From: legg@nospam.magma.ca   
      
   On Fri, 20 Dec 2024 07:16:35 -0000 (UTC), Wu Ming  wrote:   
      
   >legg  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   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)