Re: Water kettle energy consumption logging

On Fri, 20 Dec 2024 07:16:35 -0000 (UTC), Wu Ming <[email protected]
> wrote:

>legg <[email protected]> wrote:
>> The common solution with 'flash heaters' is to turn the system 
>> on x minutes prior to use and turn it off when the demand ceases.
>
>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. 

If it makes things any easier for you, already given energy loss 
from the container, you are probably aware that it takes a fixed 
amount of energy to raise 1cc of water by one degree. 

This is a characteristic of most momogenous materials, referred 
to as 'Specific heat'. Metric system units of neasurement are 
fairly simple.

Needless to say, the same thing goes when the material is cooling; 
it will loose the same energy per volume to produce similar 
temperature changes.

Knowing material volumes, delta T and timing gives energy gain 
or loss.  .  .  . either intentional or inevitable.

RL