Heating water power calculation
Heating water power calculation
(OP)
Dear Reader,
I know this is a chemical engineering forum and thought this might be my best bet, I seem do hav displaced all my physics lectures from varsity ..
I warm up 1m ³ water from 12 °C to 16°C within 15 min…
Does anybody have the formula to work out the required power in kilo-watts (kw)
I would appreciate any help on this
Kind Regards
Rheinhardt.
I know this is a chemical engineering forum and thought this might be my best bet, I seem do hav displaced all my physics lectures from varsity ..
I warm up 1m ³ water from 12 °C to 16°C within 15 min…
Does anybody have the formula to work out the required power in kilo-watts (kw)
I would appreciate any help on this
Kind Regards
Rheinhardt.
--Off all the things i've lost , i miss my mind the most--





RE: Heating water power calculation
Take it from 1 kW = 1 kJ/s = 3600 kJ/h = 860 kcal/h
RE: Heating water power calculation
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C == 4186 j/kg °C
Q=c*m*deltaT
c = 4186 j/kg°C
m = 1000kg = 1000l = 1m³ of water
deltaT = 60°C-12°C >> I have an error in first post it is 60°C deltaT=48°C
Q=200.928 Million Joules
Since a kilowatt-hour is 3.6 million Joules, this energy amounts to about 55.81 kw/h
Now 15 min of an hour is 4 times quicker
Result is 223.24 kW not taking into account thermal losses
Regards
Rheinhardt
--Off all the things i've lost , i miss my mind the most--
RE: Heating water power calculation
Rheinhardt, excellent thinking. One thing you didn't lose is your intellectual power.