Prove Bernoulli's Principle.

(N/A) Consider an incompressible,non-viscous fluid flowing through a pipe of varying cross-section and height.
At the inlet (point $B$):
- Cross-sectional area $= A_1$
- Fluid speed $= v_1$
- Pressure $= P_1$
At the outlet (point $D$):
- Cross-sectional area $= A_2$
- Fluid speed $= v_2$
- Pressure $= P_2$
In a small time interval $\Delta t$,the fluid at the inlet moves a distance $v_1 \Delta t$. The volume of fluid entering is $\Delta V = A_1 v_1 \Delta t$. The work done by the pressure force at the inlet is $W_1 = F_1 \times (v_1 \Delta t) = P_1 A_1 v_1 \Delta t = P_1 \Delta V$.
Similarly,at the outlet,the work done by the fluid against the pressure is $W_2 = P_2 A_2 v_2 \Delta t = P_2 \Delta V$. The net work done by pressure is $W = W_1 - W_2 = (P_1 - P_2) \Delta V$.
According to the work-energy theorem,this net work equals the change in kinetic energy plus the change in potential energy of the fluid mass $\Delta m = \rho \Delta V$ moving from the inlet to the outlet:
$W = \Delta K + \Delta U$
$(P_1 - P_2) \Delta V = \frac{1}{2} \Delta m (v_2^2 - v_1^2) + \Delta m g (h_2 - h_1)$
Dividing by $\Delta V$ and substituting $\Delta m / \Delta V = \rho$:
$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g (h_2 - h_1)$
Rearranging gives Bernoulli's equation:
$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$
Thus,$P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$.