Question 16.1: Air at a stagnation temperature of 27º C enters the impeller...

(i) From Eq (16.7), we can write

again with the help of Eq (16.5) and $T_{2 t}=T_{3 t}$ it becomes

$w=\Psi \sigma U_{2}^{2}=c_{p}\left(T_{2 t}-T_{1 t}\right)$                (16.5)

Here,         $p_{3 t} / p_{1 t}=4$

Therefore, $U_{2}^{2}=\frac{1005 \times 300 \times\left(4^{\frac{0.4}{1.4}}-1\right)}{0.85 \times 0.867 \times 1.05}$

which gives         $U_{2}=435 \mathrm{~m} / \mathrm{s}$

Thus the impeller tip radius is

(ii) Power input to the air $=\frac{2 \times 1.05 \times 0.867 \times(435)^{2}}{1000} \mathrm{~kW}$