Anova Sous Vide Circulator (Part 2)

[gfweb]

How much lag is there with UPS turning on? Even a little would trigger a shut down.

[paulraphael]

It would make sense to check the specs of the individual UPS, and then see what the circulator manufacturer says. But I suspect most UPSs click over in a tiny fraction of a second.

 

How much does a decent UPS cost than supply a kilowatt? Granted, these circulators are usually asking for much less than that, but you don't want to trip a breaker when the circulator's doing its initial heating. And I don't know what happens if the circulator does ask for more than the UPS batter can provide.

[alanz]

gfweb,

 

UPS stands for Uninterruptible Power Supply.  There should be no time lag at all.  As another member mentioned, the issue is that the battery in the UPS would likely only sustain the heating element for a few minutes, if at all (depending on the capacity of the UPS).  When heating, the coil likely draws about the same as the copy machine mentioned.

Edited June 17, 2014 by alanz (log)

[dcarch]

A UPS system can be any power, including a generator. In a battery (portable) system, I am not sure it always mean instantaneous response.

 

A battery UPS trickle charges the internal batteries, typically lead acid sealed 12VDC batteries. When the circuit detect a power interruption,  an inverter circuit kicks in to convert the 12vdc to 120vac (or 240vac) at 60hz (or 50hz). All these may take a fraction of a second.

 

A surge protector, OTOH, by definition, acts instantaneously.

 

dcarch

[alanz]

dcarch,

 

Some UPS unit route their power through the battery (and often voltage conditioning circuit), so in effect the battery is always supplying the power, but is also in a constant state of recharge.

 

In any event, even if there is a switchover, it's internal to the UPS, the attached appliance never sees any downtime.

[dcarch]

dcarch,

 

Some UPS unit route their power through the battery (and often voltage conditioning circuit), so in effect the battery is always supplying the power, but is also in a constant state of recharge.

 

In any event, even if there is a switchover, it's internal to the UPS, the attached appliance never sees any downtime.

 

There must be a thousand different designs for a UPS, but I have a problem understanding what you are saying, which does not mean your are wrong.

 

A battery is a 12vdc device, you can't send 120vac thru it.

 

A UPS cannot (I think) always be supplying power even when the line power in not out. The only way to do that would be parallel the UPS and line power. But to do that you end up with a difficult situation. UPS is a regulated constant voltage device, let's say at 120vac, with a sine wave or square wave timing that is non-synchronous to line wave's wave form output. Line voltage varies in voltage, when that happens, if the voltage is higher then the UPS, it can burn out the UPS, and if the line voltage is lower than the UPS, the UPS will be powering up your entire house and the whole neighborhood thru the main panel.

 

dcarch

[paulraphael]

The ones I've had in the past mentioned a lag of some number of milliseconds. It never made my Mac blink. But, mine used a kind of power supply which I think is common in less expensive units .... one that doesn't generate a true sine wave pattern alternating current. It makes a square wave, which may be hard on the power supplies of some electronics.

 

My monitors would buzz and flicker whenever the UPS kicked in, and I had two expensive professional graphics monitors die early deaths in the years I had the UPS. I can't prove causation, but it was suspicious enough that I eventually got rid of the thing. It turns out that most of the power outages/lags around here are so brief that they don't shut down my computer even without the UPS.

 

Someday I might consider a UPS for for the cooker, but I'd be inclined to check with Anova about the specs of any model before buying it.

 

BTW, my UPS was by APC ... a pretty good brand. But it was not one of their higher end models.

[dcarch]

Things can look very good for electronic units to filter out peaks and valleys in supply line power.

 

With nano carbon technology, there is a new generation of super capacitors which can go into the power section of electronic gadgets.

 

dcarch

[alanz]

dcarch,

 

Many devices use 12volts DC to drive an inverter that produces 120 volts AC (and provide very clean sine wave power).

So a UPS can use the incoming AC power to step down to 12 volts (or some other DC voltage, I think mine use two 12v batteries in series, producing 24 volts) and have that power feed the inverter and keep the batter charged.

If the battery is wired in parallel to the incoming DC power, in the event of power loss, the battery continues to power the inverter. 

 

I'm not suggesting that this is how all UPS systems work, but think of it similar to a laptop computer that's plugged into the wall.  When you remove the AC, it doesn't 'switch over' to an alternate power source... it simply lets the battery continue to power the laptop power circuitry, uninterrupted.

Edited June 17, 2014 by alanz (log)

[Chris Hennes]

How much power does it really need to supply, anyway? Once the bath is at temp, the heater should never be actually drawing 1000 watts, right?

[paulraphael]

How much power does it really need to supply, anyway? Once the bath is at temp, the heater should never be actually drawing 1000 watts, right?

I assume so. But I can imagine a couple of issues.

 

1) if power goes out during that initial heating, will it trip a fuse and shut the whole thing down?

 

2) if the the UPS is limited to, say, 500 watts, will it have a fuse or breaker that will trip if you go beyond this, even when you still have a.c.?

[KennethT]

Most UPSs don't have a breaker limited to the UPS limit. In fact, the limit is really only in reference to how much power it can continuously provide during an outage for about 10 minutes. So a 1000W UPS has a much bigger battery which can store and deliver more energy than the 500W UPS.

As you drain the battery, it's voltage decreases. There is typically a voltage regulator circuit in the UPS either before or in the inverter stage. So, as the battery voltage decreases, your output voltage stays constant, until it reaches a threshold where it will just shut off.

So, you can run 1000W off of a 500W UPS, but it will only last a minute or two. Or you can run 100W off of a 1000W UPS for 20-30 minutes, maybe more.

Once the bath reaches temp, the heater draws very little power (maybe 100W) on average. Very short bursts of 1000W look like the average power to the battery.

[scubadoo97]

As a trial I plugged the Anova into my UPS started it and pulled the plug to the UPS unit. The Anova kept chugging along. That what I need it to do

[paulraphael]

Has anyone plugged their circulator into a power meter over the course of a long cook? It would be interesting to see how efficient they are, and to compare different water containers.

[KennethT]

Years ago, I made a few of my own circulators. The PID thermostat I used could display average percent power. Once my bath was up to temperature, depending on temperature, the heater used between 8% and 20% power (8% for a bath temp of 131degF, 20% for 185degF). My bath was just a covered stockpot with no additional insulation, and I used a 1000W heater.

[dcarch]

Has anyone plugged their circulator into a power meter over the course of a long cook? It would be interesting to see how efficient they are, and to compare different water containers.

 

By definition, electric heating is 100% efficient. 

 

I think what you mean is how much $ it would cost to do a long cook.

 

Do remember that also in the calculation, all the BTUs in the cooking is 100% used to heat your house in the winter.

 

dcarch

[dcarch]

All storage/rechargeable batteries have limited recharge cycles, and they age every time you use them. The system which charges the battery and invert the battery power full time to power line voltage devices may have a shorter life. Also, The charger is not 100% efficient and the inverter is even less efficient. You will be wasting energy if you use a ups to power your other devices.

 

Except super capacitors, which operates not by chemical reactions, and is 100% efficient.

 

dcarch 

Edited June 18, 2014 by dcarch (log)

[paulraphael]

By definition, electric heating is 100% efficient. 

 

I think what you mean is how much $ it would cost to do a long cook.

 

Do remember that also in the calculation, all the BTUs in the cooking is 100% used to heat your house in the winter.

 

dcarch

I mean the efficiency of the whole system, which comes down to how much of the electric power ends up heating the food vs. heating the house. I imagine the biggest factor is the insulation of the container, but even so, I have no idea if a circulator maintaining temperature in a cooler averaging 10 watts or 200.

[KennethT]

In my testing, I've found that most energy loss was through evaporation, so keeping the bath as covered as possible is important. After that, you're right, it comes down to insulation. A well insulated container is not as important when the differential between bath and ambient temp is low. It is more important when cooking at higher temps - like for veggies or confit that mimics traditional confit.

[paulraphael]

Kenneth, can you elaborate a bit on your testing and what you found?

 

I don't have a meter and haven't done any. I was a bit surprised when doing a long cook in a cooler at 60°C that the outsides of the cooler were pretty warm. I keep a layer of reflectix on the water, and have a lid that's cut with a pretty tight fit. The lid's also insulated with spray foam. I have no idea how good the insulation in the cooler itself is. It's just a $30 coleman.

[pbear]

I was a bit surprised when doing a long cook in a cooler at 60°C that the outsides of the cooler were pretty warm..

 

Can't speak to the rest, but this isn't surprising.  Insulation isn't magic and it's never 100% efficient.  Eventually, heat from the water bath will saturate the insulation (figuratively, not literally) and warm the outside surface, where it then dissipates into the air.  But the rate of transmission through an insulated container will be much less than that of an uninsulated one, which is what you're after.  The easiest way to measure this would be to put a watt meter (available from any hardware store) on the input, i.e., the circulator.

[Chimo]

Has anyone plugged their circulator into a power meter over the course of a long cook? It would be interesting to see how efficient they are, and to compare different water containers.

I have a "Kill-a-Watt" meter that I plugged the Anova into on one of my first cooks. I recall calculating the energy cost of the cook, but I don't recall the numbers. I will try to remember to give it another go this weekend. I do recall being a bit surprised by the steady state power that the circulation motor drew.

To comment on the UPS discussion, most consumer models are of the off-line type, meaning the mains provide the power until it is lost and then battery inverter circuit kicks in after a percentage of a cycle. As far as the output waveform, some use a modified square wave, some use pulse width output fed through a smoothing filter to replicate a sine wave. Very few consumer models actually output a pure sine wave.

[dcarch]

I have a "Kill-a-Watt" meter that I plugged the Anova into on one of my first cooks. I recall calculating the energy cost of the cook, but I don't recall the numbers. I will try to remember to give it another go this weekend. I do recall being a bit surprised by the steady state power that the circulation motor drew.

To comment on the UPS discussion, most consumer models are of the off-line type, meaning the mains provide the power until it is lost and then battery inverter circuit kicks in after a percentage of a cycle. As far as the output waveform, some use a modified square wave, some use pulse width output fed through a smoothing filter to replicate a sine wave. Very few consumer models actually output a pure sine wave.

 

Most circulators use simple shaded pole motors, which are somewhat efficient. To circulate water, you need very little power. However, they just don't make special shaded pole motors small enough just for sous vide use.

 

Inverters invert by switching (pulsing, chopping) DC to transform to higher voltage, switching means on and off. on and off means square wave. Pure square wave can do damage to some electrical devices. So they have to reshape square waves to more like sine waves.

 

dcarch

[paulraphael]

Can't speak to the rest, but this isn't surprising.  Insulation isn't magic and it's never 100% efficient.  Eventually, heat from the water bath will saturate the insulation (figuratively, not literally) and warm the outside surface, where it then dissipates into the air

Yes, but the question is how much. You can calculate the R-value of insulation by measuring the surface temperature of the vessel relative to the ambient air temperature, and comparing to the temperature differential between inside and outside the vessel.

 

The higher the R-value, and the lower the temp. variation between in and out, the less warm you'd find the outside of the vessel. If the cooler were filled with 1/2" closed cell foam, which has very high r-value, I'd expect the warmth of the cooler on the outside to barely perceptibly warm to the touch. At least with a 140F cook, which is only about 70F warmer than the air in the room.

 

I haven't done the math, so I could be wrong about it. Maybe it's well insulated and I should expect it to get as warm as it does. If that's the case, though, a 30 quart cooler with that much warm surface area makes for a pretty big radiator. It's definitely burning up some watts. Imagine the light bulb it would take to warm up a big cooler that much.

[KennethT]

At the time, my vessel was just a large stockpot (All-Clad anodized aluminum (a wedding gift)), uninsulated. Like I said before, my PID controller has a function that displays percent output. So, while the bath is heating and the heating element is on full power, the controller shows 100% - which is 1000W since I was using a 1000W element. As the bath reaches temperature, the controller begins cycling the power to the element. I was most interested in power used once the bath reaches steady state, comparing the covered vs. non-covered values at various temperatures.

So, obviously, the covered pot was much more efficient, since it didn't constantly lose temperature due to evaporative cooling - which is much more substantial (depending on bath temperature) than the loss due to radiation/convection with the air around the pot - especially if there is no breeze.

Also, keep in mind that in my system, the heater is not submerged in the water, but instead, heats the pot like an electric burner. This type of heating will be less efficient than the submerged type.

Anyway, at a bath temp of 185F, the covered pot required an on-cycle of about 20% (200W). An uncovered pot used more than double that amount at that temperature. At 140F, the covered pot required an on-cycle of about 10% (100W), while the uncovered version used about 50% more (150W). At my preferred salmon cooking temp. of 115F (I gradient cook my salmon using SV dash), the covered and uncovered values are practically identical, at about 4-5% (40-50W).

I never tested an insulated bath because, at the time, I was designing the circulator with an eye towards production for the mainstream. So, it was designed to look less like lab equipment, and more like a kitchen appliance. How many mainstream people can you imagine wrapping their pot with insulation? And, once I saw the steady state values and compared them to traditional cooking methods, I figured that an extra few percent (meaning maybe 100W max.) would be inconsequential to the eventual marketing. SV, by its nature, is already orders of magnitude more efficient than using your oven.

Like pbear said, given enough time, even the best insulation will eventually equilibrate to the bath temperature. But, even though the outside may be warm to the touch, I'd guess that the power outputs would be less than an uninsulated bath. If you think of an analogy to a wetsuit when diving, the outside of the wetsuit will still feel warm, however, the wearer is still significantly warmer wearing it than not. The wetsuit traps water between it and the wearer, and the dead water, once warmed to body temp, doesn't require much more heat to stay at equilibrium. So, let's say you were diving in 50F water - with the suit, you might get uncomfortably cold after about an hour or two (I found this out by experience years ago), but without the suit, you wouldn't last 5 minutes!

Paul - in your situation, I wonder if the Reflectix is doing very much, other than limiting evaporative cooling. It might be just as well to use the tight fitting lid without it. Once the air space between the water and lid becomes saturated at 100% humidity, evaporation will stop, assuming there are no leaks where humidity can escape. At a certain point, increasing side insulation is "gilding the lily" so to speak - a lot more effort for not much gain.

ETA: Paul - Maybe a way to save a bit more energy would be to wrap the cooler with a reflective surface - like aluminum foil or something to reduce the radiation losses. Having a surface with lower emissivity should help. So then, the primary heat loss would be through convection, which, unless there's a breeze, should be minimal.

Edited June 20, 2014 by KennethT (log)