Answer:
B.197 gpm and 12.4 L/s
Explanation:
Given that
Load Q = 404.5 KW
Water inlet temperature= 6.1 °C
Water outlet temperature= 13.9°C
We know that specific heat for water
[tex]C_p=4.187\ \frac{KJ}{kg.K}[/tex]
Now from energy balance
[tex]Q=\dot{m}C_p\Delta T[/tex]
by putting the values
[tex]Q=\dot{m}C_p\Delta T[/tex]
[tex]404.5=\dot{m}\times 4.187(13.9-6.1)[/tex]
[tex]\dot{m}=12.38\ \frac{kg}{s}[/tex] (1 Kg/s = 15.85 gal/min)
We can say that
[tex]\dot{m}=196.31\ \frac{gal}{min}[/tex]
We know that
[tex]\dot{m}=\rho\times volume\ flow\ rate[/tex]
12.38=1000 x volume flow rate
[tex]volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}[/tex]
So
volume flow rate = 12.38 L/s
So the option B is correct.