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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /405/CH3/EX3.7 | |
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initial commit / add all books
Diffstat (limited to '405/CH3/EX3.7')
| -rwxr-xr-x | 405/CH3/EX3.7/3_7.sce | 111 |
1 files changed, 111 insertions, 0 deletions
diff --git a/405/CH3/EX3.7/3_7.sce b/405/CH3/EX3.7/3_7.sce new file mode 100755 index 000000000..e12e206cb --- /dev/null +++ b/405/CH3/EX3.7/3_7.sce @@ -0,0 +1,111 @@ +clear;
+clc;
+printf("\t\t\tExample Number 3.7\n\n\n");
+// numerical formulation with heat generation
+// Example 3.7 (page no.-99-100)
+// solution
+
+d = 4;// [mm] diameter of wire
+Q = 500;// [MW/cubic meter] heat generation
+Tos = 200;// [degree celsius] outside surface temperature of wire
+k = 19;// [W/m degree celsius] thermal conductivity
+// we shall make the calculations per unit length
+dz = 1;
+// because the system is one-dimensional, we take
+dphai = 2*%pi;
+dr = 0.5;// [mm]
+// a summary of values for different nodes are following
+
+// node 1.
+
+rm1 = 0.25;// [mm]
+Rmplus1 = (dr/2)/((rm1+dr/4)*dphai*dz*k);// [degree celsius/W]
+// Rmminus1 = infinity
+dV1 = rm1*dr*dphai*dz;// [cubic micro meter]
+q1 = Q*dV1;// [W]
+
+// node 2.
+
+rm2 = 0.75;// [mm]
+Rmplus2 = (dr/2)/((rm2+dr/4)*dphai*dz*k);// [degree celsius/W]
+// Rmminus2 = infinity
+dV2 = rm2*dr*dphai*dz;// [cubic micro meter]
+q2 = Q*dV2;// [W]
+
+// node 3.
+
+rm3 = 1.25;// [mm]
+Rmplus3 = (dr/2)/((rm3+dr/4)*dphai*dz*k);// [degree celsius/W]
+// Rmminus3 = infinity
+dV3 = rm3*dr*dphai*dz;// [cubic micro meter]
+q3 = Q*dV3;// [W]
+
+// node 4.
+
+rm4 = 1.75;// [mm]
+Rmplus4 = (dr/2)/((rm4+dr/4)*dphai*dz*k);// [degree celsius/W]
+// Rmminus1 = infinity
+dV4 = rm4*dr*dphai*dz;// [cubic micro meter]
+q4 = Q*dV4;// [W]
+
+// a summary of values of sum_one_by_Rij and Ti according to equation (3-32) is now given to be used in gauss seidal iteration scheme
+
+// node 1
+
+sum_one_by_Rij1 = (1/Rmplus1);// [degree celsius/W]
+// the equations formed after putting values are
+// T1 = 3.288+T2
+
+// node 2
+
+sum_one_by_Rij2 = (1/Rmplus2);// [degree celsius/W]
+// the equations formed after putting values are
+// T2 = 3.289+(1/3)*T1+(2/3)*T3
+
+// node 3
+
+sum_one_by_Rij3 = (1/Rmplus3);// [degree celsius/W]
+// the equations formed after putting values are
+// T3 = 3.290+ 0.4*T2+06*T4
+
+// node 4
+
+sum_one_by_Rij4 = (1/Rmplus4);// [degree celsius/W]
+// the equations formed after putting values are
+// T4 = 2.193+(2/7)*T3+142.857
+
+// now we will solve these equations by iteration
+A=[1 -1 0 0;-(1/3) 1 -(2/3) 0;0 -0.4 1 -0.6;0 0 -(2/7) 1];
+b=[3.288;3.289;3.290;142.857+2.193];
+x=[240;230;220;210];
+NumIters=13;
+D=diag(A);
+A=A-diag(D);
+n=length(x);
+x=x(:);
+y=zeros(n,NumIters);
+for j=1:NumIters
+ for z=1:n
+ x(z)=(b(z)-A(z,:)*x)*D(z);
+ end
+ y(:,j)=x;
+end
+printf("thirteen iterations are now tabulated :\n");
+disp(y);
+// the total heat loss from the wire may be calculated as the conduction through Rmplus at node 4. then
+T4 = y(4,13);// [degree celsius]
+q = (T4-Tos)/(Rmplus4);// [W/m]
+// this must equal the heat generated in the wire, or
+V = %pi*(d*10^(-3)/2)^(2);// [square meter]
+q_exact = Q*10^(6)*V;// [W/m]
+printf("\n\n the total heat loss from the wire by the conduction through Rmplus at node 4 is %f kW/m",q/1000);
+printf("\n\n heat generated in the wire is %f kW/m",q_exact/1000);
+printf("\n\n the difference between the two values results from the inaccuracy in determination of T4");
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