As an example, consider the flow from a reservoir whose water surface is 8 ft above the bottom of a trapezoidal channel with a bottom width b = 10 ft, a side slope m = 1.5, a Manning’s roughness n = .013 and a bottom slope So = 0.0008. The entrance loss coefficient for the water entering the

Question

As an example, consider the flow from a reservoir whose water surface is 8 ft above the bottom of a trapezoidal channel with a bottom width b = 10 ft, a side slope m = 1.5, a Manning’s roughness n = .013 and a bottom slope [latex]S_{o}[/latex] = 0.0008. The entrance loss coefficient for the water entering the channel equals [latex]K_{e}[/latex] = 0.12. At a distance 1200 ft downstream from the reservoir, a smooth transition occurs to a pipe with a diameter D = 10 ft. The pipe is laid on a steep slope. The bottom of the pipe and the trapezoidal channel are at the same level. Determine the discharge from the reservoir into the channel and the depth of flow throughout the length of the trapezoidal channel.

Accepted Answer

This problem involves the simultaneous solution of the ordinary differential equation for the gradually varied flow in the trapezoidal channel,

[latex]\frac{dy}{dx}=\frac{S_{o}-S_{f}}{1-F^{2}_{r}}[/latex]

and the algebraic equations that govern the flow. At the entrance to the channel, the energy equation controls, e.g.,

[latex]H= 8= y+ \frac{Q^{2}}{\left(2gA^{2} ight)}[/latex]

At the head of the steep pipe, the critical flow equation governs, e.g.,

[latex]\frac{Q^{2}T}{\left(gA^{3} ight)}+1[/latex]

and the specific energy at the beginning of the transition in the trapezoidal channel must equal the critical specific energy at the end of the transition, e.g.,

[latex]y_{2}+\frac{Q^{2}}{2gA^{2}_{2}}=E_{c}[/latex]

Since ODEs cannot be solved simultaneously with algebraic equations except by trial, or iterative techniques, the procedure might be as follows:

First, solve Manning’s equation and the energy equation simultaneously at the entrance giving [latex]Q_{o}[/latex] = 1127.1 cfs and [latex]y_{o}[/latex] = 6.845 ft. With this flow rate, the critical depth in the pipe is [latex]y_{c}[/latex] = 8.059 ft and the associated critical specific energy is [latex]E_{c}[/latex] = 12.35 ft. Solving the energy equation gives a depth of [latex]y_{2}[/latex] = 12.184 ft at the end of the trapezoidal channel. This is the depth that can be used as a boundary condition in solving the ODE for the gradually varied flow in the trapezoidal channel. The main program and the procedure for accomplishing such a solution are listed below:

Program CHANRE2.C
#include
#include
#include
#include
#include "odesols.h"
void main(void){
int rep;
float yb,fn,c,tol,q,g,x,xbeg,xend,delx,xz,a,fmom,h1,hmin=.001;
float *y,*dydx;
y=(float *)calloc(nv,sizeof(float));
dydx=(float *)calloc(nv,sizeof(float));
wk1=(float *)calloc(13*nv,sizeof(float));
h1=.01;
L1:printf("Give: TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,g ");
scanf("%f %f %f %f %f %f %f %f %f %f %f",&tol,&delx,&yb,&q,&fn,\

&so,&b,&fm,&xbeg,&xend,&g);

y[0]=yb; fms=2.*sqrt(fm*fm+1); c=1; if(g>30) c=1.486; fnq=fn*q/c;

q2=q*q/g;

x=xbeg; a=(b+fm*y[0])*y[0]; fmom=(b/2+fm*y[0]/3)*y[0]*y[0]+q2/a;
printf("%10.3f %10.3f %10.3f ",x,y[0],fmom);
do {xz=x+delx;

odesolsc(y,dydx,x,xz,tol,h1,hmin);

x=xz; a=(b+fm*y[0])*y[0];fmom=(b/2+fm*y[0]/3)*y[0]*y[0]+q2/a;

printf("%10.3f %10.3f %10.3f ",x,y[0],fmom);

if(delx>0) {if(x

else {if(x>xend) rep=1; else rep=0;}

}while (rep);
printf("Solve another problem? 1=yes, 0=no ");scanf("%d",&rep);

if(rep) goto L1;

free(y);free(dydx);free(wk1);

}
void slope(float x,float *y,float *dydx){
extern float b,fm,so,q2,fnq,fms;
float a,t,p,sf,fr2;
a=(b+fm*y[0])*y[0]; t=b+2*fm*y[0]; p=b+fms*y[0];
sf=fnq*pow(p/a,0.66666667)/a; sf=sf*sf;
fr2=q2*t/(a*a*a);
dydx[0]=(so−sf)/(1−fr2); return;
}

Listing of header file odesols.h
int odesolsc(float *y,float *dydx,float x1,float x2,float\

err,float h1,float hmin);

int nv=1;
float b,fm,so,q2,fnq,fms,dxbetw=0.000001,*wk1;

Notice that the above program listing calls on the smaller version of the ODE solver that does not allow intermediate values to be stored. To use the larger program you will need to include declarations for these added variables. The following program uses the complete ODE solver odesolc.c; with the header file odesol.h needed with it. A listing of these programs, and the header files, are on the CD.

CHANRE2O.C
#include
#include
#include
#include
#include "odesol.h"
void main(void){
int rep;
float yb,fn,c,tol,q,g,x,xbeg,xend,delx,xz,a,fmom,h1,hmin=.001;
float *y,*dydx;
y=(float *)calloc(nv,sizeof(float));
dydx=(float *)calloc(nv,sizeof(float));
wk1=(float *)calloc(13*nv,sizeof(float));
xp=(float *)calloc(nstor,sizeof(float));
yp=(float *)calloc(nv*nstor+12,sizeof(float));h1=.01;
L1: printf("Give: TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,g ");
scanf("%f %f %f %f %f %f %f %f %f %f %f",&tol,&delx,&yb,&q,&fn,\

&so,&b,&fm,&xbeg,&xend,&g);

y[0]=yb; fms=2.*sqrt(fm*fm+1); c=1; if(g>30) c=1.486; fnq=fn*q/c;

q2=q*q/g;

x=xbeg; a=(b+fm*y[0])*y[0]; fmom=(b/2+fm*y[0]/3)*y[0]*y[0]+q2/a;
printf("%10.3f %10.3f %10.3f ",x,y[0],fmom);
do {xz=x+delx;

odesolc(y,dydx,x,xz,tol,h1,hmin);

x=xz; a=(b+fm*y[0])*y[0];fmom=(b/2+fm*y[0]/3)*y[0]*y[0]+q2/a;

printf("%10.3f %10.3f %10.3f ",x,y[0],fmom);

if(delx>0) {if(x

if(x>xend) rep=1; else rep=0;}

}while (rep);
printf("Solve another problem? 1=yes, 0=no ");scanf("%d",&rep);

if(rep) goto L1;

free(y);free(dydx);free(wk1);free(xp);free(yp);

}
void slope(float x,float *y, float *dydx){
extern float b,fm,so,q2,fnq,fms;
float a,t,p,sf,fr2;
a=(b+fm*y[0])*y[0]; t=b+2*fm*y[0]; p=b+fms*y[0];
sf=fnq*pow(p/a,0.66666667)/a; sf=sf*sf;
fr2=q2*t/(a*a*a);
dydx[0]=(so−sf)/(1−fr2);
}

Using the following line of input to the prompt of this program:

.0001 −200. 12.184 1127.1 .013 .0008 10 1.5 1200. 0. 32.2

(i.e. specifying Q = 1127.1 and the beginning depth at 1200 ft of, YB = 12.184 ft) gives the following solution:

Since the depth of 11.274 ft is well above even the water surface elevation in the reservoir, it is clear that the flow rate will be considerably less than 1127.1 cfs that would exist under uniform flow conditions in a long trapezoidal channel of its size. Even though the pipe 1200 ft downstream is on a steep slope, it “chokes” the flow with the effect extending to the reservoir and reduces the flow rate. Trial flow rates must now be selected, the various algebraic equations resolved, and finally the ODE solved again. The table below shows trials until an adequate agreement is achieved between column 2 and the last column.

The last flow rate of 600 cfs gives a close enough agreement to be acceptable. This last gradually varied flow for the resulting [latex]M_{1}[/latex] profile is:

Question C.4: As an example, consider the flow from a reservoir whose wate...

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