Let fig. 118 represent a longitudinal section of the stream, A0A1 being the water surface, B0B1 the stream bed. Let A0B0, A1B1 be cross sections normal to the direction of flow. Suppose the mass of water A0B0A1B1 comes in a short time θ to C0D0C1D1, and let the work done on the mass be equated to its change of kinetic energy during that period. Let l be the length A0A1 of the portion of the stream considered, and z the fall, of surface level in that distance. Let Q be the discharge of the stream per second.
| Fig. 119. |
Change of Kinetic Energy.—At the end of the time θ there are as many particles possessing the same velocities in the space C0D0A1B1 as at the beginning. The change of kinetic energy is therefore the difference of the kinetic energies of A0B0C0D0 and A1B1C1D1.
Let fig. 119 represent the cross section A0B0, and let ω be a small element of its area at a point where the velocity is v. Let Ω0 be the whole area of the cross section and u0 the mean velocity for the whole cross section. From the definition of mean velocity we have
u0 = Σ ωv / Ω0.
Let v = u0 + w, where w is the difference between the velocity at the small element ω and the mean velocity. For the whole cross section, Σωw = 0.
The mass of fluid passing through the element of section ω, in θ seconds, is (G/g) ωvθ, and its kinetic energy is (G/2g) ωv3θ. For the whole section, the kinetic energy of the mass A0B0C0D0 passing in θ seconds is
(Gθ / 2g) Σωv3 = (Gθ/2g) Σω (u03 + 3u02w + 3u02 + w3),
= (Gθ / 2g) {u03Ω + Σωw2 (3u0 + w)}.
The factor 3u0 + w is equal to 2u0 + v, a quantity necessarily positive. Consequently Σωv3 > Ω0u03, and consequently the kinetic energy of A0B0C0D0 is greater than
(Gθ / 2g) Ω0u03 or (Gθ) / 2g) Qu02,