Bernoulli.s theorem: According to Bernoulli.s theorem, the sum of pressure energy, kinetic. a + energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically
` = P/2 + 1/2 v^2 + gh ` = constant
This is known as bernoulli.s equation
Proof: Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time i which is equal to the volume of the liquid leaving B in the same time. Let `a_A, v_A ` and `P_A` be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively. Let the force exerted by the liquid at A is
`F_A = P_A a_A`
Distance travelled by the liquid in time t is `d = v_A t`
Therefore, the work done is `W = F_A d = P_A a_A v_A t`
But `a_A v_A t = a_A d = V` volume of the liquid entering at A.
Thus, the work done is the pressure energy (at A), `W = F_A d = P_AV`
Pressure energy per unit volume at ` A = ("pressure energy ")/("volume ") = (P_A V)/(V) = P_A`
Pressure energy per unit mass at ` A = ("pressure energy")/("mass") = (P_A V)/(m) = ((P_A)/(m)) = P_A/rho`
Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
` EP_A = P_AV = P_A V xx (m/m) = m (P_A)/(rho)`
Potential energy of the liquid at A,
`PE_A = mg h_A`
Due to the flow of liquid, the kinetic energy of the liquid at A, `KE_A = 1/2 m v_A^2`
Therefore, the total energy due to the flow of liquid at A,
`E_A = EP_A + KE_A + PE_A`
`E_A = m (P_A)/(rho) + 1/2 mv_A^2 + mg h_A`
Similarly, let `a_B , v_B` and `P_B` be the area of cross section of the tube, velocity of the line and pressure exerted by the liquid at B. Calculating the total energy at `F_B` . we get
` E_B = m (P_B)/(rho) + 1/2 mv_B^2 + mgh_B`
From the law of conservation of energy,
`E_A = E_B`
`m (P_A)/(rho) + 1/2 mV_A^2 + mgh_A = m (P_B)/(rho) + 1/2 mv_B^2 + mg h_B`
` m (P_A)/(rho) + 1/2 mV_A^2 + mgh_A = (P_B)/(rho) + 1/2 v_B^2 + gh_B =` constant
Thus , the above equation can be written as
` (P)/(rho g) + 1/2 (v^2)/(g) + h` =constant
The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli.s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
`h = 0 rArr (P)/(rho g) + 1/2 (v^2)/(g) ` = constant
