Respuesta :
Answer:
Q = 30.07 kJ/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[tex]h_f[/tex] value is 2406 kj/kg
rate of heat transfer (Q) can be determine by using following relation
[tex]Q = \dot{m} h_f[/tex]
putting all value to get Q value
Q = 45 *2406
Q = 108270 kJ/h
[tex]Q = 108270 *\frac{1}{3600} kJ/s[/tex]
Q = 30.07 kJ/sec
Explanation:
According to the water table, the value of enthalpy of evaporation at a temperature of [tex]40^{o}C[/tex] is 2406.0 kJ/kg.
Hence, we will calculate the rate of heat transfer by using the formula as follows.
Q = [tex]m \times h_{fg}[/tex]
where, m = mass
[tex]h_{fg}[/tex] = enthalpy of evaporation
Putting the given values into the above formula as follows.
Q = [tex]m \times h_{fg}[/tex]
= [tex]63 kg/h \times 2406.0 kJ/kg[/tex]
= 151578 kJ/h
or, = [tex]151578 \times \frac{1}{3600} kJ/s[/tex]
= 42.105 kW
Thus, we can conclude that rate of heat transfer from the steam to the cooling water flowing through the pipe is 42.105 kW.
