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diff --git a/405/CH8/EX8.1/8_1.sce b/405/CH8/EX8.1/8_1.sce
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+clear;
+clc;
+printf("\t\t\tExample Number 8.1\n\n\n");
+// transmission and absorption in a gas plate
+// Example 8.1 (page no.-381) 
+// solution
+
+T = 2000+273;// [K] furnace temperature 
+L = 0.3;// [m] side length of glass plate
+t1 = 0.5;// transmissivity of glass between lambda1 to lambda2
+lambda1 = 0.2;// [micro m]  
+lambda2 = 3.5;// [micro m]  
+E1 = 0.3;// emissivity of glass upto lambda2 
+E2 = 0.9;// emissivity of glass above lambda2
+t2 = 0;// transmissivity of glass except in the range of lambda1 to lambda2
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+A = L^(2);// [square meter] area of glass plate
+// calculating constants to use table 8-1(page no.-379-380)
+K1 = lambda1*T;// [micro m K]
+K2 = lambda2*T;// [micro m K]
+// from table 8-1
+Eb_0_lam1_by_sigmaT4 = 0;
+Eb_0_lam2_by_sigmaT4 = 0.85443;
+Eb = sigma*T^(4);// [W/square meter]
+// total incident radiation is 
+// for 0.2 micro m to 3.5 micro m
+TIR = Eb*(Eb_0_lam2_by_sigmaT4-Eb_0_lam1_by_sigmaT4)*A;// [W]
+TRT = t1*TIR;// [W]
+RA1 = E1*TIR;// [W] for 0<lambda<3.5 micro m
+RA2 = E2*(1-Eb_0_lam2_by_sigmaT4)*Eb*A;// [W] for 3.5 micro m <lambda< infinity 
+TRA = RA1+RA2;// [W]
+printf("total energy absorbed in the glass is %f kW",TRA/1000);
+printf("\n\n total energy transmitted by the glass is %f kW",TRT/1000);
diff --git a/405/CH8/EX8.10/8_10.sce b/405/CH8/EX8.10/8_10.sce
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index 000000000..eacacf30f
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+clear;
+clc;
+printf("\t\t\tExample Number 8.10\n\n\n");
+// heat transfer reduction with parallel plate shield
+// Example 8.10 (page no.-413) 
+// solution
+
+E1 = 0.3;// emissivity of first plane
+E2 = 0.8;// emissivity of second plane
+E3 = 0.04;// emissivity of shield
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// the heat transfer without the shield is given by 
+// q_by_A = sigma*(T1^4-T2^4)/((1/E1)+(1/E2)-1) = 0.279*sigma*(T1^4-T2^4)
+// where T1 is temperature of first plane and T2 is temperature of second plane
+// the radiation network for the problem with the shield in place is shown in figure (8-32) (page no.-410). 
+// the resistances are 
+R1 = (1-E1)/E1;
+R2 = (1-E2)/E2;
+R3 = (1-E3)/E3;
+// the total resistance with the shield is 
+R = R1+R2+R3;
+// and the heat transfer is 
+// q_by_A = sigma*(T1^4-T2^4)/R = 0.01902*sigma*(T1^4-T2^4)
+printf("so the heat tranfer is reduced by %f percent",((0.279-0.01902)/0.279)*100);
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diff --git a/405/CH8/EX8.11/8_11.sce b/405/CH8/EX8.11/8_11.sce
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index 000000000..4e3034c61
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+clear;
+clc;
+printf("\t\t\tExample Number 8.11\n\n\n");
+// open cylindrical shield in large room
+// Example 8.11 (page no.-413-415) 
+// solution
+
+// two concentric cylinders of example(8.3) have 
+T1 = 1000;// [K] 
+E1 = 0.8;
+E2 = 0.2;
+T3 = 300;// [K] room temperature 
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// please refer to figure example 8-11(page no.-413) for radiation network
+// the room is designed as surface 3 and J3 = Eb3, because the room is very large,(i.e. its surface is very small) 
+// in this problem we must consider the inside and outside of surface 2 and thus have subscripts i and o to designate the respective quantities. 
+// the shape factor can be obtained from example 8-3 as
+F12 = 0.8253;
+F13 = 0.1747;
+F23i = 0.2588;
+F23o = 1.0;
+// also
+A1 = %pi*0.1*0.2;// [square meter] area of first cylinder
+A2 = %pi*0.2*0.2;// [square meter] area of second cylinder
+Eb1 = sigma*T1^4;// [W/square meter]
+Eb3 = sigma*T3^4;// [W/square meter]
+// the resistances may be calculated as 
+R1 = (1-E1)/(E1*A1);
+R2 = (1-E2)/(E2*A2);
+R3 = 1/(A1*F12);
+R4 = 1/(A2*F23i);
+R5 = 1/(A2*F23o);
+R6 = 1/(A1*F13);
+// the network could be solved as a series-parallel circuit to obtain the heat transfer, butwe will need the radiosities anyway, so we setup three nodal equations to solve for J1,J2i, and J2o.
+// we sum the currents into each node and set them equal to zero:
+
+// node J1: (Eb1-J1)/R1+(Eb3-J3)/R6+(J2i-J1)/R3 = 0
+// node J2i: (J1-J2i)/R3+(Eb3-J2i)/R4+(J2o-J2i)/(2*R2) = 0
+// node J2o: (Eb3-J2o)/R5+(J2i-J2o)/(2*R2) = 0
+// these equations can be solved by matrix method and the solution is 
+J1 = 49732;// [W/square meter]
+J2i = 26444;// [W/square meter]
+J2o = 3346;// [W/square meter]
+// the heat transfer is then calculated from
+q = (Eb1-J1)/((1-E1)/(E1*A1));// [W]
+// from the network we see that
+Eb2 = (J2i+J2o)/2;// [W/square meter]
+// and 
+T2 = (Eb2/sigma)^(1/4);// [K]
+// if the outer cylinder had not been in place acting as a "shield" the heat loss from cylinder 1 could have been calculated from equation(8-43a) as 
+q1 = E1*A1*(Eb1-Eb3);// [W]
+printf("temperature of the outer cylinder is %f K",T2);
+printf("\n\ntotal heat lost by inner cylinder is %f W",q1);
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diff --git a/405/CH8/EX8.12/8_12.sce b/405/CH8/EX8.12/8_12.sce
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index 000000000..31ffb71c4
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+clear;
+clc;
+printf("\t\t\tExample Number 8.12\n\n\n");
+// network for gas radiation between parallel plates
+// Example 8.12 (page no.-422-423) 
+// solution
+
+T1 = 800;// [K] temperature of first plate 
+E1 = 0.3;// emissivity
+T2 = 400;// [K] temperature of second plate
+E2 = 0.7;// emissivity
+Eg = 0.2;// emissivity of gray gas
+tg = 0.8;// transmissivity of gray gas 
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// the network shown in figure 8-39(page no.-419) applies to this problem. all the shape factors are unity for large planes and the various resistors can be computed on a unit area basis as 
+F12 = 1;
+F1g = 1;
+F2g = F1g;
+R1 = (1-E1)/E1;
+R2 = (1-E2)/E2;
+R3 = 1/(F12*(1-Eg));
+R4 = 1/(F1g*Eg);
+R5 = 1/(F2g*Eg);
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb2 = sigma*T2^(4);// [W/square meter]
+// the equivalent resistance of the center "triangle" is 
+R = 1/[(1/R3)+(1/(R4+R5))];
+// the total heat transfer is then 
+q_by_A = (Eb1-Eb2)/(R1+R2+R);// [W/square meter]
+// if there were no gas present the heat transfer would be given by equation (8-42):
+q_by_A1 = (Eb1-Eb2)/[(1/E1)+(1/E2)-1];// [W/square meter]
+// the radiosities may be computed from q_by_A = (Eb1-J1)*(E1/(1-E1)) = (J2-Eb2)*(E2/(1-E2))
+J1 = Eb1-q_by_A*((1-E1)/E1);// [W/square meter]
+J2 = Eb2+q_by_A*((1-E2)/E2);// [W/square meter]
+// for the network Ebg is just the mean of these values
+Ebg = (J1+J2)/2;// [W/square meter]
+// so that the temperature of the gas is
+Tg = (Ebg/sigma)^(1/4);// [K]
+printf("the heat-transfer rate between the two planes is %f W/square meter",q_by_A);
+printf("\n\n the temperature of the gas is %f K",Tg);
+printf("\n\n the ratio of heat-transfer with presence of gas to without presence of gas is %f",q_by_A/q_by_A1);
+
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diff --git a/405/CH8/EX8.13/8_13.sce b/405/CH8/EX8.13/8_13.sce
new file mode 100755
index 000000000..74ef28fb7
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+clear;
+clc;
+printf("\t\t\tExample Number 8.13\n\n\n");
+// cavity with transparent cover 
+// Example 8.13 (page no.-433-434) 
+// solution
+
+E1 = 0.5;// emissivity of rectangular cavity
+t2 = 0.5;// transmissivity
+rho2 = 0.1;// reflectivity
+E2 = 0.4;// emissivity
+// from example 8-9 we have
+// per unit depth in the z direction we have 
+A1 = 25+25+10;
+A2 = 10;
+// we may evaluate K from equation(8-96a)
+K = E1/(t2+(E2/2));
+// the value of Ea is then computed from equation (8-96) as 
+Ea = (t2+(E2/2))*K/[(A2/A1)*(1-E1)+K];
+printf("apparent emissivity of covered opening is %f ",Ea);
+// if there were no cover present, the value of Ea would be given by equation (8-47) as
+Ea1 = E1*A1/[A2+E1*(A1-A2)];
+printf("\n\n if there were no cover present, the value of Ea(apparent emissivity) would be %f",Ea1);
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diff --git a/405/CH8/EX8.14/8_14.sce b/405/CH8/EX8.14/8_14.sce
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index 000000000..d776061d8
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+clear;
+clc;
+printf("\t\t\tExample Number 8.14\n\n\n");
+// Transmitting and reflecting system for furnace opening
+// Example 8.14 (page no.-434-435) 
+// solution
+
+T1 = 1000+273;// [K] temperature of furnace
+lambda = 4.0;// [micro meter]
+//for  0 < lambda < 4 micro meter
+t1 = 0.9;
+E1 = 0.1;
+rho1 = 0;
+//for  4 micro meter < lambda < infinity  
+t2 = 0;
+E2 = 0.8;
+rho2 = 0.2;
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+T3 = 30+273;// [K] room temperature
+// the diagram of this problem is shown in figure example 8-14(page no.-434). because the room is large it may be treated as a blackbody also.
+// we shall analyze the problem by calculating the heat transfer for each wavelength band and then adding them together to obtain the total. the network for each band is a modification of figure 8-57(page no.-430), as shown here for black furnace and room. we shall make the calculation for unit area; then 
+A1 = 1.0;// [square meter]
+A2 = 1.0;// [square meter]
+A3 = 1.0;// [square meter]
+F12 = 1.0;
+F13 = 1.0;
+F32 = 1.0;
+// the total emissive powers are 
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb3 = sigma*T3^(4);// [W/square meter]
+// to determine the fraction of radiation in each wavelength band, we calculate
+lamba_into_T1 = lambda*T1;// [micro meter K]
+lamba_into_T3 = lambda*T3;// [micro meter K]
+// consulting table 8-1(page no.-379-380), we find 
+Eb1_0_to_4 = 0.6450*Eb1;// [W/square meter]
+Eb3_0_to_4 = 0.00235*Eb3;// [W/square meter]
+Eb1_4_to_inf = (1-0.6450)*Eb1;// [W/square meter]
+Eb3_4_to_inf = (1-0.00235)*Eb3;// [W/square meter]
+// we now apply these numbers to the network for the two wavelengths bands, with unit areas.
+
+// 0 < lambda < 4 micro meter band:
+R1 = 1/(F13*t1);
+R2 = 1/(F32*(1-t1));
+R3 = 1/(F12*(1-t1));
+R4 = rho1/(E1*(1-t1));
+// the net heat transfer from the network is then 
+R_equiv_1 = 1/(1/R1+1/(R2+R3+R4));
+q1 = (Eb1_0_to_4-Eb3_0_to_4)/R_equiv_1;// [W/square meter]
+
+// 4 micro meter < lambda < infinity band:
+R2 = 1/(F32*(1-t2));
+R3 = 1/(F12*(1-t2));
+R4 = rho2/(E2*(1-t2));
+// the net heat transfer from the network is then 
+// R1 is infinity
+R_equiv_2 = R2+R3+R4*2;
+q2 = (Eb1_4_to_inf-Eb3_4_to_inf)/R_equiv_2;// [W/square meter]
+
+// the total heat loss is then 
+q_total = q1+q2;// [W/square meter]
+// with no windows at all, the heat transfer would have been the difference in blackbody emissive powers,
+Q = Eb1-Eb3;// [W/square meter]
+printf("radiation lost through the quartz window to a room temperature of 30 degree celsius is %f W/square meter",q_total);
+printf("\n\n with no windows at all, the heat transfer would be %e W/square meter",Q);
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diff --git a/405/CH8/EX8.15/8_15.sce b/405/CH8/EX8.15/8_15.sce
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+clear;
+clc;
+printf("\t\t\tExample Number 8.15\n\n\n");
+// numerical solution for enclosure 
+// Example 8.15 (page no.-440) 
+// solution
+
+// the geometry of example 8-5 is used
+d = 0.6;// [m] diameter of long half-circular cylinder
+L = 0.2;// [m] length of square rod
+E2 = 0.5;
+T2 = 1000;// [K] temperature of body 2
+T3 = 300;// [K] temperature of body 3
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// for unit length we have:
+Eb2 = sigma*T2^(4);// [W/square meter]
+Eb3 = sigma*T3^(4);
+A1 = 4*L;// [square meter]
+A2 = %pi*d/2;// [square meter/meter]
+// we will use the numerical formulation. we find from example 8-5, using the nomenclature of the figure 
+F11 = 0.314;
+F12 = 0.425;
+F13 = 0.261;
+F21 = 0.5;
+F22 = 0;
+F23 = 0.5;
+// F31, F32 tends to zero so 
+F33 = 1;
+// we now write the equations. 
+// surface 1 is insulated so we use equation(8-107a):
+// J1*(1-F11)-F12*J2-F13*J3 = 0
+// surface 2 is constant temperature so we use equation (8-106a):
+// J2*(1-F22*(1-E2))-(1-E2)*[F21*J1+F23*J3] = E2*Eb2
+// because surface 3 is so large 
+J3 = Eb3;// [W/square meter]
+// rearranging the equation gives 
+// J1*(1-F11)-J2*F12 = F13*J3
+// J1*(-1)*(1-E2)*F21+J2*(1-F22*(1-E2)) = E2*Eb2+(1-E2)*(F23*J3)
+// solving the above two equations using matrix
+X = [(1-F11) -F12;(-1)*(1-E2)*F21 (1-F22*(1-E2))];
+Y = [F13*J3;E2*Eb2+(1-E2)*(F23*J3)];
+J = X^(-1)*Y;
+J1 = J(1);// [W/square meter]
+J2 = J(2);// [W/square meter]
+// the heat transfer is thus 
+q = (Eb2-J2)/((1-E2)/(E2*A1));// [W/m] length
+// because surface 1 is insulated 
+Eb1 = J1;// [W/square meter]
+// we could calculate the temperature as 
+T1 = (Eb1/sigma)^(1/4);// [K]
+printf("heat lost to the large room per unit length of surface 2 is %f W/m",q);
+printf("\n\n temperature of the insulated surface is %f K",T1);
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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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+
+
+
+
+
diff --git a/405/CH8/EX8.17/8_17.sce b/405/CH8/EX8.17/8_17.sce
new file mode 100755
index 000000000..66d23dcf2
--- /dev/null
+++ b/405/CH8/EX8.17/8_17.sce
@@ -0,0 +1,117 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.17\n\n\n");
+// radiation from a hole with variable radiosity 
+// Example 8.17 (page no.-443-446)
+// solution
+
+T1 = 1273;// [K]
+T5 = 293;// [K]
+E1 = 0.6;
+// all the shape factors can be obtained with the aid of figure 8-13(page no.-387) and the imaginary disk surfaces 6 and 7. we have
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb2 = Eb1;// [W/square meter]
+Eb3 = Eb2;// [W/square meter]
+Eb4 = Eb3;// [W/square meter]
+Eb5 = sigma*T5^(4);// [W/square meter]
+E2 = E1;
+E3 = E2;
+E4 = E3;
+E5 = 1.0;
+r = 0.01;// [m]
+A1 = %pi*r^(2);// [square m]
+A5 = A1;// [square m]
+A6 = A1;// [square m]
+A7 = A1;// [square m]
+A2 = %pi*2*r*0.01;// [square m]
+A3 = A2;// [square m]
+A4 = A2;// [square m]
+F11 = 0;
+F55 = 0;
+F16 = 0.37;
+F17 = 0.175;
+F15 = 0.1;
+F12 = 1-F16;
+F54 = F12;
+F13 = F16-F17;
+F53 = F13;
+F14 = F17-F15;
+F52 = F14;
+F21 = F16*A1/A2;
+F26 = F21;
+F45 = F21;
+F36 = F45;
+F37 = F36;
+F22 = 1-F21-F26;
+F33 = F22;
+F44 = F22;
+F31 = F13*A1/A3;
+F32 = F36-F31;
+F34 = F32;
+F43 = F34;
+F23 = F34;
+F27 = F26-F23;
+F46 = F27;
+F41 = F14*A1/A4;
+F25 = F41;
+F42 = F46-F41;
+F24 = F42;
+// the equations for the radiosities are now written in the form of equation 8-106, noting that
+F11 = 0;
+J5 = Eb5;// [W/square meter]
+//  J1 = (1-E1)*(F12*J2+F13*J3+F14*J4+F15*Eb5)+E1*Eb1
+//  J2 = [(1-E2)*(F21*J1+F23*J3+F24*J4+F25*Eb5)+E2*Eb2]/(1-F22*(1-E2))
+//  J3 = [(1-E3)*(F31*J1+F32*J2+F34*J4+F35*Eb5)+E3*Eb3]/(1-F33*(1-E3))
+//  J4 = [(1-E2)*(F41*J1+F42*J2+F43*J3+F45*Eb5)+E4*Eb4]/(1-F44*(1-E4))
+// we have 4 equations with 4 variables which can be solved by matrix method
+Z = [1 -(1-E1)*F12 -(1-E1)*F13 -(1-E1)*F14;-F21*(1-E2)/(1-F22*(1-E2)) 1 -F23*(1-E2)/(1-F22*(1-E2)) -F24*(1-E2)/(1-F22*(1-E2));-F31*(1-E3)/(1-F33*(1-E3)) -F32*(1-E3)/(1-F33*(1-E3)) 1 -F34*(1-E3)/(1-F33*(1-E3));-F41*(1-E4)/(1-F44*(1-E4)) -F42*(1-E4)/(1-F44*(1-E4)) -F43*(1-E4)/(1-F44*(1-E4)) 1];
+C = [E1*Eb1+(1-E1)*F15*Eb5;(E2*Eb2+F25*Eb5*(1-E2))/(1-F22*(1-E2));104859;(E4*Eb4+F45*Eb5*(1-E4))/(1-F44*(1-E4))];
+J = Z^(-1)*C;
+J1 = J(1);// [W/square meter]
+J2 = J(2);// [W/square meter]
+J3 = J(3);// [W/square meter]
+J4 = J(4);// [W/square meter]
+// the heat transfer can be calculated from equation(8-104):
+q1 = E1*A1*(Eb1-J1)/(1-E1);// [W]
+q2 = E2*A2*(Eb2-J2)/(1-E2);// [W]
+q3 = E3*A3*(Eb3-J3)/(1-E3);// [W]
+q4 = E4*A4*(Eb4-J4)/(1-E4);// [W]
+// THE TOTAL HEAT TRANSFER 
+q = q1+q2+q3+q4;// [W]
+printf("the heat transfer rate is %f W",q);
+// It is of interest to compare this heat transfer with the value we would obtain by assuming uniform radiosity on the hot surface. we would then have a two-body problem with
+A1 = %pi+3*(2*%pi);// [square cm]
+A5 = %pi*10^(-4);// [square cm]
+F51 = 1.0;
+E1 = 0.6;
+E5 = 1.0;
+// the heat transfer is then calculated from equation(8-43), with appropriate change of nomenclature:
+q_new = (Eb1-Eb5)*A5/((1/E5)+(A5/A1)*((1/E1)-1));// [w]
+printf("\n\nthus the assumption of uniform radiosity gives a heat transfer that is %f percent below the value obtained by breaking the hot surface into four parts for the calculations",(q-q_new)*100/q);
+// let us now consider the case where surface 1 is still radiating at 1000 degree celsius 
+E = 0.6;
+// the nodal equations for J1 is the same as before but now the equations for J2, J3, and J4 must be written in the form of equation(8-107). when that is done and the numerical values are inserted, we obtain
+//  J1 = 0.252*J2+0.078*J3+0.03*J4+89341
+//  J2 = 0.5*J1+0.3452*J3+0.09524*J4+24.869
+//  J3 = 0.1548*J1+0.3452*J2+0.3452*J4+64.66
+//  J4 = 0.05952*J1+0.0952*J2+0.3452*J3+208.9
+// when these equations are solved, we obtain
+J1 = 1.1532*10^(5);// [W/square meter]
+J2 = 0.81019*10^(5);// [W/square meter]
+J3 = 0.57885*10^(5);// [W/square meter]
+J4 = 0.34767*10^(5);// [W/square meter]
+// the heat transfer at surface 1 is
+A1 = %pi*10^(-4);// [square cm]
+A5 = %pi*10^(-4);// [square cm]
+A2 = %pi*10^(-4);// [square cm] 
+q1 = (E1*A1)*(Eb1-J1)/(1-E1);// [W]
+// the temperatures of the insulated surface elements are obtained from 
+T2 = 820;// [degree celsius]
+T3 = 732;// [degree celsius]
+T4 = 612;// [degree celsius]
+// it is of interest to compare the heat transfer calculated above with that obtained by assuming surfaces 2,3 and 4 uniform in temperature and radiosity. equation(8-41) applies for this case:
+q2 = A1*(Eb1-Eb5)/[((A1+A2+2*A1*F15)/(A5-A1*F15^2))+(1/E1-1)+(A1/A5)*(1/E5-1)];// [w]
+
+
+
diff --git a/405/CH8/EX8.18/8_18.sce b/405/CH8/EX8.18/8_18.sce
new file mode 100755
index 000000000..69f26eb1c
--- /dev/null
+++ b/405/CH8/EX8.18/8_18.sce
@@ -0,0 +1,170 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.18\n\n\n");
+// heater with constant heat flux and surrounding shields 
+// Example 8.18 (page no.-446-449)
+// solution
+
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+T6 = 293;// [K] temperature of room
+E1 = 0.8;
+E2 = 0.4;
+E3 = 0.4;
+E4 = 0.4;
+E5 = 0.4;
+// in reality, surfaces 2,3,4, and 5 have two surfaces each; an inside and an outside surface. we thus have nine surfaces plus the room, so a 10 body problem is involved. of course, from symmetry we can see that T2 = T4 and T3 = T5.
+// we designate the large room as surface 6 and it behaves as E6 = 1.0. 
+// the shape factors of the inside surfaces are obtained from figure 8-12 and 8-14:
+F16 = 0.285;
+F61 = F16;
+F13 = 0.24;
+F15 = 0.24;
+F31 = 0.24;
+F51 = 0.24;
+F12 = 0.115;
+F14 = 0.115;
+F24 = 0.068;
+F42 = 0.068;
+F35 = 0.285;
+F53 = 0.285;
+F32 = 0.115;
+F52 = 0.115;
+F34 = 0.115;
+F25 = 0.23;
+F23 = 0.23;
+F45 = 0.23;
+F43 = 0.23;
+F21 = 0.23;
+F41 = 0.23;
+F26 = 0.23;
+F46 = 0.23;
+F11 = 0;
+F22 = 0;
+F33 = 0;
+F44 = 0;
+F55 = 0;
+// for the outside surfaces,
+F_26 = 1;
+F_36 = 1;
+F_46 = 1;
+F_56 = 1;
+// Where the underscore indicate the outside surfaces.
+// for the room 
+Eb6 = sigma*T6^(4);// [W/square meter]
+// for surface 1 with constant heat flux, we use equation (8-108a) and write
+// J1-(F12*J2+F13*J3+F14*J4+F15*J5+F16*J6) = 1.0*10^(5)               GIVEN             (a)
+// because of the radiant balance condition we have
+// (J2-Eb2)*E2*A2/(1-E2) = (Eb2-J_2)*E2*A2/(1-E2)
+// and      Eb2 = (J2+J_2)/2                                                            (b)
+// Where underscore indicates the outside radiosity. a similar relation applies for surfaces 3,4, and 5. thus we can use equation (8-106a) for inside surface 2
+// J2-(1-E2)*(F21*J1+F23*J3+F24*J4+F25*J5+F26*J6) = E2*(J2+J_2)/2                       (c)
+// and for outside surface 2
+// J_2-(1-E2)*(F_26*J6) = E2*(J2+J_2)/2                                                 (d)
+// Equations like (c) and (d) are written for surfaces 3,4, and 5 also, and with the shape factors and emmissivities inserted the following set of equations is obtained
+// J1-0.115*J2-0.24*J3-0.115*J4-0.24*J5 = 1.0012*10^(5)                         1
+// -0.138*J1+0.8*J2-0.2*J_2-0.138*J3-0.0408*J4-0.138*J5 = 57.66                 2
+// 0.2*J2-0.8*J_2 = -250.68                                                     3
+// -0.144*J1-0.069*J2+0.8*J3-0.2*J_3-0.069*J4-0.05*J5 = 60.16                   4
+// 0.2*J3-0.8*J_3 = -250.68                                                     5
+// -0.138*J1-0.0408*J2-0.138*J3+0.8*J4-0.2*J_4-0.138*J5 = 57.66                 6
+// 0.2*J4-0.8*J_4 = -250.68                                                     7
+// -0.144*J1-0.069*J2-0.057*J3-0.069*J4+0.8*J5-0.2*J_5 = 60.16                  8
+// 0.2*J5-0.8*J_5 = -250.68                                                     9
+// We thus have nine equations and nine unknowns, which may be solved by matrix method
+Z = [1 -0.115 -0.24 -0.115 -0.24 0 0 0 0;-0.138 0.8 -0.138 -0.0408 -0.138 -0.2 0 0 0;0 0.2 0 0 0 -0.8 0 0 0;-0.144 -0.069 0.8 -0.069 -0.05 0 -0.2 0 0;0 0 0.2 0 0 0 -0.8 0 0;-0.138 -0.0408 -0.138 0.8 -0.138 0 0 -0.2 0;0 0 0 0.2 0 0 0 -0.8 0;-0.144 -0.069 -0.057 -0.069 0.8 0 0 0 -0.2;0 0 0 0 0.2 0 0 0 -0.8];
+C = [1.0012*10^(5);57.66;-250.68;60.16;-250.68;57.66;-250.68;60.16;-250.68];
+J = Z^(-1)*C;
+J1 = J(1);// [W/square meter]
+J2 = J(2);// [W/square meter]
+J3 = J(3);// [W/square meter]
+J4 = J(4);// [W/square meter]
+J5 = J(5);// [W/square meter]
+J_2 = J(6);// [W/square meter]
+J_3 = J(7);// [W/square meter]
+J_4 = J(8);// [W/square meter]
+J_5 = J(9);// [W/square meter]
+// the temperatures are thus computed from equation (b):
+Eb2 = (J2+J_2)/2;// [W/square meter]
+T2 = (Eb2/sigma)^(1/4);// [K]
+T4 = T2;// [K]
+Eb3 = (J3+J_3)/2;// [W/square meter]
+T3 = (Eb3/sigma)^(1/4);// [K]
+T5 = T3;// [K]
+// for surface 1 we observed that
+q = 1*10^(5);// [W/square meter]
+Eb1 = q*(1-E1)/E1+J1;// [W/square meter]
+// and 
+T1 = (Eb1/sigma)^(1/4);// [K]
+printf("temperature of all surfaces are following ");
+printf("\n\n\t T1 = %f K",T1);
+printf("\n\n\t T2 = %f K",T2);
+printf("\n\n\t T3 = %f K",T3);
+printf("\n\n\t T4 = %f K",T4);
+printf("\n\n\t T5 = %f K",T5);
+
+// surfaces 2,3,4, and 5 as one surface
+// we now go back and take surfaces 2,3,4, and 5 as one surface, which we choose to call surface 7. the shape factors are then
+F16 = 0.285;
+F61 = 0.285;
+F17 = 1-0.285;
+A1 = 2.0;
+A7 = 6.0;
+// THUS
+F71 = A1*F17/A7;
+F77 = 1-2*F71;
+F76 = F71;
+F_76 = 1.0;
+// then for surface 1 we use equation(8-109a) to obtain 
+// J1-(F17*J7+F16*J6) = 1.0*10^(5)
+// using Eb7 = (J7+J_7)/2, we have for the inside of surface 7 
+// J7*[1-F77*(1-E7)]-(1-E7)*(F71*J1+F76*J6) = (J7+J_7)*E7/2
+// while for the outside we have 
+// J_7-(1-E7)*F_76*J6 = (J7+J_7)*E7/2
+// when the numerical values are inserted, we obtain the set of three equations:
+// J1-0.715J7 = 1.0012*10^(5)
+// -0.143*J1+0.486*J7-0.2*J_7 = 59.74
+// 0.2*J7-0.8*J_7 = -250.68
+// Solving above three equations by matrix method
+Z = [1 -0.715 0;-0.143 0.486 -0.2;0 0.2 -0.8];
+C = [1.0012*10^(5);59.74;-250.68];
+J = Z^(-1)*C;
+J1 = J(1);// [W/square meter]
+J7 = J(2);// [W/square meter]
+J_7 = J(3);// [W/square meter]
+// the temperatures are thus computed as before
+Eb7 = (J7+J_7)/2;// [W/square meter]
+T7 = (Eb7/sigma)^(1/4);// [K]
+Eb1 = q*(1-E1)/E1+J1;// [W/square meter]
+T11 = (Eb1/sigma)^(1/4);// [K]
+printf("\n\n from second method T1 = %f K",T11);
+printf("\n\n so there is a difference of %f K between the two methods",T11-T1);
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/405/CH8/EX8.19/8_19.sce b/405/CH8/EX8.19/8_19.sce
new file mode 100755
index 000000000..46bf7b38a
--- /dev/null
+++ b/405/CH8/EX8.19/8_19.sce
@@ -0,0 +1,72 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.19\n\n\n");
+// numerical solution for combined convection and radiation(non-linear system) 
+// Example 8.19 (page no.-449-452)
+// solution
+
+l = 0.5;// [m] length of plate 
+b = 0.5;// [m] breadth of plate
+T1 = 1300;// [K] temperature of plate 
+Tinf = 300;// [K] temperature of surrounding
+T4 = Tinf;// [degree celsius]
+h = 50;// [W/square meter] convection heat transfer coefficient
+E1 = 0.8;
+E2 = 0.3;
+E3 = 0.3;
+// using figures 8-12(page no.-386) and 8-14(page no.-387), we can evaluate the shape factors as 
+F12 = 0.2;
+F13 = 0.2;
+F23 = 0.2;
+F32 = 0.2;
+F14 = 1-0.2-0.2;
+F24_L = 1;
+F34_R = 1;
+F21 = F12;
+F31 = F12;
+F24_R = 0.6;
+F34_L = 0.6;
+F11 = 0;
+F22 = 0;
+F33 = 0;
+// J2R = J3L
+// J2L = J3R      From symmetry
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+Eb4 = sigma*T4^(4);// [W/square meter]
+Eb1 = sigma*T1^(4);// [W/square meter]
+J4 = Eb4;// [W/square meter]
+// we now use equation(8-107) to obtain a relation for J1:
+// J1 = (1-E1)*[F12*J2R+F13*J3L+F14*J4]+E1*Eb1
+// but J2R = J3L and F12 = F13 so that
+// J1 = (1-E1)*[2*F13*J2R+F14*J4]+E1*Eb1                                       (a)
+// we use equation (8-108) for the overall energy balance on surface 2:
+// 2*h*(Tinf-T2) = E2*(Eb2-J2R)/(1-E2)+E2*(Eb2-J2L)/(1-E2)
+// 2*h*(Tinf-T2) = E2*(2*Eb2-J2R-J2L)/(1-E2)                                   (b)
+// equation (8-105) is used for surface J2R.
+// J2R = (1-E2)*(F21*J1+F23*J3L+F24_R*J4)+E2*Eb2
+//  But J2R = J3L so that
+// J2R = [(1-E2)*(F21*J1+F24_R*J4)+E2*Eb2]/(1-(1-E2)*F23)                       (c)
+// for surface J2L the equation is 
+// J2L = (1-E2)*(F24_L*J4)+E2*Eb2                                               (d)
+// we now have four equations with four unknowns, J1,J2R,J2L,Eb2, with T2 = (Eb2/sigma)^(1/4).
+// however equation (b) is nonlinear in Eb so we must use a special procedure to solve the set.
+for T2 = 300:0.1:400
+    Z = [1 -(1-E1)*2*F13 0 0;0 -E2/(1-E2) -E2/(1-E2) 2*E2/(1-E2);(1-E2)*F21/(1-(1-E2)*F23) -1 0 E2/(1-(1-E2)*F23);0 0 1 -E2];
+    C = [E1*Eb1;2*h*(Tinf-T2);-F24_R/(1-(1-E2)*F23);(1-E2)*F24_L];
+    S = Z^(-1)*C;
+    Eb2_E = S(4);
+    Eb2_T = sigma*T2^(4);
+    dEb2 = Eb2_E-Eb2_T;
+    if (dEb2>0 & dEb2<5) then
+        J1 = S(1);// [W/square meter]
+        J2R = S(2);// [W/square meter]
+        J2L = S(3);// [W/square meter]
+        Eb2 = S(4);// [W/square meter]
+        T2new = T2;// [K]
+    end
+end
+// the total heat flux lost by surface 1 is 
+q1_by_A1 = h*(T1-Tinf)+(Eb1-J1)*E1/(1-E1);// [W/square meter]
+// for a 0.5 by 0.5 m surface the heat lost is thus 
+q1 = q1_by_A1*l*b;// [W]
+printf("\n\n the heat lost by plate is %f W",q1);
diff --git a/405/CH8/EX8.2/8_2.sce b/405/CH8/EX8.2/8_2.sce
new file mode 100755
index 000000000..5a0a6e272
--- /dev/null
+++ b/405/CH8/EX8.2/8_2.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.2\n\n\n");
+// heat transfer between black surfaces
+// Example 8.2 (page no.-389-390) 
+// solution
+
+L = 1;// [m] length of black plate
+W = 0.5;// [m] width of black plate
+T1 = 1000+273;// [K] first plate temperature
+T2 = 500+273;// [K] second plate temperature
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// the ratios for use with figure 8-12(page no.-386) are
+Y_by_D = W/W;
+X_by_D = L/W;
+// so that 
+F12 = 0.285;// radiation shape factor 
+// the heat transfer is calculated from
+q = sigma*L*W*F12*(T1^(4)-T2^(4));
+printf("net radiant heat exchange between the two plates is %f kW",q/1000);
+
+
+
+
+
+
+
+
diff --git a/405/CH8/EX8.20/8_20.sce b/405/CH8/EX8.20/8_20.sce
new file mode 100755
index 000000000..fc869f50f
--- /dev/null
+++ b/405/CH8/EX8.20/8_20.sce
@@ -0,0 +1,36 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.20\n\n\n");
+// solar-environment equilibrium  temperatures 
+// Example 8.20 (page no.-454) 
+// solution
+
+q_by_A_sun = 700;// [W/m^(2)] solar flux
+T_surr = 25+273;// [K] surrounding temperature
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// at radiation equilibrium the netenergy absorbed from sun must equal the long-wavelength radiation exchange with the surroundings,or
+// (q_by_A_sun)*alpha_sun = alpha_low_temp*sigma*(T^4-T_surr^4)         (a)
+
+// case (a) for white paint
+
+// for white paint we obtain from table 8-4
+alpha_sun = 0.12;
+alpha_low_temp = 0.9;
+// so that equation (a) becomes
+T = [(q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr^(4)]^(1/4);// [K]
+printf("radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with white paint is %f degree celsius",T-273);
+
+// case (b) for flat black lacquer we obtain
+
+alpha_sun = 0.96;
+alpha_low_temp = 0.95;
+// so that equation (a) becomes
+T = [(q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr^(4)]^(1/4);// [K]
+printf("\n\nradiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with flat black lacquer is %f degree celsius",T-273);
+
+
+
+
+
+
+
diff --git a/405/CH8/EX8.21/8_21.sce b/405/CH8/EX8.21/8_21.sce
new file mode 100755
index 000000000..47994474b
--- /dev/null
+++ b/405/CH8/EX8.21/8_21.sce
@@ -0,0 +1,35 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.21\n\n\n");
+// influence of convection on solar equilibrium temperature 
+// Example 8.21 (page no.-455) 
+// solution
+
+T_surr = 25+273;// [K] surrounding temperature
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+h = 10;// [W/square meter] heat transfer coefficient
+// in this case the solar energy absorbed must equal the sum of the radiation and convection transfers to the surroundings
+// (q_by_A_sun)*alpha_sun = alpha_low_temp*sigma*(T^4-T_surr^4)+h*(T-T_surr)         (a)
+q_by_A_sun = 700;// [W/m^(2)] solar flux
+
+// for the white paint, using the same surface properties as in example 8-20 gives 
+
+alpha_sun = 0.12;
+alpha_low_temp = 0.9;
+// so that equation (a) becomes
+deff('[y] = f(T)','y = (q_by_A_sun)*alpha_sun-alpha_low_temp*sigma*(T^4-T_surr^4)-h*(T-T_surr)');
+T = fsolve(1,f);
+printf("the radiation-convection equillibrium temperatures for case (a) is %f degree celsius",T-273);
+
+//for flat black lacquer we obtain
+
+alpha_sun = 0.96;
+alpha_low_temp = 0.95;
+// so that equation (a) becomes
+deff('[y] = f2(T1)','y = (q_by_A_sun)*alpha_sun - alpha_low_temp*sigma*(T1^4-T_surr^4)-h*(T1-T_surr)');
+T1 = fsolve(1,f2);
+printf("\n\n the radiation-convection equillibrium temperatures for case (b) is %f degree celsius",T1-273);
+printf("\n\n where case (a)     surface is coated with white paint");
+printf("\n\n       case (b)     surface is coated with flat black lacquer");
+
+
diff --git a/405/CH8/EX8.23/8_23.sce b/405/CH8/EX8.23/8_23.sce
new file mode 100755
index 000000000..b31ec3922
--- /dev/null
+++ b/405/CH8/EX8.23/8_23.sce
@@ -0,0 +1,17 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.23\n\n\n");
+// temperature measurement error caused by radiation 
+// Example 8.23 (page no.-460) 
+// solution
+
+E = 0.9;// emissivity of mercury-in-glass thermometer 
+Tt = 20+273;// [K] temperature indicated by thermometer 
+Ts = 5+273;// [K] temperature of walls
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+h = 8.3;// [W/square meter] heat transfer coefficient for thermometer
+// we employ equation(8-113) for the solution: h*(Tinf-Tt) = sigma*E*(Tt^4-Ts^4)
+// inserting the values in above equation 
+Tinf = sigma*E*(Tt^4-Ts^4)/h+Tt;// [K]
+printf("the true air temperature is %f degree celsius",Tinf-273);
+
diff --git a/405/CH8/EX8.3/8_3.sce b/405/CH8/EX8.3/8_3.sce
new file mode 100755
index 000000000..1c318d868
--- /dev/null
+++ b/405/CH8/EX8.3/8_3.sce
@@ -0,0 +1,45 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.3\n\n\n");
+// shape-factor algebra for open ends of cylinder
+// Example 8.3 (page no.-395) 
+// solution
+
+d1 = 0.1;// [m] diameter of first cylinder
+d2 = 0.2;// [m] diameter of second cylinder
+L = 0.2;// [m] length of cylinder
+// we use the nomenclature of figure 8-15(page no.-388) for this problem and designate the open ends as surfaces 3 and 4.
+// we have 
+L_by_r2 = L/(d2/2);
+r1_by_r2 = 0.5;
+// so from figure 8-15 or table 8-2(page no.-389) we obtain
+F21 = 0.4126;
+F22 = 0.3286;
+// using the reciprocity relation (equation 8-18) we have
+F12 = (d2/d1)*F21;
+// for surface 2 we have F12+F22+F23+F24 = 1.0
+// and from symmetry F23 = F24 so that
+F23 = (1-F21-F22)/2;
+F24 = F23;
+// using reciprocity again,
+A2 = %pi*d2*L;// [m^2]
+A3 = %pi*(d2^2-d1^2)/4;// [m^2]
+F32 = A2*F23/A3;
+// we observe that F11 = F33 = F44 = 0 and for surface 3 F31+F32+F34 = 1.0
+// so, if F31 can be determined, we can calculate the desired quantity F34. for surface 1 F12+F13+F14 = 1.0
+// and from symmetry F13 = F14 so that
+F13 = (1-F12)/2;
+F14 = F13;
+// using reciprocity gives
+A1 = %pi*d1*L;// [square meter]
+F31 = (A1/A3)*F13;
+// then 
+F34 = 1-F31-F32;
+printf("shape factor between the open ends of the cylinder is %f ",F34);
+
+
+
+
+
+
+
diff --git a/405/CH8/EX8.4/8_4.sce b/405/CH8/EX8.4/8_4.sce
new file mode 100755
index 000000000..622b551d6
--- /dev/null
+++ b/405/CH8/EX8.4/8_4.sce
@@ -0,0 +1,58 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.4\n\n\n");
+// shape-factor algebra for truncated cone
+// Example 8.4 (page no.-396) 
+// solution
+
+d1 = 0.1;// [m] diameter of top of cone
+d2 = 0.2;// [m] diameter of bottom of cone
+L = 0.1;// [m] height of cone
+// we employ figure 8-16(page no.-390) for solution of this problem and take the nomenclature as shown, designating the top as surface 2,
+// the bottom as surface 1, and the side as surface 3. thus the desired quantities are F23 and F33. we have 
+Z = L/(d2/2);
+Y = (d1/2)/L;
+// thus from figure 8-16(page no.-390)  
+F12 = 0.12;
+// from reciprcity(equatin 8-18)
+A1 = %pi*d2^(2)/4;// [square meter]
+A2 = %pi*d1^(2)/4;// [square meter]
+F21 = A1*F12/A2;
+//and
+F22 = 0;
+// so that 
+F23 = 1-F21;
+// for surface 3 F31+F32+F33 = 1,  so we must find F31 and F32 in order to evaluate F33. since F11 = 0 we have
+F13 = 1-F12;
+// and from reciprocity 
+A3 = %pi*((d1+d2)/2)*[(d1/2-d2/2)^(2)+L^(2)]^(1/2);// [square meter]
+// so from above equation
+F31 = A1*F13/A3;
+// a similar procedure is applies with surface 2 so that 
+F32 = A2*F23/A3;
+// finally from above equation 
+F33 = 1-F32-F31;
+printf("shape factor between the top surface and the side is %f ",F23);
+printf("\nshape factor between the side and itself is %f ",F33);
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/405/CH8/EX8.5/8_5.sce b/405/CH8/EX8.5/8_5.sce
new file mode 100755
index 000000000..db472ddd8
--- /dev/null
+++ b/405/CH8/EX8.5/8_5.sce
@@ -0,0 +1,38 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.5\n\n\n");
+// shape-factor algebra for cylindrical reflactor
+// Example 8.5 (page no.-397-398) 
+// solution
+
+d = 0.6;// [m] diameter of long half-circular cylinder
+L = 0.2;// [m] length of square rod
+// we have given figure example 8-5(page no.-397) for solution of this problem and take the nomenclature as shown, 
+// from symmetry we have 
+F21 = 0.5;
+F23 = F21;
+// in general, F11+F12+F13 = 1. to aid in the analysis we create the fictious surface 4 shown in figure example 8-5 as dashed line.
+// for this surface 
+F41 = 1.0;
+// now, all radiation leaving surface 1 will arrive either at 2 or at 3. likewise,this radiation will arrive at the imaginary surface 4, so that F41 = F12+F13 say eqn a
+// from reciprocity
+A1 = %pi*d/2;// [square meter]
+A4 = L+2*sqrt(0.1^(2)+L^(2));// [square meter]
+A2 = 4*L;// [square meter]
+// so that 
+F14 = A4*F41/A1;// say eqn b
+// we also have from reciprocity
+F12 = A2*F21/A1;// say eqn c
+// combining a,b,c, gives
+F13 = F14-F12;
+// finally
+F11 = 1-F12-F13;
+printf("value of F12 is %f ",F12);
+printf("\nvalue of F13 is %f ",F13);
+printf("\nvalue of F11 is %f ",F11);
+
+
+
+
+
+ 
+\ No newline at end of file
diff --git a/405/CH8/EX8.6/8_6.sce b/405/CH8/EX8.6/8_6.sce
new file mode 100755
index 000000000..b28e5242c
--- /dev/null
+++ b/405/CH8/EX8.6/8_6.sce
@@ -0,0 +1,65 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.6\n\n\n");
+// hot plates enclosed by a room
+// Example 8.6 (page no.-402-404) 
+// solution
+
+w = 0.5;// [m] width of plate 
+L = 1;// [m] length of plate
+t = 0.5;// [m] seperation between two plates
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// this is a three-body problem, the two plates and the room, so the radiation network is shown in figure(8-27) page no.-401.
+// from the data of the problem 
+T1 = 1000+273;// [K] temperature of first plate
+T2 = 500+273;// [K] temperature of second plate
+T3 = 27+273;// [K] temperature of walls of plates
+A1 = w*L;// [square meter] area of plate
+A2 = A1;// [square meter] area of plate
+E1 = 0.2;// emissivity of plate 1
+E2 = 0.5;// emissivity of plate 2
+// because the area of the room A3 is very large, the resistance (1-E3)/(E3*A3) may be taken as zero and we obtain Eb3 = J3.
+// the shape factor F12 was given in example 8-2:
+F12 = 0.285;
+F21 = F12;
+F13 = 1-F12;
+F23 = 1-F21;
+// the resistance in the network are calculated as 
+R1 = (1-E1)/(E1*A1);
+R2 = (1-E2)/(E2*A2);
+R3 = 1/(A1*F12);
+R4 = 1/(A1*F13);
+R5 = 1/(A2*F23);
+// taking the resistance (1-E3)/(E3*A3) as zero, we have the network (as shown in figure example 8-6(page no.-403)).
+// to calculate the heat flows at each surface we must determine the radiosities J1 and J2. the network is solved by setting the sum of the heat currents entering nodes J1 and J2 to zero 
+
+// node J1:
+// (Eb1-J1)/R1+(J2-J1)/R3+(Eb3-J1)/R4 = 0                            (a)
+
+// node J2:
+// (J1-J2)/R3+(Eb3-J2)/R5+(Eb2-J2)/R2 = 0                            (b)
+
+// now
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb2 = sigma*T2^(4);// [W/square meter]
+Eb3 = sigma*T3^(4);// [W/square meter]
+J3 = Eb3;// [W/square meter]
+// inserting the values of Eb1,Eb2, and Eb3 into equations (a) and (b), we have two equations and two unknowns J1 and J2 that may be solved simultaneously to give 
+// on simplifying we get J1 = (J2-R3*[(Eb3-J2)/R5+(Eb2-J2)/R2])
+// putting this value in equation (a) and solve for J2
+deff('[y] = f3(J2)','y = (Eb1-(J2-R3*[(Eb3-J2)/R5+(Eb2-J2)/R2]))/R1+(J2-(J2-R3*[(Eb3-J2)/R5+(Eb2-J2)/R2]))/R3+(Eb3-(J2-R3*[(Eb3-J2)/R5+(Eb2-J2)/R2]))/R4');
+J2 = fsolve(1,f3);// [W/square meter]
+J1 = (J2-R3*[(Eb3-J2)/R5+(Eb2-J2)/R2]);// [W/square meter]
+// the total heat lost by plate 1 is 
+q1 = (Eb1-J1)/[(1-E1)/(E1*A1)];// [W]
+// and the heat lost by plate 2 is 
+q2 = (Eb2-J2)/[(1-E2)/(E2*A2)];// [W]
+// the total heat received by the room is 
+q3 = [(J1-J3)/(1/(A1*F13))]+[(J2-J3)/(1/(A2*F23))];// [W]
+printf("the net heat transfer for plate 1 is %f kW",q1/1000)
+printf("\n\n the net heat transfer for plate 2 is %f kW",q2/1000)
+printf("\n\n the net heat transfer to the room  is %f kW",q3/1000)
+
+
+
+
diff --git a/405/CH8/EX8.7/8_7.sce b/405/CH8/EX8.7/8_7.sce
new file mode 100755
index 000000000..a8c69eca9
--- /dev/null
+++ b/405/CH8/EX8.7/8_7.sce
@@ -0,0 +1,52 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.7\n\n\n");
+// surface in radiant balance
+// Example 8.7 (page no.-404-405) 
+// solution
+
+w = 0.5;// [m] width of plate 
+L = 0.5;// [m] length of plate
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// from the data of the problem 
+T1 = 1000;// [K] temperature of first surface
+T2 = 27+273;// [K] temperature of room
+A1 = w*L;// [square meter] area of rectangle
+A2 = A1;// [square meter] area of rectangle
+E1 = 0.6;// emissivity of surface 1
+// although this problems involves two surfaces which exchange heat and one which is insulated or re-radiating, equation (8-41) may not be used for the calculation because one of the heat-exchanging surfaces(the room) is not convex. The radiation network is shown in figure example 8-7(page no.-404) where surface 3 is the room and surface 2 is the insulated surface. note that J3 = Eb3 because the room is large and (1-E3)/(E3*A3) approaches zero.Because surface 2 is insulated it has zero heat transfer and J2 = Eb2. J2 "floats" in the network and is determined from the overall radiant balance. 
+// from figure 8-14(page no.-387) the shape factors are 
+F12 = 0.2;
+F21 = F12;
+// because
+F11 = 0;
+F22 = 0;
+F13 = 1-F12;
+F23 = F13;
+// the resistances are 
+R1 = (1-E1)/(E1*A1);
+R2 = 1/(A1*F13);
+R3 = 1/(A2*F23);
+R4 = 1/(A1*F12);
+// we also have
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb3 = sigma*T2^(4);// [W/square meter]
+J3 = Eb3;// [W/square meter]
+// the overall circuit is a series parallel arrangement and the heat transfer is 
+R_equiv = R1+(1/[(1/R2)+1/(R3+R4)]);
+q = (Eb1-Eb3)/R_equiv;// [W]
+// this heat transfer can also be written as q = (Eb1-J1)/((1-E1)/(E1*A1))
+// inserting the values 
+J1 = Eb1-q*((1-E1)/(E1*A1));// [W/square meter]
+// the value of J2 is determined from proportioning the resistances between J1 and J3, so that 
+// (J1-J2)/R4 = (J1-J3)/(R4+R2)
+J2 = J1-((J1-J3)/(R4+R2))*R4;// [W/square meter]
+Eb2 = J2;// [W/square meter]
+// finally, we obtain the temperature of the insulated surface as
+T2 = (Eb2/sigma)^(1/4);// [K]
+printf("temperature of the insulated surface is %f K",T2);
+printf("\n\n heat lost by the surface at 1000K is %f kW",q/1000);
+
+
+
+
diff --git a/405/CH8/EX8.8/8_8.sce b/405/CH8/EX8.8/8_8.sce
new file mode 100755
index 000000000..6f9544872
--- /dev/null
+++ b/405/CH8/EX8.8/8_8.sce
@@ -0,0 +1,38 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.8\n\n\n");
+// open hemisphere in large room
+// Example 8.8 (page no.-406-408) 
+// solution
+
+d = 0.3;// [m] diameter of hemisphere
+T1 = 500+273;// [degree celsius] temperature of hemisphere
+T2 = 30+273;// [degree celsius] temperature of enclosure 
+E = 0.4;// surface emissivity of hemisphere
+sigma = 5.669*10^(-8);// [W/square meter K^(4)] constant
+// the object is completely surrounded by a large enclosure but the inside surface of the sphere is not convex.
+// in the given figure example 8-8(page no.-407) we take the inside of the sphere as surface 1 and the enclosure as surface 2.
+// we also create an imaginary surface 3 covering the opening.
+// then the heat transfer is given by
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb2 = sigma*T2^(4);// [W/square meter]
+A1 = 2*%pi*(d/2)^(2);// [square meter] area of surface 1
+// calculating the surface resistance 
+R1 = (1-E)/(E*A1); 
+// since A2 tends to 0 so R2 also tends to 0
+R2 = 0; 
+// now at this point we recognize that all of the radiation leaving surface 1 which will eventually arrive at enclosure 2 will also hit the imaginary   surface 3(F12 = F13). we also recognize that A1*F13 = A3*F31. but 
+F31 = 1.0;
+A3 = %pi*(d/2)^(2);// [square meter]
+F13 = (A3/A1)*F31;
+F12 = F13; 
+// then calculating space resistance 
+R3 = 1/(A1*F12);
+// we can claculate heat transfer by inserting the quantities in equation (8-40):
+q = (Eb1-Eb2)/(R1+R2+R3);// [W]
+printf("net radiant exchange is %f W",q);
+
+
+
+
+
diff --git a/405/CH8/EX8.9/8_9.sce b/405/CH8/EX8.9/8_9.sce
new file mode 100755
index 000000000..9ba635333
--- /dev/null
+++ b/405/CH8/EX8.9/8_9.sce
@@ -0,0 +1,45 @@
+clear;
+clc;
+printf("\t\t\tExample Number 8.9\n\n\n");
+// effective emissivity of finned surface
+// Example 8.9 (page no.-409-410) 
+// solution
+
+// for unit depth in the z-dimension we have 
+A1 = 10;// [square meter]
+A2 = 5;// [square meter]
+A3 = 60;// [square meter]
+// the apparent emissivity of the open cavity area A1 is given by equation(8-47) as 
+// Ea1 = E*A3/[A1+E*(A3-A1)]
+// for constant surface emissivity the emitted energy from the total area A1+A2 is
+// e1 = Ea1*A1+E*A2*Eb
+// and the energy emitted per unit area for that total area is 
+// e_t = [(Ea1*A1+E*A2)/(A1+A2)]*Eb
+// the coefficient of Eb is the effective emissivity, E_eff of the combination of the surface and open cavity. inserting 
+// above equations gives the following values
+
+// for E = 0.2
+
+E = 0.2;
+Ea1 = E*A3/[A1+E*(A3-A1)];
+E_eff = [(Ea1*A1+E*A2)/(A1+A2)];
+printf("For emissivity of 0.2 the value of effective emissivity is %f ",E_eff);
+
+// for E = 0.5
+
+E = 0.5;
+Ea1 = E*A3/[A1+E*(A3-A1)];
+E_eff = [(Ea1*A1+E*A2)/(A1+A2)];
+printf("\n\n For emissivity of 0.5 the value of effective emissivity is %f ",E_eff);
+
+// for E = 0.8
+
+E = 0.8;
+Ea1 = E*A3/[A1+E*(A3-A1)];
+E_eff = [(Ea1*A1+E*A2)/(A1+A2)];
+printf("\n\n For emissivity of 0.8 the value of effective emissivity is %f ",E_eff);
+
+
+
+
+
author	priyanka	2015-06-24 15:03:17 +0530
committer	priyanka	2015-06-24 15:03:17 +0530
commit	b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree	ab291cffc65280e58ac82470ba63fbcca7805165 /405/CH8
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