Bernoulli's Theorem:According to Bernoulli's theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flow in streamline.Potential energy + Kinetic energy + Pressure energy = Constant\\\\(P + \\frac{1} {2}pv^{2} + pgh = Constant\\\\)\\\\(gh + \\frac{1} {2}v^{2} + \\frac{P} {p} = C\\\\) ............(11.11)Where C is a constant.This relation is called Bernoulli's theorem.Dividing eqn. (11.11) by g, we get\\\\(h + \\frac{P} {pg} + \\frac{1} {2} \\frac{v^{2}} {g} = C'\\\\) ............(11.12)Where C is another constant.For horizontal flow, h remains same throughout.So, \\\\( \\frac{P} {pg} + \\frac{v^{2}} {2g} = Constant\\\\) or; \\\\( P + \\frac{1} {2} pv^{2} = Constant\\\\) P is static pressure of the liquid and \\\\(\\frac{1} {2} pv^{2}\\\\) is its dynamic and velocity pressure.Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If \\\\(p_1v_1\\\\) and \\\\(p_2v_2\\\\) represent pressure and velocities at two points. Then\\\\( P_1 + \\frac{1} {2} pv^{2}_1 = P_2 + \\frac{1} {2} pv^{2}_2 \\\\) Concepts : 1. In Bernoulli's theorem \\\\( P + pgh + \\frac{1} {2} ev^{2}\\\\)= constant. The term \\\\(\\left(p + pgh \\right)\\\\) is called static pressure and the term \\\\(\\frac{1} {2} pv^{2} \\\\) is the dynamic pressure of the fluid.2. Bernoulli theorem is fundamental principle of the energy.3. The equation \\\\(\\frac{P} {pg} + \\frac{1} {2} \\frac{v^{2}} {g} + h = constant\\\\)the term \\\\(\\frac{P} {pg}\\\\) = pressure headthe term \\\\(\\frac{v^{2}} {2g}\\\\) = velocity headh = gravitational head.Derivation of Bernoulli's Theorem :The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli's theorem using the work-energy theorem.Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid \\\\(A_1\\\\) and \\\\(A_2\\\\). In time interval \\\\(\\Delta t\\\\), the liquid displaces from \\\\(A_1\\\\) by \\\\(\\Delta x_1 = v_1\\Delta t\\\\) and displaces from \\\\(A_2\\\\) by \\\\(\\Delta x_2 = v_2\\Delta t\\\\). Here \\\\(v_1\\\\) and \\\\(v_2\\\\) are the velocities of the liquid at \\\\(A_1\\\\) and \\\\(A_2\\\\).The work done on the liquid is \\\\(P_1 A_1 \\Delta x_1\\\\) by the force and \\\\(P_2 A_2 \\Delta x_2\\\\) against the force respectively.Net work done,\\\\(W = P_1 A_1 \\Delta x_1 - P_2 A_2 \\Delta x_2\\\\)\\\\(\\Rightarrow W = P_1 A_1 v_1 \\Delta t - P_2 A_2 v_2 \\Delta t\\\\)\\\\(= \\left (P_1 - P_2 \\right) \\Delta V\\\\) .........(11.13)Here, \\\\(\\Delta V \\rightarrow \\\\) the volume of liquid that flows through a cross-section is same (from equation of continuity).But, the work done is equal to net change in energy (K.E. + P.E.) of the liquid, and \\\\(\\Delta K = \\frac{1} {2} p \\Delta V \\left (v^{2}_1 - v^{2}_2 \\right)\\\\) ........ (11.14)and \\\\(\\Delta U = p \\Delta V \\left (h_2 - h_1 \\right)\\\\) ........ (11.15)\\\\(\ herefore \\left (P_1 - P_2 \\right) \\Delta V = \\frac{1} {2} p \\Delta V \\left (v^{2}_1 - v^{2}_2 \\right) + pg \\Delta V \\left (h_2 - h_1 \\right)\\\\) \\\\(P_1 + \\frac{1} {2} v^{2}_1 + pgh_1 = P_2 + \\frac{1} {2} v^{2}_2 + pgh_2\\\\)or \\\\( P + \\frac{1} {2} pv^{2} + pgh = constant\\\\) .......... (11.16)This is the required relation for Bernoulli's theorem.\\\\(\ herefore A_1 v_1 = A_2 v_2\\\\)So, more the cross-sectional area, lesser is the velocity and vice-versa.So, Bernoulli's theorem,\\\\( P_1 + \\frac{1} {2} p_1 v^{2}_1 = P_2 + \\frac{1} {2} p_2 v^{2}_2\\\\)