Watts, volts, amps, theory (please be gentle)

[Pausim]

I am very disappointed that the big hitters have not brought Peukert’s exponent into the bun fight.

 

[Cush]

I am very disappointed that the big hitters have not brought Peukert’s exponent into the bun fight.

That might be more precision than I need (edited, more than I can cope with ) for my OP

 

[Pausim]

Without wishing to blow anyone’s brain cells I feel it is worth mentioning that the Ah capacity of a battery is usually measured at a very slow rate of discharge. A 100Ah battery will often have been measured at only a 5A discharge rate. If you then drag 50A out of it you will get less than the original 100Ah out of it. This affects lead batteries far more than lithium ones.

 

[Raul]

12v, 800/12 x 3.7/60 = 4.111 ah, so in conclusion and accounting for a few losses and maybe the mugs were a bit full you could call it 7ah

You are spot on there, our kettle boils 800ml of water using 7Ah out of the battery at 12.8V. .

 

[Cush]

Without wishing to blow anyone’s brain cells I feel it is worth mentioning that the Ah capacity of a battery is usually measured at a very slow rate of discharge. A 100Ah battery will often have been measured at only a 5A discharge rate. If you then drag 50A out of it you will get less than the original 100Ah out of it. This affects lead batteries far more than lithium ones.

You put that very well, even I understood. Not something you hear much about and something to bear in mind. Is that the Peukert thingy (exponent) that applies less to Lifepo than Pb/ acid?

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[Pausim]

You put that very well, even I understood. Not something you hear much about and something to bear in mind. Is that the Peukert thingy (exponent) that applies less to Lifepo than Pb/ acid?

Yes Peukert devised a formula and exponent for calculating how much battery capacity is lost to heavier use. The effect on Li is much less than on Pb.

 

[BouncyBoatFolk]

I am very disappointed that the big hitters have not brought Peukert’s exponent into the bun fight.

Feel free to "lead" the way.
Mike.

 

[Rog NotJane]

If you go back to first principles Lenny

The specific enthalpy of water is 4.184 J/gK so that requires 4.184 Joules (1 calorie if you want to use gas) to raise one gramme of water through one degree K (same as deg C), assuming 2 mugs of water is half a litre so 500g, 500x4.184x85 (assuming the water tank to be at 15 deg C) = 177,820 J or watts per second as that is what a Joule is, gives the kettle is 800watts 177,820/800 = 222.275 seconds or 3.7 minutes, so energy required is 49.33 watt hours or with our obsession with ah at 12v, 800/12 x 3.7/60 = 4.111 ah, so in conclusion and accounting for a few losses and maybe the mugs were a bit full you could call it 7ah

Thank you getting some rigorous thinking into MHF.

Isn’t the slightly low figure because enthalpy increases as you heat it? 4.148 J/kg is the enthalpy of water straight out of Lenny HB ‘s fridge.

 

[funflair]

Thank you getting some rigorous thinking into MHF.

Isn’t the slightly low figure because enthalpy increases as you heat it? 4.148 J/kg is the enthalpy of water straight out of Lenny HB ‘s fridge.

I think I should really have said "specific heat capacity" rather that enthalpy, yes I believe the enthalpy increases as you heat it.

 

[Pausim]

Enthalpy, entropy and the other laws of thermodynamics make my brain hurt. I am off to make a cup of tea and don’t care how many Joules it takes.

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[Rog NotJane]

I think I should really have said "specific heat capacity" rather that enthalpy, yes I believe the enthalpy increases as you heat it.

The MHF Physics committee have voted and will forgive you. THIS ONCE. Don’t make the mistake again or I’ll have Jim ban you for bad physics.