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diff --git a/926/CH8/EX8.4/Chapter8_Example4.sce b/926/CH8/EX8.4/Chapter8_Example4.sce new file mode 100644 index 000000000..787479ca4 --- /dev/null +++ b/926/CH8/EX8.4/Chapter8_Example4.sce @@ -0,0 +1,33 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-8, Illustration 4, Page 281
+//Title: Calculation of heat of vaporization
+//=============================================================================
+clear
+clc
+
+//INPUT
+Tb = 78; //Normal boiling point of ethyl alcohol in degree C
+Tc = 243; //Critical temperature of ethyl alcohol in degree C
+T = 180; //Given temperature of ethyl alcohol in degree C
+Lamda_1 = 204; //latent heat of vaporisation of ethyl alcohol at normal boiling point in cal per gram
+
+//CALCULATION
+TB = 273.15+Tb; //Normal boiling point of ethyl alcohol in K
+TC = 273.15+Tc; //Critical temperature of ethyl alcohol in K
+T1 = 273.15+T; //Given temperature of ethyl alcohol in K
+
+Tr1 = TB/TC; //Reduced temperature with reference to boiling point
+Tr2 = T1/TC; //Reduced temperature with reference to temperature at which heat of vaporization is to be estimated
+Lamda_2 = Lamda_1*((1-Tr2)/(1-Tr1))^0.38; //Heat of vaporization at given temperature in cal per gram
+
+//OUTPUT
+// Console Output
+mprintf('\n Heat of vaporization at a temperature of %3.0f degree C = %3.0f cal per gram',Tb,Lamda_2);
+
+// File Output
+fd= mopen('.\Chapter8_Example4_Output.txt','w');
+mfprintf(fd,'\n Heat of vaporization at a temperature of %3.0f degree C = %3.0f cal per gram',Tb,Lamda_2);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
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