An open water tank in air springs a leak at position- 2 in Fig. $14-2$, where the pressure due to the water at position- 1 is $500 \mathrm{kPa}$. What is the velocity of escape of the water through the hole? The pressure at position- 2 in the free jet is atmospheric. We use Bernoulli's Equation with $P_{1}-P_{2}=5.00 \times 10^{5}$ $\mathrm{N} / \mathrm{m}^{2}, h_{1}=h_{2}$, and the approximation $v_{1}=0 .$ Then $$\left(P_{1}-P_{2}\right)+\left(h_{1}-h_{2}\right) \rho g=\frac{1}{2} \rho v_{2}^{2}$$
whence
$$v_{2}=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}}=\sqrt{\frac{2\left(5.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)}{1000 \mathrm{~kg} / \mathrm{m}^{3}}}=31.6 \mathrm{~m} / \mathrm{s}.
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