(d) In any bounded plane section traversed normally by streams which are rectilinear for a certain distance on either side of the section, the distribution of pressure is the same as in a fluid at rest.
Distribution of Energy in Incompressible Fluids.
§ 29. Application of the Principle of the Conservation of Energy to Cases of Stream Line Motion.—The external and internal work done on a mass is equal to the change of kinetic energy produced. In many hydraulic questions this principle is difficult to apply, because from the complicated nature of the motion produced it is difficult to estimate the total kinetic energy generated, and because in some cases the internal work done in overcoming frictional or viscous resistances cannot be ascertained; but in the case of stream line motion it furnishes a simple and important result known as Bernoulli’s theorem.
| Fig. 25. |
Let AB (fig. 25) be any one elementary stream, in a steadily moving fluid mass. Then, from the steadiness of the motion, AB is a fixed path in space through which a stream of fluid is constantly flowing. Let OO be the free surface and XX any horizontal datum line. Let ω be the area of a normal cross section, v the velocity, p the intensity of pressure, and z the elevation above XX, of the elementary stream AB at A, and ω1, p1, v1, z1 the same quantities at B. Suppose that in a short time t the mass of fluid initially occupying AB comes to A′B′. Then AA′, BB′ are equal to vt, v1t, and the volumes of fluid AA′, BB′ are the equal inflow and outflow = Qt = ωvt = ω1v1t, in the given time. If we suppose the filament AB surrounded by other filaments moving with not very different velocities, the frictional or viscous resistance on its surface will be small enough to be neglected, and if the fluid is incompressible no internal work is done in change of volume. Then the work done by external forces will be equal to the kinetic energy produced in the time considered.
The normal pressures on the surface of the mass (excluding the ends A, B) are at each point normal to the direction of motion, and do no work. Hence the only external forces to be reckoned are gravity and the pressures on the ends of the stream.
The work of gravity when AB falls to A′B′ is the same as that of transferring AA′ to BB′; that is, GQt (z − z1). The work of the pressures on the ends, reckoning that at B negative, because it is opposite to the direction of motion, is (pω × vt) − (p1ω1 × v1t) = Qt(p − p1). The change of kinetic energy in the time t is the difference of the kinetic energy originally possessed by AA′ and that finally acquired by BB′, for in the intermediate part A′B there is no change of kinetic energy, in consequence of the steadiness of the motion. But the mass of AA′ and BB′ is GQt/g, and the change of kinetic energy is therefore (GQt/g) (v12/2 − v2/2). Equating this to the work done on the mass AB,
GQt (z − z1) + Qt (p − p1) = (GQt/g) (v12/2 − v2/2).
Dividing by GQt and rearranging the terms,
v2/2g + p/G + z = v12/2g + p1/G + z1;