Low density element's

[2much2spend]

I'm looking for a new heating element but I'm curious to know what's the difference between a low density and ultra low density element's ?
These are the ones I'm looking at.
http://www.stilldragon.com.au/4800-watt-element-ultra-low-density/
http://www.stilldragon.com.au/4800-watt-super-low-density-element
http://www.stilldragon.com.au/3600-watt-element-special-runout-low-density/
http://www.stilldragon.com.au/3600-watt-incolloy-element-ultra-low-density/
/

The KK element has just given up the ghost .
So I want something better!

 

[sp0rk]

Ultra low is usually below around 60W per square inch (I think)
I know both 2400W elements from Five Star Distilling are ULWD, not sure about the stilldragon ones
http://www.5stardistilling.net/element-guard-2400w/
The lower the watt density, the lower the stressing on materials, so it theoretically last longer
Also it will have a lower surface temp and scorch less
MHB once told me the Braumeister element is cool enough to touch during mash temps, because it's watt density is so low (it is a very long element)

 

[2much2spend]

So the elements that are like a solid rod compared with the thinner one that look like a bent tube are more efficient ?
I'm thinking the less surface area is more efficient and less power load?

 

[A3k]

it's not an efficiency thing. With the element, the density is referring to the output energy per area. A high density element will apply the same amount of energy as a low density one (if the same power), but in a much smaller area. This causes the element to heat up much more and potentially burn the beer. if used in mashing, this is worse as convection is reduced.
Lower density ones are better for wort as they apply the energy over a larger area, and reduce hot spots / burning.

cheers
Al

 

[sp0rk]

More surface area has a lower surface wattage,
There's no difference in efficiency (total input wise)
The rod type elements (the weldless ones) don't necessarily have lower watt density that normal tube elements, it all depends on the length and diameter
For example the Five Star 2400W tube element has slightly lower watt density than their 2400W rod element
To calculate your watt density, you divide the element wattage by the effective surface area (diameter x heated length x 3.14)

EG; a 2400W element with .496" diameter and 11.25" heated length (just random figures here)

.496 x 11.25 x 3.14 = 17.52
2400 / 17.52 = 136.98

The watt density of that element would be 136.98 W/in2, which would be a fairly high watt density

 

[Online Brewing Supplies]

This causes the element to heat up much more and potentially burn the beer. if used in mashing, this is worse as convection is reduced.


cheers
Al

I tend to agree but in all my time it doesnt appear to happen often. I have seen burning on the elements during the early part of the mash (protein stage) only when the mash is not kept moving and this was from an element produced specifically for a mash bucket back in the 90,s. So dont be too concerned and pay a heap extra if using just in a kettle.
Nev

 

[dave_h]

"For example the Five Star 2400W tube element has slightly lower watt density than their 2400W rod element
To calculate your watt density, you divide the element wattage by the effective surface area (diameter x heated length x 3.14)

EG; a 2400W element with .496" diameter and 11.25" heated length (just random figures here)

.496 x 11.25 x 3.14 = 17.52
2400 / 17.52 = 136.98"

Would you not need to multiply your 17.52 by two as there are the element does not just have one rod, ie it comes back again?

 

[sp0rk]

"For example the Five Star 2400W tube element has slightly lower watt density than their 2400W rod element
To calculate your watt density, you divide the element wattage by the effective surface area (diameter x heated length x 3.14)

EG; a 2400W element with .496" diameter and 11.25" heated length (just random figures here)

.496 x 11.25 x 3.14 = 17.52
2400 / 17.52 = 136.98"

Would you not need to multiply your 17.52 by two as there are the element does not just have one rod, ie it comes back again?

Oops, you're right there
but if we're talking about the rod type weldless element, we wouldn't times that by 2

 

[dave_h]

It might be because Im full of a cold but my calcs dont seem right....


A picture is a thousand words so...



With this element


Watts / (Diameter x total length of element x Pi)

in mm

2500 /(8 x (260+260+220+220) x 3.14)

in inchs

2500/ (0.3149 x 373.7 x 3.14)

This comes out as 6.7W/in2 which just cant be right, am I missing something?

 

[GibboQLD]

It might be because Im full of a cold but my calcs dont seem right....

Watts / (Diameter x total length of element x Pi)

in mm

2500 /(8 x (260+260+220+220) x 3.14)

in inchs

2500/ (0.3149 x 373.7 x 3.14)

This comes out as 6.7W/in2 which just cant be right, am I missing something?

Yep, decimal place in the wrong spot -- (260+260+220+220)mm = 37.795in.

Works out to 66.85 W/in².

 

[dave_h]

Sweet thanks GibboQLD