READERS'FORUM 149 of reference 4 for details.) Actually, there are only two. Loh's analogy is one; it is useful only for "inside" problems which can be considered one-dimensional. When external--i.e., tw r odimensional, flow is considered there is just one other possible analogy; and this involves the use of a rectangular tank. The same applies to two-dimensional internal flows. To demonstrate the above statement, consider pseudo twodimensional shallow water flow in a channel having a horizontal bottom, (y, z) = (0, 0), with a vertical cross section defined by Z -cy n . (Symbols used are those of Loh, 1 except as noted above.)Following the usual assumptions we can write x, y, z, t) andboth of negligible magnitude. For a channel of fixed shape, it is clear that we can also write bZ/bt = 0, bZ/bx = 0, bZ/by = nZ/y for n > 0, bh/bx =
b8/bx t bh/by = (b8/by) -(bZ/by), bh/bt = bd/btNow, by continuity, we have for any (x, y, t),into which we can introduce the channel boundary conditions to findFurthermore, the existence of a gravitational body force potential U = pgz enables us to write Eulerian equations of motion:We now ask under what conditions will Eqs. (2), (5), and (6) be analogous to the equations of two-dimensional flow of a perfect gas. For such flow we know that
~{bp/by) = p[(bv/bt) + u(bv/bx) + v(bv/by)]where P = P(P, s)) , m
P=P(p,s)j { 'By putting Eq. (7) into the same form as Eq.(2)--viz.,
p(bu/bx) + u(bp/bx) -f-
P (bv/by) + v(b P /by) + (bp/bt) = 0 (11)it becomes evident that we can look for an analogous correspondence between p and h or p and 8, but then only if bZ/by = 0. Thus, we could conclude immediately that Z must be independent of y and the channel rectangular with 8 = h at all points. If we consider Eqs. (5), (6), (8), and (9), the position becomes even clearer. Using Eq. (10), we can rewrite Eqs. (8) and (9) as
~(bp/bp) s (bp/bx) + (bp/bs) p (bs/bx) = P [(bu/bt) + u(bu/bx) + v(bu/by)} (12)
~(bp/bp) s (bp/by) + (bp/bs) p (bs/by)These can be compared with Eqs. (5) and (6) rewritten:) -gh[(bh/by) + (bZ/by)] = h[(bv/bt) + u(bv/bx) + z;(dz;/dy)j (6a) For p and & to be analogous, it is then quite clear that (i) the gas flow must be isentropic bs/bx = 0, bs/by = 0 and (ii) the water channel must be of rectangular cross section bZ/by = 0 Under these conditions, the equations reduce to -C\b P /bx) = P [(bu/bt) + u(bu/bx) + »(d«/d^)] (14) -C\b P /by) = p[(bv/bt) + u(bv/bx) + K^/d;y)] (15) -gh(bh/bx) = h[(bu/bt) + u(bu/bx) + »(d«/dy)] (16) -gh(bk/by) = h[(bv/bt) + u{bv/bx) + p(d»/dy)] (17) and we can sa3^ that p/po = AAo (18) CVCo 2 = T/To = h/ho (19) and easily demonstrate that p/p 0 = (h/hY (20) with K = C P /Cv = 2. It will be noticed that no restriction is placed on this analogy as far as vorticity is concerned. Consequently, for rotational flows the analogous prediction equations, Eqs. (18)~(20), can be applied along corresponding streamlines as long as the two flows are kinematically similar. Generally, it is necessary to assume a priori the existence of kinematic similarity. Loh's onedimensional...
