The reading of pressure metre attached with a closed pipe is $$4.5 \times 10^4$$ N m$$^{-2}$$. On opening the valve, water starts flowing and the reading of pressure metre falls to $$2.0 \times 10^4$$ N m$$^{-2}$$. The velocity of water is found to be $$\sqrt{V}$$ m s$$^{-1}$$. The value of $$V$$ is _____.

Using Bernoulli's equation (the pipe is horizontal and both points are at the same height):

$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

When the valve is closed, $$v_1 = 0$$, $$P_1 = 4.5 \times 10^4$$ N/m$$^2$$.

When opened, $$P_2 = 2.0 \times 10^4$$ N/m$$^2$$, $$v_2 = v$$.

$$4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2}(1000)v^2$$

$$2.5 \times 10^4 = 500v^2$$

$$v^2 = 50$$

$$v = \sqrt{50}$$ m/s

So $$V = 50$$.

The answer is $$\boxed{50}$$.