5500w ULD element taking too long to heat

[ajdelange]

As heating sets up currents the problem with pot configuration is not that the heat can't get to the sides but that once it gets there it is lost through the sides. The larger the surface area the more heat passes through the surface and the less is available to heat the water. It should be clear that the ideal situation would be one in which the surface area of the volume of water were the absolute minimum required to contain the volume. For this to be the case the 'pot' would need to be a sphere - not a very practical configuration for a home brewer. Practicality dictates that we use cylindrical HLTs, kettles, mashtuns.... For a cylindrical vessel the height of the liquid should be the same as the diameter of the vessel. This results in a minimum surface area for the volume contained.

The attached graph shows the ratio of the surface area of a cylindrical volume to the surface area of a sphere of equal volume as a function of the ratio of the diameter of the pot to the height to which it is filled. As the plot shows if the height of fill is equal to the diameter the surface area is only 14.5% greater than that of the ideal spherical container and, all else being equal one would lose 14.5% more heat from a kettle filled to the height of its diameter than from the sphere. If, conversely, one filled to 3 times the diameter or 1/3 the diameter the surface area would be 29% greater than that of the equivalent sphere or twice that of a cylindrical kettle filled to the diameter. Loss in this case would be twice the loss of the optimally filled cylindrical vessel.

Thus a tall skinny kettle or a short squat one isn't as efficient as a 'square' one but it takes quite a bit of 'aspect ratio' to appreciably change the surface area ratio. A 3:1 kettle loses twice as much heat as a 1:1 kettle. If the heat loss is 2% of the total in the 1:1 kettle it is 4% in the 3:1 kettle. Not a big deal. But if the loss is 40% in the 1:1 kettle it is 80% in the 3:1 and it is.

If you want to do an interesting experiment run your kettles as is and then repeat with them wrapped in a quilt or blanket. Be sure the lid is on and is insulated too. Try it with the lid on and off.

 

[Yambor44]

Interesting ajdelange. My kettle is 17.5" tall and 16" in diameter. I fill to about 15.5-16" so my height and diameter of my volume is almost identical. Here's a short video of my first brew.

 

[passedpawn]

Watched your video. Shiney!

Regarding the wort not boiling in your hop screen: don't worry about that. The boil only happens at the surface of the heating element. It is there that the water is converted to vapor. Nowhere else. That's where the energy is entering the pot.

The agitation you see at the surface of the pot is just the buoyant vapors rushing up from the element. Those vapors aren't going to enter your screen since the only force they are seeing is straight up, and the surface tension of the bubbles keep them from going through the small holes. I've been using one of these for a long time and I don't have a utilization problem (AFAIK).

 

[Yambor44]

Thanks passedpawn! That makes perfect sense.

 

[ajdelange]

The boil only happens at the surface of the heating element. It is there that the water is converted to vapor. Nowhere else.

Look at the video carefully (or observe your own pot next time you brew). You will see a lot of turbulence at the surface right over the element but you will also see small bubbles all the way out to the edges of the pot. Keep in mind that the boiling point of water is 100 °C at the surface of the pot because, on a standard day at sea level, the pressure is 1013.25 mB and the vapor pressure of water is 1013.25 at 100 °C. At the bottom of the pot, under a foot of wort, the pressure is approximately 1013.25 + 1013.25/33.8 = 1043.3 mB because 33.8' of water column exerts a force of about 1 atmosphere. The temperature of water at which the vapor pressure is 1043 mB is a little over 100.8 °C thus the boiling point at the bottom of the pot (where the heater is) is higher than it is at the surface. Wort hotter than 100 °C but cooler than 100.8 °C wont flash to vapor at the bottom of the vessel but may do so as it rises (carried up by the currents that are plainly visible) to a point where the pressure is lower provided that it does not cool faster than the slope of the temperature vs. vapor pressure curve. Right or not that's my theory as to where the small bubbles come from.

 

[passedpawn]

Look at the video carefully (or observe your own pot next time you brew). You will see a lot of turbulence at the surface right over the element but you will also see small bubbles all the way out to the edges of the pot. Keep in mind that the boiling point of water is 100 °C at the surface of the pot because, on a standard day at sea level, the pressure is 1013.25 mB and the vapor pressure of water is 1013.25 at 100 °C. At the bottom of the pot, under a foot of wort, the pressure is approximately 1013.25 + 1013.25/33.8 = 1043.3 mB because 33.8' of water column exerts a force of about 1 atmosphere. The temperature of water at which the vapor pressure is 1043 mB is a little over 100.8 °C thus the boiling point at the bottom of the pot (where the heater is) is higher than it is at the surface. Wort hotter than 100 °C but cooler than 100.8 °C wont flash to vapor at the bottom of the vessel but may do so as it rises (carried up by the currents that are plainly visible) to a point where the pressure is lower provided that it does not cool faster than the slope of the temperature vs. vapor pressure curve. Right or not that's my theory as to where the small bubbles come from.

Hmm, I hadn't considered the pressure difference from the column of water. Thanks for that.

I keep a vigorous boil. There are bubbles everywhere. I haven't noted the effect you mentioned, but I'll look for it next time I brew.

 

[NiteOwlBrewing]

This may be a bit off topic (thank the white rajah), but here are my experiences with the camco 5500 ULWD that everyone uses on my first brew:
This was a 21g boil with the element unrestricted (contactor run straight to outlet for element)
I fired the element after first runnings covered the element
On second runnings (of a batch sparge) it boiled before I could get third/fourth runnings into kettle, but it didn't seem overly vigorous
Keep in mind this was my first 1/2bbl AND first all electric brew
This is a kettle that is ~19.5" in both dimensions with reflectix sitting on a towel for insulation/leakage catch
It seemed to boil about the same with full volume once temp was reached; wasn't overly vigorous, but seemed to be boiling

My questions are do you normally "throttle" back the element by 25% and what do you do about your boil off rate? Seemed I would be fine boiling ~13-14 with it at full bore. Looked like I'd be losing ~2g regardless of level of kettle volume; is this anyone's experience? So, do you adjust your boiloff percentage based on batch size to equal ~2g for a 60 min boil (in my case)?

Thanks and apologies if necessary...

 

[passedpawn]

This may be a bit off topic (thank the white rajah), but here are my experiences with the camco 5500 ULWD that everyone uses on my first brew:
This was a 21g boil with the element unrestricted (contactor run straight to outlet for element)
I fired the element after first runnings covered the element
On second runnings (of a batch sparge) it boiled before I could get third/fourth runnings into kettle, but it didn't seem overly vigorous
Keep in mind this was my first 1/2bbl AND first all electric brew
This is a kettle that is ~19.5" in both dimensions with reflectix sitting on a towel for insulation/leakage catch
It seemed to boil about the same with full volume once temp was reached; wasn't overly vigorous, but seemed to be boiling

My questions are do you normally "throttle" back the element by 25% and what do you do about your boil off rate? Seemed I would be fine boiling ~13-14 with it at full bore. Looked like I'd be losing ~2g regardless of level of kettle volume; is this anyone's experience? So, do you adjust your boiloff percentage based on batch size to equal ~2g for a 60 min boil (in my case)?

Thanks and apologies if necessary...

Boil-off percentage is solely dependent on the power being put into the pot. So, if you keep it at 75% with the same insulated pot, it will boil off at the same rate regardless how full the pot is.

Of course, AJ's comment above regarding the boil temp at the slightly higher pressure at the bottom of the pot has me worried, but that won't effect this. For every X units of energy put into the pot, a gallon of water will disappear into vapor.

 

[ajdelange]

Don't worry about the change in temperature. Its effect on heat of evaporation is small. Have a look at the diagram at http://whitebrooksolutions.com/documents/P-HDiagram_Water.pdf. Note that as one goes up in temperature the liquid saturation lines and steam saturation lines do move closer together but slowly. At 200 °F they are about 970 BTU/lb apart while at 250°F they are about 940 BTU/lb apart. Thus a change in temperature of 50 ° results in a 30 BTU/lb difference in heat of vaporization. This is 100*30/970/50 = 0.06% per °F. Also keep in mind that the boiling point is higher only at the bottom of the kettle - at the surface it is whatever the local atmospheric pressure dictates and is intermediate at intermediate depths.

 

[NiteOwlBrewing]

Thanks fellas for the input. My question on boil off rate centers more around whether it is a % of volume of the liquid in the kettle or more an absolute volume. In my test with 18 gallons I boiled off ~2 gallons in 60 minutes of boiling. This is around 11%. So, if I'm boiling 21 gallons would my boil off be the 11% of the 21 gallons (~2.3g) or the 2 gallons (~9.5% of the 21 gallons)? Seems to me the surface area of the liquid is the same, so the evaporation amount in gallons would be roughly the same as well.

Also of note, I'm not sure I see a reason to dial back the 5500 with boil sizes in the range of 15-21 gallons. It may be the shape of my pot driving this mostly though. Are other people getting violent boils with a 5500 ULWD? I may try another 10 gallon batch to see if the boil seems to vigorous.

 

[Dan]

What size is your conductor?

 

[NiteOwlBrewing]

That seems a rather personal question...

I'm using #10 for all wiring in the box that is 240 as it is all protected by 25a breakers. Could it be that I'm using the flexible silicone sheathed stuff? It only gets mildly warm with current flowing. The elements are run from the box on 30a dryer cords.

 

[ajdelange]

The rate of evaporation depends on the rate at which energy is supplied to the liquid. It takes 2260 kJ (kw-sec) to vaporize 1 kg of water whether there are 40 kilograms in the kettle or 14. As long as the power delivered to the water is the same the same mass will be vaporized per unit time. Note, however, that 14 kg of water has less surface area than 40 so that there will be more heat lost to the surroundings in the larger volume and less of what the element produces is delivered to the water. Thus the boil off rate will be slightly higher later in the boil than earlier.

 

[NiteOwlBrewing]

The surface heat lost through the sides is mildly mitigated by reflectix. So, it sounds like roughly the same volume assumption should be used for boil off. Wondered why beersmith has it as a % to volume.

 

[passedpawn]

The surface heat lost through the sides is mildly mitigated by reflectix. So, it sounds like roughly the same volume assumption should be used for boil off. Wondered why beersmith has it as a % to volume.

Beersmith 2 fixed this. Enter a Boil Off amount (gallons) in the Equipment Profile for your equipment. You can either select a fixed amount for the entire session, or an hourly amount if you check the checkbox under Boil Off entry.

Beersmith shows a Evaporation Rate number below that, which is still a mistake. Two mistakes, actually, since evaporation is the wrong term here. Vaporization Rate or Boiling Rate would be correct. Evaporation Rate isn't. I guess it's a semantics, but I like to keep vaporization due to evaporation and boiling separate in my head.

 

[ajdelange]

Keep in mind that evaporation and vaporization mean the same thing: the transition from liquid to vapor phase. That occurs below the boiling point and it occurs above the boiling point and the same amount of energy (as adjusted per the discussion in #129) is required to transition a given mass whether ebullition is occurring or not.

Heat lost to vaporization/evaporation before the commencement of boiling does not remove that much water but it does lengthen the time it takes water to come to a boil as the heat which is going into vapor formation is not going into raising the temperature of the rest of the water. This graph shows two runs in which the same volume of water was heated to boiling in a 55 gal drum. For the blue curve the cover was on so that any vapor condensed and was returned, along with most of the enthalpy of evaporation. For the red curve the cover was off so the vapor could escape taking the enthalpy of evaporation with it.

 

[passedpawn]

Keep in mind that evaporation and vaporization mean the same thing

Yes, semantics (sort of - note that sublimation is a form of vaporization too). I'd say (in my own mind) that evaporation is just one form of vaporization. I don't want to get into an argument over semantics (especially with you!).

I graphed a boil session myself with my logging thermometer (in my keggle). While the non-linearity is not as exaggerated as the one you showed, it is similar. The fact that the curve is not linear is a good reason to differentiate between boiling vaporization and evaporation, a way to keep the variables separate.

Thanks for pointing out the energy loss due to the evaporation. I assumed the slight curve was heat loss to the environment through the un-insulated keggle walls. As the temp of the water increased, the loss was faster and hence the curve. I should repeat this sometime with insulation.

 

[jeffmeh]

I believe that, generally speaking, evaporation rate will be a function of the surface area of the kettle, i.e. a constant volume per time multiplied by the surface area. This is likely an oversimplification, and assumes everything else is held constant (amount of heat energy applied, ambient temperature, humidity, barometric pressure, wind, etc.). That said, it should be good enough for brewing estimates.

For laughs, at the minimum heat you need to get a good rolling boil, see how close you come to (0.00729 gal/sq in/hr * surface area sq in) for your gal/hr boil off rate.

 

[ajdelange]

Your belief flies in the face of physics as it does not include the amount of energy being delivered to the water/wort. If, in a boiling kettle, you double the energy per unit time (power) delivered to the kettle by switching on a second element, increasing the gas flow to a burner, increasing the amount of steam delivered to a jacket or whatever where can that energy go? It cannot go into increasing the temperature or pressure in a vessel open to the atmosphere (though it can in an enclosed boiler e.g. a pressure cooker). It cannot be lost through the sides of the vessel as the heat loss there is proportional to temperature and the temperature is fixed (though in an improperly skirted gas system some of the heat from the bigger flame would be lost). It can't go into the creation of mass. The only place that energy can go is to increased rate of vaporization. I don't see how you could have failed to observe that if you put a pot on the stove it boils more vigorously when you turn up the heat. Same surface area - more steam production.

 

[ajdelange]

I assumed the slight curve was heat loss to the environment through the un-insulated keggle walls. As the temp of the water increased, the loss was faster and hence the curve. I should repeat this sometime with insulation.

That is a factor but not the only one. If you are going to repeat the experiment try it with the lid on as well.

 

[jeffmeh]

Your belief flies in the face of physics as it does not include the amount of energy being delivered to the water/wort. If, in a boiling kettle, you double the energy per unit time (power) delivered to the kettle by switching on a second element, increasing the gas flow to a burner, increasing the amount of steam delivered to a jacket or whatever where can that energy go? It cannot go into increasing the temperature or pressure in a vessel open to the atmosphere (though it can in an enclosed boiler e.g. a pressure cooker). It cannot be lost through the sides of the vessel as the heat loss there is proportional to temperature and the temperature is fixed (though in an improperly skirted gas system some of the heat from the bigger flame would be lost). It can't go into the creation of mass. The only place that energy can go is to increased rate of vaporization. I don't see how you could have failed to observe that if you put a pot on the stove it boils more vigorously when you turn up the heat. Same surface area - more steam production.

If that was directed at me, of course you are correct. I was trying to point out that with everything else held constant, the boil-off rate is a function of the open surface area. I failed to explicitly point out that the heat applied must also be held constant. Holding that constant is much easier with electric than propane, lol. I will edit the post accordingly.

 

[Yambor44]

Thanks fellas for the input. My question on boil off rate centers more around whether it is a % of volume of the liquid in the kettle or more an absolute volume. In my test with 18 gallons I boiled off ~2 gallons in 60 minutes of boiling. This is around 11%. So, if I'm boiling 21 gallons would my boil off be the 11% of the 21 gallons (~2.3g) or the 2 gallons (~9.5% of the 21 gallons)? Seems to me the surface area of the liquid is the same, so the evaporation amount in gallons would be roughly the same as well.

Also of note, I'm not sure I see a reason to dial back the 5500 with boil sizes in the range of 15-21 gallons. It may be the shape of my pot driving this mostly though. Are other people getting violent boils with a 5500 ULWD? I may try another 10 gallon batch to see if the boil seems to vigorous.

See video in post #122 above. I have a 550W ULD and that is 75% power to the element. I later turned it down to 70% and finished boiling off about 2.25 gallons if I recall correctly starting with 13.25 gallons.