Re: Water kettle energy consumption logging

Wu Ming <[email protected]> wrote:
> Energy payback: W1*(t-tcool)+Wh =W2*t
> 
> Where:
> W1 is the average wattage at lower T
> W2 is the average wattage at 98°
> t is the number of hours 
> tcool is the time to cool down from 98° to T
> Wh is the energy required to re-heat from T to 98°
> 
> W1, W2 are known from my previous 
> tcool is unknown and I tried to guesstimate it
> Wh is calculated as:
> 
> T° C to 98° C	Wh	Wh *1.25
> 90	37.24	46.55
> 80	83.8	104.75
> 60	176.91	221.14
> After 6 hrs 21°	358.48	448.1
> 
> Second column from an online calculator. Wolfram Alpha as example. Third
> accounts for the not perfect energy transfer. 
> 
> Solving for t is simple. Then the fun begins. 
> 

Formatting appears to be a mess. On my client at least. Doesn’t know how to
keep LF.