Write a C program using one dimensional arrays to do the following: The insulati
ID: 3767031 • Letter: W
Question
Write a C program using one dimensional arrays to do the following:
The insulation thickness' of several steel pipes are to be computed (as shown). Consider the formulas listed below for determining such thickness.
Where:
QPL = heat loss per unit length of pipe to the insulation; QPL <= 500 Btu/hr-ft
k1 = coefficient of heat conduction of steel = .07
k2 = coefficient of heat conduction of the insulation = 12.5
T1 = temperature on inside of pipe; use 120°F for all pipes
T2 = temperature on the outside surface of the insulation
D1, D2, D3 = inside diameter of the pipe, outside diameter of the pipe, and outside
diameter of the insulation, respectively
T = insulation thickness required = (D3 – D2)/2
Make a flowchart and do the following steps using your desire arrays
1. Create the headings
2. Treat the tabulated values given below of pipe and diameter temperatures as a one-dimensional arrays, respectively. Read these tables into the computer.
3. Compute QPL and T for each pipe.
4. Initialize the diameter D3 as D2 + 0.5
5. Check QPL = 500. If true, write output as shown.
If false, increase D3 by 0.5 and recalculate QPL
Given Data:
Tempera
Diameters tures
________________________________
D1 D2 T2
________________________________
1 2 2.375 550
2 4 4.50 620
Pipes 3 5 5.563 700
4 6 6.625 760
____________________________________
Out- put layout:
PIPE INSULATION THICKNESSES REQUIRED
PIPE THICKNESSES REQUIRED (in)
1 x.xx
2 x.xx
3 x.xx
4 x.xx
Explanation / Answer
#include <iostream>
#include <math.h>
#include <iomanip>
#include <cmath>
using namespace std;
// Variable Meanings
// Pipe radius = a = 0.05m
//Radius of pip + 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/m cubed
//Insulation installation cost =CL=$1.50/m
//Cost of heat =Cstheat= 1.11x10^-9
int main()
{
float i, thick_increment, start, end, start_temp, end_temp, temp_increment, a, b, L, Ta, k, F, Cvol, CL, Cstheat, tenneg9, dQ, Q3, c, d, e, f, g, h, j,CI, CI_1, CI_2, tenpos8, one5sev, cf, CF, fuel_savings, tair, dQ_1, dQ_2, CF_1, CF_2, Q3_1, Q3_2, fuel_savings1, fuel_savings2, rerun, y;
float temp[3];
do {
cout<<"Enter the starting thickness"" ";
cin>>start;
cout<<"enter the end thickness"" ";
cin>>end;
cout<<"Enter the thickness increment"" ";
cin>>thick_increment;
cout<<"Enter the starting temperature"" ";
cin>>start_temp;
cout<<"Enter the ending temperature "" ";
cin>>end_temp;
cout<<"Enter the temperature increment size"" ";
cin>>temp_increment;
cout<<setw(10)<<"Thickness"<<setw(15)<<"CI"<<setw(25)<<"CF"<<setw(35)<<"fuel_savings @ "<<start_temp<<setw(25)<<"fuel_savings @ "<<start_temp+temp_increment<<setw(25)<<"fuel_savings @ "<<end_temp<<endl;
for (i=start; i<=end; i=i+thick_increment)
{ a=.05;
b= a+i;
L=100;
Ta=150;
k=.1;
F=3;
Cvol=325;
CL=1.5;
tenneg9= pow(10, -9);
Cstheat=1.11*tenneg9;
CI_1=(b*b-a*a)*(L)*(Cvol);
CI_2=L*CL;
CI=CI_1+CI_2;
for (tair=start_temp; tair<=end_temp; tair=tair+temp_increment)
c=b/a;
d=b*F;
e=d/k;
f=log(c);
g=(e*f)+1;
h=c/g;
j=1-h;
dQ=Q3*j;
dQ_1=Q3_1*j;
dQ_2=Q3_2*j;
Q3_1=2*3.14*a*F*(Ta-0)*L;
Q3_2=2*3.14*a*F*(Ta-10)*L;
Q3=2*3.14*a*F*(Ta-(-10))*L;
tenpos8= pow(10, 8);
> cf=dQ*Cstheat;
CF=dQ*one5sev*Cstheat;
CF_1=dQ_1*one5sev*Cstheat;
CF_2=dQ_2*one5sev*Cstheat;
fuel_savings=CF-CI;
fuel_savings1=CF_1-CI;
fuel_savings2=CF_2-CI;
cout<<i<<setw(25)<<CI<<setw(25)<<CF<<setw(25)<<fuel_savings<<setw(25)<<fuel_savings1<<setw(25)<<fuel_savings2<<endl;
}
cout<<"Do you want to run this program again? (y/n)"" ";
cin>>rerun;
} while (rerun=='y' ||'Y');
return 0;
}
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