ABSTRACT
If the pressure at X is P, and at Y it is P, (Figure 5.18), for the fluid to flow in the specified direction PI must be greater than S. Consider an element of fluid cross-sectional area (A,) and length (dl,) at X, and area (A2) and length (dl,) at Y. Since it is incompressible and the mass-rate at X is the same as at Y, then the net work done in taking a volume element of the fluid from X to Y is
where V is the volume. This work done is equal to the corresponding change in the kinetic and potential ener-
gies of the element. Hence:
Dividing both sides by V and separating the terms for X and Y respectively:
