! Model of flow in an uncovered stream, 
e.g., small river, sewer, irrigation canal, etc.;
! Given a desired flow in volume/second, and
  a slope in length/length, find the width and
  height of a rectangular channel to achieve
  this flow at minimum cost of materials;
! Keywords: Hydraulics, Channel flow, Water flow,
  Gauckler-Manning-Strickler, River flow,
  Irrigation channel, Sewer flow;
 S=0.001; ! Slope of the channel;
 Q=100;   ! Required flow in meters^3 or feet^3/ second;
 k = 1;   !units conversion factor, = 1 for SI units (meters)
         = 1.48592 = (3.28083499 ft/m)^(1/3), for English units (feet);
! Some bounds on b and h;
 blo = .5; bhi = 100;
 hlo = .5; hhi = 100;

 ! Parameters:
    Q = discharge rate in meters^3 or feet^3/ second,
    S = slope of water surface, or head loss in length/length,
    n = Gauckler-Manning-Strickler roughness coefficient,
         higher means a slower flow,
         concrete has a coefficient of about .014,
    k = units conversion factor, = 1 for SI units (meters)
         = 1.4859 for English units (feet).
    b = channel width,
    P = wetted perimeter of rectangular cross section,
        e.g., amount of concrete needed,
    A = area of cross section,
    R = hydraulic radius,
    h = height of fluid in channel,

 min=P; ! Min sum of sidewalls + base;

 A=b*h; ! Area;
 P*R=A; ! Compute hydraulic radius;
! The Gauckler-Manning-Strickler formula for discharge rate gives
 flow velocity = (R^(2/3))*(S^0.5)/n, so discharge rate
           Q = A*(R^(2/3))*(S^0.5)/n, or ;

! This model is hard to solve, so be sure to turn on
 either global or multi-start;
! Some reasonable bounds on b and h;
 @BND(blo, b, bhi);
 @BND(hlo, h, hhi);
CALC: @SET("TERSEO",2); ! Turn off default output; @WRITE('Channel height, h, and width b, for various coefficients, n:',@NEWLINE(1)); @WRITE(' to achieve a target flow volume of ',Q,' cubic '); @IFC( k #EQ# 1: @WRITE('meters/sec.',@NEWLINE(1)); @ELSE @WRITE('feet/sec.',@NEWLINE(1)); ); @WRITE(' n b h P ',@NEWLINE(1)); FIRST = .01; ! Smallest value of n to try; LAST = .03; ! Largest value of n to try; STEP = .01; ! Step size over n; n = FIRST; @WHILE( n #LE# LAST: @SOLVE( STREAMFLO); @WRITE( @FORMAT(n,'6.3f'),' ',@FORMAT(b,'7.3f'),' ',@FORMAT(h,'7.3f'),' ',@FORMAT(p,'7.3f'), @NEWLINE(1)); n = n + STEP; ); ENDCALC