As before, substituting the weights G1, G2 per cubic foot for the volumes per pound, we get for the work of expansion
(p1/G1) {1/(γ − 1)} {1 − (G2/G1) γ−1},
(2a)
= p1v1 {1/(γ − 1)} {1 − (p2/p1) γ−1/γ}.
(2b)
| Fig. 77. |
§ 62. Modification of the Theorem of Bernoulli for the Case of a Compressible Fluid.—In the application of the principle of work to a filament of compressible fluid, the internal work done by the expansion of the fluid, or absorbed in its compression, must be taken into account. Suppose, as before, that AB (fig. 77) comes to A′B′ in a short time t. Let p1, ω1, v1, G1 be the pressure, sectional area of stream, velocity and weight of a cubic foot at A, and p2, ω2, v2, G2 the same quantities at B. Then, from the steadiness of motion, the weight of fluid passing A in any given time must be equal to the weight passing B:
G1ω1v1t = G2ω2v2t.
Let z1, z2 be the heights of the sections A and B above any given datum. Then the work of gravity on the mass AB in t seconds is
G1ω1v1t (z1 − z2) = W (z1 − z2) t,