Answer:
The answer 30.07J/sec
Explanation:
GIVEN DATA:
temperature of steam = 40 degree celcius
mass flow rate = 63 kg/h
from saturated water tables,
from temperature 40 °, enthalpy of evaporation value is 2406 kj/kg
rate of heat transfer (Q) can be determine by using following relation
putting all value to get Q value
Q = 45 *2406
Q = 108270 kJ/h
Q = 30.07 kJ/sec