1 files changed, 41 insertions, 0 deletions
diff --git a/2705/CH5/EX5.18/Ex5_18.sce b/2705/CH5/EX5.18/Ex5_18.sce
new file mode 100755
index 000000000..3266c108e
--- /dev/null
+++ b/2705/CH5/EX5.18/Ex5_18.sce
@@ -0,0 +1,41 @@
+clear;
+clc;
+disp('Example 5.18');
+
+//  aim : To determine the
+//  (a) final pressure
+//   (b) final dryness fraction of the steam
+
+//  Given values
+P1 = 130;//  initial pressure, [kN/m^2]
+T1 = 273+75.9;//  initial temperature, [K]
+x1 = .92;//  initial dryness fraction
+T2 = 273+120.2;//  final temperature, [K]
+
+//  solution
+
+//  (a)
+//  from steam table, at 75.9 C
+Pws = 40;//  partial pressure of wet steam[kN/m^2]
+Pa = P1-Pws;//  partial pressure of air, [kN/m^2]
+vg = 3.99//  specific volume of the wet steam, [m^3/kg]
+// hence
+V1 = x1*vg;//  [m^3/kg]
+V2 = V1/5;//  [m^3/kg]
+//  for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so
+P2 = Pa*V1*T2/(V2*T1);//  final pressure,[kN/m^2]
+
+//  now for steam at 120.2 C
+Ps = 200;//  final partial pressure of steam,[kN/m^2]
+//  so by dalton's law total pressure in cylindert is
+Pt = P2+Ps;//  [kN/m^2]
+mprintf('\n (a) The final pressure in the cylinder is  =  %f  kN/m^2\n',Pt);
+
+//  (b)
+//  from steam table at 200 kN/m^2 
+vg = .885;//  [m^3/kg]
+//  hence
+x2 = V2/vg;//  final dryness fraction of the steam
+mprintf('\n (b) The final dryness fraction of the steam is  =  %f\n ',x2);
+
+//  End