Irodov Problem 1.327

Consider an infinitesimally thin section of water of width dx escaping out of and at depth x. Let the velocity of this flow be v. The reaction pressure exerted by this section of flowing water is given by . The cross section of this infinitesimally thin section of water flow is bdx. Thus, the reaction force exerted by the infinitesimally thin section of flow is given by . Similar to as solved in Problem 1.327,


Hence we have,

Irodov Problem 1.326

Suppose that v1 and v2 are the speeds with which the water flows out of the lower and the higher hole respectively.
The reactions exerted by the water flowing out by these two escaping jets is given by and . The net reaction on the container is given by,



Suppose that the depth of the lower hole is h1. The velocity of water just before the entrance of the hole is assumed to be 0. The fluid pressure here is , where Po is the atmospheric pressure. Just outside the hole, the water escapes into open atmosphere i.e. into a region of atmospheric pressure. Applying Bernoulli's equation to points just before and after the hole we have,

Irodov Problem 1.325

Eulerian equation is a very general equation. There are three basic assumptions that restrict Eulerian equation into Bernoulli's equation. i) the fluid must be non-rotational (there should be no vorticity), or mathematically , ii) the force f must be a conservative force field in other words sum of potential energy and kinetic energy is conserved (such as gravity), and iii) the fluid must be incompressible (water can be considered incompressible for most practical purposes).

In order for the flow to be non-rotational any pressure change can only occur along the direction of the flow (otherwise a rotation will be induced - this not very different from if you apply force in a direction not aligned with velocity of a particle its path will exhibit curvature), in other words the direction of the gradient of pressure must coincide with the direction of the flow of the fluid or where s is along the flow direction of the fluid.

If the only force acting is gravity then along the z-axis, the negative sign comes because gravity acts aong the negative z -axis.