Electric v Gas

[DWC]

Hi all, I'm thinking about changing from gas fired to electric kettle. 3600W lwd element in a 50L keggle.
Just looking for pros and cons for doing this. I'm a sparky so wiring is not an issue. Only cons I can think of are cleaning and whirlpool being more difficult. 3v HERMS btw.
Cheers, Dave

 

[thumbsucker]

I would never go electric, I have a Rambo burner and even if its 10ºc outside I can bring my 50L keggle to a heavy rolling boil with ease in under 20 minutes. I have to be careful with boil overs. I have a mate who used to have electric and if it was cold he struggled to get a rolling boil, he switched to gas.

Also with gas I can really do some nice maillard reactions with the wort.

The problem with gas is that you can run out and that is a pain, so I have two 9kg tanks that way I always have a spare.

Electricity is always on so thats is a plus for electricity. Maybe if you got two 3600W elements then you maybe rolling.

 

[seamad]

I've got an all electric rig, 15 amp RIMS. Preboil kettle volume about 56l. Have 2 over the side 10 amp elements, usually turn them down to 90-95% once boil reached . Kettle is fully insulated ( glued on black foam rated to 120C iirc). Don't whirlpool anymore, use brewbrite, at end of boil take out elements, put lid on kettle. Break setlles pretty solid in about 10 minutes. Discard first cup out of kettle then fill 2 cubes. Break stays put in bottom of kettle. Clean up is really easy ( get my brother to do it), the elements are easily cleaned compared to the RIMS element.
With 5kW of solar on the roof, on a sunny day the energy costs would be pretty low.
I also have a rambo burner that I did use before going all electric and insulating the kettle, no real difference in boil intensity that I can see.

 

[Ducatiboy stu]

Gas for me. Just works and gives nice control...and probably cheaper

Gas is great for caramelising first running for Scottish Ales

 

[Benn]

Are you brewing indoors or outdoors? I don't use the gas fired keggle in my single car garage anymore, the funny taste in my mouth and heavy headidness was detracting from the sheer extacy of hanging around the kettle weighing out multiple Hop additions.
Gas outside, electric inside for me.

 

[Matplat]

Electric is cheaper to run, and always available, plus its pretty much silent compared with noisy gas burners. You can use a simple temp controller to maintain mash temps too.
Only downside is slower ramp rates if you're on the limit of your wiring circuit. You don't need a raging rapid boil to make delicious beer.

Someone on here did a calc to work out how much of the energy thats in an 8.5kg bottle of gas had actually been used for boiling the wort.... was somewhere around 13% I believe....

 

[droid]

A keggle with 2 x 2200 watt elements running is too much and one is not enough - 3600 sounds about perfect - especially if you put a power controller on as well, to be sure to be sure.

^personal experience - yours may differ

 

[Tony121]

A keggle with 2 x 2200 watt elements running is too much and one is not enough - 3600 sounds about perfect - especially if you put a power controller on as well, to be sure to be sure.

^personal experience - yours may differ

I run 3600 watts in a keggle and is perfect, ramp down to about 70% once rolling boil is established.

 

[DWC]

Matplat, that' pretty much my reasoning.

 

[pirateagenda]

electric is practical if you have the amps required or do under double batch size

Gas is easier for big batches, unless you want to spend a fortune getting set up.

 

[evoo4u]

Electricity is always on

Half your luck. Where I live a generator wired into the meter box with a changeover switch is pretty much essential. At least then you can still run lights, water pump, and watch tv etc.

 

[Lyrebird_Cycles]

Someone on here did a calc to work out how much of the energy thats in an 8.5kg bottle of gas had actually been used for boiling the wort.

Easy version of that calc: Percent efficiency = (4.8 x start vol - 4.52 x finish vol) / mass of gas used.

Calculation assumes the wort was heated from 65 oC, that it has a heat capacity about 4 kJ/kg, that the vapour driven off is substantially water and that there is no condensate return.

 

[malt junkie]

Easy version of that calc: Efficiency = (2.4 x start vol - 2.26 x finish vol) / 50 x mass of gas used.

I think he meant potential stored energy vs the actual energy used vs the actual energy required to do the job. The actual energy expended by an element immersed in the liquid you are aiming to heat is as high efficiency as your going to get on a HB scale. Obviously at larger scales the speed at which energy(heat) can be exchanged without excessive milliard reactions requires steam and in commercial application with heat recovery systems in place efficiency is quite high but still not quite that of an immersed element.

 

[Lyrebird_Cycles]

I don't understand your post but to be clear the calculation gives the efficiency according to a standard engineering definition eg work done / energy expended. It should be obvious that the "energy expended" is the heat of combustion of the gas.

 

[Ducatiboy stu]

Looking at that gas energy table... I am going to go Hydrogen as it has the most heat value.

Just got to buy bulk caustic soda and aluminum foil

 

[Matplat]

Yes, I didn't look into it in detail, but my understanding of it is the energy required to convert a volume of water into steam (boil-off volume) + energy req'd to raise entire starting volume of wort to boiling / energy consumed (heat of combustion of 8.5kg LPG (butane?))

 

[Lyrebird_Cycles]

Yes, that's what my equation does.

Enthalpy of vaporisation of water is around 2.26 MJ / kg, so the energy required for boiloff =

2.26 MJ/ kg x (Vstart - Vend) (Eq 1)

Heat capacity of wort is around 4 kJ / kg / oC, (slightly less than water at 4.18 due to the sugar content), so the energy required to heat wort =

4 kJ / kg / oC x Vstart x Trise (Eq 2)

Heat of combustion of propane is around 50 MJ / kg,so the energy from burning gas =

50 MJ / kg x mass of gas burnt. (Eq 3)

Set T rise to 35 degrees, Eq 2 becomes

E = .014 MJ / kg x Vstart (Eq 2a)

Add eqns 1 and 2a to get

E = 2.4 MJ / kg x Vstart - 2.26 MJ / kg x Vend

Efficiency is thus (2.4 x Vstart - 2.26 x Vend) / 50 x mass of gas burnt*.

Multiply both sides by 100 / 50 to get

Efficiency (in percent) = (4.8 x Vstart - 4.52 x Vend) / mass of gas

which is where we started.


I presented the bare finished product above because experience says 99% of the audience hasn't followed this.






* The units are consistent so they drop out of the quotient. The vapour is substantially water so 1 litre lost = 1 kg. The heat capacity of wort is dependent on sugar content, as is the density but they move in opposite directions so assuming 1litre = 1 kg introduces a trivial error.

 

[Zorco]

Go electric. I'm a certifier for 266MW of renewable energy farms in Queensland...

Your sexual attractiveness, brewhouse mastery and enjoyment of beer are all enhanced when you brew on electric rigs.

Plus, other than fish tanks, you're blissfully breaking the rule of mixing water and electricity.

When my entire rig is electric I'll probably miss the portability (driveway brewdays at a mate's)..

 

[buckerooni]

the ability to automate an electric setup was the kicker for me. running 2 x 3600w elements in a 100L pot - it ramps 80L at 1 deg per minute. In terms of cleaning, I went the tri-clamp fittings and removable elements. while pricey to set up it's great for cleaning and portability.

 

[malt junkie]

Yes, that's what my equation does.

Enthalpy of vaporisation of water is around 2.26 MJ / kg, so the energy required for boiloff =

2.26 MJ/ kg x (Vstart - Vend) (Eq 1)

Heat capacity of wort is around 4 kJ / kg / oC, (slightly less than water at 4.18 due to the sugar content), so the energy required to heat wort =

4 kJ / kg / oC x Vstart x Trise (Eq 2)

Heat of combustion of propane is around 50 MJ / kg,so the energy from burning gas =

50 MJ / kg x mass of gas burnt. (Eq 3)

Set T rise to 35 degrees, Eq 2 becomes

E = .014 MJ / kg x Vstart (Eq 2a)

Add eqns 1 and 2a to get

E = 2.4 MJ / kg x Vstart - 2.26 MJ / kg x Vend

Efficiency is thus (2.4 x Vstart - 2.26 x Vend) / 50 x mass of gas burnt*.

Multiply both sides by 100 / 50 to get

Efficiency (in percent) = (4.8 x Vstart - 4.52 x Vend) / mass of gas

which is where we started.


I presented the bare finished product above because experience says 99% of the audience hasn't followed this.






* The units are consistent so they drop out of the quotient. The vapour is substantially water so 1 litre lost = 1 kg. The heat capacity of wort is dependent on sugar content, as is the density but they move in opposite directions so assuming 1litre = 1 kg introduces a trivial error.

Understood this much better, keep this up and one day I may have to change my sig.

the ability to automate an electric setup was the kicker for me. running 2 x 3600w elements in a 100L pot - it ramps 80L at 1 deg per minute. In terms of cleaning, I went the tri-clamp fittings and removable elements. while pricey to set up it's great for cleaning and portability.

Gas can be automated it's just painful and expensive to do.