Question

*Using C++ (no printf command)*

So, this is a basic cost vs. benefit problem.  For different thicknesses of insulation, we will need to calculate the cost of purchasing and installing that thickness of insulation and compare that to the fuel savings generated by insulating the pipe with that thickness.

DISCUSSION:  Your program will ultimately need to determine overall savings (in dollars) for each thickness and then find the thickness that corresponds to the largest overall savings.

The overall savings is found by subtracting the cost of insulation from the fuel savings:

Overall savings = (C_F– C_I).  

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 – C_I

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):

C_I = (b²-a²)(L)(Cvol) + (L)(C_L)

b.  Fuel Savings – C_F

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 (dQ) is:

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 10⁸sec).  The formula for fuel savings over this 5-year period is:

C_F= dQ*(1.578 x 10⁸) * (CstHeat)

What we want to do is maximize the difference of  C_Fand C_I.   This difference is the overall savings (in dollars) and the thickness that corresponds to this difference is the optimum thickness!

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=150 C

Insulation conductivity            =k=0.1watt/(m C)

Convection constant                =F=3.0watt/(m C)

Pipe insulation cost                 =Cvol=$325/m³

Insulation installation cost       =C_L=$1.50/m

Cost of heat                             =Cstheat = 1.11x 10^-9per watt-sec

The only two quantities that are not constant in our problem are the insulation thickness and the air temperature (Tair).  For our program, the insulation is available in thicknesses ranging from 1 to 10 cm, in 1 cm increments (note this quantity must be changed to meters).  The air temperature will range from –10 C to 10 C in increments of 10 (in other words, you must evaluate at air temperatures of –10, 0, and 10).

Your program should output a table of thicknesses, insulation costs (C_I), fuel savings (C_F), and overall savings (C_F– C_I), for each air temperature (so, you’ll have three of these tables – one for –10, one for 0, and one for 10).  Your output must be in chart form, neat, and labeled.  You must also output the constant values used.

For example, the title of your first table may look like this:

Temperature:  -10

Thickness      CF             CI       Savings

.01 m

Answer 1

#include<iostream>
using namespace std;
int main(){

float a = 0.05; //m
float L = 100; //m
float Ta = 150; // o C
float k = 0.1; // watt/(m C)
float F = 3.0; // watt/(m C)
float Cvol = 325; // $/m^3
float CL = 1.50; // $/m
float Cstheat = 0.00000000111; //1.11*10^-9 per watt-sec

// thickness R
float R;
float minR = 0.01;// 1cm to m
float maxR = 0.1; // 10cm to m
float dR = 0.01; // 1cm to m -->increments
int nR = ((maxR-minR)/dR)+1; // number of possible thickness values

float Q3,dQ;
float b;
// temperature Tair
float vTair[] = {-10,0,10}; // vector of Tair values
float Tair;
int nT = 3; // number of Tair values

// loop to traverse all Tair values
for(int j=0;j<nT;j++){
float CI[nR],CF[nR],savings[nR];
float maxSaving = 0; int i_max=0;
Tair = vTair[j];
cout<<"Temperature: "<<Tair;
cout<<endl<<"CF \t\tCI\tsavings";
R = minR;

// loop to traverse all thickness values for the current Tair value
for(int i=0;i<nR;i++){
b = a + R;
CI[i] = ((b*b)-(a*a))*L*Cvol + L*CL;
Q3 = 2*3.14*a*F*(Ta-Tair)*L;

//************ take note *******************
dQ = Q3;
//******************************************

CF[i] = dQ*1.578*108*Cstheat;
savings[i] = CF[i] - CI[i];

// computing the max savings and corresponding max index thickness
if (savings[i]>maxSaving){
maxSaving = savings[i];
i_max = i;
}

cout<<endl<<CF[i]<<"\t"<<CI[i]<<"\t"<<savings[i];

R = R + dR;
}
float thicknessAtMaxSavings = (i_max+1)*dR;
cout<<endl<<"Max saving: $ "<<maxSaving;
cout<<endl<<"Thickness at max savings: "<<thicknessAtMaxSavings<<" meters"<<endl;
cout<<"---------------------------------------"<<endl;
}

return 0;
}

Temperature: -10 savings CF CI 0.00285118 185.75 -185.747 -227.997 0.00285118 228 0.00285118 276.75 -276.747 0.00285118 332 331.997 - 393.75 0.00285118 -393.747 0.00285118 462 -461.997 0.00285118 536.75 -536.747 -617.997 0.00285118 618 0.00285118 705.75 -705.747 0.00285118 800 -799.997 Max saving: $ e Thickness at max savings: 0.01 meters Temperature: 0 savings CF сI 185.75 0.00267298 -185.747 0.00267298 228 -227.997 276.75 0.00267298 -276.747 0.00267298 332 -331.997 0.00267298 393.75 -393.747 0.00267298 462 -461.997 0.00267298 536.75 -536.747 0.00267298 618 -617.997 0.00267298 705.75 -705.747 - 799.997 0.00267298 800 Max saving: $ e Thickness at max savings: 0.01 meters Temperature: 10 savings CF CI 0.00249478 185.75 -185.748 0.00249478 228 -227.998 0.00249478 276.75 -276.747 0.00249478 332 -331.998 -393.747 0.00249478 393.75 0.00249478 462 -461.997 0.00249478 536.75 -536.747 0.00249478 618 -617.997 - 705.747 - 799.997 0.00249478 705.75 0.00249478 800 Max saving: $e Thickness at max savings: 0.01 meters

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