Suppose the mass ABCD comes in a short time t to A′B′C′D′. The resultant force parallel to the axis of the stream is
pω + p0 (ω1 − ω) − p1ω1,
where p0 is put for the unknown pressure on the annular space between AB and EF. The impulse of that force is
{ pω + p0 (ω1 − ω) − p1ω1 } t.
| Fig. 38. |
The horizontal change of momentum in the same time is the difference of the momenta of CDC′D′ and ABA′B′, because the amount of momentum between A′B′ and CD remains unchanged if the motion is steady. The volume of ABA′B′ or CDC′D′, being the inflow and outflow in the time t, is Qt = ωvt = ω1v1t, and the momentum of these masses is (G/g) Qvt and (G/g) Qv1t. The change of momentum is therefore (G/g) Qt (v1 − v). Equating this to the impulse,
{ pω + p0 (ω1 − ω) − p1ω1 } t = (G/g) Qt (v1 − v).
Assume that p0 = p, the pressure at AB extending unchanged through the portions of fluid in contact with AE, BF which lie out of the path of the stream. Then (since Q = ω1v1)
(p − p1) = (G/g) v1 (v1 − v);
p/G − p1/G = v1 (v1 − v) / g;