Euler's equation is a fundamental equation in fluid mechanics that describes the relationship between pressure, velocity, and potential energy in one-dimensional flow. It is given by:

$$ \frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} = -\frac{1}{\rho}\frac{\partial p}{\partial x} - g\sin\theta $$

where $v$ is the velocity, $p$ is the pressure, $\rho$ is the density, $g$ is the acceleration due to gravity, and $\theta$ is the angle of inclination.



Given that the diameter of the pipe is 20 cm, the radius is 10 cm or 0.1 m. The initial discharge is 25 litres per second or 0.025 m$^3$/s and the final discharge is 100 litres per second or 0.1 m$^3$/s. The time taken for this change is 3 seconds.

Using the formula for the volume flow rate, we have:

$$Q = Av$$

where $Q$ is the volume flow rate, $A$ is the cross-sectional area of the pipe, and $v$ is the velocity.

The cross-sectional area of the pipe is given by:

$$A = \pi r^2 = \pi (0.1)^2 = 0.0314 m^2$$

Using the initial discharge, we have:

$$v_1 = \frac{Q}{A} = \frac{0.025}{0.0314} = 0.795 m/s$$

Using the final discharge, we have:

$$v_2 = \frac{Q}{A} = \frac{0.1}{0.0314} = 3.183 m/s$$

The change in velocity is:

$$\Delta v = v_2 - v_1 = 2.388 m/s$$

The time taken for this change is 3 seconds, so the acceleration is:

$$a = \frac{\Delta v}{\Delta t} = \frac{2.388}{3} = 0.796 m/s^2$$

Using the density of water at room temperature, which is approximately 1000 kg/m$^3$, we have:

$$\frac{1}{\rho} = 0.001 m^3/kg$$

Substituting these values into Euler's equation, we have:

$$0.796 = -\frac{1}{1000}\frac{\partial p}{\partial x}$$

Solving for the pressure gradient, we have:

$$\frac{\partial p}{\partial x} = -796 Pa/m$$

Therefore, the pressure gradient that can sustain the flow is -796 Pa/m.