Terms on the Right Hand Side

we have on the RHS of Eqn. 3.45,

$$-V_s~\rho A V_s~~+~~(V_s~+~dV_s)~\rho~(A~+~dA)(V_s~+~dV_s)$$

$$-V_s~\rho~A~~V_s~+~~(V_s~+~dV_s)\rho A V_s~$$

$$= \rho A V_s dV_s$$ (3.48)

Now collecting terms for the LHS and RHS we have,

$$\rho A~ V_s~ dV_s~~=~~-\rho~g~A~dz~~-~A~dp$$

$$\rho ~ V_s~ dV_s~~+\rho~g~dz~~+~~dp~~=~~0$$

$$V_s~ dV_s~~+~g~dz~~+~~{dp \over \rho}~~=~~0$$

$$d\left({V_s^2 \over 2}\right)~~+~g~dz~~+~~{dp \over \rho}~~=~~0$$

The above equation is readily integrated for an incompressible flow ( =constant). As a result we have,

$$~{p \over \rho}~+~{ {V_s^2} \over 2}~+~g~z~=~\texttt{constant}$$ (3.50)

Equation 3.50 is called the Bernoulli Equation. Note that it connects pressure (p), elevation (z) and velocity (V_s). Once it is understood that the equation is valid along a streamline (i.e., within a stream tube) we can drop the subscript,s for velocity giving,

$$~{p \over \rho}~+~{ {V^2} \over 2}~+~g~z~=~\texttt{constant}$$ (3.51)

It may be pointed out that the equation is valid for steady flows only in absence of any friction such as the one due to viscosity. Further the flow is to be incompressible. We will derive the Bernoulli equation again but based on energy considerations.