How long will it take to heat 100 litres of water with a 42kW heat exchanger, presuming no heat loss?
Insufficient information. Initial and final temperature of the water must be specified.
Hello nigel : The water mass, TW1, and TW2 are needed to work this problem : ---------------------------------------... Assume : mw = 50 kg Tw1 = 25 C Tw2 = 75 C Qw = ( mw ) ( Cpw ) ( Tw2 - Tw1 ) Qw = ( 50 kg ) ( 4.184 kJ/ kg - C deg ) ( 75 C - 25 C ) = 10460 kJ t = Qw / Qwdot t = ( 10460 kJ ) / ( 42 kJ per s ) = 249 s <---------------------------------------...
Since it takes about 4.2 kJ to heat 1 kg of water by 1 degree C, the provision of 42 kJ/s would heat 100 kg (100 liters) of water by 1 degree C in about 10 seconds (or 2 degrees in 20 seconds, etc.).
It depends upon temperature rise ΔT !!!!! Lets assume ΔT = 1.0°C...then : Q = P*t = 100*4186*1 t = Q/P = 100*4.186/42 = 10 sec approx. dT/dt = 1.0/10.0 = 0.1°C/sec it means that water temperature will increase by 1.00°C every 10 seconds !!!....now you are in position to compute time as function of whichever temperature rise !!!!!
