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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+twb1 = 15.2// Wbt in degree Celsius
+twb2 = 26.7// Wbt in degree Celsius
+tw3 = 30 // Temperature at state 3 in degree Celsius
+h1 = 43 // Enthalpy at state 1 in kJ/kg
+h2 = 83.5 // Enthalpy at state 2 in kJ/kg
+hw = 84 // Enthalpy of water in kJ/kg
+mw = 1.15 // mass flow rate of water in kg/s
+W1 = 0.0088 // Humidity ratio of inlet stream
+W2 = 0.0213 // Humidity ratio of exit stream
+hw3 = 125.8 // Enthalpy of water entering tower in kJ/kg
+hm = 84 // Enthalpy of make up water in kJ/kg
+G = 1 // mass flow rate of dry air in kg/s
+hw34 = (G/mw)*((h2-h1)-(W2-W1)*hw) // Enthalpy change
+tw4 = tw3-(hw34/4.19) // Temperature of water leaving the tower
+A = tw4-twb1 //Approach of cooling water
+R = tw3-tw4 //Range of cooling water
+x = G*(W2-W1) //Fraction of water evaporated
+
+printf("\n Example 15.8\n")
+printf("\n Temperature of water leaving the tower is %f degree celcius",tw4)
+printf("\n Range of cooling water is %f degree Celsius",R)
+printf("\n Approach of cooling water is %f degree celcius",A)
+printf("\n Fraction of water evaporated is %f kg/kg dry air",x)
+//The answers vary due to round off error
+