curve.h

求解梯级水库群优化调度

#include <math.h>
double getvolume(double l,int i)
{
    double v;
	switch(i)
	{
	case 0:
		v=122676349.715378-215914.232575954*l+95.2853295840258*pow(l,2);//返回小湾电站对应库容
		break;
	case 1:
		v=280980206.114625-835626.727754842*l+826.065343863284*pow(l,2)-0.271311822174764*pow(l,3);
		break;
	case 2:
		v=34646332.7094881-96428.3812771281*l+83.2624816754715*pow(l,2)-0.020887175505279*pow(l,3);
		break;
	default:
		v=-1.0;
		break;
	}
	return v;
}
double getlevel(double v,int i)
{
	double l;
	switch(i)
	{
	case 0:
		l=1112.8713166181+0.000126492193726789*v-0.000000000026829619*pow(v,2);	//返回小湾电站水位
		break;
	case 1:
		l=906.29751307191+0.00202380197622316*v-0.000000017602491186*pow(v,2)+0.000000000000064832*pow(v,3);
        break;
	case 2:
		l=854.428957650126+0.000936041749652965*v-0.000000005915661845*pow(v,2)+0.000000000000019115*pow(v,3);
		break;
	default:
		l=-1;
		break;
	}
	return l;
}
double gettail(double d,int i)
{
	double tail;
	switch(i)
	{
	case 0://返回小湾尾水位
		tail=993.499999999979+0.000804761904787449*d+0.00000009642857142*pow(d,2)-0.000000000008333333*pow(d,3);
		break;
	case 1:
		tail=898.032075609325+0.00427878222891111*d-0.000000938614328573*pow(d,2);
		break;
	case 2:
		tail=817.327018626232+0.00337411474829784*d-0.000000199356911976*pow(d,2);
		break;
	default:
		tail=-1;
		break;
	}
	return tail;
}
double getlost(double p,int i)
{
	double lost;
	switch(i)
	{
	case 0://返回小湾水头损失
		if(p<1600)lost=2.214e-6*pow(p,2);
		else lost=1.978e-6*pow(p,2);
		break;
	case 1:
		lost=2.7;//漫湾水头损失，要改
		break;
	case 2:
		lost=4.62;
		break;
	default:
		lost=-1;
		break;
	}
	return lost;
}