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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/CH8/EX8.16 | |
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| -rwxr-xr-x | 405/CH8/EX8.16/8_16.sce | 100 |
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diff --git a/405/CH8/EX8.16/8_16.sce b/405/CH8/EX8.16/8_16.sce new file mode 100755 index 000000000..8adabca8d --- /dev/null +++ b/405/CH8/EX8.16/8_16.sce @@ -0,0 +1,100 @@ +clear;
+clc;
+printf("\t\t\tExample Number 8.16\n\n\n");
+// numerical solution for parallel plates
+// Example 8.16 (page no.-440-443)
+// solution
+
+T1 = 1000;// [K]
+T2 = 400;// [K]
+E1 = 0.8;//
+E2 = 0.5;//
+// consulting figure 8-12, we obtain
+F12 = 0.2;
+F21 = 0.2;
+F11 = 0;
+F22 = 0;
+F13 = 0.8;
+F23 = 0.8;
+A1 = 1;// [square meter]
+A2 = 1;// [square meter]
+// surface 3 is the surrounding or insulated surface. For part A(the plates are surrounded by a large room at 300K)
+T3 = 300;// [K]
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb2 = sigma*T2^(4);// [W/square meter]
+Eb3 = sigma*T3^(4);// [W/square meter]
+// because A3 tends to infinity, F31 and F32 must approach zero since A1*F13 = A3*F31 and A2*F23 = A3*F32. the nodal equations are written in the form of equations (8-107):
+// surface 1 J1-(1-E1)*(F11*J1+F12*J2+F13*J3) = E1*Eb1
+// surface 2 J2-(1-E2)*(F21*J1+F22*J2+F23*J3) = E2*Eb2
+// surface 3 J3-(1-E3)*(F31*J1+F32*J2+F33*J3) = E3*Eb3
+
+// because F31 and F32 approach zero, F33 must be 1.0.
+F33 = 1;
+// inserting the various numerical values for the various terms and solving the third equation we get
+// the third equation as: J3*E3 = E3*Eb3 so we get the value of J3 as
+J3 = Eb3;// [W/square meter]
+// finally the equations are written in compact form after getting the value of J3 we solve for J2 and J1 by matrix method
+Z = [1-(1-E1)*F11 -(1-E1)*F12;-(1-E2)*F21 1-(1-E2)*F22];
+C = [E1*Eb1+(1-E1)*F13*J3;E2*Eb2+(1-E2)*F23*J3];
+J = Z^(-1)*C;
+J1 = J(1);// [W/square meter]
+J2 = J(2);// [W/square meter]
+// the heat transfers are obtained from equation (8-104):
+q1 = A1*E1*(Eb1-J1)/(1-E1);// [W]
+q2 = A2*E2*(Eb2-J2)/(1-E2);// [W]
+// the net heat absorbed by the room is algebric sum of q1 and q2
+q3_absorbed = q1+q2;// [W]
+printf("\t\t CASE(A)");
+printf("\n\n the heat transfers are \n\n\t\t q1 = %f kW",q1/1000);
+printf("\n\t\t q1 = %f kW",q2/1000)
+printf("\n\n the net heat absorbed by the room in part (a) is %f kW",q3_absorbed/1000);
+
+// for part(b), A3 for the enclosing wall is 4.0 square meter
+
+A3 = 4;// [square meter]
+// and we set
+J3 = Eb3;// [W/square meter], because surface 3 is insulated.
+// from reciprocity we have
+F31 = A1*F13/A3;
+F32 = A2*F23/A3;
+// then, we have
+F33 = 1-F31-F32;
+// the set of equations are same with J3 = Eb3
+// surface 1 J1-(1-E1)*(F11*J1+F12*J2+F13*J3) = E1*Eb1
+// surface 2 J2-(1-E2)*(F21*J1+F22*J2+F23*J3) = E2*Eb2
+// surface 3 J3-(1-E3)*(F31*J1+F32*J2+F33*J3) = E3*J3
+// the third equation of set can be written as
+// J3(1-E3)-(1-E3)*(F31*J1+F32*J2+F33*J3) = 0
+// so that (1-E3) term drops out, and we obtain three equation in three variable which can be solved by matrix method
+Z = [1-(1-E1)*F11 -(1-E1)*F12 -(1-E1)*F13;-(1-E2)*F21 1-(1-E2)*F22 -(1-E2)*F23;-F31 -F32 1-F33];
+C = [E1*Eb1;E2*Eb2;0];
+J = Z^(-1)*C;
+J1n = J(1);// [W/square meter]
+J2n = J(2);// [W/square meter]
+J3n = J(3);// [W/square meter]
+// the heat transfers are
+q1n = A1*E1*(Eb1-J1n)/(1-E1);// [W]
+q2n = A2*E2*(Eb2-J2n)/(1-E2);// [W]
+// of course these heat transfers should be equal in magnitude with opposite sign because the insulated wall exchanges no heat.
+// the temperature of the insulated wall is obtained from
+T3 = (J3n/sigma)^(1/4);// [degree celsius]
+printf("\n\n \t\tCASE(B)");
+printf("\n\n the heat transfers are \n\n\t\t q1 = %f kW",q1n/1000);
+printf("\n\t\t q2 = %f kW",q2n/1000);
+printf("\n\n the temperature of the insulated wall is %f K",T3);
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