Answer
$\dot{m}_{coal}=2.893\times10^6\ kg/day$
$\dot{m}_{air}=401.8\ kg/s$
Work Step by Step
$\eta = \dot{W}_e/\dot{Q}_H$
Given $\eta = 0.32,\ \dot{W}_e=300\ MW$:
$\dot{Q}_H=937.5\ MW=8.1\times10^7\ MJ/day$
$\dot{Q}_H=\dot{m}q$
With $q=28\ MJ/kg$
$\dot{m}=2.893\times10^6\ kg/day=33.48\ kg/s$
Since $r=12\ kg_{air}/kg_{coal}=\dot{m}_{air}/\dot{m}_{coal}$
$\dot{m}_{air}=401.8\ kg/s$
