貯水池

最大流出率

推定

dt=Q (t)- q (z), where z = elevation of water level in reservoir, f (z) = surface area of reservoir, t=time, Q(t)=infIow rate, q(z)=outf1ow rate. Integrating between the limits, t = 0 and t = h (the instant of maximum outflow rate) we have F(z_h) - F(z_0)=int_0^h Q( t)dt -int_0^h q( z )dt, where F’(z)=f(z), and z_h, z_0 are the values of z at t=h, t = 0 respectively. On the assumption that q (z) is representable with sufficient closeness by a quartic in t, having 2nd-order contact with q (z) at t = 0 and 1st-order contact at t = h, we get int 0^h q( z)=( h/5)( 3q( z_0)+2Q(h)) +(h^3/60)q’(z_0)Q’(o)/f(z_0), provided Q( a)=q( z_0).Hence F( Z_h) - F(z_0) = int 0^h Q( t )dt - (h/5)(3q(z_0) +2Q(h))- (h^3/(60)q'(z_0)Q'( o )/f(z_0) and Q(h)=q(z_h).If the last term in the first of these equations is neglected, they can conveniently be solved for h with the aid of a chart. Take a pair of rectangular axes and draw the inflow mass curve with the horizontal axis as time axis. Graduate the vertical axis so that the graduation zon it is at a distance F(z)-F(z_0) from the origin, and draw through z two straight lines with angular coefficients q(z) and 0.6q(z_0)+0.4q(z)respectively. The forn1er set of lines shall be called the first outflow lines and the latter set the second outflow lines. Then the abscissa of the point where one of the first outflow lines touches the mass curve gives an upper bound to the required root of the equations, while the abscissa of the point where one of the second _outflow lines cuts the mass curve so that the corresponding first outflow line is parallel to the tangent to the curve at the point gives the required root (Fig. 5) In practice it will be convenient to draw the mass curve on a sheet of transparent paper and place it on the outflow-line diagram, which, for a given reservoir, can be constructed once for all. A numerical example in which is retained the term neglected in the graphical method of solution is given on p. 121.