Solution:
Temperature of air in the tank,
$
T_{1}=20+273=293 \mathrm{~K} \\
$
Diameter at the outlet of the nozzle,
$
D_{2}=25 \mathrm{~mm}=0.025 \mathrm{~m} \\
$
Area,
$\quad A_{2}=\pi / 4 \times 0.025^{2}=0.0004908 \mathrm{~m}^{2} \\
$
$
R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K}, \gamma=1.4 \\
$
(i) Mass rate of flow of air when pressure in the tank is,
$
140 \mathrm{kN} / \mathrm{m}^{2}(abs.) \\
$ :
$$
\rho_{1}=\frac{p_{1}}{R T_{1}}=\frac{140 \times 10^{3}}{287 \times 293}=1.665 \mathrm{~kg} / \mathrm{m}^{3} \\
$$
$$
p_{1}=140 \mathrm{kN} / \mathrm{m}^{2} \text { (abs.) } \\
$$
Pressure at the nozzle,
$\quad p_{2}=$,
atmospheric pressure,
$
=100 \mathrm{kN} / \mathrm{m}^{2} \\
$
Pressure ratio,
$$
\quad \frac{p_{2}}{p_{1}}=\frac{100}{140}=0.7143 \\
$$
Since the pressure ratio is more than the critical value, flow in the nozzle will be subsonic, hence mass rate of flow of air is given by,
$$
m=A_{2} \sqrt{\frac{2 \gamma}{\gamma-1} p_{1} \rho_{1}\left[\left(\frac{p_{2}}{p_{1}}\right)^{\frac{2}{\gamma}}-\left(\frac{p_{2}}{p_{1}}\right)^{\frac{\gamma+1}{\gamma}}\right]} \\\\
$$
$$
=0.0004908 \sqrt{\left(\frac{2 \times 1.4}{1.4-1}\right) \times 140 \times 10^{3} \times 1.665\left[(0.7143)^{\frac{2}{1.4}}-(0.7143)^{\frac{1.4+1}{1.4}}\right]} \\\\
$$
$$
=0.0004908 \sqrt{1631700(0.7143)^{1.4285}-(0.7143)^{1.7142}} \\\\
$$
$$or$$
$$
m=0.0004908 \sqrt{1631700(0.6184-0.5617)}=0.1493 \mathrm{~kg} / \mathbf{s} \quad \text { (Ans.) } \\
$$