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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/EX8.11 | |
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| -rwxr-xr-x | 405/CH8/EX8.11/8_11.sce | 76 |
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diff --git a/405/CH8/EX8.11/8_11.sce b/405/CH8/EX8.11/8_11.sce new file mode 100755 index 000000000..4e3034c61 --- /dev/null +++ b/405/CH8/EX8.11/8_11.sce @@ -0,0 +1,76 @@ +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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