Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P₁. The reading of the pressure gauge falls to P₂ when the valve is opened. The speed of water flowing in the pipe is proportional to

When the valve is opened, water starts flowing through the horizontal pipe, and the pressure decreases from $P_1$ to $P_2$. To determine the speed of water flow, we use Bernoulli’s equation, which states:

$$P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}$$

Since the pipe is horizontal, the gravitational potential energy term $\rho gh$ cancels out, reducing Bernoulli’s equation to:

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

Assuming that the water was initially at rest ($v_1 = 0$), we get:

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

Rearranging for $v$:

$$v = \sqrt{\frac{2 (P_1 - P_2)}{\rho}}$$

Conclusion:

The speed of water flow in the pipe is proportional to the square root of the pressure difference:

$$v \propto \sqrt{P_1 - P_2}$$