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diff --git a/926/CH8/EX8.7/Chapter8_Example7.sce b/926/CH8/EX8.7/Chapter8_Example7.sce new file mode 100644 index 000000000..98f908a51 --- /dev/null +++ b/926/CH8/EX8.7/Chapter8_Example7.sce @@ -0,0 +1,35 @@ +//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 7, Page 283
+//Title: Calculation of total enthalpy
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [100 32]; //Given temperature and standard temperature of dry air in degree F
+MW = [18 29]; //Molecular weight of water and air respectively
+
+//DATA FROM GRAPH
+H = 0.0215; //humidity in terms of lb of water per lb of dry air at given conditions obtained from Fig 20, Page 122
+DP = 79; //Dew point corresponding to above humidity in degree F obtained from Fig 20, Page 122
+CP = [8.02 6.95]; //Molal heat capacities of water vapor between 79-100 degree F and 32-100 degree F in BTU per lb-mole degree F obtained from Fig 62, Page 259
+lamda = 1046; //Heat of vaporization at 79 degree F in BTU per lb obtained from Fig 19, Page 120
+
+//CALCULATION
+Lamda1 = (T(1)-T(2))*CP(2)/MW(2); //Sensible enthalpy of air in BTU per lb
+Lamda2 = (DP-T(2))*H; //Sensible enthalpy of liquid water in BTU per lb
+Lamda3 = lamda*H; //Latent heat of water in BTU per lb
+Lamda4 = (T(1)-DP)*H*CP(1)/MW(1); //Superheat of water vapor in BTU per lb
+Lamda = Lamda1+Lamda2+Lamda3+Lamda4; //Total enthalpy in BTU per lb of dry air
+
+//OUTPUT
+// Console Output
+mprintf('\n Total enthalpy of dry air is %2.0f BTU per lb',Lamda);
+
+// File Output
+fd= mopen('.\Chapter8_Example7_Output.txt','w');
+mfprintf(fd,'\n Total enthalpy of dry air is %2.0f BTU per lb',Lamda);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
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