Question

Write a C++ program to solve:

DISCUSSION: Your program will ultimately need to determine overall savings (in dollars) for each thickness. The savings is found by subtracting the cost of insulation from the fuel savings (CF-C). The insulation thickness that corresponds to the largest overall savings will be our answer. To find the overall savings, your program will need to calculate two quantities: a. Insulation cost-G. This quantity has two components: the cost of purchasing the insulation and the cost of installing the insulation. The formula for this calculation is (see below for identification of variables): G=(b2-a2)(L)(Cyol) + (L)(CL) b. Fuel Savings - CF The fuel savings is found by taking the difference in heat lost, dQ, and then multiplying this quantity by a dollar amount to get the amount of money we save using the insulation. The formula for finding the difference in heat lost (Q) is: dQ = Q3[1 - - 1 + b *ln() Q3 represents the quantity of heat lost with no insulation and is found using: Q3 = 2*3.14*a*F*(Ta - Tair)*L Now, once we have dQ, we can calculate the fuel savings. We'll use a five year period (5 years=1.578 x 108 sec). The formula for fuel savings over this 5-year period is: Cp=dQ*(1.578 x 108) * (CstHeat) What we want to do now is maximize the difference of these two values. That is, we want to know the insulation thickness that maximizes the quantity, CF-C1. This difference is actually the overall savings (in dollars)! Here are the constant variables for your program: Pipe radius =a=0.05m Radius of pipe + insulation =b=(a + insulation thickness) Pipe length =L=100m Pipe temperature =Ta=150C Insulation conductivity =k=0.1 watt/(m C) Convection constant =F=3.0watt/(m C) Pipe insulation cost =Cyol=$325/m3 Insulation installation cost =Cu=$1.50/m Cost of heat =Cstheat= 1.11x 10-9 per watt-sec

The completed program should include the following:

—Program input and output
—Allow the user to input the starting, stopping and increment values for the insulation thickness and the air temperature. Also, at the end of the program, provide an opportunity for the user to use the program again without having to re-execute the program (ie. Ask them if they want to use the program again).
—your program must find and output the value of the optimum thickness for each air temperature (you must have the program sort through the savings column for each air temperature and find the maximum value). Also, you must store the values you calculate for CF, CI, and overall savings in an array(s). You may store additional values in array(s) as well if you’d like.

Answer 1

/*Following program contains all the required functions. All of them are self explainatory, Comments have been added at relevent lines for better understanding. Still if there is any doubt, feel free to ask in the comment section. Hope it helps :)*/

#include <bits/stdc++.h>

using namespace std;

// printTable function to print the output.
void printTable(double Ts, double Ti, vector<double> &maxSavings,
vector<double> &CFs, vector<double> &CIs, vector<double> &thickness) {
  
cout << "Temperatue Thickness CF CI Savings" << endl;
  
for(int i = 0; i < maxSavings.size(); i++) {
cout << (Ts + Ti*i) << " " << (thickness[i] * 100) << "cm $" << CFs[i];
cout << " $" << CIs[i] << " $" << maxSavings[i] << endl;
}
  
}

//CalculateCost function to find the savings at all the Temperatue and thicknesses and store them in an array.
void CalculateCost(double Ts, double Te, double Ti, double ts, double te, double ti) {
  
// set values to all the given constants
double a = 0.05;
double L = 100;
double Ta = 150;
double k = 0.1;
double F = 3.0;
double Cvol = 325;
double Cl = 1.50;
double Cstheat = 1.11 * pow(10, -9);
  
// convert thickness from cm to m.
ts = ts * 0.01;
te = te * 0.01;
ti = ti * 0.01;
  
// create vectors(dynamic arrays, functions same as that of arrays, but thier size is not fixed)
// (Size of vector increases as the requirement increases)
// these vectors store the optimal value of Savings, CIs, CFs at every Temperatue.
double Tair = Ts;
vector<double> maxSavings;
vector<double> CFs;
vector<double> CIs;
vector<double> thickness;
while(Tair <= Te) {
// these vectors store the optimal value of Savings, CIs, CFs at every Thickness.
vector<double> Savings;
vector<double> CF1s;
vector<double> CI1s;
  
double t = ts;
while(t <= te) {
double b = a + t;
double CI = (b*b - a*a)*L*Cvol + L*Cl;
double Q3 = 2*3.14*a*F*(Ta - Tair)*L;
double dQ = Q3*(1 - (b/a)/(1 + (b*F/k)*log(b/a)));
double CF = dQ*(1.578*pow(10, 8))*Cstheat;
CI1s.push_back(CI);
CF1s.push_back(CF);
Savings.push_back(CF - CI);
t = t + ti;
}
  
// we find the max saving from all the thicknesses of pipe available
double max = -9999999;
double max_index = -1;
for(int i = 0; i < Savings.size(); i++) {
double diff = CF1s[i] - CI1s[i];
Savings[i] = diff;
if(diff > max) {
max = diff;
max_index = i;
}
}
  
// We put the max values in the Temperatue vectors.
maxSavings.push_back(Savings[max_index]);
CFs.push_back(CF1s[max_index]);
CIs.push_back(CI1s[max_index]);
thickness.push_back(ts + ti*(max_index));
Tair = Tair + Ti;
}
  
// prints the table.
printTable(Ts, Ti, maxSavings, CFs, CIs, thickness);
}

int main()
{
// ask the user what to do ?
while(1) {
cout<<"Do you wish to use the system ? (1)Yes (2)No : ";
int choice;
cin >> choice;
if(choice == 1) {
// take input
double Ts, Te, Ti;
double ts, te, ti;
cout << "Enter starting, ending and incremental value for the Temperatue(C): " << endl;
cin >> Ts;
cin >> Te;
cin >> Ti;
cout << "Enter starting, ending and incremental value for the Thickness(cm): " << endl;
cin >> ts;
cin >> te;
cin >> ti;
// call the function
CalculateCost(Ts, Te, Ti, ts, te, ti);
} else if(choice == 2) {
// terminate program
cout << "Program Treminated. Have a nice day!" << endl;
break;
} else {
cout << "Invalid Choice." << endl;
}
}

return 0;
}

Sample Output:-

Do you wish to use the system ? (1) Yes (2) No : 1 Enter starting, ending and incremental value for the Temperatue (C): -20 20 10 Enter starting, ending and incremental value for the Thickness (cm) : 10 7 cm 7cm Temperatue Thickness CF CI Savings -20 $1183.48 $536.75 $646.732 7 cm $1113.87 $536.75 $577.116 0 7 cm $1044.25 $536.75 $507.499 10 $974.632 $536.75 $437.882 6cm $834.856 $462 $372.856 Do you wish to use the system ? (1) Yes (2) No : 1 Enter starting, ending and incremental value for the Temperatue (C): -30 30 20 10 Enter starting, ending and incremental value for the Thickness (cm) : CF -20 10 0.5 Temperatue Thickness CI -30 7.5cm $1296.22 $576.562 7.5cm $1224.21 $576.562 -10 7cm $1113.87 $536.75 7 cm $1044.25 $536.75 10 6.5cm $938.379 $498.562 20 $834.856 $462 30 $770.636 $462 Do you wish to use the system ? (1) Yes (2) No : 2 Program Treminated. Have a nice day! Savings $719.657 $647.645 $577.116 $507.499 $439.816 $372.856 $308.636 бcm 6cm

For Indentation:-

#include <bits/stdc++.h> using namespace std; // printTable function to print the output. void printTable double Ts, double Ti, vector double> &maxSavings, vector double> &CFs, vector double> <CIS, vector double> &thickness) { cout << "Temperatue Thickness CF CI Savings" << endl; cout << (Ts + Ti i) << " " «< (thickness[i] * 100) << "cm $" << CFs[i]; cout << " $" << CIS[i] << " $" << maxSavings[i] << endl; //CalculateCost function to find the savings at all the Temperatue and thicknesses and store them in an array. void CalculateCost(double Ts, double Te, double Ti, double ts, double te, double ti) { // set values to all the given constants double a = 0.05; double L = 100; double Ta = 150; double k = 0.1; double F = 3.0; double Cvol = 325; double cl = 1.50; double Cstheat = 1.11 * pow(10, -9);

Il convert thickness from cm to m. ts = ts * 0.01; te = te * 0.01; ti = ti * 0.01; // create vectors (dynamic arrays, functions same as that of arrays, but thier size is not fixed), // (Size of vector increases as the requirement increases) // these vectors store the optimal value of Savings, CIS, CFs at every Temperatue. double Tair = Ts; vector double maxSavings; vector double> CFs; vector double> CIS; vector double> thickness; while(Tair <= Te) { // these vectors store the optimal value of Savings, CIS, CFs at every Thickness. vector double> Savings; vector double> CF1s; vector double> CI1s; double t = ts; while(t <= te) { double b = a + t; double CI = (b b - a a) L Cvol + L*c1; double 03 = 2*3.14 a F (Ta - Tair) L; double d = Q3 (1 - (ba)/(1 + (b F/k) *log (b/a))); double CF = dQ (1.578*pow(10, 8)) Cstheat; CI1s.push_back(CI); CF1s.push_back(CF); Savings.push_back (CF - CI); t = t + ti;

// we find the max saving from all the thicknesses of pipe available double max = -9999999; double max_index = -1; for(int i = 0; i < Savings.size(); i++) { double diff = CF1s[i] - CI1s[i]; Savings[i] = diff; if(diff > max) { max = diff; max_index = i; // We put the max values in the Temperatue vectors. maxSavings.push_back(Savings[max_index]); CFs.push_back(CF1s[max_index]); CIS.push back(CI1s[max_index]); thickness.push_back(ts + ti (max_index)); Tair = Tair + Ti; // prints the table. printTable(Ts, Ti, maxSavings, CFS, CIS, thickness);

int main() // ask the user what to do ? while(1) { cout<<"Do you wish to use the system ? (1) Yes (2)No : "; int choice; cin >> choice; if (choice == 1) { // take input double Ts, Te, Ti; double ts, te, ti; cout << "Enter starting, ending and incremental value for the Temperatue(C): " << endl; cin >> Ts; cin >> Te; cin >> Ti; cout << "Enter starting, ending and incremental value for the Thickness (cm): " << endl; cin >> ts; cin >> te; cin >> ti; // call the function CalculateCost(Ts, Te, Ti, ts, te, ti); } else if(choice == 2) { Il terminate program cout << "Program Treminated. Have a nice day!" << endl; break; } else { cout << "Invalid Choice." << endl; return 0;