inlet valve, and a heating element.
Lucía tells me the target is for a year’s lifetime.
I’m thinking that probably the control solution for these is to hook up an STM32 to a ULN2003 or something similar, then use the ULN2003 to run some relays. None of them needs subsecond response times; I think they are each turned on and off less than ten times during a cycle, which lasts roughly a day, so a lifetime of 20000 openings/closings and an MTBF of 100 000 openings/closings is probably adequate. Most of them will run on line power rather than a lower voltage, so the ULN2003 can’t handle them directly.
I back-of-the-envelope estimated the heating element at 300 W, but maybe 600 W is a safer level (in the sense of “we are sure it will work”, not in the sense of “less likely to cause injury in case of malfunction”.) 600 W is about 2.6 amps at 240 VAC (RMS), so a 3-amp relay is probably fine.
But wait, this needs to be PWM-controlled. PWM through a relay is not a good idea; if the PWM cycle is 20 seconds in length (50 mHz), 100 000 openings and closings only gets us to 555 hours of dehydration time. An SSR would achieve higher levels of reliability, but 500 hours is probably already a year, since the overall cycle might have an hour or two of dehydration in it.
The most challenging objective is dehydrating 5 kg dry weight of soaked, activated beans to more or less their original dry weight. They will have absorbed roughly 5 kg of water. Water’s enthalpy of vaporization is about 2.4 MJ/kg, so this is 12 MJ which get turned into latent heat. Doing this in one hour would require 3000 watts, which would be 12 amps. At my previous guess of 600 watts it would take five hours, which might be long enough for the beans to rot.
If we could use an air-countercurrent recuperative condenser like the ones in modern condensing clothes dryers, we could perhaps reduce that somewhat, but I’m not sure how much. (There are also heat-pump dryers, basically air-conditioner-based dehydrators in a box, which cost an extra US$1200 or so.) Traditional vented clothes dryers use 4–6 kW and vent 100–225 cfm, or more broadly 3–9 kW, while non-heat-pump condenser dryers typically use 2 kW, and heat-pump condensers use ≈ 1 kW.
Let’s say we don’t have a recuperator, but 2 hours is adequate, so we can get by with 1500 watts. And let’s say the air stream is limited to 40° in order to make sure we don’t heat the food past 42°. My psychrometric chart doesn’t go to 100% humidity at 40°, but it says that 30 mg H₂O / g air is 64% humidity at 40°, so I guess 100% is about 47 mg H₂O / g air, and 95% would be about 44 mg H₂O / g air. This means that removing 5 kg of water vapor would require passing about 114 kg of dry air through the beans, or a bit more, since the air in Buenos Aires already contains something like 10 mg H₂O / g air.
Air is about 1.23 kg/m³, so 114 kg of air is 92 m³. This is enough air to fill a 35 m² room of 2.6 meters height, and we’re proposing to turn it all into the kind of fog that fills your bathroom after a shower. So you really, really need to vent it outside or run it through a condenser.
Either way, those 92 m³ of air need to be blown through the dehydrator during those 2 hours, which works out to about 12.8 ℓ/s, 770 ℓ/minute, or 27 cfm. This is the kind of airflow associated with bathroom fans with 8 cm ducts, or with small or quiet fans for computer cases — definitely feasible.
You don’t want to blow too much new air in because you have to heat it all up — air’s heat capacity is 1.01 kJ/kg/K, so each additional ℓ/s is 1.23 g/s and 1.24 W/K, and with ΔT ≈ 20 K that works out to about 24 W of extra heater power, plus whatever the fan uses. This isn’t very much extra energy use but it’s energy use that doesn’t evaporate much extra water.
(Or is it?)
Here in Buenos Aires, we have abundant municipal water which we could use to chill a condenser, and the Nutra already needs access to both a water intake and a drain. How much water would it waste? If we drop the dewpoint of the air to 25° (thus removing more than half the water) by running it through a countercurrent heat exchanger cooled by tap water starting at 15°, well, we would need 290 ℓ of water to dump those 12 MJ of heat. This might be okay in some kind of objective sense but people who run dietéticas are not going to tolerate it.
Lucía previously calculated that she would need a 2000-watt heater.
I have no idea what the water inlet valve is like. I’m guessing it’s a solenoid driven by line power (through a diode) at under an amp.
The pump needs to drain 6 liters (I think) of water within, I don’t know, 20 minutes. 5 mℓ/s, 0.08 gallons per minute, 5 gallons per hour. This is a very small pump; a US$11 aquarium pump on Amazon is 227 GPH, 45 times as big, and it’s 18 watts. So I think this is more like a 1-watt pump. The intrinsic work being done is very small, possibly even negative, but pumping 5 milliliters per second against a pressure of 1 meter column of water would be 50 milliwatts, plus whatever mechanical inefficiencies are encountered.
It might be feasible to run this directly off the ULN2003.
This is 190 watts output, perhaps 220 watts input — 1 amp at 240 volts.
A typical PC fan is like 100 to 300 milliamps at 12 volts. This could possibly be run directly off the ULN2003, which can handle 500 mA on a given pin at up to 50 V, though only one pin at a time.
So we probably need a couple of 1-amp relays, a 3-amp relay, and maybe another sub-1-amp relay for the pump. If we simplify this to four 3-amp relays, well, the usual suspect seems to be something like the Omron G5LE-1A4 DC12, which costs US$1.30 in quantity 1 and switches 10 amps at up to 250 VAC when you apply 12 VDC and 33.3mA to the coil. It’s SPST, normally open. Or the TE Connectivity/Brumfield OJE-SH-112HM,000, apparently identical in every way except for costing US$1.32 and drawing 37.5 mA.
However, the Omron datasheet has a chart that shows durability of 200 000 operations at 3 amps and a 250 VAC resistive load or 120 VAC inductive load. With a 120 VAC resistive load, it’s rated for twice the lifetime. The TE relay is only rated for 10 000 ops, though there’s an “LM” variant rated to 100k.
All four relays at once would only be 133 milliamps, well within the ULN2003’s limits.
There do exist mercury-wetted reed relays with lifetimes of a million operations, but these probably aren’t necessary for a lifetime of one year.
Some kind of triac-optoisolator thing might be a reasonable alternative, but a typical triac drops 1.5 volts, so 3 amps would dissipate 5 watts, requiring a hefty heatsink and posing potential reliability problems of its own.
The ULN2003 only has a guaranteed β of 1000, so getting 33 milliamps on its output requires 33 microamps on its input. Except that actually the curves in the datasheet show that you need over 100 microamps to get anything to happen. This means that, at 3.3 V, the base resistor shouldn’t be over 33 kilohms, which means you can’t quite use the pullup resistors integrated in the STM32 I/O pins, since they could be as high as 50 kΩ according to the datasheet. You could maybe parallel two I/O pins, but it would probably be better to use external resistors, thus avoiding the chance of a software error burning up the STM32 or the ULN2003.
The ULN2003’s maximum base current is 25 mA; 1 mA per pin should be plenty, so anything between 3.3kΩ and 33kΩ is adequate. 10kΩ, everybody’s favorite resistor, is probably optimal, giving 330 μA, and thus 330 mA or better on the output pin.
We can hook up a speaker to an extra output of the ULN2003 and control it through the built-in PWM of the STM32. If it’s a ¼W 8Ω speaker, we need to keep current through it under about 100mA; a dc-blocking capacitor is probably worthwhile here, so that if the ULN2003 is on we don’t have a constant current through the speaker with no sound, but also it would be a good idea to limit the current of ac signals and also perhaps filter out inaudible high frequencies. More precisely, with 8Ω, to keep I²R < ¼W we need |I| < 177 mA, so a series resistance of 68Ω or greater is called for with 12V. Except with dc-blocking that’s 12V P-t-P, so 6V RMS with a perfect square wave, so 33Ω would be adequate. The resistor is going to have most of that 6V RMS across it, like about 5V, so it needs to be pretty hefty, nearly a watt in the worst case. Probably prudent to limit the output power further, using 100Ω or so.
At 100Ω series resistance, an instantaneous 12V spike would push 120 mA, and 6V RMS would dissipate 360 mW in the resistor, which requires a largish resistor but nothing ridiculous; it would experience spikes to 1.3 watts. The 60 mA RMS would output 29 mW from the 8Ω speaker. You can see why people prefer to drive these damn things through transformers. Still, 29 mW is probably still audible; it’s less than 10 dB below what I guessed was the maximum output for the speaker.
Now, what about filtering out ultrasound? If we shunt the speaker with a capacitor, we’d like its RC time constant to be in the tens of microseconds somewhere, so that 64 kHz (2.5 μs per radian) is strongly attenuated but 10 kHz (15.9 μs per radian) is minimally attenuated, if at all. But the relevant R here is not the 100Ω that’s being used to limit the current, but rather the parallel combination of that 100Ω (plus the dc-blocking capacitor, which is hopefully negligible at the relevant frequencies) and the 8Ω of the speaker, which works out to be 7.4Ω. So we need something like 1.3 μF, not the 100 nF you would expect. At 1 μF, we get 7.4 μs, which puts the knee around 21 kHz, which means that we only have about 3× attenuation at 64 kHz (like 10 dB). If we use 2.2 μF, the knee is at 9.8 kHz, which gives us an extra 6 dB.
The dc-blocking capacitor needs to have a sufficiently long time constant with the series resistance to not mess up the bass too bad. Say we want its time constant to be below 40 Hz (4 ms per radian), then we need at least 40 μF. 100 μF is probably fine; at 12 volts this holds 7.2 mJ, which is what gets dissipated in that 1.3-W spike in the series resistance.
Perhaps a second RC section is worthwhile in this case: 10 ohms before
the 100 ohms, then a 1μF to ground. So the whole output circuit from
the ULN2003 pin would be { 100 μF 10 Ω { 1 μF gnd } 100 Ω (1 μF ||
speaker) 12 V }, where concatenation is series combination, {} is a
branch, () are grouping, and || is parallel combination. In a
Falstad simulation, this configuration gives us about -4 dB at 40 Hz,
-0 dB from 80 Hz to 5 kHz, -1 dB at 7.5 kHz, -2 dB at 10 kHz, -4 dB at
15 kHz, -5 dB at 20 kHz, -15 dB at 40 kHz, and -22 dB at 65 kHz.
However, at high frequencies, this puts the whole 6 V RMS across this
new 10Ω resistor through the capacitor shunt to ground, which means it
will be dissipating 3.6 watts! And of course the PWM output from the
ULN2003 is always close to 6 V RMS, except when it’s close to dc.
So the second RC section is a bad idea. Probably much less hassle to use an inductor in series with the resistor, one whose impedance becomes large relative to 100 ohms at somewhere around 10 kHz, like a 4.7mH inductor. In a Falstad simulation, this gives about 3 dB attenuation at 3.5 kHz, 6 dB at 6.3 kHz, 10 dB at 10 kHz, 17 dB at 20 kHz, 27 dB at 40 kHz, 30 dB at 48 kHz, 38 dB at 80 kHz, and 42 dB at 99 kHz. Simulation also suggests about a 13.7 mV peak amplitude part of a 48 kHz PWM modulating square wave would remain across the terminals of the speaker, compared to some 440 mV peak for the actual signal. (Maybe it works even better with L speaker || C instead of L (speaker || C), with a +6dB peak around 6kHz.)
Wait, actually the whole thing is bad in that form. I need a 220Ω pullup resistor on the ULN2003 to get current. And then maybe I don’t really need an inductor. Here’s the Falstad design:
$ 1 3.0000000000000004E-7 26.59566520631553 50 12.0 50
c 960 432 1088 432 0 9.999999999999999E-5 3.5513458288348803
r 1168 432 1264 432 0 8.0
R 816 304 816 224 0 0 40.0 12.0 0.0 0.0 0.5
w 1264 480 1264 432 0
r 816 304 816 416 0 220.0
a 592 432 672 432 0 12.0 0.0 1000000.0
174 272 384 368 480 0 1000.0 0.005 Resistance
g 272 480 272 496 0
R 272 384 272 352 0 0 40.0 5.0 0.0 0.0 0.5
R 592 416 560 416 0 4 65000.0 5.0 0.0 0.0 0.5
O 672 432 672 368 0
S 592 448 368 448 0 0 false 0
170 368 464 320 464 3 20.0 100000.0 5.0 0.3
l 1088 432 1168 432 0 0.001 0.02828594427035853
t 736 432 768 432 0 1 -5.526250662994705 -0.11573671215990365 100.0
t 768 448 816 448 0 1 -5.410513950834801 0.11562175983470305 100.0
w 736 432 736 496 0
r 736 496 816 496 0 10000.0
w 816 464 816 496 0
g 816 496 816 528 0
w 768 416 816 416 0
w 816 416 816 432 0
w 816 432 848 432 0
r 672 432 736 432 0 10000.0
g 1264 480 1264 528 0
r 848 432 960 432 0 100.0
c 960 432 960 496 0 1.0E-6 2.5834701256300843
g 960 496 960 528 0
o 1 64 0 35 0.5846006549323611 0.09353610478917779 0 -1
o 10 64 0 34 20.0 9.765625E-5 0 -1
o 15 4 6 35 10.0 9.765625E-5 1 -1
o 13 4 0 35 2.5 0.1 2 -1
Oh, I think I have a reasonable approach that eliminates most of the
inefficiencies. {100Ω (L || 100μF speaker) 12V}, where the L
provides a dc path around the speaker and capacitor, and the capacitor
basically just protects the speaker from dc. The resistor in series
with the whole shebang limits the current, but maybe it could be
bypassed with a capacitor so that it can be bigger than would be
desirable for ac. And you can still shunt the speaker alone with a
capacitor to reduce ultrasound.
A potential problem with this is that it could introduce a kind of distortion when the conduction switches from being through the Darlington to being through the freewheel diode, just because of the forward voltage drop of the freewheel diode.
How much inductance does the shunt inductor need? Ideally it shouldn’t steal much current from the speaker at audio frequencies, which means its impedance at audio frequencies should be large compared to the speaker’s 8Ω, although maybe the capacitor can help compensate for this. But let’s say we want its impedance at 40 Hz to be 16Ω. This requires a fairly hefty 68-mH inductor (like, 8 mm × 12 mm.)
Or you could just drive the speaker through a little 4:1 100Ω-ESR audio-frequency transformer, which will probably filter out the ultrasound as a side effect.
What is the PWM frequency? The STM32F030x4 etc. reference manual RM0360 says, “If the APB prescaler is 1, the timer clock frequencies are set to the same frequency as that of the APB domain,” which is normally 48MHz except for power saving. I think that means that if you want 8-bit PWM, you can get 187.5 kHz PWM. At such a speed, 1mH of inductance should be plenty — and how much does the speaker itself have? 1mH series and 1μF in parallel reduces the 187.5 kHz square wave to 5.6 mV peak, while the AF signal is like 440 mV peak.
A Nokia 5110 84×48 display ought to be adequate and might fit in well with the overall feeling of the device, although OLEDs are nicer on several axes nowadays.
Monarca sells a DHT-22 temperature and humidity sensor for $220. Nubbeo has the cheaper DHT-11 for $60, but its precision is ±2°, which would be terrible for the Nutra, restricting it to a temperature of 40° or below. By contrast, the DHT-22’s precision is ±0.1°.
Apart from cost and precision, the sensors are otherwise quite similar.