If you look at an instruction set for cooking, you will most likely see a message like this: ‘boil the water and add the ‘tastysoopynotyetfood’ into it for 5 mins’. The blog is on how I learned to set the heating equipment (I use an electric hob) to keep the temperature of the water unchanged (approx) so that the food gets cooked well and the water doesn’t completely evaporated(approx).
To understand how to maintain the temperature first I needed to know how temperature of something increases ! Answer was when we supply heat. When it decreases? you guessed right, when the heat is removed.
Typical heat rating of a hob is 2KW maximum and mine is 1 KW in the medium setting. So this is the maximum energy that I can pump into the food. Now I needed to know how much heat is removed from the container when the water is heated and not boiling, this is very difficult to calculate but approximately this comes around 0.6KW to 1.6KW to 6KW (when boiling I don’t do this) (I Assumed the ambient temperature to be at 20 degrees and water to be at 100 degrees, area of the container to be 0.025 to 0.045 m^2).
- Heat out = A hard to find constant * Area of cooking vessel * (Temperature of food – Ambient temperature)
The interesting thing to note was that heat transfer from the container is proportional to the square of the container radius and this is the most important variable to consider as the ambient temperature and heat transfer coefficient of water doesn’t change much.
So here was my action plan. Firstly I need to find out approximately what is the watt rating of the setting that I use. For this I boiled 1L water in a container and saw the time it takes for it.
I found it to be 4800*DT/(Time to boil). Typical DT is 70 to 80. Took 80 for safety (No over cooking). So it is 4800*80/(Time to boil in secs). In my case it was 1 KW in the medium setting. Now things were easy. So the net heat I was pumping into the vessel was 1+0.6KW = 1.6KW in the medium setting.
- Energy needed to raise temperature of something= mass of something*an easy to find constant(4800)*temperature that you need to raise
I did this for all the settings. Found the pumped in heat, then to maintain the temperature I chose the appropriate container dia such that heat removed matched the heat that goes in.
- Example: Medium setting implies a vessel diameter of 24cm (A = 0.045 m^2)
The main take away was the following: The heat removed is proportional to the area of the container , so a 50% increase in diameter doubles the amount of heat transfer and I have to change the setting from lower medium (3) to upper medium(4) to have the same effect.
To keep it simple. boil in highest setting . move it to medium and see if it input output matching happens (water level don’t decrease a lot over time) . If it does then when you use a bigger container use the next higher setting and vice versa. (See how I found something with such an effort which you all already know)