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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2705/CH4/EX4.19/Ex4_19.sce | |
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Diffstat (limited to '2705/CH4/EX4.19/Ex4_19.sce')
| -rwxr-xr-x | 2705/CH4/EX4.19/Ex4_19.sce | 50 |
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diff --git a/2705/CH4/EX4.19/Ex4_19.sce b/2705/CH4/EX4.19/Ex4_19.sce new file mode 100755 index 000000000..edc6f9d44 --- /dev/null +++ b/2705/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,50 @@ +clear;
+clc;
+disp('Example 4.19');
+
+// aim : To determine the condition of the steam after
+// (a) isothermal compression to half its initial volume,heat rejected
+// (b) hyperbolic compression to half its initial volume
+
+// Given values
+V1 = .3951;// initial volume,[m^3]
+P1 = 1.5;// initial pressure,[MN/m^2]
+
+// solution
+
+// (a)
+// from steam table, at 1.5 MN/m^2
+hf1 = 844.7;// [kJ/kg]
+hfg1 = 1945.2;// [kJ/kg]
+hg1 = 2789.9;// [kJ/kg]
+vg1 = .1317;// [m^3/kg]
+
+// calculation
+m = V1/vg1;// mass of steam,[kg]
+vg2b = vg1/2;// given,[m^3/kg](vg2b is actual specific volume before compression)
+x1 = vg2b/vg1;// dryness fraction
+h1 = m*(hf1+x1*hfg1);// [kJ]
+Q = m*x1*hfg1;// heat loss,[kJ]
+mprintf('\n (a) The Quantity of steam present is = %f kg \n',m);
+mprintf('\n Dryness fraction is = %f \n',x1);
+mprintf('\n The enthalpy is = %f kJ \n',h1);
+mprintf('\n The heat loss is = %f kJ \n',Q);
+
+// (b)
+V2 = V1/2;
+// Given compression is according to the law PV=Constant,so
+P2 = P1*V1/V2;// [MN/m^2]
+// from steam table at P2
+hf2 = 1008.4;// [kJ/kg]
+hfg2 = 1793.9;// [kJ/kg]
+hg2 = 2802.3;// [kJ/kg]
+vg2 = .0666;// [m^3/kg]
+
+// calculation
+x2 = vg2b/vg2;// dryness fraction
+h2 = m*(hf2+x2*hfg2);// [kJ]
+
+mprintf('\n (b) The dryness fraction is = %f \n',x2);
+mprintf('\n The enthalpy is = %f kJ\n',h2);
+
+// End
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